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3 years ago

6) A pyramid with a TRIANGULAR base has a volume of 240 cubic inches and a height of 15 inches. The base of

Mathematics

1 answer:

Misha Larkins [42]3 years ago

8 0

Answer: 12 inches

Step-by-step explanation:

Vol = 1/3 x b x h

240 = 1/3 x b x 15

b = 240/5 = 48 in = area of triangle base

Area = 1/2 x b x h

48 = 1/2 x 8 x h

h = 48/4 = 12 inches

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Speedy crawled 5/9 yards. Slimy crawled 4 times as far as speedy. How far did slimy crawl?

ELEN [110]

Answer:

2 2/9 yards or 20/9 yards

Step-by-step explanation:

Slimy went 4 times as much as Speedy did so that means 5/9 x 4 yards which is the same thing as 5 x 4/9 so 5 x 4/9 equals 20/9 or 2 2/9 so Slimy crawled 2 2/9 yards.

5 0

3 years ago

Assume Y=1+X+u, where X, Y, and u=v+X are random variables, v is independent of X; E(v)=0, Var(v)=1, E(X)=1, and

kobusy [5.1K]

Answer:

a) $E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1+$

b) $E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2$

c) $E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2$

d) $E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6$

e) $E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1$

f) $E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3$

g) $E(u) = E(v) +E(X) = 0+1=1$

h) E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

Step-by-step explanation:

For this case we know this:

$Y = 1+X +u$

$u = v+X$

with both Y and u random variables, we also know that:

$[tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2$

And we want to calculate this:

Part a

$E(u|X=1)= E(v+X|X=1)$

Using properties for the conditional expected value we have this:

$E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1$

Because we assume that v and X are independent

Part b

$E(Y| X=1) = E(1+X+u|X=1)$

If we distribute the expected value we got:

$E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2$

Part c

$E(u|X=2)= E(v+X|X=2)$

Using properties for the conditional expected value we have this:

$E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2$

Because we assume that v and X are independent

Part d

$E(Y| X=2) = E(1+X+u|X=2)$

If we distribute the expected value we got:

$E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6$

Part e

$E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1$

Part f

$E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3$

Part g

$E(u) = E(v) +E(X) = 0+1=1$

Part h

E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

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3 years ago

How do you do function machines?

timurjin [86]

Well you are going to need a rule or formula like x + 6 or 3 (x) + 2

Example:

Input x + 6 Output
2 2+ 6 8
4 4+6 10
6 6+6 12

Hope this helps sorry if it doesn’t

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3 years ago

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kvasek [131]

Answer:

I think it is the third one

Step-by-step explanation:

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3 years ago

Select the correct answer from each drop-down menu. How many solutions does each polynomial have? Equation y=-6x-9, y=2^5-x^4+2x

Svetlanka [38]

Answer:

Step-by-step explanation:

Y=-6x-9

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3 years ago

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