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PilotLPTM [1.2K]
3 years ago
7

Solve for X and show work

Mathematics
1 answer:
Len [333]3 years ago
7 0
I don’t know how to show how I got it but the answer is X=12.
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A circle has a sector with area 27 and central angle of 120
Alex_Xolod [135]

Answer:

81π

Step-by-step explanation:

-The sum of central angles in a  circle add up to 360°.

-Let x be the area of the circle.

-Given the area of a 120° is 27π, the circles area can be calculated as:

A=27\pi\\\\120\textdegree=27\pi\\360\textdegree=x\\\\\therefore x=\frac{360\textdegree\times 27\pi}{120\textdegree}\\\\=3\times 27\pi\\\\=81\pi

Hence, the area of the circle is 81π

6 0
3 years ago
» What is the product?<br> 0.9 x 100 =
Zepler [3.9K]
The product of this is 90
4 0
3 years ago
Read 2 more answers
Which pair of ratios are equal?6/15 and 3/17 3/4 and 9/11 2/5 and 12/35 or 2/3 and 8/12
Elanso [62]
The last pair.
Because 2*4/3*4=8/12
7 0
4 years ago
The mean MCAT score 29.5. Suppose that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.
Paul [167]

Answer:

We conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

Step-by-step explanation:

We are given that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.2 with a standard deviation of 4.2.

Let \mu = <u><em>population mean score</em></u>

So, Null Hypothesis, H_0 : \mu \leq 29.5      {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}

Alternate Hypothesis, H_A : \mu > 29.5      {means that the students that took the Kaplan tutoring have a mean score greater than 29.5}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because 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 MCAT score = 32.2

            s = sample standard deviation = 4.2

            n = sample of students = 40

So, <u><em>the test statistics</em></u> =  \frac{32.2-29.5}{\frac{4.2}{\sqrt{40} } }  ~  t_3_9

                                    =  4.066

The value of t-test statistics is 4.066.

Now, at 0.05 level of significance, the t table gives a critical value of 1.685 at 39 degrees of freedom for the right-tailed test.

Since the value of our test statistics is more than the critical value of t as 4.066 > 1.685, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

3 0
3 years ago
Simplify (please help)
valkas [14]

Answer:

\frac{8x^{3}y^{6}  }{z^{3} }

Step-by-step explanation:

4 0
3 years ago
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