June 2023 Algebra 2, Part II

This exam was adminstered in June 2023.

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Algebra 2 June 2023

Part II: Each correct answer will receive 2 credits. Partial credit can be earned. One mistake (computational or conceptual) will lose 1 point. A second mistake will lose the other point. It is sometimes possible to get 1 point for a correct answer with no correct work shown.


25. The business office of a local college wishes to determine the methods of payment that will be used by students when buying books at the beginning of a semester. Explain how the office can gather an appropriate sample that minimizes bias.

Answer:

The college should take a random sample survey from the list of students.

Your answer should imply sort of randomization and include an appropriate sample.

26. Determine the solution of √(3x + 7) = x - 1 algebraically.

Answer:

Square both sides of the equation to get rid of the radical and then solve the equation that results from the exponents being equal.

√(3x + 7) = x - 1
3x + 7 = x² - 2x + 1
0 = x² - 5x - 6
0 = (x - 6)(x + 1)
x = 6 or x = -1

Check both answers:

√(3(6) + 7) = √(25) = 5; 6 - 1 = 5, Check.
√(3(-1) + 7) = √(4) = 2; -1 - 1 = -2. discard as extraneous.

x = 6 is the only answer.



27. The population of bacteria, P(t), in hundreds, after t hours can be modeled by the function P(t) = 37e^0.0532t . Determine whether the population is increasing or decreasing over time. Explain your reasoning.

Answer:

You can graph this with your calculator and state that the graph is rising so the population is increasing. Or you can evaluate e^0.0532t.

Since e^0.0532t = 1.0546..., which is greater than 1, the population is increasing.

28. The polynomial function g(x) = x³ + ax² - 5x + 6 has a factor of (x - 3). Determine the value of a.

Answer:

Set the function equal to 0, substitute x = 3, and solve for a.

g(3) = (3)³ + a(3)² - 5(3) + 6 = 0
27 + a(9) - 15 + 6 = 0
9a = - 18
a = -2

29. Write a recursive formula for the sequence 189, 63, 21, 7, … .

Answer:

A recursive function needs and a₁ value followed by a_n defined in terms of a_n-1.

a₁ = 189
a_n = (1/3) a_n-1

30. Solve algebraically for x to the nearest thousandth:

2e^0.49x = 15

Answer:

Solve using inverse operations, including natural logs.

2e^0.49x = 15
e^0.49x = 15/2
ln(e^0.49x) = ln(15/2)
0.49x = 2.01490...
x = 4.112

31. For all values of x for which the expression is defined, write the expression below in simplest form.

(2x³ + x² - 18x - 9) / (3x - x²)

Answer:

Factor the numerator and the denominator and cancel out the common factors.

(2x³ + x² - 8x - 9) / (3x - x²)

( x²(2x + 1) - 9(2x + 1) ) / ( (x)(3 - x) )

( (x²- 9)(2x + 1) ) / ( (x)(3 - x) )

( (x + 3)(x - 3)(2x + 1) ) / ( (x)(-1)(x - 3) )

( (x + 3)(2x + 1) ) / ( (-x) )

32. An app design company believes that the proportion of high school students who have purchased apps on their smartphones in the past 3 months is 0.85. A simulation of 500 samples of 150 students was run based on this proportion and the results are shown below.

Suppose a sample of 150 students from your high school showed that 88% of students had purchased apps on their smartphones in the past 3 months. Based on the simulation, would the results from your high school give the app design company reason to believe their assumption is incorrect? Explain.

Answer:

Find the confidence interval by multiplying the standard deviation (SD) by 2. Subtract that from the mean to get the lower bound of the interval and add it to the mean to get the upper bound. See if 88% falls within that interval.

(2)(0.029) = 0.058

Lower bound = 0.852 - 0.058 = 0.794. Upper bound = 0.852 + 0.058 = 0.910.

Since 88% falls inside the confidence interval, the results would not give the company reason to believe that their assumption is incorrect.

End of Part II

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