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GalinKa [24]
3 years ago
9

Find four consecutive odd integers whose sum is 8

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
8 0

Answer:

4x + 12 = 8; 4x = -4; x = -1. Therefore the four consecutive odd integers are -1, 1, 3, 5.

Step-by-step explanation:

Svetradugi [14.3K]3 years ago
5 0

Answer:

<h3>             -1, 1, 3, 5</h3>

Step-by-step explanation:

k  - an integer

so:

2k+1  - an odd integer

2k+1+2 = 2k+3  - an odd integer consecutive to 2k+1

2k+3+2 = 2k+5  - next consecutive odd integer

2k+5+2 = 2k+7  - the fourth consecutive odd integer

2k+1 + 2k+3 + 2k+5 + 2k+7 = 8

               8k + 16 = 8

                    - 16      -16

                  8k = - 8

                  ÷8     ÷8

                  k = -1

2k+1 = 2(-1) + 1 = -2 + 1 = -1

2k+3 = 2(-1) + 3 = -2 + 3 = 1

2k+5 = 2(-1) + 5 = -2 + 5 = 3

2k+7 = 2(-1) + 7 = -2 + 7 = 5

Check:  -1+1+3+5 = 3+5 = 8

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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

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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.

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3 years ago
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