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zloy xaker [14]
3 years ago
6

Please help with this AP Calculus question!

Mathematics
1 answer:
mario62 [17]3 years ago
5 0

9514 1404 393

Answer:

  B.  -1

Step-by-step explanation:

A graph can show you that the expression approaches -1 as x approaches 2π.

The limit is -1.

__

The expression evaluated at x=2π is 0/0, an indeterminate form. This suggests that L'Hopital's rule is applicable. Applying that, we have ...

  lim = (f'(2π) -f'(x))/(x-2π)' = -cos(x)/1

Evaluated at x=2π, this is ...

  lim = -1

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Using the normal distribution, it is found that there was a 0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

\mu = 700, \sigma = 50.

The probability of a month having a PCE between $575 and $790 is the <u>p-value of Z when X = 790 subtracted by the p-value of Z when X = 575</u>, hence:

X = 790:

Z = \frac{X - \mu}{\sigma}

Z = \frac{790 - 700}{50}

Z = 1.8

Z = 1.8 has a p-value of 0.9641.

X = 575:

Z = \frac{X - \mu}{\sigma}

Z = \frac{575 - 700}{50}

Z = -2.5

Z = -2.5 has a p-value of 0.0062.

0.9641 - 0.0062 = 0.9579.

0.9579 = 95.79% probability of a month having a PCE between $575 and $790.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

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The coordinates of the vertices of triangle ABC are A(9.-10), B(-15, 0), and C(15, 10). A dilation centered at (0, 0) is perfome
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Answer: 1.5 (D)
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