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2 years ago

How many digits are there in the repeating block for the decimal equivalent of 3 over 7?

Mathematics

1 answer:

Sindrei [870]2 years ago

7 0

Step-by-step explanation:

$\text{By long division, we know that}$

$\frac37=0.425871425871425871\ldots$

$\text{Observing the pattern, we know the repeating block is }\overline{425871}\text{. Therefore, there are 6 digits in the repeating block}$

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3 years ago

The radius of a circle is 4 centimeters.what is the area

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3 years ago

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A lumber company is making boards that are 2920.0 millimeters tall. If the boards are too long they must be trimmed, and if the

Effectus [21]

Answer:

The value of the test statistic is of 0.22.

Step-by-step explanation:

Our test statistic is:

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the expected mean, $\sigma$ is the standard deviation(square root of the variance) and n is the size of the sample.

A lumber company is making boards that are 2920.0 millimeters tall.

This means that $\mu = 2920$ .

A sample of 12 is made, and it is found that they have a mean of 2922.7 millimeters with a variance of 121.00.

This means that $X = 2922.7, n = 12, \sigma = \sqrt{121} = 11$ . So

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$t = 0.22$

The value of the test statistic is of 0.22.

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2 years ago

PLEASE HELP the formula m=12,000+12,000rt/12t gives keri's monthly loan payment where t is the annual interest rate and t is the

tatiyna

Answer:

The answer is below

Step-by-step explanation:

The formula m = (12,000 + 12,000rt)/12t gives Keri's monthly loan payment, where r is the annual interest rate and t is the length of the loan, in years. Keri decides that she can afford, at most, a $275 monthly car payment. Give an example of an interest rate greater than 0% and a loan length that would result in a car payment Keri could afford. Provide support for your answer.

Answer: Let us assume an annual interest rate (r) = 10% = 0.1. The maximum monthly payment (m) Keri can afford is $275. i.e. m ≤ $275. Using the monthly loan payment formula, we can calculate a loan length that would result in a car payment Keri could afford.

$m=\frac{12000+12000rt}{12t}\\ but\ m\leq275, \ and \ r=10\%=0.1\\275= \frac{12000+12000(0.1)t}{12t}\\275= \frac{12000}{12t} +\frac{12000(0.1)}{12t}\\275= \frac{1000}{t} + 100\\275-100= \frac{1000}{t} \\175= \frac{1000}{t} \\175t = 1000\\t= \frac{1000}{175}\\ t=5.72\ years$

The loan must be at least for 5.72 years for an annual interest rate (r) of 10%

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3 years ago