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Digiron [165]
3 years ago
11

Which value of x is a solution to x2 = 16?

Mathematics
1 answer:
Helga [31]3 years ago
8 0

Answer:

8

Step-by-step explanation:

if we take the 2 that is in the R.H.S and put it in L.H.S

it becomes 16÷2=8

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How do the values of the 1s in 14.937 and 6.1225 compare?
Darina [25.2K]

Answer:

I would say the answer is b

Step-by-step explanation:

8 0
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Find the surface area of the rectangular and triangular prisms
harina [27]
Is there a picture to this????

if your just asking how to find  the area of something all you need to do is multiply all the numbers together and the total is the area.
For example if its 8cm,3cm,2cm all you need to do is multiply them together
8x3x2=48  so your area is 48cm
7 0
3 years ago
Miguel wants to buy a condominium. He has the choice of buying it now or renting it with the option to buy at the end of 3 years
kolbaska11 [484]
Total amount paid if he bought = 6000 + 726(12 x 3) = 6000 + 726(36) = 6000 + 26,136 = 32,136
Total amount paid if he rents = 2000 + 37P, where P is the monthly rent

For renting to be cheaper
2000 + 37P < 32,136
37P < 32136 - 2000 = 30136
P < 30136 / 37 = 814.49

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5 0
3 years ago
Read 2 more answers
Which expression is equivalent to (8x – 20) + (-4x + 12)?
lana66690 [7]

Answer:

B. 4x-8

Step-by-step explanation:

6 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
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