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dem82 [27]
3 years ago
9

WILL GIVE BRAINLIEST George is building a fence around a rectangular dog run. He is using his house as one side of the run. The

area of the dog run will be 240 square feet. The length of the run is 30 feet, and the width is (30 minus x) feet. The diagram below shows his plan. Recall the formulas for area and perimeter: A = lw and P = 2l + 2w. A rectangle labeled Dog run has a length of 30 feet and width of (30 minus x) feet. How many feet of fencing will George need for the dog run? 22 feet 68 feet 76 feet 82 feet
Mathematics
2 answers:
sdas [7]3 years ago
6 0

Answer:

b68

Step-by-step explanation:

koban [17]3 years ago
4 0

Answer:

68 feet

Step-by-step explanation:

May i plz get brainliest i really need it

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Answer: 100

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bazaltina [42]

Answer:

P(B|W)=0.5

Step-by-step explanation:

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P(B|W)=\frac{P(W)\times P(B)}{P(W)}\\\\=\frac{0.50\times0.29}{0.29}\\\\=0.5

Hence, the conditional probability that a degree is earned by a woman, given that the degree is a bachelor's degree is 0.5

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3 years ago
A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. ​(a) What must be true
valentina_108 [34]

Answer:

(a) The distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b) The value of P(\bar X is 0.7642.

(c) The value of P(\bar X\geq 69.1) is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

\mu=68\\\sigma=15

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

\mu_{\bar x}=\mu=68

And the standard deviation of the distribution of the sample mean is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74

Thus, the distribution of the sample mean (\bar x) is <em>N</em> (68, 4.74²).

(b)

Compute the value of P(\bar X as follows:

P(\bar X

                    =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the value of P(\bar X is 0.7642.

(c)

Compute the value of P(\bar X\geq 69.1) as follows:

Apply continuity correction as follows:

P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)

                    =P(\bar X>69.6)

                    =P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})

                    =P(Z>0.34)\\=1-P(Z

Thus, the value of P(\bar X\geq 69.1) is 0.3670.

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