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3 years ago

Plz help this is due today NO LINKS OR GROSS PICTURES OR I Will REPORT And please show work.

Mathematics

2 answers:

Darya [45]3 years ago

8 0

Answer:

224 m^3

Step-by-step explanation:

Volume = l x w x h

V = 8 x 4 x 7

V = 32 x 7

V = 224 m^3

Have a great day! :D

Salsk061 [2.6K]3 years ago

3 0

Answer:

224m³

Step-by-step explanation:

volume =surface area ×height

thus surface area=l×w (since its a rectangle)

V=l×w×h

V=8×4×7

V=224m³

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Urgent!!!! The scores of a high school entrance exam are approximately normally distributed with a given mean Mu = 82.4 and stan

Effectus [21]

Answer:95%

Step-by-step explanation:

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3 years ago

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The distribution of heights for adult men in a certain population is approximately normal with mean 70 inches and standard devia

KengaRu [80]

Answer:

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12]

Step-by-step explanation:

I beleive those options corresponds to another question, i will ignore them. We want to know an interval in which the probability that a height falls there is 0.8.

In such interval, the probability that a value is higher than the right end of the interval is (1-0.8)/2 = 0.1

If X is the distribuition of heights, then we want z such that P(X > z) = 0.1. We will take W, the standarization of X, wth distribution N(0,1)

$W = \frac{X-\mu}{\sigma} = \frac{X-70}{4}$

The values of the cumulative distribution function of W, denoted by $\phi$ , can be found in the attached file. Lets call $y = \frac{z-70}{4}$ . We have

$0.1 = P(X > z) = P(\frac{X-70}{4} > \frac{z-70}{4}) = P(W > y) = 1-\phi(y)$

Thus

$\phi(y) = 1-0.1 = 0.9$

by looking at the table, we find that y = 1.28, therefore

$\frac{z-70}{4} = 1.28\\z = 1.28*4+70 = 75.12$

The other end of the interval is the symmetrical of 75.12 respect to 70, hence it is 70- (75.12-70) = 64.88.

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12] .

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8 0

3 years ago

Shureka Washburn has scores of 67, 68, 76, and 63 on her algebra tests. a. Use an inequality to find the scores she must make

mash [69]

Average=(total number)/(number of items)
given that the final exam counts as two test, let the final exam be x. The weight of the final exams on the average is 2, thus the final exam can be written as 2x because any score Shureka gets will be doubled before the averaging.
Hence our inequality will be as follows:
(67+68+76+63+2x)/6≥71
(274+2x)/6≥71
solving the above we get:
274+2x≥71×6
274+2x≥426
2x≥426-274
2x≥152
x≥76

b] The above answer is x≥76, the mean of this is that if Shureka is aiming at getting an average of 71 or above, then she should be able to get a minimum score of 76 or above. Anything less than 76 will drop her average lower than 71.

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3 years ago

In the following exercises, solve

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Answer:

When the volume of a gas in a container varies inversely with the pressure on the gas and a container of nitrogen has a volume of 29.5 litres with 2000 psi. So, if the tank has a 14.7 psi pressure, then the volume of the tank would be 4013.6 litres.

Step-by-step explanation:

The volume of a gas in a container varies inversely with the pressure on the gas. If a container of nitrogen has a volume of litres with psi, what is the volume if the tank has a psi pressure.