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AleksAgata [21]

3 years ago

10

In 2012 Obama receiver 65,899,660 votes and Romney received 60,932,152 votes. In this election, if a voter for either Obama or R

omney is chosen at random what is the probability of the event that the voter opted for Romney?

2 answers:

creativ13 [48]3 years ago

5 0

Answer:

Social media is now a key tool candidates use to reach the public.

Step-by-step explanation:

On edg.

Contact [7]3 years ago

3 0

For the sake of simplicity
obama = 66 mil
romney = 61 mil

66+61 = 127mil total votes

there are 66mil chances to pick an obama vote from 127mil -- 66/127

there are 61mil chance to pick a romney vote from 127 mil -- 61/127

see where im going with this?

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ella [17]

Answer:

x=50 so ABD = 175

Step-by-step explanation:

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3 years ago

I Need to find the Function Operations

Alinara [238K]

Answer:

1) $(f+g)(x)=2x^2+3x-6$

2) $(f-g)(x)=2x^2+x$

3) $(f \cdot g)(x)=2x^3-4x^2-9x+9$

Step-by-step explanation:

So we have the two functions:

$f(x)=2x^2+2x-3\text{ and } g(x)=x-3$

And we want to find (f+g)(x), (f-g)(x), and (f*g)(x).

1)

(f+g)(x) is the same to f(x)+g(x). Substitute:

$(f+g)(x)=f(x)+g(x)\\=(2x^2+2x-3)+(x-3)$

Combine like terms:

$=(2x^2)+(2x+x)+(-3-3)$

Add:

$=2x^2+3x-6$

So:

$(f+g)(x)=2x^2+3x-6$

2)

(f-g)(x) is the same to f(x)-g(x). So:

$(f-g)(x)=f(x)-g(x)\\=(2x^2+2x-3)-(x-3)$

Distribute:

$=(2x^2+2x-3)+(-x+3)$

Combine like terms:

$=(2x^2)+(2x-x)+(-3+3)$

Simplify:

$=2x^2+x$

So:

$(f-g)(x)=2x^2+x$

3)

(f*g)(x) is the same to f(x)*g(x). Thus:

$(f\cdot g)(x)=f(x)\cdot g(x)\\=(2x^2+2x-3)(x-3)$

Distribute:

$=(2x^2+2x-3)(x)+(2x^2+2x-3)(-3)$

Distribute:

$=(2x^3+2x^2-3x)+(-6x^2-6x+9)$

Combine like terms:

$=(2x^3)+(2x^2-6x^2)+(-3x-6x)+(9)$

Simplify:

$=2x^3-4x^2-9x+9$

So:

$(f \cdot g)(x)=2x^3-4x^2-9x+9$

4 0

3 years ago

Dustin has a bag of marbles with 4 blue marbles, 4 white marbles, and 3 red marbles.

diamong [38]

Answer:

A.

$\dfrac{6}{55}$

B.

$\dfrac{6}{55}$

C.

$\dfrac{4}{165}$

Step-by-step explanation:

Dustin has a bag of marbles with 4 blue marbles, 4 white marbles, and 3 red marbles. There are 11 marbles in total.

A. The probability that the first marble drawn is blue marble is

$\dfrac{4}{11},$

the probability that the second marble drawn is red marble is

$\dfrac{3}{10}$

The probability that the first marble drawn is blue and the second is red is

$\dfrac{4}{11}\cdot \dfrac{3}{10}=\dfrac{12}{110}=\dfrac{6}{55}$

B. The probability that the first marble drawn is red marble is

$\dfrac{3}{11},$

the probability that the second marble drawn is white marble is

$\dfrac{4}{10}$

The probability that the first marble drawn is red and the second is white is

$\dfrac{3}{11}\cdot \dfrac{4}{10}=\dfrac{12}{110}=\dfrac{6}{55}$

C. The probability that the first marble drawn is blue marble is

$\dfrac{4}{11},$

the probability that the second marble drawn is blue marble is

$\dfrac{3}{10},$

the probability that the third marble drawn is blue marble is

$\dfrac{2}{9},$

The probability that the first, second and third marbles drawn are blue is

$\dfrac{4}{11}\cdot \dfrac{3}{10}\cdot \dfrac{2}{9}=\dfrac{24}{990}=\dfrac{4}{165}$

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3 years ago

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3 years ago

If you solve this please show me how!

Veronika [31]

By similarity theorem, I will send you a picture when I'm done.

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3 years ago