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Helga [31]
3 years ago
5

A logo is made of a square and a semicircle. The square has side lengths of 12 inches. One side of the square is also the diamet

er of the semicircle. What is the approximate total area of the logo?
Mathematics
1 answer:
boyakko [2]3 years ago
6 0

Answer:

The square has side lengths of 22 inches. One side of the square is also the diameter of the semicircle

Step-by-step explanation:

Hopefully its right!

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Help!!!
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Answer:

(8, -5)

Step-by-step explanation:

When you reflect across the line y = x, you switch the two numbers in the coordinate. The point (a, b) turns into (b, a).

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How many 1/7 inch pieces can be cut from a piece of ribbon 5/21 of an inch long
yawa3891 [41]

Hello there! :D

6 in' pieces can be cut! :D

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3 years ago
A car originally priced at $15,000 is discounted 26%<br> What is the sale price?
Viktor [21]
26% of 15,000 is $3,900
Which means the sale price is 15000-3900
= $11, 100
8 0
3 years ago
Read 2 more answers
What is the volume of the cylinder shown below?
pav-90 [236]

Answer:

2258 cu. units

Step-by-step explanation:

you take 22/7 multiply by radius squared and multiply that by 5

7 0
3 years ago
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A candy company claims that 20% of the candies in its bags are colored green. Steve buys 30 bags of 30 candies, randomly selects
Allisa [31]

Answer:

The probability of Steve agreeing with the company’s claim is 0.50502.

Step-by-step explanation:

Let <em>X</em> denote the number of green candies.

The probability of green candies is, <em>p</em> = 0.20.

Steve buys 30 bags of 30 candies, randomly selects one candy from each, and counts the number of green candies.

So, <em>n</em> = 30 candies are randomly selected.

All the candies are independent of each other.

The random variable <em>X</em> follows a binomial distribution with parameter <em>n</em> = 30 and <em>p</em> = 0.20.

It is provided that if there are 5, 6, or 7 green candies, Steve will conclude that the company’s claim is correct.

Compute the probability of 5, 6 and 7 green candies as follows:

P(X=5)={30\choose 5}(0.20)^{5}(1-0.20)^{30-5}=0.17228\\\\P(X=6)={30\choose 6}(0.20)^{6}(1-0.20)^{30-6}=0.17946\\\\P(X=7)={30\choose 7}(0.20)^{7}(1-0.20)^{30-7}=0.15328

Then the probability of Steve agreeing with the company’s claim is:

P (Accepting the claim) = P (X = 5) + P (X = 6) + P (X = 7)

                                       = 0.17228 + 0.17946 + 0.15328

                                       = 0.50502

Thus, the probability of Steve agreeing with the company’s claim is 0.50502.

7 0
3 years ago
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