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madam [21]
2 years ago
5

Which equation has a constant of proportionality equal to 8?

Mathematics
1 answer:
tiny-mole [99]2 years ago
8 0
B is the right answer for you question
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Write the equation for the perpendicular bisector of a segment that has endpoints (–1, 10) and (3, 6).
BaLLatris [955]

Answer: I believe it’s y=x+7

Step-by-step explanation:

7 0
3 years ago
Help me pls i have been stuck on this for the past hour
Sedbober [7]

Answer:

-4, -6, -3, -5, -1. The inequality solved for n is n ≥ -6.

Step-by-step explanation:

Substitute all the values in the equation.

n/2 ≥ -3

-10/2 ≥ -3

-5 is not ≥ -3.

n/2 ≥ -3

-7/2 ≥ -3

-3.5 is not ≥ -3.

n/2 ≥ -3

-4/2 ≥ -3

-2 is ≥ -3.

n/2 ≥ -3

-9/2 ≥ -3

-4.5 is not ≥ -3.

n/2 ≥ -3

-6/2 ≥ -3

-3 is ≥ -3.

n/2 ≥ -3

-3/2 ≥ -3

-1.5 is ≥ -3.

n/2 ≥ -3

-8/2 ≥ -3

-4 is not ≥ -3.

n/2 ≥ -3

-5/2 ≥ -3

-2.5 is ≥ -3.

n/2 ≥ -3

-2/2 ≥ -3

-1 is ≥ -3.

To solve the inequality n/2 ≥ -3 for n, do these steps.

n/2 ≥ -3

Multiply by 2.

n ≥ -6.

3 0
3 years ago
John drew a trapezoid ABCD on a coordinate plane. The coordinates are as follows
sleet_krkn [62]

Answer: 5

Step-by-step explanation:

If the sides of A is (8,4) and the coordinates of D is (8-1), the same value in both coordinates are 8 and the different values are 4 (A) and -1 (D). We will take the absolute values (AV) of both when added: |4 + -1|.

The answer will be<u><em> 5 grids</em></u> (or whatever measurements the question uses).

7 0
1 year ago
Read 2 more answers
Can someone please help with 15?
natima [27]
The answer would be c
8 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

            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
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