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2 years ago

PLEASE ANSWER What is the area of minor sector AOB?

Mathematics

1 answer:

scoray [572]2 years ago

3 0

<h3>Answer: Approximately 13 square units (choice B)</h3>

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Explanation:

The given reflex angle is 215 degrees. A reflex angle is anything over 180 degrees, but less than 360. Subtract 215 from 360 to get the measure of angle AOB

angle AOB = 360 - 215 = 145

angle AOB = 145 degrees

We'll use this later.

Now find the area of the full circle. Use the formula A = pi*r^2. The radius is r = sqrt(10) which can be found through the distance formula or the pythagorean theorem. You want to find the length of either OA or OB to get the radius.

The area of the circle is

A = pi*r^2

A = pi*(sqrt(10))^2

A = 10pi

This is the exact area of the full circle, but we want just a fractional portion of it. Specifically we want the pie slice that is formed by angle AOB

area of sector AOB = [ (angle AOB)/360 ] * (area of full circle)

area of sector AOB = (145/360)*10pi

area of sector AOB = 145pi/36

area of sector AOB = 145*3.14/36

area of sector AOB = 12.647 approximately

area of sector AOB = 13 square units approximately, after rounding to the nearest whole number

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Need help on this problems!

Fofino [41]

For six you can write 1/4 because there is 4 pennies and 1 represents each penny

for number 5 you can put 2/10 because there is 10 coins and 2 quarters

and for number 7 is that Bradley has 10 coins not 9 so it would be 1/10

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2 years ago

The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of m

UkoKoshka [18]

Answer:

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

$X \sim N(14.7,3.7)$

Where $\mu=14.7$ and $\sigma=3.7$

Since the distribution of X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we have;

$\mu_{\bar X}= 14.70$

$\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59$

Step-by-step explanation:

Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of $\bar X$ Round your answers to two decimal places.

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

$X \sim N(14.7,3.7)$

Where $\mu=14.7$ and $\sigma=3.7$

Since the distribution of X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we have;

$\mu_{\bar X}= 14.70$

$\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59$

4 0

2 years ago

Jermaine has just painted a large mural. It is 8 1/2 feet wide. The wall he wants to place it on is 11 3/4 feet wide. About how

PolarNik [594]

First turn 1/2 to 2/4 so you have a common denominator.
Next turn the mixed numbers into improper fractions:
11 3/4 --> 47/4 and 8 2/4 --> 34/4
Subtract:
47/4 - 34/4 = 13/4
Divide:
13/4 = 3.25
3.25/2 --> 1.625 or 1 5/8
Therefore, he will have to place it 1 5/8 feet from each side

Hope this helps!

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2 years ago

Find the value of n .<br> 6/n = 24/28<br> y = ___.

il63 [147K]

Answer:

$n=7$