**Algebra II**exam, but I've received enough requests that I've decided to post some of the open-ended questions.

Some questions were posted earlier here.

### August 2016, Algebra II Part 2 (continued)

**28.***Using the identity sin ^{2} O + cos^{2}O = 1, find the value of tan O, to the
nearest hundredth, if cos O is –0.7 and O is in Quadrant II.*

Copy the identity. Substitute the value for cosine and then square it. Solve for sine. Remember, because it is in Quadrant II where the y values are positive, we need a positive answer for sine. Once we know sine and cosine, we can divide to find tangent.

sin^{2} ~~O~~ + cos^{2}~~O~~ = 1

sin^{2} ~~O~~ + (-0.7)^{2} = 1

sin^{2} ~~O~~ + .49 = 1

sin^{2} ~~O~~ = 0.51

sin~~O~~ = SQRT(0.51) = .7141...

tan~~O~~ = sin~~O~~ / cos~~O~~ = 0.714 / -0.7 = -1.02

You could also have gotten this answer by converting the identity, by dividing all the terms by cos^{2}~~0~~, which would then give you an equation with tan ^{2}~~O~~ in it. Again, remember that because it is Quadrant II where x is negative and y is positive, that the tangent must be a negative value.

**29.** *Elizabeth waited for 6 minutes at the drive thru at her favorite fast-food restaurant the last time she visited. She was upset about having to wait that long and notified the manager. The manager assured her that her experience was very unusual and that it would not happen again.
*

*A study of customers commissioned by this restaurant found an approximately normal distribution of results. The mean wait time was 226 seconds and the standard deviation was 38 seconds. Given these data, and using a 95% level of confidence, was Elizabeth’s wait time unusual? Justify your answer.
*

A 95% level of confidence means that the value should be within 2 standard deviations from the mean.

The mean is 226. Plus 1 deviation = 226 + 38 = 264. Plus another deviation 264 + 38 = 302 seconds.

Six minutes is 360 seconds, which would be unusual because it is more than two standard deviations away from the mean.

**30.** *The x-value of which function’s x-intercept is larger, f or h? Justify your answer.
*

The x-intercept of h(x) = 2, as shown in the table.

To find the x-intercept of f(x), set f(x) = 0.

log(x - 4) = 0

10^{0} = x - 4

1 = x - 4

5 = x

f(x) has the higher x-intercept.

to be continued.

Any questions?

Comments, corrections welcome.