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3 years ago

Chords and intersect in circle O as shown in the diagram. Two of the angle measurements are marked. What is mCED?

Mathematics

2 answers:

jenyasd209 [6]3 years ago

8 0

we know that

The measure of the interior angle is the half sum of the arcs comprising it and its opposite

m∠CED= $\frac{1}{2} *(arc\ AB+arc\ CD)$

see the attached figure with letters to better understand the problem

$arc\ AB=82$ ° ------> by central angle

$arc\ CD=48$ ° ------> by central angle

Substitute the values in the formula above

m∠CED= $\frac{1}{2} *(82+48)$

m∠CED= $\frac{1}{2} *130$

m∠CED= $65$ °

therefore

<u>the answer is</u>

m∠CED= $65$ °

blsea [12.9K]3 years ago

3 0

Most of the information's required to find mCED is already given in the diagram attached with the question. It has a very simple formula.
mCED = (48 + 82)/2
= 130/2
= 65
I hope that this is the answer that you were looking for and it has actually come to your great help.

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$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ 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 check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

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