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3 years ago

please help asap!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Phantasy [73]3 years ago

5 0

Answer:

Area = 7.853981 in^2 (2.5 Pi)

Perimeter = 11.42478 in (3.63Pi)

Step-by-step explanation:

Area

Lg semi circle area = Pi(2^2)(0.5) = 3.141593(4)0.5) = 6.283185 in^2

Sm semi circle area = Pi(1^2)(0.5) = 3.141593(2)(0.5) = 1.570796 in^2

added together = 7.853981 in^2

Perimeter

Lg semi circle perimeter = Pi(4)(0.5) + 2 = 8.283185 in

Sm semi circle perimeter = Pi(2)(0.5) = 3.141593 in

added together = 11.42478 in

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Answer:

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$

$p_v =P(t_9>10.349)=1.34x10^{-6}$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=51.6$ represent the sample mean

$s=1.1$ represent the standard deviation for the sample

$n=10$ sample size

$\mu_o =48$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean score is higher than 48, the system of hypothesis would be:

Null hypothesis: $\mu \geq 48$

Alternative hypothesis: $\mu > 48$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$

Calculate the P-value

First we find the degrees of freedom:

$df=n-1=10-1=9$

Since is a one-side upper test the p value would be:

$p_v =P(t_9>10.349)=1.34x10^{-6}$

Conclusion

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3 years ago