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3 years ago

Find the area and round to the nearest hundredth??

Mathematics

1 answer:

bagirrra123 [75]3 years ago

8 0

Answer:

total area = 193.31 m²

Step-by-step explanation:

rectangle area = 7 x 15 = 105 m²

semi circle area = (3.14)(7.5²)/2 = 88.31 m²

total area = 105 m² + 88.31 m² = 193.31 m²

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1) If you borrow $120,000 at an APR of 7% for 25 years, you will pay $848.13 per month. If you

Arlecino [84]

Step-by-step explanation:

If you borrow $120,000 at an APR of 7% for 25 years, you will pay $848.13 per month. If you

borrow the same amount at the same APR for 30 years, you will pay $798.36 per month.

a. What is the total interest paid on the 25-year mortgage? $134,439.

5 0

3 years ago

8+x/-9=9 Please solve this equation and show your work!

Oduvanchick [21]

8+(x/-9)=9 subtract 8

x/-9= 1 multiply by -9

x= -9

7 0

3 years ago

Read 2 more answers

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babymother [125]

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2 years ago

An assembly line produces 20 items per hour. If the line runs 24 hours per day, 7 days week, how many items are made by this mac

Margaret [11]

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7 0

3 years ago

Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0

3 years ago