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Dmitriy789 [7]
3 years ago
8

What is (4.81 times 10 Superscript 16 Baseline) (1.1 times 10 Superscript negative 4 Baseline) in scientific notation?

Mathematics
2 answers:
torisob [31]3 years ago
4 0

Answer:

c

Step-by-step explanation:

justtook that test yesterday trust me its c

viktelen [127]3 years ago
3 0

Answer:

5.291 times 10 Superscript 12

Step-by-step explanation:

(4.81 × 10¹⁶)(1.1 × 10^-4)

(4.81 × 1.1) × 10^(16-4)

5.291 × 10¹²

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How to factor out xsquared+6x-27
Aleks04 [339]
X^2 + 6x - 27
(x - 3)(x + 9)

The answer is: (x - 3)(x + 9).
5 0
3 years ago
Eddie deposited $1200 into an account that earns 3% interest compounded 2 times per year. How much money will Eddie have in his
nirvana33 [79]
Total = Principal * (1+(rate/n))^n*years
Total = 1,200 * (1.015)^14
Total = 1,200 * <span><span><span>1.2317557307 </span> </span> </span>
<span>Total = 1,478.11</span>
Source:
http://www.1728.org/compint3.htm




8 0
3 years ago
Find two consecutive integers whose sum is -29
LuckyWell [14K]

Example:

The sum of two consecutive integers is -15.  What is the product of the two integers?

x = 1st consecutive integer

x + 1 = 2nd consecutive integer   {consecutive integers increase by 1}


x + x + 1 = -15   {the sum of two consecutive integers is -15}

2x + 1 = -15   {combined like terms}

2x = -16   {subtracted 1 from each side}

x = -8   {divided each side by 2}

x + 1 = -7   {substituted -8, in for x, into x + 1}


-8 and -7 are the two consecutive integers


Their product is 56.

​


8 0
3 years ago
If X and Y are independent continuous positive random
Leni [432]

a) Z=\frac XY has CDF

F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy

F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy

where the last equality follows from independence of X,Y. In terms of the distribution and density functions of X,Y, this is

F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy

Then the density is obtained by differentiating with respect to z,

f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy

b) Z=XY can be computed in the same way; it has CDF

F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy

F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy

Differentiating gives the associated PDF,

f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy

Assuming X\sim\mathrm{Exp}(\lambda_x) and Y\sim\mathrm{Exp}(\lambda_y), we have

f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy

\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}

and

f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy

\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0
3 years ago
Lloyd was paid a 6% commission on sales of $95,292. How much money was he paid in commissions?
alexgriva [62]
$5717.52

You’d just take the 6 percent and make it 0.06 then multiply it by 95,292
4 0
3 years ago
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