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Stells [14]
2 years ago
12

What is the length of PR​

Mathematics
2 answers:
Sergeeva-Olga [200]2 years ago
7 0
The answer is 13 hope this helps
guajiro [1.7K]2 years ago
3 0

Answer:

9

Step-by-step explanation:

This is because 21/7 is 3. So, you use the scale factor of 3 to divide 27/3 to get x which is 9.

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I really need help with this no sum answers or else. You will get 20 points and brain list
nordsb [41]

Answer:

if their asking for % then this is the answers

1. 41% of 900 is 369

2. 24%

3. 46%

4. 23%

5. 95%

6. 23%

7. 75%

8. 22%

9. 92%

10. 45%

5 0
2 years ago
Read 2 more answers
I’ll love u forever if u help me with this ASAP PLZZZ I don’t get it plzzz
laiz [17]

Answer:

v= 13

w= -32

Step-by-step explanation:

2v+6w=-36

5v+2w=1

2w= 1-5v

2v+ (3)(1-5v)= -36

2v+3-15v= -36

3-13v =-36

3--36= 13v

39= 13v

13 = V

5×13+2w=1

65+2w=1

1-65= 2w

-64=2w

-32=w

3 0
2 years ago
Can someone please help I will give you 30 points and brainliest if the answer is right also no links please
Zolol [24]

Answer:

B: BC ≅EC

Step-by-step explanation:

You know that the two angles starting in C are congruent (they're opposite by vertex C).

Given that the B angle and the E angle are congruent, in order for the two triangles to be congruent by A(ngle)S(ide)A(ngle) you want the side inbetween to be congruent. That is BC and EC. Option B

5 0
2 years ago
CAN SOMEONE HELP ME WITH THIS PLEASE AND THANK YOU.
Aloiza [94]
If you convert the fractions to improper, you get 9/2 times 7/4. you can then just multiply the tops and bottoms and simplify! hope this helps
6 0
3 years ago
Read 2 more answers
For a population with µ = 80 and σ = 20, the Sampling distribution of the Mean, based on n = 16 will have an expected value of t
allsm [11]

Answer:

Will have an expected value of the mean = 80 and a standard error of the mean = 5

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 80, \sigma = 20

n = 16

So the mean is 80 and the standard deviation is s = \frac{20}{\sqrt{16}} = 5

Will have an expected value of the mean = 80 and a standard error of the mean = 5

6 0
3 years ago
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