June 2024 Algebra 2 Regents Part III

This exam was adminstered in January 2024 .

June 2024 Algebra 2, Part III

Each correct answer is worth up to 4 credits. Partial credit can be given. Work must be shown or explained.

33. Solve the following system of equations algebraically for all values of x, y, and z:

3x - 8y + 2z = -60
2x - 7y - 5z = -31
-6x + 2y - 4z = 36

Answer:

You can solve a system of three equations with three variables by eliminating one or more variables through multiplication and elimination.

In this example, if you look at the first and third equations, you will see that -6x is 3x * -2 and -4z is 2x * -2, so if you multiply the first equation by 2, you can eliminate two variables at once.

3x - 8y + 2z = -60
-6x + 2y - 4z = 36

6x - 16y + 4z = -120
-6x + 2y - 4z = 36

-14y = -84
y = 6

Once you know that y = 6, you can substitute into the original equations and solve for the other two variables.

3x - 8(6) + 2z = -60 --> 3x + 2z = -12
2x - 7(6) - 5z = -31 --> 2x - 5z = 11
-6x + 2(6) - 4z = 36 --> -6x - 4z = 24

3x + 2z = -12
2x - 5z = 11
6x + 4z = -24
-6x + 15z = -33

19z = -57
z = -3

3x - 8(6) + 2(-3) = -60
3x - 48 - 6 = -60
3x = -6
x = -2

You can check your answers by substituting your solutions into the other equations and getting true statements.

34. In the town of Skaneateles, New York, house prices since 2008 have changed based on the function H(t) = 200,000(1.045)^t, where t is the number of years since 2008 and H(t) is the median house price. Determine the average rate of change for the median house price in Skaneateles from 2010 to 2018 to the nearest dollar per year.

Explain what this rate of change means as it relates to median house prices.

Answer:

THe average rate of change is the difference between H(10) and H(2) divided by 8 years (2018 - 2010).

(200000(1.045)¹⁰ - 200000(1.045)²) / 8 = 11523.610...

To the nearest dollar, the average rate of change is $11524. (Make sure you round UP.)

This means that the median home price rose about $11524 every year from 2010 to 2018.

35. IA sporting goods manufacturer is trying to determine if they should continue to produce multiple types of hockey pucks. The company surveyed 50 randomly chosen customers and asked them if they purchased both game regulation pucks and lighter training pucks. Of those surveyed, 40 of them said that they purchase both types of pucks. A simulation that was run 100 times based on the survey results produced the approximately normal results below

a) Determine an interval containing the middle 95% of plausible values that estimates the proportion of all customers who would purchase both types of pucks from the company.

b) The company will continue to manufacture both types of hockey pucks if it is reasonable to assume that the true proportion of customers who buy both types of hockey pucks is above 0.60. Using the interval from part a, explain whether or not the company should continue to produce both types of hockey pucks.

Answer:

The middle 95% of the data will fall within two Standard Deviations (SD) from the mean. This will leave 2.5% of the data on the low end and 2.5% of the data on the high end. The Mean is 0.795 and the SD is 0.085.

This means that 95% of the data falls within the range:

0.795 - 2(0.085) = 0.625
0.795 + 2(0.085) = 0.965

The range is 0.625 - 0.965

The company should continue to make both types because 0.625, which is the bottom of the confidence interval, is greater than .60.

36. Graph y = f(x), where f(x) = log₂(x - 1) + 3 on the set of axes below.

State the equation of the asymptote of f(x).

When f(x) is reflected over the line y = x, a new function is formed: g(x) = 2^{(x - 3)} + 1. State the equation of the asymptote of g(x).

Answer:

Put the equation intop your graphing calculator and check the table of values. You will get the points (2,3), (3,4), (5,5), (9,6). You will see that there is a vertical asymptote at x = 1. You will see that the graph crosses the x-axis at a number a little more than x = 1 (at x = 1.125).

The asymptote is x = 1.

If the function is reflection over the line y = x, then the asymptote of g(x) will be y = 1 because g(x) is an exponential function, which would have an asymptote of y = 0 but there is a constant of +1 which raises it up to y = 1.

End of Part III

How did you do?

Questions, comments and corrections welcome.

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