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antoniya [11.8K]
3 years ago
12

Which of the following polygons can be circumscribed by a circle?

Mathematics
1 answer:
atroni [7]3 years ago
8 0

Answer:

The correct answer is C

Step-by-step explanation:

If you put a circle over a rectangle you can touch all four corners

If you put a circle over polygon A then you could not touch all four verticies.

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Can someone help me plz!!! I would appreciate it ❤️
kolezko [41]

Answer:

25% because 75% receive anywhere between $10-$20. If 75% receive this amount than 25% don't.

Step-by-step explanation:

100% is the whole if you subtract 75% from 100% you get 25% left over. Hope this helps

7 0
2 years ago
Emily has $19 more dollars than Juan. Juan has $65. How much money does Emily have? Type your answer in the box below​
kozerog [31]

Answer:

65+19=84

Step-by-step explanation:

Emily has $84

6 0
2 years ago
Read 2 more answers
If ⅓y = 9; then y =?​
Svetach [21]

<u>Answer:</u>

<h2>y = 27</h2>

<u>Eexplanation:</u>

1/3y = 9

y/3 = 9

y = 9 × 3

y = 27

4 0
2 years ago
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PLZ ANSWER FAST WILL GIVE BRAINLIEST Find the value of x.<br> 148
ludmilkaskok [199]
74 degrees

An angles value is 1/2 of the value of its corresponding arc. Angle x has the corresponding arc of 148, so it’s 148/2 or 74 degrees
3 0
3 years ago
According to a survey, high school girls average 100 text messages daily (The Boston Globe, April 21, 2010). Assume the populati
Ghella [55]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: number of daily text messages a high school girl sends.

This variable has a population standard deviation of 20 text messages.

A sample of 50 high school girls is taken.

The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;δ²/n)

This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:

Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)

a.

P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836

b.

P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)

P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328

I hope you have a SUPER day!

3 0
2 years ago
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