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2 years ago

I just have one question left you don’t need to simplify if you give the answer

Mathematics

2 answers:

Lemur [1.5K]2 years ago

8 0

Photomath will help u

Ugo [173]2 years ago

7 0

It is either 400 or negative 400 because if you devide 12 by .03 it is 400

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Find the missing number in this proportion.<br><br> 7 / 35 = ? / 28

Phantasy [73]

using proportional law 5.6

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3 years ago

How to simplify y = 34 - 0.5x

Tems11 [23]

Answer:

y = 0.5

Step-by-step explanation:

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2 years ago

Anyone knows the answer to these questions?

11111nata11111 [884]

So take the given value for question one, 210, and realize that it increases by 10%. What is 10% of 210? If you don't know, basically 10% = 0.1 so multiply 210 by .1 and add that to 210. leave a comment if you can't figure it the answer to either of them and I'll help more.

7 0

3 years ago

Read 2 more answers

Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and

Naddika [18.5K]

Answer:

Using a 90% confidence level

A. A sample size of 68 should be used.

B. A sample size of 98 should be used.

Step-by-step explanation:

I think there was a small typing mistake and the confidence level was left out. I will use a 90% confidence level.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find the margin of error M as such

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

A. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 72 seconds, what sample size should be used?

We have the standard deviation in minutes, so the margin of error should be in minutes.

72 seconds is 72/60 = 1.2 minutes.

So we need a sample size of n, and n is found when M = 1.2. We have that $\sigma = 6$ . So

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

$1.2 = 1.645*\frac{6}{\sqrt{n}}$

$1.2\sqrt{n} = 6*1.645$

$\sqrt{n} = \frac{6*1.645}{1.2}$

$(\sqrt{n})^{2} = (\frac{6*1.645}{1.2})^{2}$

$n = 67.65$

Rounding up.

A sample size of 68 should be used.

B. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 1 minute, what sample size should be used?

Same logic as above, just use M = 1.

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

$1 = 1.645*\frac{6}{\sqrt{n}}$

$\sqrt{n} = 6*1.645$

$(\sqrt{n})^{2} = (6*1.645)^2$

$n = 97.42$

Rounding up

A sample size of 98 should be used.

3 0

3 years ago

CORRECT ANSWER GETS BRAINLIEST

Kisachek [45]

Answer:

$\frac{1}{18}$

Step-by-step explanation:

The problem is asking for the fraction to be divided by a whole number. This is equal to multiplying a fraction by the reciprocal of the whole number.

$\frac{1}{3} /6$

$\frac{1}{3} *\frac{1}{6}$

Then you can multiply like a regular multiplication problem

$\frac{1}{18}$

The final answer would be $\frac{1}{18}$

8 0

3 years ago