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rewona [7]
3 years ago
12

What is the word name for 20,004

Mathematics
1 answer:
Fofino [41]3 years ago
6 0
<span>twenty thousand and four</span>
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A rabbit population doubles every 4 weeks.There are currently five rabbits in a restricted area. If t represents the time in wee
sertanlavr [38]
Doubling formula is this:

P(t)=P(2)^{\frac{t}{d}}
where P=initial number of rabbits
t=time
d=time it takes to doulbe

ok, so 4 weeks is the doubling time so that is 4*7=28 days

we wawnt time=98
and oroiginal number of rabbits is 5 so
P(98)=5(2)^{\frac{98}{28}}
P(98)=5(2)^{3.5}
P(98)=5(2^3)(\sqrt{2})
P(98)=5(8)\sqrt{2}
P(98)=40\sqrt{2}
so P(98)≈56.56
we can't have .56 rabbit so round down or up
about 56 or 57 rabbits in 98 days
8 0
4 years ago
C6.2 PA4 Special Pricing Ken Owens Construction specializes in small additions and repairs. His normal charge is $400/day plus m
postnew [5]
Not reading all that LOL
3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20-%204u%20%2B%204x%20%3D%2012%20%5C%5C%20%20-%20u%20%3D%20%20-%202" id="TexFormula1" title="
AleksAgata [21]
When u=-2, put that in where you see "u". use parentheses.
-4(-2)+4x=12

-4*-2=8 ➡positive 8 since a negative times a negative is always positive
一一一一一
now look for x
8+4x=12
subtract 8 from both sides. get x by itself.

-8+8+4x=12-8
4x=4➡ divide both sides by 4. a number divided by itself is 1 (unless one side is negative, then the answer is negative)
4/4=1
一一一一一

x=1
6 0
4 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]

8 0
3 years ago
PLEASE HELP
Alexus [3.1K]

Answer:

x=1

y=0

Step-by-step explanation:

(x-1)=-2x+2

3x-1=2

3x=3

x=1

-2(1)+2

-2+2=0

6 0
3 years ago
Read 2 more answers
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