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3 years ago

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Mathematics

2 answers:

Rus_ich [418]3 years ago

8 0

Answer:

thank you

Step-by-step explanation:

I appreciate it

FrozenT [24]3 years ago

4 0

Painpinaaiisnxns jsjznsjxjamxje sisnxjznx

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You want to put $8,000 in a simple interest account. It has a 2% annual interest rate. How long must you invest that money to ea

Nutka1998 [239]

Answer:

2.5 years

Step-by-step explanation:

Interest= P×R×T÷100

T= 100I÷P×R

T= 100×400÷8000×2

T= 40000÷16000

T= 5÷2

T= 2.5 years

5 0

3 years ago

A cup has a circumference of 28 cm. Approximately what is the diameter?

Ainat [17]

Answer:

The diameter is approximately 8.9 cm

Step-by-step explanation:

The circumference is given by

C = pi *d

28 = pi *d

Divide each side by pi

28/pi = d

We can approximate pi by 3.14

28 /3.14 =d

8.917 = d

The diameter is approximately 8.9 cm

7 0

3 years ago

Calculate the area of the regular pentagon below:

djverab [1.8K]

Assuming the vertex of the triangle shown is the center of the pentagon, and the line segment shown is an altitude of the triangle:

If we join the center of (the circumscribed circle and of) the pentagon to the 5 vertices, 5 isosceles triangles are formed, all congruent to the one shown in the figure. It is clear that these triangles are congruent, so to find the area of the pentagon, we find the area of one of these triangles and multiply by 5.

The base of the triangle is 22.3 in, and the height is 15.4 ins, thus the area of the pentagon is:

5(Area triangle)=5*[(22.3*15.4)/2]=<span>858.55 (square inches).

Answer: </span>858.55 (square inches).

7 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$