## Tuesday, March 12, 2024

### January 2024 Algebra 2 Regents, Part III

This exam was adminstered in January 2024.

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### Algebra 2 January 2024

Part III: Each correct answer will receive 4 credits. Partial credit can be earned. One computational mistake will lose 1 point. A conceptual error will generally lose 2 points (unless the rubric states otherwise). It is sometimes possible to get 1 point for a correct answer with no correct work shown.

33. A researcher wants to determine if nut allergies and milk allergies are related to each other. The researcher surveyed 1500 people and asked them if they are allergic to nuts or milk. The survey results are summarized in the table below.

Determine the probability that a randomly selected survey respondent is allergic to milk
Determine the probability that a randomly selected survey respondent is allergic to milk, given that the person is allergic to nuts.
Based on the survey data, determine whether nut allergies and milk allergies are independent events. Justify your answer.

Add up the rows and columns. You will see that that it adds to 1500, as stated in the problem.

There are 45 respondents out of 1500 who are allergic to milk, so the probability is 45/1500. (You don't need to simplify the fraction.)

There are 15 people who are allergic to nuts, and of those, 3 are also allergic to milk, so the probability is 3/15.

If the nut allergies and milk allergies are independent, then the previous two answers would be the same because P(A) would have to be equal to P(A|B). However, 45/1500 = 0.03 and 3/15 = 0.2. So the events are not independent.

34. Algebraically solve for x: 2x = 6 + 2√(x - 1)

Isolate the radical. Then square both sides. Finally, solve the quadratic equation.

2x = 6 + 2√(x - 1)
2x - 6 = 2√(x - 1)
x - 3 = √(x - 1)
(x - 3)2 = (√(x - 1))2
x2 - 6x + 9 = x - 1
x2 - 7x + 10 = 0
(x - 5)(x - 2) = 0
x - 5 = 0 or x - 2 = 0
x = 5 or x = 2

Throw out x = 2 as extraneous because 2(2) =/= 6 + 2√(2-1).

The only solution is x = 5.

35. During the summer, Adam saved \$4000 and Betty saved \$3500. Adam deposited his money in Bank A at an annual rate of 2.4% compounded monthly. Betty deposited her money in Bank B at an annual rate of 4% compounded quarterly. Write two functions that represent the value of each account after t years if no other deposits or withdrawals are made, where Adam’s account value is represented by A(t), and Betty’s by B(t).

Using technology, determine, to the nearest tenth of a year, how long it will take for the two accounts to have the same amount of money in them. Justify your answer.

Write the functions A(t) and B(t) using the given initial amounts and rates. Note that that A is compounded monthly, so 0.024 will be divided by 12 and the exponent will be multiplied by 12. Likewise, B is compounded quarterly, so 0.04 will be divided by 4 and the exponent will be multiplied by 4.

A(t) = 4000(1 + 0.024/12)12t
or A(t) = 4000(1.002)12t

B(t) = 3500(1 + 0.04/4)12t
or B(t) = 3500(1.01)4t

"Using technology" means that you can graph the two functions to see when they intersect rather than writing an equation and solving. Put the two equations into your graphing calculator. Graph and trace the functions, or look at the table of values, setting the calculator to show every 0.1 value of x.

At t=8.4, A(8.4) = 4892.40 and B(8.4) = 4889.50, a difference of \$3.10, which is the smallest difference to the nearest tenth of year.

36.
On the graph below, draw at least one complete cycle of a sine graph passing through point (0,2) that has an amplitude of 3, a period of p, and a midline at y = 2.
Based on your graph, state an interval in which the graph is increasing.

The period of π instead of 2π means that means we need sin(2x) instead of sin(x). The amplitude of 3 is a mulitplier in front of the function and the midline of 2 is an addition after the function.

y = 3 sin(2x) + 2

The graph would look like the one below. The y-intercept is at (0,2). The maximum value is 5 and the minimum is -1.

One interval where it is increasing would be from 3π/4 to 5π/4.

End of Part III

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