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STatiana [176]
3 years ago
13

URGENT PLEASE convert y = -5/6x -9 to standard slope form

Mathematics
1 answer:
SpyIntel [72]3 years ago
3 0

Answer:

6y + 5x = -54  

Step-by-step explanation:

y = m x + b -> slope-intercept form

ax + by = c -> standard form

y = - \frac{5}{6} x - 9

6y = -5x - 54

6y + 5x = -54  

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Consider the following hypothesis test:
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Answer:

a. P-value = 0.039.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. P-value = 0.013.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. P-value = 0.130.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean significantly differs from 100.

Then, the null and alternative hypothesis are:

H_0: \mu=100\\\\H_a:\mu\neq 100

The significance level is 0.05.

The sample has a size n=65.

The degrees of freedom for this sample size are:

df=n-1=65-1=64

a. The sample mean is M=103.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.5.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.5}{\sqrt{65}}=1.4264

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{103-100}{1.4264}=\dfrac{3}{1.4264}=2.103

 

This test is a two-tailed test, with 64 degrees of freedom and t=2.103, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>2.103)=0.039

As the P-value (0.039) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

b. The sample mean is M=96.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11}{\sqrt{65}}=1.3644

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{96.5-100}{1.3644}=\dfrac{-3.5}{1.3644}=-2.565

This test is a two-tailed test, with 64 degrees of freedom and t=-2.565, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t

As the P-value (0.013) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.

c. The sample mean is M=102.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=10.5.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{10.5}{\sqrt{65}}=1.3024

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{102-100}{1.3024}=\dfrac{2}{1.3024}=1.536

This test is a two-tailed test, with 64 degrees of freedom and t=1.536, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>1.536)=0.130

As the P-value (0.13) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.

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Mrs. Myles wants to know if her students prefer multiple choice or short answer tests. What question would be MOST appropriate t
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Answer: The answer is multiple choice questions.

Step-by-step explanation:  Given that Mrs. Myles wants to know if her students prefer multiple choice or short answer tests.

According to me, Mrs. Myles should ask her students multiple choice questions, because in these type of questions, we get information not only about the correct option but it also reminds us what the incorrect options are related to.

So, these questions are better than short answers which are related to only one topic rather than three or four in the queue for multiple choice questions.

Thus, Mrs. Myles should ask her students multiple choice questions.

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