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vivado [14]
3 years ago
10

Please help me with this question!

Mathematics
1 answer:
andrey2020 [161]3 years ago
5 0

Answer:

180 = 4b + 24

4b = 180 - 24

4b = 156

b = 156/4

b = 39

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What is the measure of each angle in any equilateral triangle?
miskamm [114]
60°; all of of the angles of a triangle add up to 180° so 180 divided by 3 equals 60°.
3 0
3 years ago
Read 2 more answers
I need help with this please thank you very much
Hitman42 [59]

So,

We can notice that the graph of g, is translated 2 units to the left and 4 units up. We can express these changes with the following equation:

g(x)=(x+2)^2+4

4 0
1 year ago
The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm.
devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

4 0
3 years ago
Is this statement true or false -7 -2 > 6 - 14
exis [7]

Answer:

false

Step-by-step explanation:

(-7-2) > (6-14)

-7-2 = -9

6-14 = -8

-9 is not > -8, because -9 is farther to the left on a number line. So, this statement would be false.

6 0
3 years ago
What is the value of x<br> 2x+3y=45 <br> x+y=10
Volgvan
Assuming this is a system of equations, here is how to find x. 

2x + 3y = 45
x + y = 10 

Multiply x + y = 10 by 2 so you are able to use the elimination method.

2x + 3y = 45
2x + 2y = 20

Subtract.

y = 25

Now that we've found y, we can plug it in to find x. 

x + 25 = 10

Subtract 25 from both sides.

x = -15
4 0
3 years ago
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