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RUDIKE [14]
2 years ago
14

Do you guys know the answers??

Mathematics
2 answers:
Mkey [24]2 years ago
7 0

Answer:

Hope this helps!

Step-by-step explanation:

1.  9.38

2. 9.80

3. 11

4. 9.85

5.  15.23

6.  15

7. 9.95

8. 27

9. 26.96

10. 14.14

sergij07 [2.7K]2 years ago
5 0

Step-by-step explanation:

on the picture

if it's helpful ❤❤❤❤❤

THANK YOU.

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What is the solution to the equation 9 cubed x ≈ 7?
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<u>Answer</u>

7/729


<u>Explanation</u>

The question requires to you to find the value of x.

9³x = 7

729 x = 7

x = 7/729

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Solve for x<br> (2x - 3)(x - 2) = 0
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Answer:

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Step-by-step explanation:

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How do you solve this?
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A professor's son, having made the wise decision to drop out of college, has been finding his way in life taking one job or anot
Trava [24]

Answer:

88.88% probability that it endures for less than a year and a half

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 66, \sigma = 10

The next career begins on Monday; what is the likelihood that it endures for less than a year and a half?

One year has 52.14 weeks. So a year and a half has 1.5*52.14 = 78.21 weeks.

So this probability is the pvalue of Z when X = 78.21.

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

Z = \frac{78.21 - 66}{10}

Z = 1.22

Z = 1.22 has a pvalue of 0.8888

88.88% probability that it endures for less than a year and a half

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Step-by-step explanation:

A proper noun is a specific name for a particular person, place, or thing.

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