All categories

3 years ago

WILL GIVE BEST RESPONSE

1 answer:

4 0

We will use demonstration of recurrences1) for n=1, 10= 5*1(1+1)=5*2=10, it is just
2) assume that the equation 10 + 30 + 60 + ... + 10n = 5n(n + 1) is true, for all positive integers n>=1
3) let's show that the equation is also true for n+1, n>=1
10 + 30 + 60 + ... + 10(n+1) = 5(n+1)(n + 2)
let be N=n+1, N is integer because of n+1, so we have
10 + 30 + 60 + ... + 10N = 5N(N+1), it is true according 2)
so the equation is also true for n+1,
finally, 10 + 30 + 60 + ... + 10n = 5n(n + 1) is always true for all positive integers n.

You might be interested in

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total s

garri49 [273]

Answer

a. The expected total service time for customers = 70 minutes

b. The variance for the total service time = 700 minutes

c. It is not likely that the total service time will exceed 2.5 hours

Step-by-step explanation:

This question is incomplete. I will give the complete version below and proceed with my solution.

Refer to Exercise 3.122. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) Is it likely that the total service time will exceed 2.5 hours?

Reference

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour.

From the information supplied, we denote that

$X=$ Customers that arrive within the hour

and since $X$ follows a Poisson distribution with mean $\alpha$ = 7

Therefore,

$E(X)= 7$

$& V(X)=7$

Let $Y =$ the total service time for customers arriving during the 1 hour period.

Now, since it takes approximately ten minutes to serve each customer,

$Y=10X$

For a random variable $X$ and a constant $c$ ,

$E(cX)=cE(X)\\V(cX)=c^2V(X)$

Thus,

$E(Y)=E(10X)=10E(X)=10*7=70\\V(Y)=V(10X)=100V(X)=100*7=700$

Therefore the expected total service time for customers = 70 minutes

and the variance for serving time = 700 minutes

Also, the probability of the distribution $Y$ is,

$p_Y(y)=p_x(\frac{y}{10} )\frac{dx}{dy} =\frac{\alpha^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-\alpha } \frac{1}{10}\\ =\frac{7^{\frac{y}{10} } }{(\frac{y}{10})! }e^{-7 } \frac{1}{10}$

So the probability that the total service time exceeds 2.5 hrs or 150 minutes is,

$P(Y>150)=\sum^{\infty}_{k=150} {p_Y} (k) =\sum^{\infty}_{k=150} \frac{7^{\frac{k}{10} }}{(\frac{k}{10})! }.e^{-7} .\frac{1}{10} \\=\frac{7^{\frac{150}{10} }}{(\frac{150}{10})! } .e^{-7}.\frac{1}{10} =0.002$

0.002 is small enough, and the function $\frac{7^{\frac{k}{10} }}{(\frac{k}{10} )!} .e^{-7}.\frac{1}{10}$ gets even smaller when $k$ increases. Hence the probability that the total service time exceeds 2.5 hours is not likely to happen.

3 0

3 years ago

Solve the equation 8(4- x) = 7x + 2 x=

Jet001 [13]

Answer: x=2

Step-by-step explanation:

Expand.

32−8x=7x+2

Add 8x to both sides.

32=7x+2+8x

Simplify 7x+2+8x7x+2+8x to 15x+215x+2.

32=15x+2

Subtract 22 from both sides.

32−2=15x

Simplify 32−2 to 30.

30=15x

Divide both sides by 15.

30/15=x

Simplify 30/15 to 2

2=x

Switch sides.

x=2

7 0

3 years ago

Solve the system of equations y = x + 5 and y = 5x - 11 using a graphical method.

tresset_1 [31]

You can use substitution for this.

x+5 = 5x -11

Combine like terms

16=4x

Divide 4 on both sides to isolate the x

x=4

Plug x=4 to one of the equations

y= 4 +5

Combine like terms

y=9

So the answer is A. (4,9)

I hope this helps you out a bunch hun :)

4 0

3 years ago

) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

Fynjy0 [20]

The surface (call it $S$ ) is a triangle with vertices at the points

$x=0,y=0\implies z=2\implies(0,0,2)$

$x=0,y=2\implies z=-2\implies(0,2,-2)$

$x=2,y=0\implies z=-8\implies(2,0,-8)$

Parameterize $S$ by

$\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=(20v,8v,4v)$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv$

$=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}$

8 0

4 years ago

B) Tyler swam 32 laps in the pool. Kevin swam 3 times as many laps as Tyler. Lets the number of

evablogger [386]

If Tyler did 32 laps and Kevin did 3 times that amount; its quite simple, you would find three times the amount of Tylers laps which is 96.

4 0

2 years ago

Read 2 more answers