Mixed practice: Calculator and Digital Testing Skills

2.

Group both parts: (11-3)/(2+2) gives 8/4 = 2. Without the denominator parentheses, part of the intended denominator could become a separate added term. Check the exact arithmetic before trusting the entered structure. Review Example 1.01.

3.

Substitute −2 using parentheses: 3(−2)² + 2(−2) − 5 = 12 − 4 − 5 = 3. A definition f(x)=3*x^2+2*x-5 followed by f(-2) is also valid. The squared negative is positive; the linear term remains negative. Review Example 1.04.

4.

The equation factors as (x − 2)(x − 4) = 0, giving intercept inputs 2 and 4. Thus the greater is 4. The vertex at (3, −1) is a different feature and does not answer the intercept question. A graph should be used to inspect the zeros, not simply the lowest point. Review Example 2.03.

9.

There are four ordered values, so average the middle two: (8 + 10)/2 = 9. A median command agrees. The mean is 40/4 = 10, which is different and illustrates why a statistics menu does not choose the requested statistic for you. Review Example 3.10.

56.5 m².

The exact area is π($\sqrt{\text{18}}$)² = 18π ≈ 56.5486678. Rounding to the nearest tenth gives 56.5. Enter 18*pi; rounding the radius before squaring would introduce unnecessary error. Review Example 4.08.

2.

The bounds on y require 2x + 5 ≤ 15 − 3x, so 5x ≤ 10 and x ≤ 2. At x = 2, choose y = 9; both boundary equations hold, so the upper bound is attainable. Graphing the overlap gives the same rightmost point (2, 9). Review Example 2.13.

608.

Use 500*(1.04)^5 to obtain approximately 608.3264512, then round to 608. The amount must exceed 500 because the factor is greater than 1. Replacing the exponential calculation by five additive increases of 4 percent of the original would represent a different model. Review Example 5.08.

36.9°. The angle is sin⁻¹(12/20) = sin⁻¹(0.6) ≈ 36.86989765°. Set degrees and use the inverse-sine function, not the reciprocal of sine. The answer rounds to 36.9° and is acute as required. Review Example 1.11.

5.

Set x² − 1 = 2x + 2, giving (x − 3)(x + 1) = 0. The positive input is x = 3, and then y = 8. Therefore y − x = 8 − 3 = 5. The other intersection is (−1, 0) and violates the positive-input condition. Review Example 4.12.

A. 0.

The original expression excludes x = 5. Cancelling for allowed inputs gives x + 5 = 10, whose only candidate is the forbidden input 5. Therefore there are no solutions. A line-shaped display after cancellation does not make the original rational expression defined at the hole. Review Example 4.03.

3.3.

The weighted total is 0(2) + 3(5) + 6(3) = 33 and the count is 10. Thus the mean is 33/10 = 3.3. Averaging the three distinct values would give 3, but would discard their different frequencies. Review Example 3.11.

32.

Adding the entire equations directly gives 6x + 4y = 14 + 18 = 32. Neither variable needs to be isolated. This is an example in which recognizing the requested combination is shorter than graphing both lines and reconstructing the expression from their intersection. Review Example 5.03.

20.58.

At t = 6, the exponent is 6/2 = 3. Thus P(6) = 60(0.7)³ = 60(0.343) = 20.58. The factor 0.7 is applied once every two time units, so using exponent 6 would double the number of decay periods. Review Example 3.08.

8.

294/40 = 7.35. Seven buses provide 280 seats, which is insufficient; eight provide 320. The capacity inequality requires increasing to the next integer, not ordinary rounding to 7. Review Example 4.10.

2.

Taking both square-root signs gives x − 4 = ±0.02, so the roots are 3.98 and 4.02. They are distinct even if a broad graph window makes them appear merged. The exact positive right side guarantees two roots; replacing it with zero would change the question. Review Example 4.13.

23.

Each input increase of 3 corresponds to an output increase of 6, giving slope 2. Moving from input 7 to 10 adds another 6 to the output: 17 + 6 = 23. Equivalently the model is y = 2x + 3. A table makes the input spacing visible and prevents mistaking 6 for the slope. Review Example 3.04.

B.

A positive fitted slope describes an association in the observed data. The design does not establish a causal effect, a universal rule about every student, or an absence of confounding. Random sampling addresses selection for population inference; it is not the same as random assignment of a treatment. Review Example 5.12.

10.

The budget condition is 25 + 17h ≤ 200, so h ≤ 175/17 ≈ 10.2941. The greatest allowable whole number is 10. Ten hours cost $195; eleven cost $212 and exceed the budget. Here the inequality requires moving down to an integer, unlike a minimum-capacity problem. Review Example 4.10.

3.

Substitute y = x² − 2 and let u = x² ≥ 0. Then u + (u − 2)² = 4, which simplifies to u(u − 3) = 0. For u = 0, there is the point (0, −2). For u = 3, the points are ($\sqrt{\text{3}}$, 1) and (−$\sqrt{\text{3}}$, 1). Count ordered pairs, not just the two possible values of u. Review Example 2.14.

28.

Write y = ax² + bx + c. The first point gives c = 1; the others give a + b = 5 and 4a + 2b = 14. Subtraction yields a = 2, then b = 3. Thus f(3) = 2(9) + 3(3) + 1 = 28. Quadratic regression on these exact pairs provides the same model. Review Example 3.07.

6.

V(5) = 53.1441 > 50 and V(6) = 47.82969 < 50. The factor 0.9 makes the model decrease as n increases. These adjacent checks show that 6 is the first whole-number input meeting the strict inequality. A decimal crossing alone is not the requested integer. Review Example 5.15.

18.

Because (x − 5)² ≥ 0, the term −2(x − 5)² is at most zero. Therefore g(x) ≤ 18, with equality at x = 5. Vertex form supplies the answer directly. The maximum input is not the question; report the output 18 rather than 5. Review Example 5.05.

1.7.

The prediction at input 4 is 3.2(4) + 0.5 = 13.3. The residual is 15 − 13.3 = 1.7, positive because the observation lies above the fitted line. Subtracting in the reverse order would reverse the sign and change the interpretation. Review Example 3.06.