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Sergio039 [100]

2 years ago

8

If a complete graph represents the road network in a country, it would be advantageous for that country in the even that an infe

ctious disease hits one of the towns on the road network. Yes, it would be advantageous No, it would not be advantageous

1 answer:

Contact [7]2 years ago

6 0

Answer:

everyone

Step-by-step explanation:

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c because no x value is repeated

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Simplify or compute: a·a·a/3a

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Answer:

. I I wish I could help. <em>Sorry</em>

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A juggler is performing her act using several balls. She throws the balls up at an initial height of 4 feet, with a speed of 15

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3 years ago

Write an equation for the line passing through the given pair of points. give the final answer in standard form (-4,-2) and (3,-

Andrew [12]

Answer:

y = -2/7x - 22/7

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

<u>Algebra I</u>

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Slope-Intercept Form: y = mx + b

m - slope

b - y-intercept

Step-by-step explanation:

<u>Step 1: Define</u>

Point (-4, -2)

Point (3, -4)

<u>Step 2: Find slope </u><em><u>m</u></em>

Substitute: $m=\frac{-4+2}{3+4}$
Add: $m=\frac{-2}{7}$

<u>Step 3: Find y-intercept </u><em><u>b</u></em>

Define: y = -2/7x + b
Substitute: -4 = -2/7(3) + b
Multiply: -4 = -6/7 + b
Isolate <em>b</em>: -22/7 = b
Rewrite: b = -22/7

<u>Step 4: Write linear equation</u>

y = -2/7x - 22/7

7 0

2 years ago

An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of mov- ing violatio

olga2289 [7]

Answer:

a) $E(Y) = \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60$

b) $E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1$

And then the expected value would be:

$E(100Y^2) = 100*1.1= 110$

Step-by-step explanation:

We assume the following distribution given:

Y 0 1 2 3

P(Y) 0.60 0.25 0.10 0.05

Part a

We can find the expected value with this formula:

$E(Y) = \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) =0*0.60 +1*0.25 +2*0.1 +3*0.05 = 0.60$

Part b

If we want to find the expected value of $100 Y^2$ we need to find the expected value of Y^2 and we have:

$E(Y^2) = \sum_{i=1}^n Y^2_i P(Y_i)$

And replacing we got:

$E(Y^2) =0^2*0.60 +1^2*0.25 +2^2*0.1 +3^2*0.05 = 1.1$

And then the expected value would be:

$E(100Y^2) = 100*1.1= 110$

4 0

2 years ago