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klio [65]
3 years ago
12

Please help! How would you find the relative frequency of being female and attending an action movie?

Mathematics
2 answers:
denpristay [2]3 years ago
6 0
Is this a STARR test? 
A. Divide 99 by 250
katen-ka-za [31]3 years ago
3 0
A)250÷9

hope this help
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A man jogs from his home to the nearst market 25km away jogs at an average speed 5km 1h. how long does it taken him to reach the
mr_godi [17]

Answer:

1/5

Step-by-step explanation:

first write the formula of average speed

  • 2) then equate the values of given question
  • 3) you will find the solution

8 0
2 years ago
A or B or C or D <br> ???????????
mrs_skeptik [129]

Answer:

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3 0
3 years ago
What is the area of this figure?<br> 6 mi<br> 5 mi<br> 2 mi<br> 2 mi<br> 11 mi<br> 9 mi<br> ורן 7
Alborosie
I can’t see the image maybe if you redo it
5 0
3 years ago
How large a sample should be selected to provide a 95% confidence interval with a margin of error of 6? Assume that the populati
marshall27 [118]

Using the z-distribution, it is found that a sample of 171 should be selected.

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

The confidence interval is:

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

The margin of error is:

M = 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.

For this problem, the parameters are:

z = 1.96, \sigma = 40, M = 6

Hence we solve for n to find the needed sample size.

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

6 = 1.96\frac{40}{\sqrt{n}}

6\sqrt{n} = 40 \times 1.96

\sqrt{n} = \frac{40 \times 1.96}{6}

(\sqrt{n})^2 = \left(\frac{40 \times 1.96}{6}\right)^2

n = 170.7.

Rounding up, a sample of 171 should be selected.

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

#SPJ1

4 0
2 years ago
Write this number in expanded form 73,489
mart [117]

the answer is 70,000+3,000+400+80+9

3 0
3 years ago
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