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otez555 [7]
3 years ago
12

HELPPP ASAPPPPPPPPPPA triangle has sides that measure 4 units, 6 units, and 7.21 units. What is the area of a circle with a circ

umference that equals the perimeter of the triangle? Use 3.14 for pie, and round your answer to the nearest whole number.
17 units ^2
24 units ^2
54 units^2
94 units^ 2
Mathematics
1 answer:
Zepler [3.9K]3 years ago
7 0

Answer:

24 units

i took the test and got it right

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the answer is C

Step-by-step explanation:

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The equation of a line is X+3Y=14<br> what is the y-intercept of the line?
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X+3y=14
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Find m (2x+14)<br> (X+7)
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Answer:

=

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Ⓗⓘ ⓣⓗⓔⓡⓔ

Well, assuming m of angle ABD is 90 degrees, then you would add and then solve.

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4 years ago
How many centimeters are in 5 inches? 1in. = 2.54 cm<br><br> 5 in. = ?
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6 0
3 years ago
The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The follow
kipiarov [429]

Answer:

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

Step-by-step explanation:

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.

At the null hypothesis, we test if the mean cost is still the same, that is:

H_0: \mu = 13252

At the alternative hypothesis, we test if the mean cost has increased, that is:

H_1: \mu > 13252

The test statistic is:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

13252 is tested at the null hypothesis:

This means that \mu = 13252

The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.

This means that n = 20, X = 15560, s = 3500

Value of the test statistic:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{15560 - 13252}{\frac{3500}{\sqrt{20}}}

t = 2.95

P-value of the test and decision:

The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

7 0
3 years ago
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