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.

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.

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.

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 y-value for a given x-value.

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

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.

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.

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.

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.

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.

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.