Given: ABCD is a trapezoid, AC ⊥ CD , AB = CD, AC= square root of 75 , AB = 5 Find: Area of ABCD

What is the length of the radius? startfraction 6 over 85 endfraction startfraction 85 over 12 endfraction startfraction 121 ove

melomori [17]

The length of the radius will be 22 units.

<h3>What is the radius?</h3>

The radius of a circle is defined as the distance of the center of the circle to its outer layer. It is a locus of a point present at some distance from the center point.

Now from the question, we have an expression:

$r=(\dfrac{6}{85} )\times(\dfrac{85}{12} )\times (\dfrac{121}{36} )\times (\dfrac{157}{12} )$

By solving the above equation we get

$r=(\dfrac{18997}{864} )$

$r=22$

Hence the length of the radius will be 22 units.

To know more about radius follow

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Need help with these type of problems & if you answer could you explain

Nuetrik [128]

Answer:

c?

Step-by-step explanation:

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The table below shows the different options available at Kreate-A-Koala Workshop. You decide to randomly select one animal, one

fredd [130]

The probability that you will receive a Monkey with a red shirt & white pants is 3/8

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A bag contains 28 blue and green tiles. The bag contains 2 blue tiles for every 5 green tiles. Which

Alchen [17]

Answer:

I think the answer is 8 blue tiles and 20 green

Step-by-step explanation:

To keep the porportion multiply 2 blue tiles *4

then you need to take 5 green tiles *4

You get 8 blue tiles and 20 Green tiles, and added together you get 28 tiles

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3 years ago

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

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

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

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

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.

6 0

3 years ago