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liubo4ka [24]
3 years ago
7

} = 12^{x} " alt=" {5}^{2x - 1} = 12^{x} " align="absmiddle" class="latex-formula">
good luck
use indices nooooo log​
Mathematics
1 answer:
nirvana33 [79]3 years ago
3 0

step by step explanation:

Add 1 to both sides

52x−1+1=12x+1

Simplify

52x=12x+1

Subtract 12x from both sides

52x−12x=12x+1−12x

Simplify

40x=1

Divide both sides by 40

40x40 =140 

Simplify

x=140 

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A new phone system was installed last year to help reduce the expense of personal calls that were being made by employees. Befor
Leya [2.2K]

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

3 0
2 years ago
Could someone please help? They don’t need to be solved I just need help with the translating.
Lorico [155]

Answer:

I won't solve i'll translate like you asked

Step-by-step explanation:

13) x - 5 = 21

14) 12 × y = 48

15) b + 10 = 13

If you want I can solve.

HOPE THIS HELPED

5 0
3 years ago
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