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motikmotik
3 years ago
7

Select the correct answer.

Mathematics
1 answer:
slavikrds [6]3 years ago
5 0
I think the best answer is B
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Nathan read 2 books in 10 months. What was his rate of reading in books per month
Novay_Z [31]
<h2>Answer:</h2><h2>The rate of reading a book per month = \frac{1}{5} = 20%</h2>

Step-by-step explanation:

Nathan read 2 books in 10 months.

The time taken to complete 1 book = \frac{10}{2} = 5 months

To complete 1 book it takes 5 months.

The rate of reading a book per month = \frac{1}{5} = 20%

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
Simplify the polynomial by combining like terms. 3.63x2 - 4.54x + 7.96x2 +9.85 -3.12x​
Brums [2.3K]

Answer:

11.59x² -7.66x + 9.85

Step-by-step explanation:

3.63x² - 4.54x + 7.96x² +9.85 -3.12x​

11.59x² -7.66x + 9.85

7 0
2 years ago
Read 2 more answers
Distance formula for cars coming from perpendicular roads
Serggg [28]
The answer to the question is
6 0
2 years ago
<img src="https://tex.z-dn.net/?f=1%2F3%20%2812x%20-%2024%29%20%3D%2016" id="TexFormula1" title="1/3 (12x - 24) = 16" alt="1/3 (
coldgirl [10]
1/3(12x-24)=16

3/1[1/3(12x-24)=16]

12x-24=48
+24 +24
12x=72

12x/12=72/12

X=6
4 0
2 years ago
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