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Which ratio is equivalent to 5:9? 9:45 25:81 27:15 30:54

barxatty [35]

Sorry but none are correct

3 0

3 years ago

Read 2 more answers

2. Find each missing term. p:4 = 9:12

sveta [45]

Answer:

27

Step-by-step explanation:

4×3=12

9×3=27 if a fraction

4/ would be on the bottom 9/12

7 0

3 years ago

What has the same value as 30%, percent of 18?

Ne4ueva [31]

Answer:

Answer: 60.

Step-by-step explanation:

30 percent (calculated percentage %) of what number equals 18?

7 0

3 years ago

please help me with a math question. Do not report please. I need a good grade for chirstmas. 50 points.

Sliva [168]

Answer: I would think

6-0

12-1

30-2

Step-by-step explanation:

8 0

3 years ago

Consider the following hypothesis test:

postnew [5]

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.

8 0

3 years ago