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Step2247 [10]
3 years ago
10

Which of these division problems has a quotient of -7 2/9?

Mathematics
1 answer:
Lapatulllka [165]3 years ago
3 0

Answer:

- 7 \frac{2}{9}  =   \frac{ - 65}{9}

it's answer B

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Jake tutors students. He charges $10 for the first session and $5 for each additional session. Let x
SIZIF [17.4K]

Answer: y = 5x + 10

Step-by-step explanation:

y= 5x + 10

THINK ABOUT IT! if 5 is for EACH additional session and x is equal to the NUMBER OF SESSIONS...... then 5 times x will give you the cost for each additional session, but NOT INCLUDING THE 10 FOR THE FIRST SESSION. so simply just do 5x + 10 and y is the total cost so

y = 5x + 10

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Solve the system of equations.<br> 3x 2 y = 2<br> x+7y=4
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3 years ago
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Answer:

Step-by-step explanation:

We can obtain that the dimensions of the rectangle are 7 by 5.

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5 0
2 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.

7 0
3 years ago
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