Number 5. Please help

In the United States, voters who are neither Democrat nor Republican are called Independent. It is believed that 11% of voters a

Radda [10]

Answer:

a) 0.0214 = 2.14% probability that none of the people are Independent.

b) 0.8516 = 85.16% probability that fewer than 6 are Independent.

c) 0.8914 = 89.14% probability that more than 2 people are Independent.

Step-by-step explanation:

For each people, there are only two possible outcomes. Either they are independent, or they are not. For each person asked, the probability of them being Independent voters is the same. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

It is believed that 11% of voters are Independent.

This means that $p = 0.11$

A survey asked 33 people to identify themselves as Democrat, Republican, or Independent.

This means that $n = 33$

A. What is the probability that none of the people are Independent?

This is P(X = 0). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

0.0214 = 2.14% probability that none of the people are Independent.

B. What is the probability that fewer than 6 are Independent?

This is

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

$P(X = 1) = C_{33,1}.(0.11)^{1}.(0.89)^{32} = 0.0872$

$P(X = 2) = C_{33,2}.(0.11)^{2}.(0.89)^{31} = 0.1724$

$P(X = 3) = C_{33,3}.(0.11)^{3}.(0.89)^{30} = 0.2202$

$P(X = 4) = C_{33,4}.(0.11)^{4}.(0.89)^{29} = 0.2041$

$P(X = 5) = C_{33,5}.(0.11)^{5}.(0.89)^{28} = 0.1463$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0214 + 0.0872 + 0.1724 + 0.2202 + 0.2041 + 0.1463 = 0.8516$

0.8516 = 85.16% probability that fewer than 6 are Independent.

C. What is the probability that more than 2 people are Independent?

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{33,0}.(0.11)^{0}.(0.89)^{33} = 0.0214$

$P(X = 1) = C_{33,1}.(0.11)^{1}.(0.89)^{32} = 0.0872$

$P(X < 2) = 0.0214 + 0.0872 = 0.1086$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1086 = 0.8914$

0.8914 = 89.14% probability that more than 2 people are Independent.

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sergiy2304 [10]

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sladkih [1.3K]

Answer:

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1/3 cup of flour is used to make 5 dinner rolls.

Radda [10]

Divide.

1/3 / 5

Or multiply by the reciprocal of 5:

1/3 * 1/5 = 1/15

Therefore 1/15 cup of flour is used to make 1 dinner roll.

We can set up a proportion:

5/1/3 = x/5 2/3

Multiply 5 2/3 to both sides:

28 1/3 / 1/3 = x

85 = x

So 85 rolls can be made with 5 2/3 cups of flour.

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enyata [817]

Answer:

The center is (2,8)

Step-by-step explanation:

The equation of a circle is written as

(x-h)^2+ (y-k)^2 = r^2

where ( h,k) is the center and r is the radius

(x-2)^2+(y-8)^2=33

The center is (2,8) and the radius is sqrt(33)

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