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Lubov Fominskaja [6]
3 years ago
12

Point Y is the circumcenter of ΔDEF. Find FY. 5 11 17 22

Mathematics
2 answers:
RideAnS [48]3 years ago
6 0

Answer:

The answer is 22.

Step-by-step explanation:

I was confident the other person was correct, I took the test. Answer turned out to be D. 22.

Leona [35]3 years ago
6 0

Answer:

D.22

Step-by-step explanation:

Put FY and YE on opposite sides and solve for X, then substitute in X

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What is 8 times as many as 3 is
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Step-by-step explanation:

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Someone please explain I have zero clue how to answer this! ​
UkoKoshka [18]

Answer:

x=62

Step-by-step explanation:

152-90(because of the right angle)

the 28 is a red herring

4 0
3 years ago
ASAP help I don’t know the answer
Nady [450]

Answer:

6

Step-by-step explanation:

It appears you have vertical angles. The rule with angles is that no matter what, vertical angles are always congruent (or the same). This means you would have to set up the equation to solve for "b" as:

70 - 2b = 28 + 5b

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4 0
3 years ago
Read 2 more answers
A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye
Katena32 [7]

Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we select appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{x}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

 \sigma_{x}=\sqrt{n}\sigma

The information provided is:

<em>μ</em> = $970

<em>σ</em> = $129

<em>n</em> = 102

Since the sample size is quite large, i.e. <em>n</em> = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)

Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

P (\sum X \leq  100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}

                              =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

6 0
3 years ago
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