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fredd [130]
3 years ago
12

A high rise building has a sink on the fourth floor located 2-feet above the floor. Each floor is 10-feet above the other. Calcu

late the psi at the sink where when the street pressure is 80 psi. The building does not have a pressure reducing device due to fire codes.
Mathematics
1 answer:
kap26 [50]3 years ago
8 0
The pressure at a certain height is the pressure at sea level plus the hydrostatic pressure.

P = 80 psi + ΔP
where
ΔP = ρgh
ρ is the density of water equal to 62.4 lbm/ft³
ΔP = (62.4 lbm/ft³)(1 lbf/lbm)(4*10 + 2 ft)(1 ft/12 in)² = 18.2 psi

Thus,
P = 80 + 18.2
<em>P = 98.2 psi</em>
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Daniel ran ¾ of a mile every day for 7 days. How many miles did David run after seven days?
VLD [36.1K]

Answer:

5.25 miles

Step-by-step explanation:

Convert 3/4 to a decimal=

0.75

Then, multiply 0.75 with 7

0.75×7=5.25

David ran 5.25 miles in 7 days

5 0
2 years ago
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A tree initially measured 18 feet tall. Over the next 312 years, it grew to a final height of 3512 feet. During those 312 years,
Leni [432]

Answer:

Average growth= 11.20 feet

Step-by-step explanation:

<u>First, we need to calculate the total growth during those 312 years:</u>

<u></u>

Total growth= 3,512 - 18

Total growth= 3,494 feet

<u>Now, the average growth:</u>

<u>Average growth= total growth / number of years</u>

Average growth= 3,494 / 312

Average growth= 11.20 feet

8 0
2 years ago
Please help. 10 points
mestny [16]
It's either A or C but correct me if I'm wrong :)
5 0
3 years ago
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elixir [45]
The first Venn diagram would be the best because 18 students play baseball and one side has 13 13+5=18 and for basketball 12+5=17
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7 0
2 years ago
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Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
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