what is the answer to this question and how??

Mathematics

1 answer:

Lesechka [4]2 years ago

4 0

Answer:

maybe there's a typo, that dot usually represents the multiplication, and I got 5.52

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An English professor assigns letter grades on a test according to the following scheme. A: Top 9% of scores B: Scores below the

sergeinik [125]

Answer:

The minimum score required for an A grade is 83.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 72.3 and a standard deviation of 8.

This means that $\mu = 72.3, \sigma = 8$

Find the minimum score required for an A grade.

This is the 100 - 9 = 91th percentile, which is X when Z has a pvalue of 0.91, so X when Z = 1.34.

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

$1.34 = \frac{X - 72.3}{8}$

$X - 72.3 = 1.34*8$

$X = 83$

The minimum score required for an A grade is 83.

5 0

2 years ago

Check attachment for picture but please help with extraneous solution:

Firlakuza [10]

It's C.!!!!!!!!!!!!!

6 0

2 years ago

Read 2 more answers

If z^2+1/z^2=11 , find the value of z^3-1/z^3,using only the positive value of z-1/z

vredina [299]

Hello,

Let's assume

$t=z- \dfrac{1}{z} \\\\

t^2=z^2-2+ \dfrac{1}{z^2}= 11-2=9==\ \textgreater \ t=3\\\\

t^3=z^3-3z^2* \dfrac{1}{z} +3z* \dfrac{1}{z^2} - \dfrac{1}{z^3} \\\\
=z^3 - \dfrac{1}{z^3} -3(z-\dfrac{1}{z})\\\\
z^3 - \dfrac{1}{z^3}=t^3+3t=3^3+9=27+9=36


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