1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Elina [12.6K]
3 years ago
7

Solve x^2+10x+7=0 by completing the square.

Mathematics
1 answer:
melisa1 [442]3 years ago
8 0
The correct answer is number 2 I do believe.
You might be interested in
the points on the number line represent the values of four different numbers which point best represents the value of 3
AveGali [126]

Answer:

its point b

Step-by-step explanation:

the square root of 3 rounds off to 1.7 so looking at the chart point b is closest to that

7 0
3 years ago
How do you solve 2 times (-3) times 5
Sever21 [200]

Answer:2 x -3 x 5 = - 30

Step-by-step explanation:

Go from left to right

3 0
3 years ago
Katya has several sources of income, listed in the table. Which are considered earned income? *****Select all that apply.******
Lady_Fox [76]

Answer: $1300 per month salary

2% commission on all sales

3 0
3 years ago
Free answers if u want
vladimir2022 [97]

Answer:

Thank you so much!!!

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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
Other questions:
  • Please Hurry!
    13·2 answers
  • What is the answer to 7 × 2/5
    7·2 answers
  • A farmer has rights to pump 1,500 gallons of water a minute from a river that runs through his property. He uses 0.9
    13·1 answer
  • 87/100 in simplest form
    7·1 answer
  • Y varies inversely with x. If y = 7 when x = -4, find y when x = 5.
    12·1 answer
  • Mr. Jones has $410,000 in a retirement account that earns 3.85% simple interest each year. Find the amount of interest earned by
    15·1 answer
  • I can’t figure it out<br> 5 = 5 ( x + 6 )
    15·2 answers
  • What is 30 ÷ 14 written as a fraction?
    11·1 answer
  • Ronnie took a survey of eight of his classmates about the number of siblings they have and the number of pets they have. A table
    5·1 answer
  • What is x and Y please answer with your work shown... this type of math is bogus smh ‍♀️
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!