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4 years ago

15

during the _____ stage, the analyst will look for patterns of responses to the survey questions. a. data editing b. data collect

ion c. data analysis d. sampling

1 answer:

Masja [62]4 years ago

8 0

C. data analysis. he would be analysing the patterns of responses.

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A circle with radius 3 has a sector with a central angle of 1/9 pi radians

marissa [1.9K]

Complete question:

A circle with radius 3 has a sector with a central angle of 1/9 pi radians

what is the area of the sector?

Answer:

The area of the sector = $\frac{\pi}{2}$ square units

Step-by-step explanation:

To find the area of the sector of a circle, let's use the formula:

$A = \frac{1}{2} r^2 \theta$

Where, A = area

r = radius = 3

$\theta = \frac{1}{9}\pi$

Substituting values in the formula, we have:

$A = \frac{1}{2}*3^2* \frac{1}{9}\pi$

$A = \frac{1}{2}*9* \frac{1}{9}\pi$

$A = 4.5 * \frac{1}{9}\pi$

$A = \frac{\pi}{2}$

The area of the sector = $\frac{\pi}{2}$ square units

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4 years ago

For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with df

rjkz [21]

Answer:

a) For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

b) For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

$t_{\alpha/2} =\pm 2.35$

Part b

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

$t_{\alpha/2} =\pm 2.62$

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4 years ago

Explain how to draw a line segment that measures 72 millimeters.

andrew-mc [135]

Answer:

You can use a ruler

Step-by-step explanation:

brainliest?

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3 years ago

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Answer:

Step-by-step explanation:

Divide the circle into 10 parts and color 2 parts black and 8 parts red.

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3 years ago

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KonstantinChe [14]

Answer: True

Explanation:
ABCD is a face on the figure

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3 years ago