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ycow [4]
3 years ago
11

How many times larger is 3 x 1015 than 6 x 1010

Mathematics
2 answers:
Deffense [45]3 years ago
8 0

it is 3,000 times larger! that is the difference.

Klio2033 [76]3 years ago
7 0

(3 x 10^15) / (6 x 10^10)

= (3/6) x 10^(15 - 10)

= 0.5 x 10^5

= 5 x 10^4

= 50,000

Answer

50,000 times larger

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The ratio of the length of a rectangle to its width is 8 to 5. If the longer side of the rectangle is 20ft, what is the length o
choli [55]

Answer: The answer is 12.5 ft.


Step-by-step explanation:  Given that there is a rectangle with ratio of its length to breadth 8 : 5. Also, the longer side of the rectangle is 20 feet. We are to find the length of the shorter side.

Let, '8x' and '5x' be the length and breadth of the rectangle respectively. Since length is the longer side, so we have

8x=20\\\\\Rightarrow x=\dfrac{20}{8}\\\\\\\Rightarrow x=2.5.

Therefore, width, length of the shorter side will be

W=5x=5\times 2.5=12.5~\textup{ft.}

Thus, the length of the shorter side is 12.5 ft.


6 0
3 years ago
Read 2 more answers
Round 939,515 to the nearest ten
Radda [10]

Answer:

939,520 would be the answer. You're rounding the 15 part up to the next ten which would be 20, making the answer 939,520.

6 0
3 years ago
Read 2 more answers
If f (x)=3^5-x + 6, what is the value of f (5), to the nearest tenth?
Amiraneli [1.4K]

Answer:

240

Step-by-step explanation:

3^5 - 5 + 6

243 - 5 = 238+6= 244

240?

8 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
Suppose that y varies directly as the square of x, and that y=567 when x=9. What is y when x=4
Gwar [14]

Answer:

y  = 112

Step-by-step explanation:

If  y varies directly as the square of x

mathematically;

y ∝ x²

y = kx² where 'k' is the constant of proportionality

let u know the value of k for this equation by making k the subject of the formular from y = kx²

divide both sides by x²

k = y / x²

given that  y = 567 x = 9

k = 567/(9)²

k = 567/81

k = 7

now to get the value of  y when x = 4

from the equation conneting x, y,

y = kx²

y = 7 × (4)²

y = 7 × 16

y  = 112

6 0
3 years ago
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