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natali 33 [55]
3 years ago
6

Find the value of y. PLS I NEED HELP WITH THIS FAST

Mathematics
1 answer:
Ugo [173]3 years ago
8 0

Answer:

y =  \sqrt{6}

or

y = 2.4494897428

Step-by-step explanation:

We know value of x is

2 \sqrt{3}

or 3.46410161

Using pythagorean theorem,

{a}^{2} +  {b}^{2}  =  {c}^{2}

{2 \sqrt{3} }^{2} +  {y}^{2} =  {4 \sqrt{3} }^{2}

6 +  {y}^{2}  = 12

Subtract 6 from 12

{y}^{2} = 6

Square root 6 to get value of y

\sqrt{6}

Answer:

y =  \sqrt{6}

or

y = 2.4494897428

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Ben also bought a package of 3 T-shirts for $14.25. What is the unit price, per T-shirt
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The perimeter of a rectangle is 46 cm,<br> The length is 13 cm. What is the width?
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Answer:

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Step-by-step explanation:

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Read 2 more answers
A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.
oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

H_0: \mu=2.4\\\\H_a:\mu> 2.4

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

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

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58

The degrees of freedom for this sample size are:

df=n-1=45-1=44

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>2.5801)=0.0066

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0
3 years ago
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