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bonufazy [111]
3 years ago
12

Evaluate 8 - 4.7y - x, when y = -1 and x = 2. A. -10.7 B. -9.7 C. 9.7 D. 10.7

Mathematics
1 answer:
zysi [14]3 years ago
5 0

The value of the equation is 10.7

Step-by-step explanation:

  • Step 1: Solve the equation when x = 2 and y = -1

⇒ 8 - 4.7y - x = 8 - 4.7 × -1 - 2 = 8 + 4.7 - 2

                     = 12.7 - 2 = 10.7

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A college entrance exam company determined that a score of 2323 on the mathematics portion of the exam suggests that a student i
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a) The null and alternative hypothesis are:

H_0: \mu=23\\\\H_a:\mu> 23

c) Test statistic t=1.53

P-value=0.064

d) The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that students who complete the core curriculum are ready for​ college-level mathematics.  That is that the true score for the group is not significantly higher than 23.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>a) State the appropriate null and alternative hypotheses.</em>

<em>c) Use the P-value approach at the 0.05 level of significance to test the hypotheses in part (a). ldentify the test statistic. (Round to two decimal places as needed.) Identfy the P-value. P-value (Round to three decimal places as needed.) </em>

<em>d) Write a conclusion based on the results. Choose the correct answer below. ? the null hypothesis and claim that there ? sufficient evidence to conclude that the population mean is ? than 20.</em>

This is a hypothesis test for the population mean.

The claim is that students who complete the core curriculum are ready for​ college-level mathematics.

Then, the null and alternative hypothesis are:

H_0: \mu=23\\\\H_a:\mu> 23

The significance level is assumed to be 0.05.

The sample has a size n=150.

The sample mean is M=23.4.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=3.2.

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

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{3.2}{\sqrt{150}}=0.261

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{23.4-23}{0.261}=\dfrac{0.4}{0.261}=1.531

The degrees of freedom for this sample size are:

df=n-1=150-1=149

This test is a right-tailed test, with 149 degrees of freedom and t=1.531, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>1.531)=0.064

As the P-value (0.064) is bigger than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that students who complete the core curriculum are ready for​ college-level mathematics.

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3 years ago
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