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3 years ago

0.6 3/5 jjkiijgthksjdj

Mathematics

1 answer:

olasank [31]3 years ago

3 0

Answer:

0.36

Step-by-step explanation:

0.6*3/5= 0.36

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Distance between 20 and -11

Mama L [17]

$20-(-11)=20+11=31$

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3 years ago

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2. Mr. Frank tyses 12 emails per<br> day. How long will it take him to<br> write 60 emails?

Nezavi [6.7K]

Answer:

5 days

Step-by-step explanation:

We can use a ratio to solve

12 emails 60 emails

--------------- = --------------

1 day x days

Using cross products

12*x = 1 * 60

12x = 60

Divide each side by 12

12x/12 = 60/12

x =5

It will take him 5 days

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3 years ago

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iren [92.7K]

Answer:

-48

Step-by-step explanation:

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3 years ago

Michael and Imani go out to eat for lunch. Part A If their food and beverages cost $25.30 and there is an 8% meals tax, how much

bagirrra123 [75]

a: 27.324 (rounded to the nearest cent is $27.32)

b: 18.0088% (rounded to the nearest percent is 18.01% or 18%)

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3 years ago

There is no prior information about the proportion of Americans who support Medicare-for-all in 2019. If we want to estimate 95%

Kay [80]

Answer:

32 randomly selected Americans must be surveyed

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

If we want to estimate 95% confidence interval for the true proportion of Americans who support Medicare-for-all in 2019 with a 0.175 margin of error, how many randomly selected Americans must be surveyed?

n randomly selected Americans must be surveyed. n is found when M = 0.175.

We have no prior estimate, so we work with the worst case scenario, which is $\pi = 0.5$ .