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Schach [20]
3 years ago
14

1/2 with an exponent of 4

Mathematics
2 answers:
Pani-rosa [81]3 years ago
7 0
The answer is 0.0625
Helen [10]3 years ago
6 0

Answer: \frac{1}{16}

1/2 to the 4th power = 1/2^4

<u>Multiply</u>

\frac{1}{2} *\frac{1}{2} *\frac{1}{2} *\frac{1}{2}

=\frac{1}{16}

<u>Decimal</u>

\frac{1}{16} =0.0625

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find the volume of cubical box . if the cost of painting of outer surface is rs 1440 at the rate rs 15 per mcube
bearhunter [10]
Surface of a cubical box=6(side²)

1)We have to calculate the surface of this cubical box.
Rate=cost of painting / surface  ⇒surface=cost of painting/rate

Data:
Rate=$15/m²
cost of painting=$1440

Surface=$1440/($15/m²)=96 m²

2)We find out the length of the side:

Surface of a cubical box=6(side²)

Data:
Surface of a cubical box=96 m2

Therefore:
96m²=6 (side²)
side²=96 m²/6
side²=16 m²
side=√(16 m²)=4 m

3) We find the volume of a cubical box.
volume=(side³)
volume=(4 m)³
volume=64 m³

Answer: the volume of this cubical box would be 64 m³.

3 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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777dan777 [17]

Answer:

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Step-by-step explanation:

(-4y-x)-(y-9x)\\\\-4y-x-(y-9x)\\\\\rightarrow-(y-9x)*-1=-y+9x\\\\-4y-x-y+9x\\\\-4y-y+9x-x\\\\\boxed{-5y+8x}

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fiasKO [112]

Answer:

A. 2CO + O2 ---> 2CO2

Step-by-step explanation:

2CO + O2 ---> 2CO2

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Answer:

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Step-by-step explanation:

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