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3 years ago

10

Which is the graph of f(x)= 4[1/2]^x ?

1 answer:

vivado [14]3 years ago

4 0

Answer:

B

Step-by-step explanation:

A

f(1) = 4*(1/2)^1 = 4/2 = 2

(1,1) is not the same thing.

B

B is the answer.

C

f(0)=4 not 2 C is incorrect. C is not the answer

D

Same problem as C

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Eric and Mark were at a ball game and visited the concession stand. They each got a hamburger, and Mark got a lemonade and some

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Write an out an equation using the information given. So:

8 + 5 + x - 2 = 14.5
<span>13 + x - 2 = 14.5 </span>
<span>11 + x = 14.5 </span>
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This means that the answer is D.

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3 years ago

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How many inches are in a foot then converted into millimeters

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There are 12 inches in a foot. When you convert 12 into millimeters, it becomes 304.8.

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A company is studying the number of monthly absences among its 125 employees. The following probability distribution shows the l

Angelina_Jolie [31]

Answer:

(1)

$E(x) = 0.72$

$Var(x) = 1.1616$

$\sigma = 1.078$

(2)

$E(x) = 12$

$Var(x) = 1.3$

$\sigma = 1.14$

Step-by-step explanation:

Solving (1):

$\begin{array}{ccccccc}{Days} & {0} & {1} & {2} & {3} & {4}& {5} \ \\ {Probability} & {0.60} & {0.20} & {0.12} & {0.04} & {0.04} & {0.00} \ \end{array}$

(a): Mean

This is calculated as:

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 0 * 0.60 + 1 * 0.20 + 2 * 0.12 + 3 * 0.04 + 4 * 0.04 + 5 * 0.00$

$E(x) = 0.72$

Solving (b): The variance

This is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Where:

$E(x) = 0.72$

and $E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 0^2 * 0.60 + 1^2 * 0.20 + 2^2 * 0.12 + 3^2 * 0.04 + 4^2 * 0.04 + 5^2 * 0.00$

$E(x^2) = 1.68$

So, we have:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = 1.68 - 0.72^2$

$Var(x) = 1.1616$

Solving (c): The standard deviation.

This is calculated as:

$\sigma = \sqrt{Var(x)}$

$\sigma = \sqrt{1.1616}$

$\sigma = 1.078$

Solving (2):

$\begin{array}{ccccccc}{x} & {10} & {11} & {12} & {13} & {14}& { } \ \\ {P(x)} & {0.1} & {0.25} & {0.3} & {0.25} & {0.1} & { } \ \end{array}$

(a): Mean

This is calculated as:

$E(x) = \sum x * P(x)$

So, we have:

$E(x) = 10 * 0.10 + 11 * 0.25 + 12 * 0.3 + 13 * 0.25 + 14 * 0.1$

$E(x) = 12$

Solving (b): The variance

This is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Where:

$E(x) = 12$

and $E(x^2) = \sum x^2 * P(x)$

So, we have:

$E(x^2) = 10^2 * 0.10 + 11^2 * 0.25 + 12^2 * 0.3 + 13^2 * 0.25 + 14^2 * 0.1$

$E(x^2) = 145.3$

So, we have:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = 145.3 - 12^2$

$Var(x) = 1.3$

Solving (c): The standard deviation.

This is calculated as:

$\sigma = \sqrt{Var(x)}$

$\sigma = \sqrt{1.3}$

$\sigma = 1.14$

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777dan777 [17]

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vivado [14]

Answer:

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Step-by-step explanation:

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4 years ago

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