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AleksandrR [38]
3 years ago
14

BRAINLIEST ✨

Mathematics
1 answer:
inna [77]3 years ago
7 0

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, or</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, or</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, or</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.Check: (2 + 2/3), (2 + 3*2/3), (2 + 6*2/3)</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.Check: (2 + 2/3), (2 + 3*2/3), (2 + 6*2/3)= 8/3, 4, 6</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.Check: (2 + 2/3), (2 + 3*2/3), (2 + 6*2/3)= 8/3, 4, 6Common ratio: 4/(8/3) = 3/2 same as 6/4 = 3/2. Correct.</em>

Step-by-step explanation:

Am I right? just commet if I'm wrong

then <em>TANKYOU>^_^<!</em>

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A sample of 900900 computer chips revealed that 66f% of the chips fail in the first 10001000 hours of their use. The company's p
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No, there is not enough evidence at the 0.02 level to support the manager's claim.

Step-by-step explanation:

We are given that the company's promotional literature states that 68% of the chips fail in the first 1000 hours of their use. Also, a sample of 900 computer chips revealed that 66% of the chips fail in the first 1000 hours of their use.

And the quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage, i.e;

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Alternate Hypothesis, H_1 : p  0.68 {means that the actual percentage that fail is different from the stated percentage}

The test statistics we will use here is;

     T.S. = \frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } } ~ N(0,1)

where, p = actual percentage of chip fail = 0.68

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           n = sample size = 900

So, Test statistics = \frac{0.66 -0.68}{\sqrt{\frac{0.66(1- 0.66)}{900} } }

                             = -1.267

Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have sufficient evidence to accept null hypothesis.

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