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frosja888 [35]
3 years ago
5

Round 14.603 round to the nearest hundredth

Mathematics
1 answer:
Studentka2010 [4]3 years ago
4 0

Answer:

I believe its 14.60

Step-by-step explanation:


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One number is 8 more than another number, and their sum is 20. Find the numbers.
melomori [17]

Answer:

First number=6

Second number=14

Step-by-step explanation:

Let the first number be x

So the second number is x+8

So the sum of the two is

X+x+8=20

2x+8=20

2x=20-8

2x=12

X=12/2

=6

The second number will x+8

So 6+8=14

First number=6

Second number=14

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3 years ago
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DAY 2 please help i dont get it
MrRa [10]
G because she need to sell 20 vases because the cost is 450 and if you divide it by 23 for the price the vases will be sold for it comes to 19.57 and you can't sell .57 of a vase you have to round up to make more than the money spent
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3 years ago
What is 37.025 written in word format
egoroff_w [7]
Thirty seven and twenty five thousandths

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3 years ago
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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
Need help :((<br><br> plss help i took a screenshot so you can see
Mashcka [7]

Answer: 5:4

Step-by-step explanation: you divide them i believe :)

7 0
3 years ago
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