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julia-pushkina [17]
3 years ago
6

PLEASE HELP MEEEEEEE I WILL GIVE BRAINLIEST!!!

Mathematics
2 answers:
Marina CMI [18]3 years ago
7 0
The y-intercept is 60
the slope is 30
the equation of the line is y=30x+60
the ordered pair (4,120) represents and increase of 30 minutes per hour
Wewaii [24]3 years ago
4 0
THIS IS VERY MUCH A SCAM

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What is the value of thi expression when d=3?
Leni [432]

Answer:

73

Step-by-step explanation:

Givens

d = 3

Equation 7d^2 + 10 = ?

Solution

7*3*3 + 10                

7*9 + 10

63 + 10 = 73

4 0
3 years ago
Read 2 more answers
If f(7)=22, then f-1(22)=
nignag [31]

Answer:

f^{-1} (22) = 7

Step-by-step explanation:

The inverse function maps the output of the function back to its input

f(7) = 22 means when input is 7 the output is 22

The inverse maps the output 22 back to the input 7 , that is

f^{-1} (22) = 7

4 0
3 years ago
Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
Help me solve for z, step by step if you can
zheka24 [161]

Answer:

z=6

Step-by-step explanation:

z/3 + 1/2 = 5/2

Subtract 1/2 from each side

z/3 + 1/2-1/2 = 5/2-1/2

z/3 = 4/2

z/3 =2

Multiply each side by 3

z/3*3 = 2*3

z = 6


5 0
3 years ago
First find the value of x the find the measure of each angle. need help !!!!​
lisabon 2012 [21]

Answer:

I cant see the photo

Step-by-step explanation:

3 0
3 years ago
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