This exam was adminstered in January 2023.

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__Algebra 2 January 2023__

__Algebra 2 January 2023__

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. Solve the equation √(49 - 10x) + 5 = 2x algebraically.
*

**Answer: **

Subtract 5 from both sides to isolate the radical. Square both sides. Then solve the quadratic equation. Finally, discard extraneous solutions that were added when you squared the equation.

You do this by checking the answers that you got.

√(49 - 10x) + 5 = 2x

√(49 - 10x) = 2x - 5

49 - 10x = 4x^{2} - 20x + 25

0 = 4x^{2} - 10x - 24

0 = 2x^{2} - 5x - 12

0 = (2x + 3)(x - 4)

2x + 3 = 0 or x - 4 = 0

x = -3/2 or x = 4

Check -3/2: √(49 - 10(-3/2)) + 5 ?= 2(-3/2)

√(49 - 10(-3/2) ?= -3 - 5 = -8

The square root cannot be negative, so **reject** this answer.

Check 42: √(49 - 10(4)) + 5 ?= 2(4)

√(9) + 5 ?= 8

3 + 5 = 8 (check!)

x = 4 is the only solution.

*34. Joette is playing a carnival game. To win a prize, one has to correctly guess which of five equally sized regions a spinner will land on, as shown in the diagram below.
*

She complains that the game is unfair because her favorite number, 2, has only been spun once in ten times she played the game.

State the proportion of 2’s that were spun.

State the theoretical probability of spinning a 2.

The simulation output below shows the results of simulating ten spins of a fair spinner, repeated 100 times.

Does the output indicate that the carnival game was unfair? Explain your answer.

**Answer: **

Empirical probability is

**1/10**because there was 1 positive result in 10 tries.

Theoretical probability is **1/5** because the number 2 covers one-fifth of the area of the wheel. There are five possible outcome and each should be as likely to occur as the others.

The chart shows that in 21 out of 100 simulations a result of 1/10, or 0.10 occured. This is not an unusual outcome.

If you think about it, with 5 possible outcomes, in 10 spins, a result of 2 should happen two times. If a result of 2 only happened one time, or even 3 times, in 10 spins that wouldn't be an unlikely occurrence because variations can be expected.

*35. Graph c(x) = -9(3)*

Describe the end behavior of c(x) as x approaches positive infinity.

Describe the end behavior of c(x) as x approaches negative infinity.

^{x - 4}+ 2 on the axes below.Describe the end behavior of c(x) as x approaches positive infinity.

Describe the end behavior of c(x) as x approaches negative infinity.

**Answer: **

You can use your graphing calculator to get the table of values for where to plot the points. Your graph should look like this one below.

The end behavior of c(x): as x approaches positive infinity, c(x) approaches negative infinity. As x approaches negative infinity, c(x) approaches 2.

*36. The monthly high temperature (°F) in Buffalo, New York can be modeled by B(m) = 24.9sin(0.5m - 2.05) + 55.25, where m is the number of the month and January = 1.*

Find the average rate of change in the monthly high temperature between June and October, to the nearest hundredth.

Explain what this value represents in the given context.

Find the average rate of change in the monthly high temperature between June and October, to the nearest hundredth.

Explain what this value represents in the given context.

**Answer: **

June is the 6th month and October is the 10th month. Find B(6) and B(10). Subtract B(10) - B(6) and divide the difference by (10 - 6), which is 4 to find the average monthly change.

B(6) = 24.9*sin(0.5(6) - 2.05) + 55.25 = 75.5040

B(10) = 24.9*sin(0.5(10) - 2.05) + 55.25 = 59.9915

(B(10) - B(6)) / 4 = (59.9915 - 75.5040) / 4 = -3.878 = -3.88

This means that for each month, the temperature is decreasing by 3.88 degrees.

End of Part III

How did you do?

More to come. Comments and questions welcome.

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