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expeople1 [14]
3 years ago
5

Solve for k: 3(k+8) = 21​

Mathematics
1 answer:
timurjin [86]3 years ago
5 0
3(k+8)=21
3k+24=21
-24 -24
3k=-3
-------- = k=-1 is your answer
3 3
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Set up the correct type of variation equation. <br>V=r^2h<br>kV=r^2h<br>V=kr^2h<br>V=k(r+h)<br><br>​
Ede4ka [16]

Answer:

V = kr²h

Step-by-step explanation:

Given that V varies jointly with r² and h, then the equation relating them is

V = kr²h ← k is the constant of variation

Note that joint variation involves the product of the joint quantities

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What is the sum of the first 26 terms of the arithmetic series?
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What would be the final location of point (2,6) after a rotation 90° counterclockwise about the origin. Write the location of th
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3 years ago
(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
3 years ago
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