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melomori [17]
2 years ago
6

Simplify The Following​

Mathematics
1 answer:
shusha [124]2 years ago
6 0

Answer:

i think this would help

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2.33333(repeating)

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Step-by-step explanation:

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Anna wants to take fitness classes. She compares two gyms to determine which would be the best deal for her. Fit Fast charges a
lutik1710 [3]

The number of classes Anna can take so the total cost for the month will be the same is 5.

The monthly cost would be  $37.50.

<h3>When would the total cost be the same?</h3>

When the monthly cost is equal, both equations would be equal: 7.5x = 5.5x + 10

In order to determine the value of x, take the following steps:

  • Combine similar terms: 7.5x - 5.5x = 10
  • Add similar terms: 2x = 10
  • Divide both sides by 2 : 10 /2 = 5

Monthly cost when the cost is the same: 7.5 x 5 = $37.50

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7 0
1 year ago
How do you solve for LMN? I keep getting 15
Marina86 [1]

Answer:

60°

Step-by-step explanation:

90° = 6x

15° = x

60° = 4x

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In 2008 the Better Business Bureau settled 75% of complaints they received (USA Today, March 2, 2009). Suppose you have been hir
Ede4ka [16]

Answer:

Explained below.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 \mu_{\hat p}= p

The standard deviation of this sampling distribution of sample proportion is:

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

(a)

The sample selected is of size <em>n</em> = 450 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0204^{2}).

(b)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

P(p-0.04

                                          =P(-1.96

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.95.

(c)

The sample selected is of size <em>n</em> = 200 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{200}}=0.0306

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0306^{2}).

(d)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

P(p-0.04

                                          =P(-1.31

Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.81.

(e)

The probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 450 is 0.95.

And the probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 200 is 0.81.

So, there is a gain in precision on increasing the sample size.

6 0
2 years ago
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