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lesya [120]
3 years ago
10

Aaron bought a new TV for $1,224. He is paying it off over a period of 12 months. How much will Aaron pay per month for his new

TV?
Mathematics
2 answers:
evablogger [386]3 years ago
7 0

Aaron will have to pay $102 a month. Just divide $1,224 by 12!

boyakko [2]3 years ago
3 0

1224 / 12 = 102

He will be paying $102 a month for his TV

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Arc length and area
Kipish [7]

Answer:

Step-by-step explanation:

4 0
3 years ago
Which expression represents the sum of (2x-5y and (x+y) ?
gayaneshka [121]
I have no clue you have to do something like x2-x1 over y1-y2 and you get yoyu answer from that. 
8 0
3 years ago
In 1999, there were 9860 Great Danes registered with the American Kennel Club. The number of
dybincka [34]

Answer:

154,897

Step-by-step explanation:

Labradors were

6997 MORE than 15 TIME Danes

So, first, what is 15 times Danes??

Given 9860 Danes, fifteen times Danes would be:

9860 * 15 = 147,900

Now, 6997 MORE to that, it would be:

147,900 + 6997 = 154,897

Thus,

There were 154,897 registered Labrador Retrievers

8 0
3 years ago
4
frutty [35]

Answer:

The answer is B

Step-by-step explanation:

3 0
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.

7 0
3 years ago
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