Data Analysis and Probability — Free Algebra 1 Review Games.

Points going up left to right indicate a positive correlation.

A fair coin has 2 equally likely outcomes, so P(heads) = 1/2.

There is 1 favorable outcome out of 6 total: 1/6.

No pattern means no correlation between the variables.

Probability 0 means the event cannot happen.

Total = 10, P(blue) = 2/10 = 1/5.

r close to -1 indicates a strong negative linear relationship.

y = 2(5) + 3 = 13.

P(HH) = 1/2 * 1/2 = 1/4.

P(not A) = 1 - P(A) = 1 - 0.3 = 0.7.

There are 6 ways to get sum 7 out of 36 outcomes: 6/36 = 1/6.

For independent events: P(A and B) = P(A)*P(B) = 0.4*0.5 = 0.2.

r^2 represents the proportion of variance explained: 0.81 = 81%.

P = (4/10)(3/9) = 12/90 = 2/15.

Curved patterns suggest a nonlinear relationship, not a straight line.

When points trend downward from left to right, as x increases, y decreases — this is negative correlation. Positive correlation occurs when points trend upward from left to right. 'Perfect correlation' is not a direction; it describes strength.

A probability of 1 means the event will definitely occur — it is certain. By contrast, a probability of 0 means the event is impossible. Probabilities always fall between 0 and 1 inclusive.

A line of best fit models the overall trend in a scatter plot and is used to make predictions about values not in the data set. It does not pass through every point — it minimizes the overall distances between the line and all data points.

The strength of a linear correlation is determined by how close the absolute value of r is to 1. Since |-0.95| = 0.95, this is closest to 1 among the choices, indicating the strongest linear relationship. The negative sign shows direction, not weakness.

Even numbers on a standard die are 2, 4, and 6 — that is 3 favorable outcomes out of 6 total. P(even) = 3/6 = 1/2. Choice A (1/6) would be the probability of rolling one specific number, and choice B (1/3) would be correct for two favorable outcomes.

An outlier is a data point that falls far from the general trend or cluster of the other points in a scatter plot. It does not necessarily have the highest x-value — it simply does not fit the pattern of the rest of the data.

r = 0 means there is no linear relationship between the two variables — knowing x gives no information about predicting y using a line. Note that a non-linear relationship (such as a curve) could still exist even when r = 0.

There are 5 red and 2 yellow marbles, giving 7 favorable outcomes out of 10 total. P(red or yellow) = 7/10. Since red and yellow are mutually exclusive events (a marble cannot be both), we add: 5/10 + 2/10 = 7/10.

The slope is -3, meaning for every 1-unit increase in x, y decreases by 3. The y-intercept (12) gives the predicted value of y when x = 0 — not the slope. The correlation coefficient r is a separate value not directly shown in the equation.

For mutually exclusive events, P(A or B) = P(A) + P(B) = 0.4 + 0.25 = 0.65. Because the events cannot both happen at the same time, there is no overlap to subtract. Choice A (0.10) is the product P(A) times P(B), which applies to independent events, not the 'or' rule.

In y = 4x - 1, the y-intercept is -1, which represents the predicted value of y when x = 0. The slope (4) tells us the rate of change. Choice D is wrong because -1 is the constant term, not the slope.

If a point is above the line of best fit, the actual y-value is greater than what the line predicts for that x-value. This is called a positive residual (actual minus predicted is greater than zero). Points on the line have a residual of zero.

Numbers greater than 5 are 6, 7, and 8 — that is 3 favorable outcomes out of 8 equally likely outcomes. P(greater than 5) = 3/8. Choice A (5/8) is the probability of spinning a number that is 5 or less, which is a common reversal error.

Substitute x = 10 into the equation: y = 1.5(10) + 4 = 15 + 4 = 19. A common error is to forget adding the y-intercept of 4, which gives the incorrect answer of 15.

A strong positive correlation means r is close to +1. r = 0.88 is close to +1, indicating a strong positive linear relationship. Negative values of r indicate negative correlation, and r = 0.05 indicates nearly no linear relationship.

There are 4 kings in a 52-card deck (one per suit). P(king) = 4/52 = 1/13. Choice A (1/52) is the probability of drawing one specific card. Choice C (1/4) is the probability of drawing a card of any specific suit, not a specific rank.

Each flip is independent with P(heads) = 1/2. For all three flips to be heads: P(HHH) = (1/2)(1/2)(1/2) = 1/8. Choice C (3/8) is the probability of getting exactly 2 heads out of 3 flips — a different event and a common mix-up.

Extrapolation means using the line of best fit to predict y for x-values that fall outside the range of the original data. This is less reliable than interpolation, which predicts within the data range, because the trend may not continue beyond the observed values.

With replacement, each draw is independent. P(red on first draw) = 5/8 and P(red on second draw) = 5/8 since the marble is returned. P(both red) = (5/8)(5/8) = 25/64. Without replacement the answer would be (5/8)(4/7) = 20/56 — a common error when students forget the 'with replacement' condition.

y = 6(8) + 45 = 48 + 45 = 93, so the predicted score is correct. Since x = 8 falls outside the original data range of 1 to 7, this is extrapolation. Even a strong r value does not guarantee the trend continues beyond the observed data.

P(first roll is even) = 3/6 = 1/2 (even numbers are 2, 4, 6). P(second roll greater than 4) = 2/6 = 1/3 (numbers 5 and 6). Since the two rolls are independent events, P(both) = (1/2)(1/3) = 1/6.

Using the addition rule: P(A or B) = P(A) + P(B) - P(A and B) = 18/30 + 12/30 - 6/30 = 24/30. We subtract the 6 students who do both to avoid counting them twice. Choice C (30/30) is wrong because it ignores the overlap entirely.

r squared = 0.64 means 64% of the variation in y is explained by the linear relationship with x. The negative slope (-2) confirms a negative relationship. Note that r = -0.8 (not 0.64), because we take the negative square root of r squared when the slope is negative.

P(first is red) = 4/9. After removing one red marble, 8 marbles remain, of which 3 are blue. P(second is blue | first was red) = 3/8. Since the draws are dependent: P(both) = (4/9)(3/8) = 12/72 = 1/6. Using 9 in the denominator for both draws is a common error that ignores the lack of replacement.

Use the complement: P(at least one win) = 1 - P(winning neither game) = 1 - P(not A) times P(not B) = 1 - (0.7)(0.6) = 1 - 0.42 = 0.58. Choice A (0.12) is P(A and B) = (0.3)(0.4), the probability of winning both. Simply adding 0.3 + 0.4 = 0.70 overcounts the case where both are won.

A positive correlation means that as the x-variable increases, the y-variable also increases. The data points trend upward from left to right. Choice A describes a negative correlation. Choice D describes a perfect correlation (r = 1), which is a special case, not the general definition.

The line of best fit (also called a trend line or regression line) is drawn to best represent the general trend of all the data points — it minimizes the distances between the line and the points. It does not connect all points (Choice A) or specifically pass through the endpoints (Choice B).

The probability of the complement of an event is 1 minus the probability of the event: P(not A) = 1 - 0.25 = 0.75. Choice A is the probability of A itself. Choice D is impossible since probabilities cannot exceed 1.

There are 4 blue marbles out of a total of 3 + 4 + 3 = 10 marbles. P(blue) = 4/10 = 2/5. Choice B (3/10) is the probability of selecting red or green. Choice A (1/4) is incorrect because the total is 10, not 16.

No correlation means there is no consistent relationship between the variables, so the data points appear scattered randomly with no upward or downward trend. Choice A describes positive correlation, Choice B describes negative correlation, and Choice D describes a perfect linear correlation.

There is exactly 1 favorable outcome (rolling a 4) out of 6 equally likely outcomes, so P(4) = 1/6. Choice A (1/3) would apply if there were 2 favorable outcomes. Choice D (2/6) simplifies to 1/3 and is also incorrect.

An outlier is a data point that does not fit the general pattern of the rest of the data — it lies noticeably far from the trend. Choice A describes a point that fits the trend well. Choice D describes the y-intercept, which is a feature of the line, not a data point.

In any linear equation y = mx + b, the y-intercept (b) is the value of y when x = 0. In a line of best fit, it represents the model's predicted starting value. Choice A describes the slope. Choice B describes the correlation coefficient r.

The slope of -3 means that for each 1-unit increase in x (hours of exercise), the predicted y (resting heart rate) decreases by 3 bpm. A negative slope indicates an inverse relationship. Choice A reverses the direction. Choice C misinterprets the slope as the y-intercept.

Substitute x = 7 into the equation: y = 2(7) + 5 = 14 + 5 = 19. Choice A (14) is the result of computing 2(7) without adding 5. Choice B (17) may come from computing 2(6) + 5. Always substitute the full equation carefully.

The strength of a linear relationship is determined by the absolute value of r, not its sign. |0.45| = 0.45, |-0.87| = 0.87, |0.12| = 0.12, |-0.30| = 0.30. Since 0.87 is the largest absolute value, r = -0.87 indicates the strongest relationship. The negative sign means the relationship is negative, not weak.

There are 4 white (non-red) candies out of 10 total: P(not red) = 4/10 = 2/5. Alternatively, P(not red) = 1 - P(red) = 1 - 6/10 = 4/10 = 2/5. Choice A (3/5) is the probability of selecting a red candy, not a non-red candy.

The sample space is {HH, HT, TH, TT}. Outcomes with at least one tail: HT, TH, TT = 3 outcomes. P(at least one tail) = 3/4. An easier method: P(at least one tail) = 1 - P(no tails) = 1 - P(HH) = 1 - 1/4 = 3/4. Choice A (1/4) is P(no tails), a common error when using the complement incorrectly.

When data points form a random cloud with no upward or downward trend, this indicates no correlation — knowing a student's shoe size gives no useful information about their expected math score. Choices A and B both describe correlations that the data does not support. Choice C contradicts the observed pattern.

In the slope-intercept form y = mx + b, the coefficient m is the slope, representing how much y changes per one-unit increase in x. Here, m = 3 means y increases by 3 for each unit x increases. Choice A (2) is the y-intercept, not the slope. Choice D describes r, which is a separate statistic.

STATISTICS has 10 letters: S-T-A-T-I-S-T-I-C-S. The letter T appears at positions 2, 4, and 7, giving 3 occurrences. P(T) = 3/10. Choice A (1/10) assumes only one T. Choice B (1/5 = 2/10) assumes two T's. Always count each letter occurrence carefully.

When one variable increases while the other decreases, the relationship is a negative correlation. The scatter plot would show a downward trend from left to right. Choice A (positive) would mean both variables increase together. Choice D (perfect) means all points fall exactly on the line, which is not stated here.

A residual = actual y - predicted y. First find the predicted y: 2.5(6) + 10 = 15 + 10 = 25. Then residual = 22 - 25 = -3. A negative residual means the actual value is below the line of best fit. Choice B (3) is the result of subtracting in reverse order (predicted - actual).

Since chips are drawn without replacement, the events are dependent. P(first red) = 5/8. After removing one red chip, 4 red and 3 blue remain (7 total). P(second red | first red) = 4/7. P(both red) = 5/8 times 4/7 = 20/56 = 5/14. Choice A (25/64) incorrectly uses 5/8 times 5/8, treating the draws as independent (with replacement).

Set y = 18 and solve: 18 = -1.5x + 45. Subtract 45 from both sides: -27 = -1.5x. Divide both sides by -1.5: x = 18. Choice A (12) may result from subtracting 45 incorrectly. Choice C (27) may come from not dividing by the slope correctly.

The strength of a linear relationship is measured by |r|, the absolute value of the correlation coefficient. |0.72| = 0.72 and |-0.89| = 0.89. Since 0.89 > 0.72, Data Set B has the stronger linear relationship. The sign of r indicates direction (positive or negative), not strength. Choices A and D confuse sign with strength.

Since the marble is replaced, the draws are independent. Total marbles = 9. P(first red) = 4/9. P(second yellow) = 2/9. P(red then yellow) = 4/9 times 2/9 = 8/81. Choice C (2/9) is only the probability of the second event. Choice D (8/72) is incorrect because the denominator should be 81 after replacement, not 72.

Experimental probability = 94/200 = 0.47. Theoretical probability of a fair coin = 0.50. Since 0.47 < 0.50, the experimental probability is less than the theoretical probability. Choice C is wrong — experimental and theoretical probabilities are rarely exactly equal. Choice D confuses the definitions: theoretical probability is based on theory (1/2), not results.

For mutually exclusive events (events that cannot both occur at the same time), P(A or B) = P(A) + P(B) = 0.35 + 0.45 = 0.80. Choice D (0.1575) is the result of multiplying P(A) times P(B), which applies to independent events using the multiplication rule, not the addition rule for mutually exclusive events.

r-squared (the coefficient of determination) represents the proportion of variation in y explained by x. r-squared = 0.81 means temperature explains 81% of the variation in lemonade sales — a fairly strong relationship. Choice B confuses r-squared with r; the actual r would be approximately 0.90. Choice C misreads the slope — the positive slope of 4 means sales increase, not decrease, with temperature.

When data points trend upward from left to right, as x increases y also increases — this is a positive correlation. Negative correlation shows a downward trend. No correlation shows no discernible pattern. Although a correlation exists, correlation alone does not establish causation, so 'causal relationship' is not the correct description.

Numbers greater than 4 on a standard die are 5 and 6 — that is 2 favorable outcomes out of 6 total equally likely outcomes. P = 2/6 = 1/3. A common mistake is counting only one number (giving 1/6) or including 4 itself as qualifying (giving 3/6 = 1/2).

In slope-intercept form y = mx + b, the y-intercept is the constant b. Here b = 7. The value 4 is the slope, not the y-intercept. Students often confuse the slope (4) with the y-intercept (7), or substitute x = 1 to get 11, which is a point on the line but not the y-intercept.

r = 0 means there is no linear correlation between the two variables. r = 1 indicates a perfect positive linear relationship and r = -1 indicates a perfect negative one. Importantly, r = 0 does not rule out a non-linear (curved) relationship — it only tells us there is no linear pattern.

A standard deck has 4 suits, each with 13 cards. There are 13 hearts, so P(heart) = 13/52 = 1/4. The value 1/52 is the probability of drawing one specific card. The value 1/13 is the probability of drawing a card of a specific rank (such as an Ace), not a specific suit.

An outlier is a data point that does not follow the general trend — it lies far from the pattern formed by the other points. A point on the line of best fit fits the trend perfectly. Having the highest x-value or the lowest y-value does not automatically make a point an outlier; it depends on whether that point deviates from the overall pattern.

Complementary events have probabilities that sum to 1. P(not on time) = 1 - 0.65 = 0.35. A common error is subtracting 0.65 from 0.80 (giving 0.15) or doubling 0.65. The complement rule simply requires subtracting the given probability from 1.

Substitute x = 8 into the equation: y = 5(8) + 55 = 40 + 55 = 95. A common error is forgetting to add the y-intercept (55), which gives only 40. Another error is adding 5 and 8 before multiplying, producing an incorrect result.

Even numbers on this spinner are 2 and 4, giving P(even) = 2/4 = 1/2 on each spin. Because the spins are independent, P(even and even) = 1/2 times 1/2 = 1/4. A common mistake is adding probabilities instead of multiplying: 1/2 + 1/2 = 1, which is impossible as a probability for a specific outcome.

The slope represents the rate of change of y per one-unit increase in x. Here, slope = -0.4 means each 1 mph increase in speed is associated with a 0.4 mpg decrease in fuel efficiency. The y-intercept (42) represents the predicted fuel efficiency at zero speed — not the slope. The x and y variables cannot be swapped in the interpretation.

With replacement, each draw is independent and the bag always contains 10 marbles. P(green) = 3/10 each time. P(green and green) = 3/10 times 3/10 = 9/100. Choice D (1/15) is the without-replacement answer: 3/10 times 2/9 = 6/90 = 1/15, a common error when students forget to replace the marble.

Use the complement: P(at least one passes) = 1 - P(neither passes). P(one student fails) = 1 - 0.8 = 0.2. P(both fail) = 0.2 times 0.2 = 0.04. P(at least one passes) = 1 - 0.04 = 0.96. Choice B (0.64) is P(both pass), and choice A only accounts for a single student passing.

Interpolation means predicting a value within the range of the observed data. Since x = 50 falls between 20 and 80, this is interpolation. Extrapolation involves predicting outside the observed data range (for example, x = 90 or x = 10), which carries greater uncertainty because the trend may not continue beyond the collected data.

Substitute x = 5: y = -0.3(5) + 4.0 = -1.5 + 4.0 = 2.5. Choice D (3.5) results from ignoring the negative sign on the slope, treating it as +0.3. Choice A (2.0) comes from incorrectly computing -0.4(5) + 4.0. The negative slope confirms that more TV time is associated with a lower predicted GPA.

Students who play sports but not an instrument = 18 - 5 = 13 (subtract the 5 who do both from the 18 who play sports). P = 13/40. Choice C (18/40) incorrectly includes students who also play instruments. Choice A (5/40) is the probability of doing both activities, not sports only.

By the Fundamental Counting Principle, multiply the number of independent choices: 3 shirts times 2 pants = 6 outfits. A common error is adding the options (3 + 2 = 5) instead of multiplying. The principle states that when making one choice from each of two groups, multiply the sizes of those groups.

A cluster is a group of data points that are closely grouped together in a region of the scatter plot. A gap is an interval where no data points appear. A residual is the vertical distance between a data point and the line of best fit. A trend line (or line of best fit) is the line drawn to model the data — it is not a region of data points.

P(first red) = 5/8. After removing one red marble, 7 marbles remain with 3 white ones, so P(second white | first red) = 3/7. Multiply: 5/8 times 3/7 = 15/56. Choice A (15/64) is the with-replacement answer (5/8 times 3/8), a common error when students forget the total decreases after the first draw.

The strength of a linear relationship is determined by the absolute value of r, not its sign. |-0.91| = 0.91 is greater than |0.65| = 0.65, so the first data set has the stronger linear relationship. The sign of r only indicates direction (positive or negative slope) and has no effect on strength.

Residual = actual value - predicted value. Predicted value: y = 2(7) + 3 = 17. Residual = 20 - 17 = 3. A positive residual means the actual value is above the line. Choice B (-3) would mean the predicted value exceeds the actual, which is the reverse. Choice C (17) is the predicted value, not the residual.

P(pass first) = 0.7. P(fail second) = 1 - 0.6 = 0.4. Since the tests are independent: P(pass first and fail second) = 0.7 times 0.4 = 0.28. Choice B (0.42) is P(pass both = 0.7 times 0.6). Choice C (0.18) incorrectly uses P(fail first) times P(fail second) = 0.3 times 0.6.

Set y = 76 and solve for x: 76 = 6x + 40, so 36 = 6x, giving x = 6. Exactly 6 hours predicts a score of 76. With 5 hours: y = 6(5) + 40 = 70, which is below 76. Since the question requires at least 76 and x = 6 gives exactly 76, 6 whole hours is the minimum.

P(first girl) = 8/20 = 2/5. After selecting one girl, 19 students remain with 7 girls. P(second girl | first girl) = 7/19. P(both girls) = 8/20 times 7/19 = 56/380 = 14/95. Choice C (8/25) is the with-replacement answer (8/20 times 8/20), a common error. Choice D (7/19) accounts for only the second draw.

This is a classic lurking variable problem. The size of the fire drives both the number of firefighters dispatched and the amount of damage — sending more firefighters does not cause the damage. Correlation only shows that two variables tend to move together; it does not establish that one causes the other. A strong r-value (close to 1) only confirms the strength of the association, not causation.

Predicted value at x = 8: y = -2(8) + 30 = -16 + 30 = 14. Actual value = 16. Residual = actual - predicted = 16 - 14 = +2. A positive residual means the actual point is above the line. Choice B would apply if the actual value were 12 (below the predicted 14). Choice D would require an actual value of exactly 14.

A positive correlation means both variables move in the same direction — when x increases, y also increases. Choice A describes a negative correlation. Choice C describes no correlation (r near 0). Choice D is never true for real scatter plot data.

Probability is always a value from 0 (impossible event) to 1 (certain event). The range -1 to 1 describes a correlation coefficient, not probability. Expressing probability as 0 to 100 requires a percent sign; as a decimal or fraction it must stay between 0 and 1.

r = 0 means there is no linear relationship between the two variables. A value of r = 1 indicates a perfect positive correlation and r = -1 indicates a perfect negative correlation. Strong relationships are shown by r values close to 1 or -1, not close to 0.

A scatter plot shows how two numerical variables relate by plotting each pair of values as a single point. Choice A describes a line graph, Choice B describes a bar chart, and Choice D describes a pie chart.

There is one favorable outcome (rolling a 4) out of six equally likely outcomes, so P(4) = 1/6. Choice A incorrectly uses the face value 4 as the numerator. Choice D doubles the correct probability without justification. Choice B uses 4 as the total number of outcomes, which is wrong.

Extrapolation means using a model to make predictions beyond the observed data range, which can be unreliable. Predicting within the data range is called interpolation (Choice A). Choices B and D describe other procedures related to lines of best fit but are not extrapolation.

An outlier is a data point that does not follow the general pattern and sits far from the line of best fit. An intercept is where a line crosses an axis. A cluster is a group of data points that are close together. The mode is the most frequently occurring value in a data set.

An impossible event has a probability of 0. A probability of 1 means the event is certain to occur. A probability of 0.5 indicates equal chances of occurring or not occurring. Probability is always a defined number for any event, never undefined.

Substitute x = 8 into the equation: y = 3(8) - 5 = 24 - 5 = 19. Choice C (24) results from multiplying 3 times 8 but forgetting to subtract 5. Choice D (29) adds 5 instead of subtracting. Choice A (16) comes from subtracting x from the product instead of subtracting the constant 5.

There are 6 blue marbles out of 10 total marbles, so P(blue) = 6/10 = 3/5. Choice A (2/5) is the probability of drawing a red marble. Choice C (1/3) is incorrect. Choice D misuses only the number of blue marbles as the denominator, ignoring the red marbles.

For mutually exclusive events, P(A or B) = P(A) + P(B) = 0.3 + 0.4 = 0.7. Mutually exclusive events cannot both occur at the same time, so there is no overlap to subtract. Choice A (0.12) results from multiplying the two probabilities, which would give P(A and B) for independent events — not P(A or B). Choice D incorrectly assumes the probabilities must sum to 1.

r = 0.87 is close to 1, indicating a strong positive linear relationship — as practice hours increase, performance scores tend to increase as well. Values of r close to 0 indicate weak or no correlation. Negative r values indicate an inverse relationship. Weak positive correlations typically have r values between 0 and about 0.5.

The slope represents the rate of change. A slope of -1.5 means y decreases by 1.5 for each one-unit increase in x. Choice A describes the y-intercept, which is 12 in this equation. Choice C incorrectly describes a positive slope. Choice D is unrelated to slope.

Numbers greater than 5 on this spinner are 6, 7, and 8 — exactly 3 favorable outcomes out of 8 total. P(greater than 5) = 3/8. Choice A (5/8) incorrectly counts the numbers less than or equal to 5. Choice D (4/8) counts numbers greater than or equal to 5, including 5 itself.

A weak negative correlation has r close to 0 but negative, indicating a slight inverse relationship. r = -0.30 fits this description. r = 0.85 is a strong positive correlation. r = -0.92 is a strong negative correlation, not weak. r = 0.10 is a weak positive correlation.

Every student belongs to exactly one region of the two-way table, so the total is the sum of all four groups: 15 + 5 + 8 + 12 = 40. Choice A (28) omits the 12 who passed neither test. Choice B (35) contains an arithmetic error. Choice D (32) also results from an incomplete sum.

First calculate the slope: m = (22 - 10) / (6 - 2) = 12 / 4 = 3. Then substitute point (2, 10): 10 = 3(2) + b, so b = 4. The equation is y = 3x + 4. Choice A passes through (2, 10) but fails at (6, 22) since 2(6) + 6 = 18, not 22. Choice D gives 3(2) + 1 = 7, not 10.

The complement rule states P(not A) = 1 - P(A). So P(no rain) = 1 - 0.6 = 0.4. Choice A simply repeats the original probability. Choice B adds 1 and 0.6 instead of subtracting, producing a value greater than 1, which is impossible for a probability. Choice D misplaces the decimal.

The predicted value is y = 4(5) + 1 = 21. The residual is observed minus predicted: 18 - 21 = -3. A negative residual means the actual data point falls below the line of best fit. Choice A has the correct magnitude but wrong sign. Choice D gets the sign right but misinterprets it — a negative residual means below the line, not above it.

Because the order is fixed (first 2 correct, last 3 wrong), multiply the probabilities for each specific outcome: (1/4)^2 * (3/4)^3. Choice B gives the probability of answering all 5 correctly. Choice C ignores the first two questions entirely. Choice D uses the combination formula, which counts all possible arrangements of 2 correct among 5 — but this question specifies a fixed order, so no combination factor is needed.

For the first draw, P(white) = 6/10. Because the marble is not replaced, only 5 white marbles remain among 9 total, so P(white on second draw) = 5/9. Combined probability: 6/10 * 5/9 = 30/90 = 1/3. Choice A assumes replacement, which is incorrect here. Choice C multiplies by the probability of drawing a red marble on the second draw instead of a white one.

Using a linear model beyond the range of the data used to create it — called extrapolation — can be unreliable. The relationship may not remain linear past 12 hours; for example, test scores may plateau at 100. The fact that you can still compute a value (Choice B) does not mean the result is trustworthy. The direction of the slope (Choice D) has nothing to do with reliability.

For independent events, P(A or B) = P(A) + P(B) - P(A)*P(B). Substituting: 0.8 = 0.5 + P(B) - 0.5*P(B), so 0.3 = 0.5*P(B), giving P(B) = 0.6. Choice A (0.3) results from computing 0.8 - 0.5 without accounting for the overlap subtraction. Choice B (0.4) is the complement of 0.6 and reflects a common sign error.

Calculate residual (observed - predicted) for each choice: (1,4): 4 - [2(1)+2] = 0; (3,9): 9 - [2(3)+2] = 9 - 8 = 1; (5,12): 12 - [2(5)+2] = 12 - 12 = 0; (4,8): 8 - [2(4)+2] = 8 - 10 = -2. The largest absolute residual is 2, belonging to point (4, 8). Points with residual = 0 lie exactly on the line.

This is conditional probability: P(science | math) = P(both) / P(math) = (8/30) / (18/30) = 8/18. The denominator must be the number of math passers (18), since we are told the student already passed math. Choice A gives P(both) without any conditioning. Choice B incorrectly uses the science passers as the denominator. Choice D is simply P(science).

A positive correlation means that as x-values increase, y-values also tend to increase. This appears in a scatter plot as points trending upward from left to right. Choice A describes a negative correlation, where higher x-values are associated with lower y-values.

In slope-intercept form y = mx + b, the value b is the y-intercept. For y = 4x - 6, b = -6, so the y-intercept is -6. The value 4 is the slope m, not the y-intercept.

Numbers greater than 4 on a six-sided die are 5 and 6, giving 2 favorable outcomes out of 6 total. P = 2/6 = 1/3. Choice A (1/6) is the probability of rolling any single specific number, such as exactly a 5.

No correlation means there is no linear relationship between the two variables, so the points appear scattered randomly without any upward or downward trend. Choices A and B describe strong positive and strong negative correlations respectively, both of which show clear directional patterns.

The slope m represents how much y changes for every one-unit increase in x. Choice A describes the y-intercept b, not the slope. The correlation coefficient r is a separate statistic that measures the strength and direction of the linear relationship.

There are 5 blue marbles out of a total of 3 + 5 + 2 = 10 marbles. P(blue) = 5/10 = 1/2. Choice C (3/10) is the probability of selecting a red marble, which has only 3 marbles out of 10.

Complementary events cannot both occur at the same time and together they cover all possible outcomes, so their probabilities must add to 1. Choice B is incorrect because complementary events do not need equal probabilities. For example, P(rain) = 0.3 and P(no rain) = 0.7 are complements with unequal values.

Substitute x = 8 into the equation: y = 3(8) + 5 = 24 + 5 = 29. Choice A (24) results from computing only 3 times 8 without adding the y-intercept of 5, which is a common error when applying a linear equation.

By the complement rule, P(not A) = 1 - P(A). Therefore P(not passing) = 1 - 0.72 = 0.28. Choice C (0.72) is the original probability of passing, not its complement. The probabilities of an event and its complement must always add to 1.

The slope represents the rate of change. Here, for every one additional week of exercise, the predicted weight loss increases by 2.5 pounds. Choice A describes the y-intercept (10 pounds), which is the baseline value when x = 0, not the slope.

For independent events, P(A and B) = P(A) times P(B) = 0.4 times 0.5 = 0.20. Choice C (0.70) incorrectly adds the two probabilities together. Adding probabilities applies to P(A or B) for mutually exclusive events, not to P(A and B).

The probability of being both a 10th grader and preferring basketball equals the joint count divided by the total: 20/60 = 1/3. Choice C (4/7) is the conditional probability P(basketball | 10th grade) = 20/35, which only applies if you already know the student is in 10th grade.

Residual = actual value minus predicted value = 14 - 11 = 3. A positive residual means the actual data point is above the line of best fit. Choice A gets the sign wrong because a negative residual would mean the actual point is below the predicted value, which is not the case here.

Substitute x = 60: y = 0.5(60) + 20 = 30 + 20 = 50. Choice A (30) results from computing only 0.5 times 60 without adding the y-intercept of 20, which is a common error when evaluating a linear model.

A negative correlation coefficient indicates that as one variable increases, the other decreases. The value r = -0.78 represents a moderately strong negative correlation, which matches the described pattern. Choice A (r = 0.85) represents a strong positive correlation, which is the opposite of the relationship described.

P(not an ace) = 1 - P(ace) = 1 - 4/52 = 1 - 1/13 = 12/13. There are 48 non-ace cards out of 52 total, which also gives 48/52 = 12/13. Choice A (1/13) is the probability of drawing an ace, which is the original event, not its complement.

Use the Addition Rule: P(A or B) = P(A) + P(B) - P(A and B) = 40% + 30% - 15% = 55%. The overlap must be subtracted to avoid double-counting students who do both. Choice C (70%) incorrectly adds the two percentages without subtracting the 15% overlap.

A correlation coefficient of r = 0.91 is close to 1, indicating a strong positive linear relationship between the variables. Choice A is incorrect because correlation does not establish causation. Even a strong r-value only shows that the variables tend to move together, not that one causes the other.

First find the slope: m = (16 - 7) / (4 - 1) = 9/3 = 3. Using point-slope form with point (1, 7): y - 7 = 3(x - 1), which gives y = 3x + 4. When x = 10: y = 3(10) + 4 = 34. Choice A (30) results from computing 3 times 10 without adding the y-intercept of 4.

Expected value = (1/6)(10) + (2/6)(4) + (3/6)(-3) = 10/6 + 8/6 - 9/6 = 9/6 = $1.50. Each outcome is multiplied by its probability and the products are summed. Choice B ($1.00) results from incorrectly weighting outcomes rather than applying the correct probabilities.

P(1st red) = 5/8. After removing one red marble, 7 remain (4 red, 3 blue), so P(2nd red | 1st red) = 4/7. P(both red) = (5/8)(4/7) = 20/56 = 5/14. Choice A (25/64) is the incorrect result of calculating (5/8)(5/8), which assumes replacement rather than the given without-replacement condition.

The predicted value at x = 8 is y = -2(8) + 30 = 14. Residual = actual - predicted = 20 - 14 = 6. A positive residual means the actual point (20) is above the line of best fit prediction (14). Choice C has the correct magnitude but an incorrect sign and interpretation — a negative residual indicates a point below the line, not above it.

This is conditional probability: P(car | bicycle) = P(car and bicycle) / P(bicycle) = (10/80) / (30/80) = 10/30 = 1/3. Choice D (5/8 = 50/80) is the overall probability of owning a car, ignoring the condition that the selected person is already known to own a bicycle.

Compute each residual (actual minus predicted): x=2: 5.5 - [1.2(2)+3] = 5.5 - 5.4 = 0.1; x=4: 8.5 - [1.2(4)+3] = 8.5 - 7.8 = 0.7; x=6: 9.0 - [1.2(6)+3] = 9.0 - 10.2 = -1.2; x=10: 16.5 - [1.2(10)+3] = 16.5 - 15.0 = 1.5. The largest absolute residual is |1.5| at x = 10. Choice C (x=6) has an absolute residual of 1.2, which is smaller than 1.5.

Use the complement: P(at least one girl) = 1 - P(both boys). P(1st boy) = 12/30. P(2nd boy | 1st boy selected) = 11/29. P(both boys) = (12/30)(11/29) = 132/870 = 22/145. P(at least one girl) = 1 - 22/145 = 123/145. Choice A (3/5) is simply the probability of selecting one girl on the first draw, not accounting for the second selection or the complement approach.

When data points trend upward from left to right, as x increases y also increases — this is a positive correlation. A negative correlation would show a downward trend from left to right, while no correlation would show a random scatter with no discernible pattern.

The numbers greater than 4 on a six-sided die are 5 and 6 — that is 2 favorable outcomes out of 6 total. P = 2/6 = 1/3. A common error is choosing 1/2, which is the probability of rolling an even number (2, 4, or 6), not a number greater than 4.

When data points show no pattern or trend, there is no linear correlation between the variables — a correlation coefficient near 0 describes this. A weak negative correlation would still show a slight downward trend, and a perfect correlation would have all points on a single line.

In y = mx + b, the constant b is the y-intercept — the value of y when x = 0. Here, it means a student who studies 0 hours is predicted to score 8 points. The slope 3 represents the rate of increase (3 points per additional hour studied), not the constant 8.

Complementary events together cover all possible outcomes and have probabilities that sum to 1. P(not passing) = 1 - P(passing) = 1 - 0.72 = 0.28. Choosing 0.72 would repeat the original probability. The value 1.72 is impossible since probabilities cannot exceed 1.

Substitute x = 6: y = 2(6) + 5 = 12 + 5 = 17. A common error is computing 2(6) = 12 and forgetting to add 5, giving 12. Another error is computing only 2 + 5 = 7 and then multiplying by 6 to get 42, which misapplies order of operations.

The strength of a linear relationship is determined by the absolute value of r, not its sign. Computing absolute values: |0.30| = 0.30, |-0.90| = 0.90, |0.50| = 0.50, |-0.20| = 0.20. The largest absolute value is 0.90, so r = -0.90 indicates the strongest relationship. The negative sign shows direction (negative slope), not weakness.

There are 3 red marbles out of 3 + 7 = 10 total marbles. P(red) = 3/10. A common mistake is choosing 3/7, which compares red marbles to blue marbles rather than to the total. The value 7/3 is greater than 1 and therefore cannot be a valid probability.

The slope represents the rate of change. For every 1-unit increase in x (1 week), y increases by 4 (4 followers). The y-intercept 12 represents the starting number of followers. 'Doubling every 4 weeks' would describe exponential growth, not linear growth represented by a constant slope.

For mutually exclusive events (they cannot both occur at the same time), P(A or B) = P(A) + P(B) = 0.30 + 0.25 = 0.55. The value 0.075 is P(A) times P(B) = 0.075, which applies when calculating the probability that both independent events occur — not the 'or' case for mutually exclusive events.

Substitute x = 8: y = 1.5(8) + 10 = 12 + 10 = 22. The value 18 is a common error from computing 8 + 10 = 18 without multiplying by the slope first. The value 20 may arise from computing 1.5 times 10 instead of using the y-intercept correctly.

There are 2 green marbles out of 12 total, so P(green) = 2/12 = 1/6. Using the complement rule: P(not green) = 1 - 1/6 = 5/6. Alternatively, non-green marbles = 4 + 6 = 10, giving 10/12 = 5/6. Choosing 2/3 is an error from computing 8/12 (perhaps by mistakenly subtracting 4 instead of 2 from the total).

An outlier is a data point that falls far from the general pattern. Since the overall trend shows higher scores with more study hours, a student who scored very high with very few hours deviates significantly from this trend. Such a point typically weakens the correlation rather than strengthens it, and it is not the y-intercept.

Extrapolation means using a model to predict values outside the range of the original data. Since the data was collected for ages 10-16, predicting for a 50-year-old is far outside this range and may be unreliable because the linear relationship may not hold there. Interpolation — predicting within the known data range — is generally more reliable.

Residual = actual y-value minus predicted y-value. The predicted value at x = 4 is y = 3(4) + 2 = 12 + 2 = 14. The actual value is 18. Residual = 18 - 14 = 4. A residual of -4 would occur if the actual value were below the predicted value. Choosing 14 gives only the predicted value, not the residual.

The odd numbers are 1, 3, and 5 — three out of five sections, so P(odd) = 3/5 per spin. Since the two spins are independent events, P(odd on both spins) = 3/5 times 3/5 = 9/25. Choosing 6/25 is a common error from doubling 3/5 (adding instead of multiplying), and 3/25 may come from multiplying 3/5 by 1/5 incorrectly.

Use the addition rule: P(A or B) = P(A) + P(B) - P(A and B) = 20/50 + 15/50 - 5/50 = 30/50 = 3/5. Choosing 7/10 is the error of forgetting to subtract the overlap: 20 + 15 = 35, but this double-counts the 5 students who have visited both parks, giving the incorrect 35/50 = 7/10.

First, find the slope: m = (22 - 10) / (6 - 2) = 12 / 4 = 3. Using point-slope form: y - 10 = 3(x - 2), which simplifies to y = 3x + 4. At x = 9: y = 3(9) + 4 = 27 + 4 = 31. A common error leading to 28 is using the slope correctly but making an arithmetic mistake when solving for the y-intercept.

Expected gross winnings = (1/6)($8) + (1/6)($2) + (4/6)($0) = 8/6 + 2/6 = 10/6 approximately $1.67. Subtract the $2 entry fee: $1.67 - \(2.00 = -\)0.33. The player expects to lose about $0.33 per game on average. Choosing +$1.67 ignores the cost to play. Choosing -$0.50 is a common arithmetic error when computing the expected winnings.

P(first white) = 4/10. After removing one white ball, 3 white and 6 black remain (9 total): P(second white given first was white) = 3/9. P(both white) = 4/10 times 3/9 = 12/90 = 2/15. Choosing 4/25 is the error of computing (4/10) squared = 16/100 = 4/25, which incorrectly assumes the balls are replaced between draws.

Correlation does not imply causation. Both ice cream sales and drowning incidents rise during hot weather because more people visit beaches and pools. Hot weather is a lurking variable that drives both outcomes. A high r-value only measures the strength of a linear association between two variables — it does not establish that one causes the other.

Total swimmers = 15 (adults) + 40 (seniors) = 55. Using conditional probability: P(senior given swimming) = P(senior and swimming) / P(swimming) = (40/100) / (55/100) = 40/55 = 8/11. Choosing 2/3 is a common error from computing seniors who swim divided by total seniors (40/60). The denominator must be total swimmers (55), not total seniors.

Strength of a linear relationship depends on the absolute value of r, not its sign. The absolute values are: |0.85| = 0.85, |-0.92| = 0.92, |0.40| = 0.40, |-0.55| = 0.55. The smallest absolute value is 0.40, so r = 0.40 indicates the weakest linear relationship. A negative sign indicates a negative direction (as x increases, y decreases), not a weaker association.

P(exactly 2 correct) = C(3,2) times (1/2) squared times (1/2) to the first = 3 times 1/4 times 1/2 = 3/8. P(exactly 3 correct) = (1/2) cubed = 1/8. P(at least 2 correct) = 3/8 + 1/8 = 4/8 = 1/2. Choosing 3/8 is the common error of computing only the probability of exactly 2 correct and forgetting to include the case where all 3 are correct.

The slope -0.8 describes the predicted change in y per unit change in x, and r = -0.87 indicates a strong negative linear association. Choice A is wrong because correlation does not prove causation. Choice C confuses r with r-squared: r squared = 0.757, meaning approximately 75.7% of variation is explained — not 87%. Choice D is wrong because extrapolating far outside the observed data range produces unreliable predictions.

A scatter plot uses points on a coordinate plane to show how two numerical variables relate to each other. Each point represents one observation with an x-value and a y-value. A bar graph or histogram would display frequency of a single variable, not a scatter plot.

A positive correlation means that as the x-values increase, the y-values also tend to increase. The data points slope upward from left to right. A negative correlation would have y decreasing as x increases. A perfect straight line describes a perfect correlation, which is a stronger claim than simply 'positive.'

A standard die has 6 faces numbered 1 through 6. The even numbers are 2, 4, and 6 — that is 3 favorable outcomes out of 6 total equally likely outcomes. P(even) = 3/6 = 1/2. Choosing 1/3 would only be correct if there were 2 even numbers, not 3.

A line of best fit is also known as a trend line or a regression line. It is the line that most closely follows the pattern of the data points, minimizing the overall distances between the line and each point. Median and mode are measures of center for a single data set, not terms for lines on scatter plots.

The strength of a linear correlation is determined by how close |r| is to 1. Since |-0.95| = 0.95, which is very close to 1, the relationship is strong. The negative sign indicates the direction: as x increases, y decreases. A weak correlation would have |r| close to 0, such as r = -0.15.

There are 3 red marbles out of a total of 3 + 7 = 10 marbles. P(red) = 3/10. Choosing 3/7 is a common error — it compares red marbles to blue marbles rather than to the total number of marbles. The denominator must always be the total number of possible outcomes.

The strength of a linear relationship depends on the absolute value of r, not its sign. |0.05| = 0.05, |0.75| = 0.75, |-0.80| = 0.80, |-0.95| = 0.95. Since 0.05 is closest to 0, r = 0.05 represents the weakest linear association. The negative sign in r = -0.80 and r = -0.95 indicates direction only, not weakness.

Substitute x = 8 into the equation: y = 2(8) + 5 = 16 + 5 = 21. A common mistake is to calculate 2(8) = 16 and forget to add 5, giving 16. Another error is multiplying 2 by 13 instead of 8. Always substitute carefully and follow the order of operations.

The sign of r is negative, so the relationship is negative (as x increases, y decreases). The absolute value |-0.55| = 0.55 falls roughly in the middle range — not close enough to 1 to be strong, not close enough to 0 to be weak — making it a moderate correlation. Strong would require |r| closer to 0.8 or above; weak would be |r| below about 0.3.

The slope of a line of best fit represents the rate of change: for every one-unit increase in x, y changes by the slope value. Here, for each extra hour studied (x increases by 1), the predicted quiz score (y) increases by 0.5 points. The y-intercept (12, not 0.5) is the predicted score when x = 0. The slope does not represent a maximum value.

List the sample space: HH, HT, TH, TT — four equally likely outcomes. The outcomes with exactly one head are HT and TH, so there are 2 favorable outcomes. P(exactly one head) = 2/4 = 1/2. Choosing 1/4 would be correct only for P(HH) or P(TT). Choosing 3/4 would be correct for 'at least one head,' which includes HT, TH, and HH.

The residual is calculated as actual y minus predicted y. The predicted value is y = 5(4) + 8 = 20 + 8 = 28. The actual value is 30. Residual = 30 - 28 = 2. A positive residual means the actual point is above the line of best fit. Choosing -2 would mean the point is below the line, which is incorrect here.

Use the addition rule for probability: P(A or B) = P(A) + P(B) - P(A and B). Here, 18 + 12 - 6 = 24 students like math or science. Subtracting 6 prevents double-counting the students who like both. Choosing 30 assumes every student likes at least one subject, which is not stated. Choosing 36 forgets to subtract the overlap.

First, find the slope: m = (14 - 4) / (5 - 0) = 10 / 5 = 2. Since one of the points is (0, 4), the y-intercept is 4. The equation is y = 2x + 4. Choosing y = 2x + 5 uses the wrong y-intercept. Choosing y = 3x + 4 uses an incorrect slope of 3 instead of 2.

Use the inclusion-exclusion principle: P(king or heart) = P(king) + P(heart) - P(king and heart). There are 4 kings, 13 hearts, and 1 king of hearts. So P = 4/52 + 13/52 - 1/52 = 16/52. Choosing 17/52 is the most common error — it forgets to subtract the king of hearts, which was counted in both groups.

Since value decreases as age increases, the correlation is negative, eliminating r = 0.88 and r = 0.05. The relationship between age and car value is well-known and strong, so r should have a large absolute value. r = -0.88 indicates a strong negative correlation, which fits the context. r = -0.15 would indicate a very weak negative correlation — unlikely for such a predictable relationship.

Substitute x = 7 into the equation: y = 4(7) + 50 = 28 + 50 = 78. A common error is computing 4(8) + 50 = 82 by using the wrong x-value. Another error is computing 4(7) = 28 and then forgetting to add 50, giving only 28. Carefully substitute the given value and follow the order of operations.

Mutually exclusive events cannot occur at the same time, so P(A and B) = 0. Therefore, P(A or B) = P(A) + P(B) = 0.3 + 0.4 = 0.7. Choosing 0.12 results from multiplying the probabilities, which gives P(A and B) for independent events — not the probability of A or B. Choosing 0.58 applies the inclusion-exclusion formula incorrectly by subtracting 0.12.

The predicted value is y = -2(15) + 60 = -30 + 60 = 30. The residual = actual - predicted = 28 - 30 = -2. A negative residual means the actual value (28) is less than the predicted value (30), so the point falls below the line of best fit. Confusing the sign leads to thinking the point is above the line, which is incorrect here.

Substituting x = 5 into y = 3x + 1 gives y = 3(5) + 1 = 15 + 1 = 16. Since the result matches the mean y-value of 16, the mean point (5, 16) does lie on the line. In fact, a key property of least-squares regression lines is that they always pass through the point (x-bar, y-bar), the means of both variables. This is not a coincidence — it is a guaranteed property.

The even numbers from 1 to 10 are 2, 4, 6, 8, and 10 — that is 5 even cards out of 10 total. P(first card is even) = 5/10. After drawing one even card, 4 even cards remain out of 9 total. P(second card is even | first was even) = 4/9. Multiply: P(both even) = (5/10)(4/9) = 20/90 = 2/9. Choosing 1/4 incorrectly treats the draws as independent.

The strength of a linear relationship is determined by the absolute value of r, regardless of sign. |0.90| = 0.90, |-0.85| = 0.85, |0.40| = 0.40. Ranking from largest to smallest absolute value gives 0.90, then -0.85, then 0.40. The negative sign in r = -0.85 describes direction only, not weakness. Placing r = 0.40 before r = -0.85 is the most common error.

After drawing one red marble, 4 red marbles remain in the bag along with 3 blue marbles, for a total of 7 remaining marbles. P(second red | first was red) = 4/7. Choosing 5/8 uses the original count of 5 red out of 8 total — this ignores the fact that one marble has already been removed. Choosing 5/7 incorrectly keeps the red count at 5 instead of reducing it to 4.

Set y = 90 and solve for x: 90 = 3x - 120. Add 120 to both sides: 210 = 3x. Divide by 3: x = 70. The predicted temperature is 70°F. Choosing 150°F comes from computing 90 + 60 instead of 90 + 120. Choosing 210°F stops after computing 90 + 120 = 210 without dividing by the slope coefficient 3.

The predicted score is y = -4(8) + 92 = -32 + 92 = 60. The residual = actual - predicted = 56 - 60 = -4. A negative residual means the actual score (56) is below the model's prediction (60), so the student scored lower than expected given their number of absences. Choosing 4 reverses the subtraction order. The residual measures deviation from the model's prediction, not from the class average.

When both variables increase together, the scatter plot shows a positive correlation. Negative correlation means y decreases as x increases. No correlation means there is no visible pattern. Perfect positive correlation specifically requires all points to lie exactly on a line with r = 1, which is not indicated here — so 'positive correlation' is the most accurate general description.

Total marbles = 4 + 6 = 10. P(red) = 4/10 = 2/5. Choice A (2/3) incorrectly uses the ratio of green to red rather than red to total. Choice C (4/6) compares red marbles only to green marbles, ignoring that the total sample space is 10. Choice D (1/4) treats the denominator as just the number of red marbles.

The slope represents the rate of change: for every 1-unit increase in x (hours of exercise), y (pounds lost) increases by 3. Choice A describes the y-intercept (7), not the slope. Choice C confuses slope with a maximum value — slope is a rate, not a cap. Choice D confuses slope with the correlation coefficient r, which is always between -1 and 1 and is a separate statistic.

Substitute x = 6 into the equation: y = 4(6) + 3 = 24 + 3 = 27. Choice A (24) is a common error — it calculates 4 times 6 but forgets to add the y-intercept of 3. Choice C (30) results from adding 3 twice (4 times 6 plus 6). Choice D (33) incorrectly adds 9 instead of 3, perhaps doubling or tripling the y-intercept.

The prime numbers from 1 to 10 are 2, 3, 5, and 7 — a total of 4 primes. Note that 1 is not a prime number. So P(prime) = 4/10. Using the complement rule: P(not prime) = 1 - 4/10 = 6/10 = 3/5. Choice A (2/5) is actually the probability of landing ON a prime. Choice B (1/2) results from the common mistake of counting 1 as prime, which gives 5 primes out of 10.

The residual is calculated as actual minus predicted: 45 - 52 = -7. A negative residual means the actual data point lies below the line of best fit. Choice A (7) results from subtracting in the wrong order (predicted minus actual). Choices B and D are the raw predicted and actual values, not the residual — the residual measures the error, not the values themselves.

Each true/false question has a 1/2 probability of being guessed correctly. Because the two guesses are independent events, multiply the probabilities: P(both correct) = 1/2 times 1/2 = 1/4. Choice A (1/2) is the probability of getting just one specific question right — a common error is forgetting to account for both questions. Choice D (3/4) is the probability of getting at least one question correct, not both.

Residual = actual minus predicted. For Student A: predicted = 5(4) + 10 = 30, so residual = 28 - 30 = -2. For Student B: predicted = 5(6) + 10 = 40, so residual = 40 - 40 = 0. The absolute residuals are 2 and 0, making Student B's score a perfect fit for the model. Choice A incorrectly assigns Student A as the better fit. Choices C and D contain arithmetic errors in the residual calculations.

Conditional probability is calculated as P(tablet and phone) divided by P(phone) = (10/40) divided by (25/40) = 10/25 = 2/5. The key is to restrict the sample space to the 25 phone owners and ask how many of them also own a tablet (10). Choice A (1/4 = 10/40) ignores the given condition and treats this as an unconditional probability. Choice C (2/3 = 10/15) incorrectly divides by the number of tablet owners rather than phone owners.

Substituting x = 5: y = 8(5) + 20 = 40 + 20 = 60. With r = 0.78, the correlation is positive (r is greater than 0) and moderately strong (r values near 0.8 indicate a strong relationship; values near 0 indicate weak). Choice A incorrectly labels the correlation as negative even though r is positive. Choice C (40) forgets to add the y-intercept of 20. Choice D (80) adds the y-intercept twice and incorrectly calls r = 0.78 a perfect correlation — perfect correlation requires r = 1 exactly.