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Vitek1552 [10]
3 years ago
8

H(a) = 3 - 2a find h(-3)

Mathematics
2 answers:
skelet666 [1.2K]3 years ago
6 0

Answer:

h(-3)=9

Step-by-step explanation:

hope this helps

Rainbow [258]3 years ago
4 0
H(-3)=9
Good luck on it
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Which of the following is an equation of the line through (3, -1) and (−2, 14)?
enot [183]
D.
Slope is rise over run and in this case it is -15/ 5 so when it is reduced you get -3 as your slope
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3 years ago
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Simplify.<br> 4/x^2-35 - 1/x+5
Oksana_A [137]
If I understand your question well, I suppose you mean \frac{4}{ x^{2} -35} -  \frac{1}{x+5}
\frac{4}{ x^{2} -35} - \frac{1}{x+5} =  \frac{4}{(x+5)(x-7)}- \frac{1}{x+5} \\ = \frac{1}{x+5}( \frac{4}{x-7}-1) \\ = \frac{1}{x+5}( \frac{4}{x-7}- \frac{x-7}{x-7}) \\ = \frac{1}{x+5}( \frac{4-x+7}{x-7}) \\ = \frac{1}{x+5}( \frac{11-x}{x-7}) \\ = \frac{11-x}{ x^{2} -35}
6 0
3 years ago
Jia is tiling a floor. The floor is a square with side
NNADVOKAT [17]

Answer:

36 tiles

Step-by-step explanation:

First, find area of the square floor knowing the formula as;

Area= s^{2}

where s= side of a square.

If one side of the floor is 12 feet, then Area= 12^{2} = 144ft^{2}

Next, find the area of each tile using the same formula and given s=2 feet;

Area(tile) = 2^{2} = 4ft^{2}

To find the number of tiles needed to cover entire floor, divide area of the floor by area of one tile; 144 / 4 = 36 tiles

5 0
3 years 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
Help picture below problem 6
r-ruslan [8.4K]

Answer:

option B is the correct answer

Step-by-step explanation:

mark me brainliest

6 0
2 years ago
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