SAT 8

PRACTICE TEST 3

28.In a geometry exercise, there is a 0.15 probability a protractor is in error of 1° or more. If 5 protractors

are used, what is the probability that all of them are in error of 1° or more?

(A)0.000076

(B)0.60

(C)0.15

(D)0.75

(E)7.6

29.For all θ, (sin θ + sin (−θ))(cos θ + cos (−θ)) =

(A)2 sin θ

(B)0

(C)1

(D)2 cos θ

(E)4 sin θ cos θ

30.If 12 sin2 θ + 5 sin θ − 2 = 0, then what is the smallest positive value of θ?

(A)8.6°

(B)14.5°

(C)20.9°

(D)41.8°

(E)83.6°

(C)1

(D)5

2

(E)11

32.If 7.8a = 3.4b, then b =

a

(A)0.361

(B)0.596

(C)1.408

(D)1.679

(E)2.294

33.A point has rectangular coordinates (6, 8). If the polar coordinates are (10, θ), then θ =

(A)21.8°

(B)36.9°

(C)53.1°

(D)60°

(E)76°

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PRACTICE TEST 3

40.The daytime high temperatures for the first 9 days in January in Boston were 19, 21 16, 22, 28, 34, 30, 28, 36 degrees Fahrenheit. What is the standard deviation of the temperatures?

(A)4.90

(B)6.48

(C)10

(D)19.44

(E)42

41.The graph in Figure 1 could be a portion of the graph of which of the following functions?

I.f (x) = x4 + ax3 + bx2 + cx + d.

II.F(x) = −x4 + ax3 + bx2 + cx + d.

III. g(x) = x6 + ax5 + bx4 + cx3 + dx2 + ex + f.

(A)I only

(B)III only

(C)I and II only

(D)I and III only

(E)I, II, and III

integer x such that f (x) > 100?

(A)51

(B)99

(C)101

(D)199

(E)201

43.If log3 (x − 9) = log9 (x − 3), then x =

(A)30

(B)7

(C)9

(D)7 or 12

(E)12

44.If n represents the greatest integer less than or

equal to n, then which of the following is the solution to −11 + 4 n = 5?

(A)n = 4

(B)4 < n < 5

(C)−2 ≤ n < −1

(D)4 ≤ n < 5

(E)4 < n ≤ 5

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y

4

2

x

–2

Figure 1

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ANSWERS AND SOLUTIONS

1.B

1 − x = 4 − 4x . x x

Raise each side of the equation to the fifth power to solve for k.

(n + 1)3.

4. D The distance between ordered triples (x1, y1, z1) and (x2, y2, z2) is given by the formula:

Distance = (x2 − x1)2 + (y2 − y1) + (z2 − z1)2.

=(11 − −2)2 + (4 − 0)2 + (−1 − 3)2

=132 + 42 + (− 4)2

=132 + 42 + (− 4)2

=201 ≈ 14.18.

5. E

If sin x = 0.45, the angle whose sine is 0.45 is:

sin−1 (0.45) = 26.74°

6. C The radicand must be greater than or equal to zero.

9 − x2 ≥ 0.

9 ≥ x2.

−3 ≤ x ≤ 3.

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7. D

7a4 = 28a2.

7a4 − 28a2 = 0. 7a2 − (a2 − 4) = 0.

7a2 − (a − 2)(a + 2) = 0. a = 0, 2, or − 2.

8. C

First, determine f (8).

f (8) = 2(8) + 3 8 = 16 + 2 = 18. Now, determine f (18).

f (8) = 2(18) + 3 18 = 36 + 2.621 ≈ 38.6 9. B

9! = 9! 4!(9 − 4)! 4!5!

=6 × 7 × 8 × 9

1 × 2 × 3 × 4

=126.

10.D The components of v are given by (3 − (−2), −5 − 7), or (5, −12). The magnitude is the length of v.

PART III / EIGHT PRACTICE TESTS

11. D Isolate the radical expressions and square both sides of the equation.

2 x − 2x + 1 − 1 = 0.

x = 2 x.

Now square both sides a second time. x2 = (2 x )2 .

x2 = 4x.

x2 − 4x = 0.

x(x − 4) = 4.

x = 0 or x = 4.

Because the radicand of a square root must be greater than zero, 4 is the only solution.

12.B Because the y-intercept of the line y = mx + 4 is positive, then the slope of the line must be negative in order for part of the line to fall in the 4th quadrant. m < 0 is the correct answer choice.

13.B

10

Now, set the two expressions equal to each other and solve for n.

5n = 5n.

5 = 5n2.

n2 = 1.

n = ±1.

Answer B is the only possible answer choice.

PRACTICE TEST 3

14. D The middle term in the binomial expansion of (x + 2)6 is given by:

6 C3 x3 23 =

3!3!6! (x3 )(8) =

20(x3 )(8) = 160x3.

15. E Try substituting the given values into the function for x to determine which results in f (x) = 0.

f (1.5) = 6(1.5)3 − 5(1.5)2 + 4(1.5) − 15 = 0.

An alternative solution is to divide the polynomial by the factor x − 32 . Doing so results in a zero remainder. You could also graph the function to determine that 32 is an x-intercept.

16. B The largest angle of a triangle is opposite its longest side. Let θ = the largest angle of the triangle. θ is opposite the side measuring 6 inches. Using the Law of Cosines:

62 = 42 + 42 − 2(4)(4)cosθ.

36 = 16 + 16 − 32 cosθ.

4 = −32 cosθ.

cosθ = − 324 .

θ= cos−1− 324 ≈ 97.2ο

17.A The midpoint of the line segment connecting (1, −4) and (3, 6) is the center of the circle.

The radius of the circle is half the length of the line segment connecting the two points.

Therefore, the standard form of the equation of the circle is (x − 2)2 + (y − 1)2 = 26.

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18. C The model uses the greatest integer function. Evaluate the model for t = 11.25. Remember that the cost of the first minute is $0.40. 12 is the least integer greater than or equal to 11.25.

C= 0.40 + 0.20 t .

C= 0.40 + 0.20 11.25 .

C= 0.40 + 0.20(12) = $2.80.

19.D For the three terms to be part of a geometric sequence there must be a common ratio between con-

secutive terms. Add c to each of the three terms to get the progression 1 + c, 13 + c, 61 + c. Now, set the common ratios equal to each other and solve for c.

20. C Think of X and Y as vertices of a right triangle with one leg measuring 6 units (the diameter of the cylinder’s base) and one leg measuring 12 units (the height).

21. C The average grade for the first class is 84%, so the sum of the grades is 18(84) or 1,512. There are 42 total students in both classes and the least possible sum for the second class is 0.

1,512 + 0 =

42

36

The greatest possible sum for the second class is 100(24) or 2,400.

1,512 + 2,400 =

42

93.1.

The correct answer choice is, therefore, 36 ≤ x ≤ 93.1.