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2 years ago

37 + 15r > or equal to 127, what is the answer?

Mathematics

1 answer:

Tpy6a [65]2 years ago

4 0

Answer:

r ≥ 6

Step-by-step explanation:

37+15r ≥ 127

-37 -37 37 cancels and you're left with 15r

15r ≥ 90

/ 15 /15 15 cancels and you're left with r

r ≥ 6

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3 years ago

This table represents a proportional relationship. Find the constant of proportionality.

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Answer:

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An animal shelter spends three dollars per day to take care for each bird and seven dollars per day take care for each cat Alexa

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Answer:

There were 16 birds at the shelter on Wednesday.

Step-by-step explanation:

Number of cats and birds on Wednesday = 34

let the number of birds on Wednesday = b

Then number of cats = (34 -b)

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The rate of taking care of each cat in shelter = $7

Total amount spent by shelter on Wednesday = $174

So, according to the question

b x ($3) + (34 -b) ($7) = $ 174

or, 3b + 238 - 7b = 174

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3 years ago

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.

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

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