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motikmotik
3 years ago
7

Your friend asks you to help him study and will pay you 5 dimes the first time you help him. You agree to help if he multiplies

your payment by 5 for each study session. After 2 study sessions, you will receive 25 dimes, and after 3 study sessions, you will receive 125 dimes.
Complete and solve the equation that finds the number of dimes he will pay you after the 7th study session.


= number of dimes for 7th study session

Payment for the 7th study session =
dimes
Mathematics
2 answers:
elena55 [62]3 years ago
8 0

Answer:

Step-by-step explanation:

The answer is 5 exponent ^7

bottom 78,125

the other answers other people answered are totally wrong

zvonat [6]3 years ago
6 0
You would have 78,125 dimes after the 7th study session. 

Which would equal out to $7,812.50
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V=length\times width \times height\\\\\dfrac{V}{length\times width}=height\\\\\dfrac{4x^{8}+3x^{6}}{x^{2}}=height\\\\height=4x^{6}+3x^{4}
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3 years ago
Can you find the sum of (-6y-2)+5(3+2.5y) ?
hjlf
(-6y-2)+5(3+2.5y) expend -6y-2+15+12.5 6.5y+13
6 0
3 years ago
Read 2 more answers
Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.
Cloud [144]

If F_Y(y) is the cumulative distribution function for Y, then

F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)

Then the probability density function for Y is f_Y(y)={F_Y}'(y):

f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}

The nth moment of Y is

E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy

Let u=\ln y, so that \mathrm du=\frac{\mathrm dy}y and y^n=e^{nu}:

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du

Complete the square in the exponent:

nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du

But \frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2} is exactly the PDF of a normal distribution with mean n and variance 1; in other words, the 0th moment of a random variable U\sim N(n,1):

E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1

so we end up with

E[Y^n]=e^{\frac12n^2}

3 0
3 years ago
How much is 14 divided by 8218
s344n2d4d5 [400]
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5 0
3 years ago
From home, Mary’s work is two thirds along the way to training. Training is 2.5km from work. Mary normally goes to work, then tr
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First thing to do is to illustrate the problem, Since it was mentioned that work was along the way to training, the order is shown in the picture. Mary's home and workplace are nearer compared to her training center. It is also mentioned that the distance between work and home, denoted as x, is 2/3 of the total distance from home to training. The total distance is (x + 2.5). Thus,

x = 2/3(x+2.5)
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Thus, the distance from home to work is 5 km. This means that Mary has to walk this distance twice to return home to get her shoes. Then, she will travel again the total distance of 5+2.5 = 7.5 km to get to her training center. So,

Total distance = 2(5km) + 7.5 km
Total distance = 17.5 km

3 0
3 years ago
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