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erma4kov [3.2K]

3 years ago

14

A jar contains 125 marbles. Given 5% of the marbles are green, 39% of the marbles are blue, and the rest are red, how many red m

arbles are in the jar?

1 answer:

scZoUnD [109]3 years ago

4 0

Answer:

70

Step-by-step explanation:

39+5=44

100-44=56

56 percent of 125 is 70

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Answer:

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Step-by-step explanation:

21 rand * ($1/7 rand) = $3

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∆ABC has A(-3, 6), B(2, 1), and C(9, 5) as its vertices. The length of side AB is units. The length of side BC is units. The len

leonid [27]

Answer:

AB = 7.07 units

BC = 8.06 units

AC = 12.04 units

Step-by-step explanation:

To find the length for each side of the triangle, apply the distance formula between each pair of vertices.

<u>AB</u>

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{(2--3)^2 + (1-6)^2} \\d = \sqrt{(5)^2 + (-5)^2} \\d = \sqrt{25 + 25} \\d = \sqrt{50} \\d=7.07$

<u />

<u>BC</u>

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{(9-2)^2 + (5-1)^2} \\d = \sqrt{(7)^2 + (4)^2} \\d = \sqrt{49 + 16} \\d = \sqrt{65} \\d=8.06$

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<u>AC</u>

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{(9--3)^2 + (5-6)^2} \\d = \sqrt{(12)^2 + (-1)^2} \\d = \sqrt{144 + 1} \\d = \sqrt{145} \\d=12.04$

4 0

3 years ago

In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz

Ahat [919]

Answer:

a) There is a 18.75% probability that the first question that she gets right is the second question.

b) There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

c) There is a 10.35% probability that she gets the majority of the questions right.

Step-by-step explanation:

Each question can have two outcomes. Either it is right, or it is wrong. So, for b) and c), we use the binomial probability distribution to solve this problem.

Binomial probability distribution

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

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

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

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

And $\pi$ is the probability of X happening.

In this problem we have that:

Each question has 4 choices. So for each question, Robin has a $\frac{1}{4} = 0.25$ probability of getting ir right. So $\pi = 0.25$ . There are five questions, so $n = 5$ .

(a) What is the probability that the first question she gets right is the second question?

There is a 75% probability of getting the first question wrong and there is a 25% probability of getting the second question right. These probabilities are independent.

So

$P = 0.75(0.25) = 0.1875$

There is a 18.75% probability that the first question that she gets right is the second question.

(b) What is the probability that she gets exactly 1 or exactly 2 questions right?

This is: $P = P(X = 1) + P(X = 2)$

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

$P(X = 1) = C_{5,1}.(0.25)^{1}.(0.75)^{4} = 0.3955$

$P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637$

$P = P(X = 1) + P(X = 2) = 0.3955 + 0.2637 = 0.6592$

There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

(c) What is the probability that she gets the majority of the questions right?

That is the probability that she gets 3, 4 or 5 questions right.

$P = P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 3) = C_{5,3}.(0.25)^{3}.(0.75)^{2} = 0.0879$

$P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146$

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

$P = P(X = 3) + P(X = 4) + P(X = 5) = 0.0879 + 0.0146 + 0.001 = 0.1035$

There is a 10.35% probability that she gets the majority of the questions right.

6 0

3 years ago

A marketing firm would like to test-market the name of a new energy drink targeted at 18- to 29-year-olds via social media. A st

Anon25 [30]

Answer:

(a) The probability that a randomly selected U.S. adult uses social media is 0.35.

(b) The probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c) The probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = an US adult who does not uses social media.

<em>Y</em> = an US adult between the ages 18 and 29.

<em>Z</em> = an US adult between the ages 30 and above.

The information provided is:

P (X) = 0.35

P (Z) = 0.78

P (Y ∪ X') = 0.672

(a)

Compute the probability that a randomly selected U.S. adult uses social media as follows:

P (US adult uses social media (<em>X'</em><em>)</em>) = 1 - P (US adult so not use social media)

$=1-P(X)\\=1-0.35\\=0.65$

Thus, the probability that a randomly selected U.S. adult uses social media is 0.35.

(b)

Compute the probability that a randomly selected U.S. adult is aged 18–29 as follows:

P (Adults between 18 - 29 (<em>Y</em>)) = 1 - P (Adults 30 or above)

$=1-P(Z)\\=1-0.78\\=0.22$

Thus, the probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c)

Compute the probability that a randomly selected U.S. adult is 18–29 and a user of social media as follows:

P (Y ∩ X') = P (Y) + P (X') - P (Y ∪ X')

$=0.22+0.65-0.672\\=0.198$

Thus, the probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

6 0

3 years ago