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nikdorinn [45]
3 years ago
11

12) Find the sum, S7, for the geometric series (1 + 5 + 25+ --- + 15625).

Mathematics
1 answer:
vfiekz [6]3 years ago
8 0

Answer:

Find the (r) between adjacent members

a2/a1=+25/5=5

a3/a2=+125/+25=5

a4/a3=+625/+125=5

a5/a4=+3125/+625=5

a6/a5=+15625/+3125=5

The ration (r) between every two adjacent members of the series is constant and equal to 5

General Form: a

n

=a

1

×r

n-1

This geometric series: a

n

=5×5

n-1

The nature of this series

Sum of finite geometric series members

sum of a geometric series

The sum of our particular series

a 1-rn = 5 1-56 = 5 1-15625 = 5 -15624 =5×3906=19530

1-r 1-5 -4 -4

Finding the n

th

element

a2 =a1×r2-1  =a1×r1 =5×51 =25

a3 =a1×r3-1  =a1×r2 =5×52 =125

a4 =a1×r4-1  =a1×r3 =5×53 =625

a5 =a1×r5-1  =a1×r4 =5×54 =3125

a6 =a1×r6-1  =a1×r5 =5×55 =15625

a7 =a1×r7-1  =a1×r6 =5×56 =78125

a8 =a1×r8-1  =a1×r7 =5×57 =390625

a9 =a1×r9-1  =a1×r8 =5×58 =1953125

a10 =a1×r10-1  =a1×r9 =5×59 =9765625

a11 =a1×r11-1  =a1×r10 =5×510 =48828125

a12 =a1×r12-1  =a1×r11 =5×511 =244140625

a13 =a1×r13-1  =a1×r12 =5×512 =1220703125

a14 =a1×r14-1  =a1×r13 =5×513 =6103515625

a15 =a1×r15-1  =a1×r14 =5×514 =30517578125

a16 =a1×r16-1  =a1×r15 =5×515 =152587890625

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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

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E(X)= 7

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

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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}

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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

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<h3><u>Solution:</u></h3>

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<em><u>The perimeter of square is given as:</u></em>

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<u>case c)</u> The function can be represented by the equation y =(1/10)x + 60

The statement is True

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