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Dominik [7]
3 years ago
12

A spool with 18 feet of ribbon will be cut into 8-inch segments. how many 8-inch segments can be cut from the spool of ribbon?

Mathematics
2 answers:
Kobotan [32]3 years ago
4 0

Answer:

27

Step-by-step explanation:

First, we convert 18 ft to inches.

18 × 12 in. = 216 in.

Now we divide 18 ft written as 216 in. by 8 in.

216 in. / 8 in. = 27

Answer: 27

Slav-nsk [51]3 years ago
4 0
27 is going to be your correct answer for this question.
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8w=120
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Answer: w=15°


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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
3 years ago
Some one factor 3x^2+7x-20
mylen [45]
Split it up
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group
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factor
(3x)(x+4)+(-5)(x+4)
undistribute (x+4) like ab+ac=a(b+c)
(3x-5)(x+4)
6 0
3 years ago
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