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Orlov [11]
3 years ago
14

Please help asap! Due tonight!

Mathematics
1 answer:
Ugo [173]3 years ago
5 0
The answer may be 10
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So im stumped on this one
AlekseyPX

Answer:

The Answer is 18 and 774 because 18 + 43 = 61 and 43 X 18 = 774

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4 years ago
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How to find the surface area of a trapezoid
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This is the formula for finding the area of a trapezoid

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Over the summer, Jackie played video games 3 hours per day. When school began in the fall, she only allowed to play video games
erik [133]
So to find this we divide 3 by 1/2 and get 6. So, she is now only at 1/6 of her original time.1/6 is equal to .1666, and to the nearest percent that would be 17%. hope this helps and its not too late!
3 0
3 years ago
Darren wants to estimate the mean age in a population of trees. He'll sample nnn trees and build a 90\%90%90, percent confidence
Alchen [17]

Using the z-distribution, as we have the standard deviation for the population, it is found that the smallest sample size required to obtain the desired margin of error is of 77.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm z\frac{\sigma}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • z is the critical value.
  • n is the sample size.
  • \sigma is the standard deviation for the population.

The margin of error is given by:

M = z\frac{\sigma}{\sqrt{n}}

In this problem, we have that the parameters are given as follows:

M = 3, z = 1.645, \sigma = 16.

Hence, solving for n, we find the sample size.

M = z\frac{\sigma}{\sqrt{n}}

3 = 1.645\frac{16}{\sqrt{n}}

3\sqrt{n} = 1.645 \times 16

\sqrt{n} = \frac{1.645 \times 16}{3}

(\sqrt{n})^2 = \left(\frac{1.645 \times 16}{3}\right)^2

n = 76.97

Rounding up, the smallest sample size required to obtain the desired margin of error is of 77.

More can be learned about the z-distribution at brainly.com/question/25890103

4 0
2 years ago
How do you the distributive property to write an equivalent expression for 2(9+5k).
Nonamiya [84]
2(9) + 2(5k)
18 + 10k
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4 years ago
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