Which of the following graphs best shows a strong negative association between d and t ?

Solution
A graph with a strong negative association between d and t would have the points on the graph closely aligned with a line that has a negative slope. The more closely the points on a graph are aligned with a line, the stronger the association between d and t, and a negative slope indicates a negative association. Of the four graphs, the points on graph D are most closely aligned with a line with a negative slope. Therefore, the graph in choice D has the strongest negative association between d and t.
Choice A is incorrect because the points are more scattered than the points in choice D, indicating a weak negative association between d and t. Choice B is incorrect because the points are aligned to either a curve or possibly a line with a small positive slope. Choice C is incorrect because the points are aligned to a line with a positive slope, indicating a positive association between d and t.
If 16 + 4x is 10 more than 14, what is the value of 8x ?

Solution
The description “16 + 4x is 10 more than 14” can be written as the equation 16 + 4x = 10 + 14, which is equivalent to 16 + 4x = 24.Subtracting 16 from each side of 16 + 4x = 24 gives 4x = 8. Since 8x is 2 times 4x, multiplying both sides of 4x = 8 by 2 gives 8x = 16. Therefore, the value of 8x is 16.
Choice A is incorrect because it is the value of x, not 8x. Choices B and D are incorrect; those choices may be a result of errors in rewriting 16 + 4x = 10 + 14. For example, choice D could be the result of subtracting 16 from the left side of the equation and adding 16 to the right side of 16 + 4x = 10 + 14, giving 4x = 40 and 8x = 80.
In the figure above, lines A and m are parallel and lines s and t are parallel. If the measure of ∠1 is 35°, what is the measure of ∠2 ?

Solution
Consider the measures of ∠3 and ∠4 in the figure below.
The measure of ∠3 is equal to the measure of ∠1 because they are corresponding angles for the parallel lines ℓ and m intersected by the transversal line t. Similarly, the measure of ∠3 is equal to the measure of ∠4 because they are corresponding angles for the parallel lines s and t intersected by the transversal line m. Since the measure of ∠1 is 35°, the measures of ∠3 and ∠4 are also 35°. Since ∠4 and ∠2 are supplementary, the sum of the measures of these two angles is 180°. Therefore, the measure of ∠2 is 180° − 35° = 145°.
Choice A is incorrect because 35° is the measure of ∠1, and ∠1 is not congruent to ∠2. Choice B is incorrect because it is the measure of the complementary angle of ∠1, and ∠1 and ∠2 are not complementary angles.Choice C is incorrect because it is double the measure of ∠1.
If y = kx, where k is a constant, and y = 24 when x = 6, what is the value of y when x = 5 ?

Solution
Substituting 6 for x and 24 for y in y = kx gives 24 = (k)(6), which gives k = 4. Hence, y = 4x. Therefore, when x = 5, the value of y is (4)(5) = 20. None of the other choices for y is correct because y is a function of x, and so there is only one yvalue for a given xvalue.
Choices A, B, and D are incorrect. Choice A is the result of using 6 for y and 5 for x when solving for k. Choice B results from using a value of 3 for k when solving for y. Choice D results from using y = k + x instead of y = kx
John runs at different speeds as part of his training program. The graph shows his target heart rate at different times during his workout. On which interval is the target heart rate strictly increasing then strictly decreasing?

Solution
On the graph, a line segment with a positive slope represents an interval over which the target heart rate is strictly increasing as time passes. A horizontal line segment represents an interval over which there is no change in the target heart rate as time passes, and a line segment with a negative slope represents an interval over which the target heart rate is strictly decreasing as time passes. Over the interval between 40 and 60 minutes, the graph consists of a line segment with a positive slope followed by a line segment with a negative slope, with no horizontal line segment in between, indicating that the target heart rate is strictly increasing then strictly decreasing.
Choice A is incorrect because the graph over the interval between 0 and 30 minutes contains a horizontal line segment, indicating a period in which there was no change in the target heart rate. Choice C is incorrect because the graph over the interval between 50 and 65 minutes consists of a line segment with a negative slope followed by a line segment with a positive slope, indicating that the target heart rate is strictly decreasing then strictly increasing. Choice D is incorrect because the graph over the interval between 70 and 90 minutes contains horizontal line segments and no segment with a negative slope.
refer to the following information.
A botanist is cultivating a rare species of plant in a controlled environment and currently has 3000 of these plants. The population of this species that the botanist expects to grow next year, N_{next year}, can be estimated from the number of plants this year, N_{this year}, by the equation below.
\(N_{next\, year}=N_{this\, year}+0.2(N_{this\, year})\left ( 1\frac{N_{this\, year}}{K} \right )\)The constant K in this formula is the number of plants the environment is able to support.
The botanist would like to increase the number of plants that the environment can support so that the population of the species will increase more rapidly. If the botanist’s goal is that the number of plants will increase from 3000 this year to 3360 next year, how many plants must the modified environment support?

Solution
The correct answer is 7500. If the number of plants is to be increased from 3000 this year to 3360 next year, then the number of plants that the environment can support, K, must satisfy the equation 3360 = 3000 + 0.2(3000)(1 − \(\frac{3000}{K}\)). Dividing both sides of this equation by 3000 gives 1.12 = 1 + 0.2( 1 − \(\frac{3000}{K}\)), and therefore, it must be true that 0.2( 1 − \(\frac{3000}{K}\)) = 0.12, or equivalently, 1 − \(\frac{3000}{K}\) = 0.6. It follows that \(\frac{3000}{K}\) = 0.4, and so K = \(\frac{3000}{0.4}\) = 7500.
refer to the following information.
A botanist is cultivating a rare species of plant in a controlled environment and currently has 3000 of these plants. The population of this species that the botanist expects to grow next year, N_{next year}, can be estimated from the number of plants this year, N_{this year}, by the equation below.
\(N_{next\, year}=N_{this\, year}+0.2(N_{this\, year})\left ( 1\frac{N_{this\, year}}{K} \right )\)The constant K in this formula is the number of plants the environment is able to support.
According to the formula, what will be the number of plants two years from now if K = 4000 ? (Round your answer to the nearest whole number.)

Solution
The correct answer is 3284. According to the formula, the number of plants one year from now will be 3000 + 0.2(3000)( 1 − \(\frac{3000}{4000}\) ), which is equal to 3150. Then, using the formula again, the number of plants two years from now will be 3150 + 0.2(3150)(1 − \(\frac{3150}{4000}\)), which is 3283.875. Rounding this value to the nearest whole number gives 3284.
In the figure above, point O is the center of the circle, line segments LM and MN are tangent to the circle at points L and N, respectively, and the segments intersect at point M as shown. If the circumference of the circle is 96, what is the length of minor arc \(\overset{\frown}{LN}\)?

Solution
The correct answer is 32. Since segments LM and MN are tangent to the circle at points L and N, respectively, angles OLM and ONM are right angles. Thus, in quadrilateral OLMN, the measure of angle O is 360° − (90° + 60° + 90°) = 120°. Thus, in the circle, central angle O cuts off \(\frac{120}{360}=\frac{1}{3}\) of the circumference; that is, minor arc \(\overset{\frown}{LN}\) is ^{1}⁄_{3} of the circumference. Since the circumference is 96, the length of minor arc \(\overset{\frown}{LN}\) is ^{1}⁄_{3} × 96 = 32.
a = 18t + 15
Jane made an initial deposit to a savings account.Each week thereafter she deposited a fixed amount to the account. The equation above models the amount a, in dollars, that Jane has deposited after t weekly deposits. According to the model, how many dollars was Jane’s initial deposit? (Disregard the $ sign when gridding your answer.)

Solution
The correct answer is 15. The amount, a, that Jane has deposited after t fixed weekly deposits is equal to the initial deposit plus the total amount of money Jane has deposited in the t fixed weekly deposits. This amount a is given to be a = 18t + 15. The amount she deposited in the t fixed weekly deposits is the amount of the weekly deposit times t; hence, this amount must be given by the term 18t in a = 18t + 15 (and so Jane must have deposited 18 dollars each week after the initial deposit). Therefore, the amount of Jane’s original deposit, in dollars, is a − 18t = 15.
In one semester, Doug and Laura spent a combined 250 hours in the tutoring lab. If Doug spent 40 morehours in the lab than Laura did, how many hours did Laura spend in the lab?

Solution
The correct answer is 105. Let D be the number of hours Doug spent in the tutoring lab, and let L be the number of hours Laura spent in the tutoring lab. Since Doug and Laura spent a combined total of 250 hours in the tutoring lab, the equation D + L = 250 holds. The number of hours Doug spent in the lab is 40 more than the number of hours Laura spent in the lab, and so the equation D = L + 40 holds. Substituting L + 40 for D in D + L = 250 gives (L + 40) + L = 250, or 40 + 2L = 250. Solving this equation gives L = 105. Therefore, Laura spent 105 hours in the tutoring lab.