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slega [8]
3 years ago
13

John spent 2/3 of his money on a pen and calculator .The calculator cost 3 times as much as the pen .If the calculator cost $24,

how much money did he have left
Mathematics
1 answer:
Helga [31]3 years ago
4 0
Calculator --- $24
Pen --- 24/3 = $8
Money spent --- 24+8 = 32
Total money --- 32 divided by 2 times 3 = $48
Money left --- 48-32 = $16
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A photograph was enlarged to a width of 15 inches. If the scale factor was 3 /2 , what was the width of the original photograph?
andre [41]
If the width is 15, and the scale factor from width to length is 3:2, then you will have to divide 15 by 3 to get your original width.
15/3 = 5.
5 is your original width.
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Right to explain how this digit 6 changes value in number 666, 666, 666
zvonat [6]
To a higher number the six changes value in this number above
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3 years ago
-2(6+s)&gt;-15-2s<br><br> Please answer!!
valkas [14]
This problem has infinitly many solutions
6 0
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Read 2 more answers
Write an equation to solve each problem and then solve it. b In a 3-digit number, the hundreds digit is four more than the ones
Misha Larkins [42]

Answer:

digits are: 0, 4, 8. The required number is: 480.

Step-by-step explanation:

→ Let ones digits is 'x', then according to the condition the hundreds digit is 'x+4', and the tens digit is '2(x+4)'.

→ According to the condition the sum of the digits is: x+(x+4)+2(x+4)=12.

→ After evaluation this equation, x=0 - this is ones digit;

→ x+4=0+4=4 - this is hundreds digit;

→ 2(x+4)=2(0+4)=8 - this is ten digit.

3 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
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