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ratelena [41]
3 years ago
8

Which shows another way to write 7^5?

Mathematics
2 answers:
Zigmanuir [339]3 years ago
8 0
Hello!

Exponents are written out as the base times itself by the number the exponent is

7^5 is the same as 7 * 7 * 7 * 7 * 7

The answer is A) 7 * 7 * 7 * 7 * 7

Hope this helps!
Anettt [7]3 years ago
7 0
Option A. Is the correct answer because you are multiplying 7, 5 times!
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A student graphs two equations on a graphing calculator. The graph of the second equation is identical to the graph of the first
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The correct option is A.

The following information should be considered:

The graph of the second equation should be the same to the first equation graph.

So here each and every solution of the first equation should be the solution to the second equation because each and every point is similar.

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X=12/11 for the first one
5 0
3 years ago
I NEED HELP ASAP!!
Oduvanchick [21]

Answer:

Step-by-step explanation:

women=85+32=117

men=70+51=121

total=117+121=238

a. P(woman)=117/238

b. P(woman or summer)=p(woman)+P(summer )-P(woman and summer)

=117/238+155/238-85/238

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7 0
4 years ago
One year Roger had the lowest ERA​ (earned-run average, mean number of runs yielded per nine innings​ pitched) of any male p
baherus [9]

Answer:

(a) The <em>z</em>-score of Roger is -1.56 and the <em>z</em>-score of Amber is -1.13.

(b) Roger had a better year relative to his peers.

Step-by-step explanation:

If X follows N (<em>µ, σ</em>₂), then z=\frac{x-\mu}{\sigma}, is a standard normal variate with mean,      E (Z) = 0 and Var (Z) = 1. That is, Z follows N (0, 1).

Let <em>X</em> = ERA of male pitchers and <em>Y</em> = ERA of ERA of female pitchers.

It is provided that the mean and standard deviation of <em>X</em> are:

\mu_{X}=4.371\\\sigma_{X}=0.787

Also, the mean and standard deviation of <em>Y</em> are:

\mu_{Y}=4.363\\\sigma_{Y}=0.869

ER of Roger is 3.14 and ERA of Amber is 3.38.

(1)

Compute the <em>z</em>-score of Roger as follows:

z=\frac{x-\mu_{X}}{\sigma_{X}}=\frac{3.14-4.371}{0.787}=-1.56

Thus, the <em>z</em>-score of Roger is -1.56.

Compute the <em>z</em>-score of Amber as follows:

z=\frac{x-\mu_{Y}}{\sigma_{Y}}=\frac{3.38-4.363}{0.869}=-1.13

Thus, the <em>z</em>-score of Amber is -1.13.

(2)

Compute the probability of ERA's that are greater than Roger's ERA as follows:

P(X>3.14)=P(\frac{X-\mu}{\sigma}>\frac{3.14-4.371}{0.787})\\=P(Z>-1.56)\\=P(Z

This implies that 94% of the other male pitchers had an ERA higher than 3.14.

Compute the probability of ERA's that are greater than Amber's ERA as follows:

P(Y>3.38)=P(\frac{Y-\mu_{Y}}{\sigma_{Y}}>\frac{3.38-4.363}{0.869})\\=P(Z>-1.13)\\=P(Z

This implies that 87% of the other female pitchers had an ERA higher than 3.38.

As it is provided that the lower the ERA the better the pitcher, then it can be concluded that Roger had a better year relative to his peers.

4 0
4 years ago
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