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3 years ago

Is 1,000 a rational number or a irrational or is it a whole number ?

Mathematics

1 answer:

ad-work [718]3 years ago

3 0

Answer:

1,000 is both a rational number and a whole number.

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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

Divide. check your answer

slava [35]

Step-by-step explanation:

:)..................

7 0

2 years ago

A store allows customers to fill their own bags of candy. Terri decides she only wants jelly beans and chocolate drops. Jelly be

NISA [10]

Jelly beans - $0.98 per pound
chocolate drops - $0.67 per pounds
Terri's bag = 2.1 pounds and $1.56

p = represents the pounds jelly beans.
c = represents the pounds in chocolate drops

p + c = 2.1
0.98p + 0.67c = 1.56

c = 2.1 - p ; number of pounds of chocolate drops

0.98p + 0.67(2.1 - p) = 1.56
0.98p + 1.41 - 0.67p = 1.56
0.98p - 0.67p = 1.56 - 1.41
0.31p = 0.15
p = 0.15/0.31
p = 0.48 pounds of jelly beans

c = 2.1 - 0.48
c = 1.62 pounds of chocolate drops.

6 0

2 years ago

Read 2 more answers

Mauricio divided two identical rectangles into equal parts. He colored one part of each rectangle. What is true about the areas

Ivenika [448]

Answer:

Areas of the colored parts are equal

Step-by-step explanation:

Given Mauricio divided two identical rectangles into equal parts.

Now given Mauricio colored one part of each rectangle. we have to tell about the true statement of colored parts.

Two identical rectangles means the rectangle having equal areas. He divides these two into equal parts and shaded one of each part

Let area of rectangle is x $unit^{2}$

∴ Colored part area = $\frac{1}{2}xunit^{2}$ → (1)

Similarly, other rectangle area identical to above rectangle is x $unit^{2}$

∴ Colored part area = $\frac{1}{2}xunit^{2}$ → (2)

From above eq (1) and (2), we get

Areas of the colored parts are equal which is equals to $\frac{1}{2}xunit^{2}$

4 0

3 years ago

Can someone help me with this geometry question????

Gnesinka [82]

3.865in^2

It's (the area of the square minus the area of the circle) divided by 2.

So, Area of the square is 6x6=36

Area of the circle is (pi)*(radius=3)^2=28.27

36-28.27=7.73

7.73/2=3.865in^2

7 0

3 years ago