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Naya [18.7K]
3 years ago
15

In a circle of radius 10 cm, a sector has an area of 40 (pi) sq. cm. What is the degree measure of the arc of the sector?

Mathematics
2 answers:
Rashid [163]3 years ago
6 0

Answer:

144°

Step-by-step explanation:

First, find the area of the circle, with the formula A = \pir²

Plug in 10 as the radius, and solve

A = \pir²

A = \pi(10²)

A = 100\pi

Using this, create a proportion that relates the area of the sector to the degree measure of the arc.

Let x represent the degree measure of the arc of the sector:

\frac{40\pi }{100\pi } = \frac{x}{360}

Cross multiply and solve for x:

100\pix = 14400\pi

x = 144

So, the degree measure of the sector arc is 144°

Semmy [17]3 years ago
3 0

Answer:

<h2><u>Solution</u><u> </u><u>:</u><u>-</u></h2>

We know that

Area = πr²

Area = 3.14 × (10)²

Area = 314/100 × 100

Area = 314 cm²

Now

40π/314 = x/360

40 × 3.14/314 = x/360

125.6/314 = x/360

0.4 = x/360

0.4 × 360 = x

144 = x

\\

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Answer:

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This is the expression we wanted.

b) Now we need to expand and simplify the expression:

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spayn [35]

Answer:

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And we can find this probability with the complement rule and we got:

P(Z>2) = 1-P(Z

And using the normal standard table or excel we got:

P(Z>2) = 1-P(Z

Step-by-step explanation:

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 prices of a population, and for this case we know the distribution for X is given by:

X \sim N(100,5)  

Where \mu=100 and \sigma=5

We are interested on this probability

P(X>110)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>110)=P(\frac{X-\mu}{\sigma}>\frac{110-\mu}{\sigma})=P(Z>\frac{110-100}{5})=P(z>2)

And we can find this probability with the complement rule and we got:

P(Z>2) = 1-P(Z

And using the normal standard table or excel we got:

P(Z>2) = 1-P(Z

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Step-by-step explanation:

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Arte-miy333 [17]

Answer:

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Step-by-step explanation:

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