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sesenic [268]
3 years ago
14

I need the answer and if you would please explain step by step

Mathematics
1 answer:
emmasim [6.3K]3 years ago
3 0

Answer:

x=17

Step-by-step explanation:

well the triangles are congruent so that equation has to equal 61, so set it equal to 61, once you do that subtract 10 with 61 giving you 51, then divide 3 with 51 and you get 17

3x + 10 = 61 \\ 3x = 51 \\ x = 17

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The expression with negative exponent is not a polynomial.

Expression in the denominator is also not a polynomial

x^2-1 has two terms without any negative exponent. so it is a polynomial.

option B,C and D has negative exponents, so it is not a polynomial.

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2 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}}

\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
2 years ago
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