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3 years ago

11 th grade algebra .

Mathematics

1 answer:

Ksju [112]3 years ago

7 0

Answer:

Step-by-step explanation:

The inverse of a matrix is the transpose of the cofactor matrix, divided by the determinant. The determinant is (4)(3) -(1)(-5) = 17, eliminating choice B.

In a 2×2 matrix, the transpose of the cofactor matrix swaps the diagonal elements, and negates the off-diagonal elements. That is, the transpose of the cofactor matrix will look like the matrix of B. When that is divided by 17, you get the matrix of answer choice C.

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3 years ago

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Answer:

The null hypothesis is $H_0: \mu = 14$

The alternate hypothesis is $H_a: \mu < 14$

The test statistic is t = -1.95.

The p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the highlighters wrote for less than 14 continuous hours.

Step-by-step explanation:

Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that their highlighters could write continuously for 14 hours.

At the null hypothesis, we test if the mean is 14 hours, that is:

$H_0: \mu = 14$

At the alternate hypothesis, we test if the mean is less than 14 hours, that is:

$H_a: \mu < 14$

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.

14 is tested at the null hypothesis:

This means that $\mu = 14$

X = 13.6 hours, s = 1.3 hours. Sample of 40:

In addition to the values of X and s given, we have that $n = 40$

Test statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{13.6 - 14}{\frac{1.3}{\sqrt{40}}}$

$t = -1.95$

The test statistic is t = -1.95.

P-value:

The p-value of the test is the probability of finding a sample mean lower than 13.6, which is a left tailed test, with t = -1.95 and 40 - 1 = 39 degrees of freedom.

Using a calculator, the p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the highlighters wrote for less than 14 continuous hours.

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3 years ago