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Gwar [14]
3 years ago
11

Name 3 planes parallel to FL

Mathematics
2 answers:
3241004551 [841]3 years ago
8 0

Answer:

17)THE PLANES ARE <em><u>BCHI</u></em><em><u>,</u></em><em><u>DJIC</u></em><em><u>,</u></em><em><u>EDJK</u></em>

JulijaS [17]3 years ago
4 0

Answer:

I have answered. it is perfect answer

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SOLVE AND JUSTIFY RIGHT ANSWER GETS BRAIN
DIA [1.3K]

Answer:

x > 14

Step-by-step explanation:

It is really easy.

8 0
3 years ago
3/8 1/7 and 1/2 in greatest to least
Alex73 [517]
1/2, 3/8, 1/7 hopefully I am right I’m not sure
3 0
3 years ago
Read 2 more answers
How are ∠1 and ∠2 related
andreyandreev [35.5K]

Answer:

  • D. They are adjacent

Step-by-step explanation:

Angles 1 and 2 have common side, so they are adjacent. They are not making a right or straight angle.

Correct choice is D

7 0
3 years ago
Subtract.
faltersainse [42]
18 1/2 -17 3/4
= 18.50 -17.75
= 0.75
= 3/4

The 4th selection is appropriate.
5 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
3 years ago
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