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Dominik [7]
3 years ago
11

In Mrs.Wilson's class, the ratio of girls to total students is 5 to 8.If there are only 24 total students, ​how many boys are in

the class.
Mathematics
1 answer:
Sergeu [11.5K]3 years ago
7 0

Answer: 15


Step-by-step explanation:  24/3=3

3(5)=15 girls


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A software company decided to conduct a survey on customer satisfaction. Out of 564 customers who participated in the online sur
nikklg [1K]

Answer:

z=\frac{0.0904 -0.1}{\sqrt{\frac{0.1(1-0.1)}{564}}}=-0.760 \approx -0.8  

p_v =2*P(z  

Step-by-step explanation:

Information given

n=564 represent the sample selected

X=51 represent the number of people who rated the overall services as poor

\hat p=\frac{51}{564}=0.0904 estimated proportion of people who rated the overall services as poor  

p_o=0.1 is the value to compare

z would represent the statistic

Hypothsis to analyze

We want to analyze if the proportion of customers who would rate the overall car rental services as poor is 0.1, so then the system of hypothesis are:  

Null hypothesis:p=0.1  

Alternative hypothesis:p \neq 0.1  

The statistic for a one z test for a proportion is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing the info given we got:

z=\frac{0.0904 -0.1}{\sqrt{\frac{0.1(1-0.1)}{564}}}=-0.760 \approx -0.8  

And the p value since we have a bilateral test is given b:

p_v =2*P(z  

3 0
3 years ago
Easy 6th grade math! Please help!
Pie

Answer:

Step-by-step explanation 10 + 2.50d = 850

David starts with $10 and each day he earns $2.50. The d represents a variable and can be any number.

To find d you must solve the equation.

10 + 2.50d = 850

2.50d = 840 (subtract 10 from 850)

d = 336 (divide 840 by 2.50)

It will take David 336 days to save $850.

3 0
3 years ago
Which steps would you use to solve this equation? Check all that apply.
Dovator [93]

Answer: The answers are

  1. Subtract 7 from both sides of the equation.
  2. Subtract 7 from 7.
  3. Substitute 5 for p to check the solution.

Step-by-step explanation:

p + 7 = 12

=   p + 7 - 7 = 12 - 7

=   p = 5

Then for the substituting part

5 + 7 = 12

Hope this helps!

3 0
3 years ago
It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

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.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So \mu = 375, \sigma = 68

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So n = 6, s = \frac{68}{\sqrt{6}} = 27.76

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 375}{27.76}

Z = 1.26

Z = 1.26 has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So \mu = 522, \sigma = 106

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So n = 6, s = \frac{106}{\sqrt{6}} = 43.27

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 523}{43.27}

Z = -2.61

Z = -2.61 has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0
3 years ago
This is an answer not a question btw if u have this same question<br> 4 2/3 divided by 1/2 = 9 1/3
Reika [66]
Thanks for the answer
7 0
3 years ago
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