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kaheart [24]
3 years ago
9

PLEASE HELP!!!

Mathematics
2 answers:
AfilCa [17]3 years ago
7 0
The answers must be B
Ivenika [448]3 years ago
5 0

Answer:

B

Step-by-step explanation:

(Standard form is automatically in integers)

Multiply by 3 to take out the fraction

3y=2x+21

-2x -2x

-2x+3y=21

Standard form means Ax+By=C with positive A (usually--some people allow A to be negative).

Multiply by (-1)

2x-3y=-21

(Technically, that's the more correct answer, but since you don't have that choices, it's just the previous one.)

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The endpoints of GH¯¯¯¯¯¯¯¯ are G(9,−6) and H(−1,8). Find the coordinates of the midpoint M. The coordinates of the midpoint M a
jok3333 [9.3K]

Answer:

( 4, 1 )

Step-by-step explanation:

midpoint formula: (x1 + x2/ 2, y1 + y2/ 2)

(9-1/ 2, -6+8/ 2) -> (4,1)

3 0
3 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

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            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

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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
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IgorLugansk [536]

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or

\sf (6x-2)^2

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3 0
3 years ago
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