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Anestetic [448]
3 years ago
6

The ratio of boys to girls is 40:56 what is the ratio in lowest terms

Mathematics
1 answer:
tiny-mole [99]3 years ago
4 0
The answer is 5:7. 
Hope this helps
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What is the value 24/25 ÷ 4/5 6th grade work.
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as a decimal 0.048 and as a fraction 6/125

Step-by-step explanation:

Hope this helps :)

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Simplify 4h + 3h + h <br> Pleaseee help asap <br> Thank youuu
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Combine like terms4ℎ+3ℎ+ℎ8ℎ

Solution8ℎ

Step-by-step explanation:

7 0
2 years ago
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solve this system of linear equations, separate the x- and y-values with a comma. -13x=-96-7y -3x=64+7y
GaryK [48]

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(2,-10)

Step-by-step explanation:

I used a graphing tool to graph the system of equations. The the two lines intercept at the point (2,-10). So, (2,-10) is the solution.

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3 years ago
9+7+(+^8+7+9 and 12 more
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3 years ago
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(1 point) Let p be the joint density function such that p(x,y)=116xy in R, the rectangle 0≤x≤4,0≤y≤2, and p(x,y)=0 outside R. Fi
baherus [9]

Answer:

The answer is \frac{7}{8} of the population.

Step-by-step explanation:

The question is wrong. The joint density function is p(x,y)=(\frac{1}{16})xy in R

and p(x,y)=0 outside R.

R is defined as the rectangle 0\leq x\leq 4 , 0\leq y\leq 2

In order to find the fraction of the population satisfying the constraint x\geq y , we will need to integrate the joint density function p(x,y) over the region defined by the constraint. It is very convenient to draw the region ''R'' and the new region define by the constraint x\geq y

I will attach a drawing with the region ''R'' and the new region where we need to apply the integral.

If we integrate outside ''R'', given that p(x,y)=0 outside ''R'', the integral will be equal to 0 (because of the joint density function).

Inside the rectangle ''R'' and given the constraint x\geq y , we define two new regions : the green region (I) and the blue region (II).

The final step is to integrate in (I) and in (II) and sum ⇒

\int\int p(x,y) dx dy ⇒

\int \int\limits_1 {p(x,y)} \, dx dy + \int \int\limits_2{p(x,y)} \, dx dy , where ''1'' is the green region and ''2'' is the blue region.

⇒ \int\limits^2_0 \int\limits^x_0 (\frac{1}{16} xy) dy dx   +  \int\limits^4_2\int\limits^2_0 (\frac{1}{16}xy) dydx  = \frac{1}{8}+\frac{3}{4}=\frac{7}{8}=0.875

We find that \frac{7}{8} of the population satisfy the constraint x\geq y.

6 0
2 years ago
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