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3 years ago

Helpppppppppppppppppppppp

Mathematics

1 answer:

Vinil7 [7]3 years ago

6 0

Answer:
7x (x^2 * y)^1/3 which is the last option

Explanation:
5x (x^2 * y)^1/3 + 2 (x^5 * y)^1/3
5x (x^2/3 * y^1/3) + 2 (x^5/3 * y^1/3)
5 * x^5/3 * y^1/3 + 2 * x^5/3 * y^1/3
(5+2) * x^5/3 * y^1/3
7* x^5/3 * y^1/3
which can be rewritten as:
7x (x^2 * y)^1/3 which is the last option

Hope this helps :)

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3 years ago

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3 years ago

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Answer:

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(b) p-value = 0.001. Null hypothesis is rejected.

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Step-by-step explanation:

To test for the significance of the population mean from a Normal population with unknown population standard deviation a t-test for single mean is used.

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If the p - value is less than the significance level then the null hypothesis will be rejected. And if the p-value is more than the value of α then the null hypothesis will not be rejected.

(a)

The alternate hypothesis is:

Hₐ: μ > μ₀

The sample size is, n = 11.

The test statistic value is, t = 1.91 ≈ 1.90.

The degrees of freedom is, (n - 1) = 11 - 1 = 10.

Use a t-table t compute the p-value.

For the test statistic value of 1.90 and degrees of freedom 10 the p-value is:

The p-value is:

P (t₁₀ > 1.91) = 0.043.

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The null hypothesis is rejected at 5% level of significance.

(b)

The alternate hypothesis is:

Hₐ: μ < μ₀

The sample size is, n = 17.

The test statistic value is, t = -3.45 ≈ 3.50.

The degrees of freedom is, (n - 1) = 17 - 1 = 16.

Use a t-table t compute the p-value.

For the test statistic value of -3.50 and degrees of freedom 16 the p-value is:

The p-value is:

P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.

The p-value = 0.001 < α = 0.05.

The null hypothesis is rejected at 5% level of significance.

(c)

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Hₐ: μ ≠ μ₀

The sample size is, n = 7.

The test statistic value is, t = 0.83 ≈ 0.82.

The degrees of freedom is, (n - 1) = 7 - 1 = 6.

Use a t-table t compute the p-value.

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The test statistic value is, t = 2.13 ≈ 2.12.

The degrees of freedom is, (n - 1) = 28 - 1 = 27.

Use a t-table t compute the p-value.

For the test statistic value of 0.82 and degrees of freedom 6 the p-value is:

The p-value is:

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The p-value = 0.444 > α = 0.05.

The null hypothesis is rejected at 5% level of significance.

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4 years ago