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Nady [450]
2 years ago
7

if 3.8 billion gallons of water are used for domestic use per day , approximately how many gallons of water are used for irrigat

ion per day

Mathematics
2 answers:
Xelga [282]2 years ago
5 0

Answer:

1.629 billion gallons of water a day are used for Irrigation.

Step-by-step explanation:

Great question, it is always good to ask away and get rid of any doubts that you may be having.

According to my research which can also be found on the USDA ERS website. The Domestic Use (indoor) amount of water accounts for 70% of the total water used per day. While the Irrigation Water (outdoor) used per day accounts for 30%.

This being the case We can use the Rule of Three technique to solve for the amount of irrigation water used per day. This rule is attached in the picture below.

3.8 billion   ⇒   0.70

   x billion   ⇒   0.30

\frac{3.8 * 0.30}{0.70} = x

1.629billion = x ... rounded to the nearest thousandth

We can see from the calculation above that approximately 1.629 billion gallons of water a day are used for Irrigation.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

dsp732 years ago
3 0
One trillion three hundred eighty seven billion
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2 years ago
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Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |
Bas_tet [7]

Answer:

M=168k

(\bar{x},\bar{y})=(5,\frac{85}{28})

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

M=\int\int \rho dA

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx

Let's solve this integral:

M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx

M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx      

M=k\int^{9}_{1}21dx

M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}

So the mass will be:

M=21k*8=168k

Now we need to find the x-coordinate of the center of mass.

\bar{x}=\frac{1}{M}\int\int x*\rho dydx

\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx

\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx

\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx

\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx

\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}

\bar{x}=\frac{21}{168}*40=5

Now we need to find the y-coordinate of the center of mass.

\bar{y}=\frac{1}{M}\int\int y*\rho dydx

\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx

\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx

\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx

\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx

\bar{y}=\frac{255}{672}\int^{9}_{1}dx

\bar{y}=\frac{255}{672}8=\frac{2040}{672}

\bar{y}=\frac{85}{28}

Therefore the center of mass is:

(\bar{x},\bar{y})=(5,\frac{85}{28})

I hope it helps you!

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2 years ago
Suppose that we've decided to test Clara, who works at the Psychic Center, to see if she really has psychic abilities. While tal
mixas84 [53]

Answer:

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Step-by-step explanation:

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q = 1-(1/4) = 3/4

So;

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Just need to check these answers
amid [387]
Great Job! they are all correct.  :)


Good luck in your next tests.
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2 years ago
Estimate the product 8.1 x 4.2
Korvikt [17]
Round 8.1 down to 8
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