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Andreas93 [3]
3 years ago
5

What are the answers for these questions 1,2,&3

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
6 0
1fyjoite yuuic hyrsd hui
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Plz Help...................
joja [24]

Answer:

55.5 u^2

Step-by-step explanation:

Split the quadrilateral into 2 triangles. There are two ways to do this, but I'm only doing one of them.

On the 5-11 triangle, use the two of the lengths and plug them into the triangle area formula. (L x W)/2 = A

5-11 triangle area: 27.5

Do the same for the 7-8 triangle. Plug it into the same formula.

7-8 triangle area: 28

Add the two areas. 27.5 + 28 = 55.5

8 0
3 years ago
100 POINTS AND BRAINLIST
Alla [95]
  • X=B

All ratios remains same

\\ \rm\hookrightarrow sinX=\dfrac{P}{H}=\dfrac{30.8}{34}=0.9

\\ \rm\hookrightarrow cosX=\dfrac{B}{H}=\dfrac{14.4}{34.4}=0.42

\\ \rm\hookrightarrow tanX=\dfrac{sinX}{cosX}=\dfrac{0.9}{0.42}=2.14

3 0
2 years ago
Read 2 more answers
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
"calculate the number of permutations possible when using the first five letters of the alphabet to create 3-letter words.
Hitman42 [59]
That is   5! / (5-3)!  =  120 / 2 = 60 answer
5 0
3 years ago
Which ones ARE in scientific notation?
Rina8888 [55]

Answer:

C

Step-by-step explanation:

Cause C met the rules of scientific notation which is:

  • (First digit of the number) followed by (the decimal point) and then (all the rest of digits of the number) times (10 to an appropriate power)
7 0
2 years ago
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