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2 years ago

Consider the series 3/2 + 3/4 + 3/8 + 3/16 + ... What expression defines Sn?

Mathematics

1 answer:

snow_lady [41]2 years ago

5 0

For the series 3/2 + 3/4 + 3/8 + 3/16 + ...the sum will be calculated by the formula $3(\dfrac{1}{2})^n$ .

<h3>What is number series?</h3>

When the numbers are arranged in a series with some logic then it is called as the number series.

here we have a series:

$\dfrac{3}{2}+\dfrac{3}{4}+\dfrac{3}{8}+\dfrac{3}{16}+............$

The sum of the series will be given as:

$S_n=3(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+............)$

$S_n=3 (\dfrac{1}{2})^n$

Hence for the series 3/2 + 3/4 + 3/8 + 3/16 + ...the sum will be calculated by the formula $3(\dfrac{1}{2})^n$

To know more about number series follow

brainly.com/question/6561461

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Answer and Step-by-step explanation:

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3 years ago

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Let X denote the number of bars of service on your cell phone whenever you are at an intersection with the following probabiliti

lorasvet [3.4K]

Answer:

Step-by-step explanation:

Hello!

Given the variable

X: number of bars of service on your cell phone.

The posible values of this variable are {0, 1, 2, 3, 4, 5}

a. F(X) is the cummulative distribution function P(X≤x₀) you can calculate it by adding all the point probabilities for each value:

X: 0; 1; 2; 3; 4; 5

P(X): 0.15; 0.25; 0.25; 0.15; 0.1; 0.1

F(X): 0.15; 0.4; 0.65; 0.8; 0.9; 1

The maximum cumulative probability of any F(X) is 1, knowing this, you can calculate the probability of X=5 as: 1 - F(4)= 1 - 0.9= 0.1

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E(X)= ∑Xi*Pi= (0*0.15)+(1*0.25)+(2*0.25)+(3*0.15)+(4*0.1)+(5*0.1)

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You can read this expression as the probability of having less than 2 bars of service. This means you can have either one or zero service bars, you can rewrite it as:

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This expression means "the probability of having at most (or less or equal to) 3.5 service bars", this variable doesn't have the value 3.5 in its definition range so you have to look for the accumulated probability until the lesser whole number. This expression includes the probabilities of X=0, X=1, X=2, and X=3, you can express it as the accumulated probability until 3, F(3):

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