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Aleksandr-060686 [28]
3 years ago
9

Someone pls help me!!! Will give brainliest! A survey of 1000 students found that 41% of the surveyed students were freshmen, 31

% were juniors, and 67% lived on the college campus. (A) If 260 of the surveyed students were freshmen living on the college cam- pus, then how many surveyed students were either freshmen or living on campus? (B) Suppose from a previous survey you know that of interviewed juniors, only 23% live on campus. What is the probability a surveyed student is a junior or someone that lives on campus?

Mathematics
1 answer:
Korvikt [17]3 years ago
8 0

Answer:

<em>(A) 560 students</em>

<em>(B) P = 0.41</em>

Step-by-step explanation:

<em>(Please check the attached picture for more details)</em>

As given, a survey of 1000 students found that:

41% of the surveyed students were freshmen

=> 1000 x 41/100 = 410 surveyed students were freshmen

31% were juniors

=> 1000 x 31/100 = 310 surveyed students were junior

67% of surveyed students lived on the college campus

=> 1000 x 67/100 = 670 surveyed students lived on the campus

<em>(A) 260 of the surveyed students were freshmen living on the campus</em>

=> A = 410 - 260 = 150 surveyed students were freshmen NOT living on the campus

=> B = 670 - 260 = 410 surveyed students were not freshmen living on the campus

=> The number of surveyed students who were either freshmen or living on the campus:

N = A + B = 150 + 410 = 560

<em>(B)  From a previous survey you know that of interviewed juniors, only 23% live on campus (we don't really need this)</em>

=>The probability a surveyed student is a junior or someone that lives on campus: P = (670 - 260)/1000 = 0.41

I hope this helps!

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A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if
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Answer:

a) The expected value is \frac{-1}{15}

b) The variance is  \frac{49}{45}

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is \binom{10}{2}.

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is \binom{5}{2}. Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is \binom{5}{1}\cdot \binom{5}{1}. So, we define the following probabilities.

Probability of winning: \frac{2\binom{5}{2}}{\binom{10}{2}}= \frac{4}{9}

Probability of losing \frac{(\binom{5}{1})^2}{\binom{10}{2}}\frac{5}{9}

Let X be the expected value of the amount you can win. Then,

E(X) = 1.10*probability of winning - 1 probability of losing =1.10\cdot  \frac{4}{9}-\frac{5}{9}=\frac{-1}{15}

Consider the expected value of the square of the amount you can win, Then

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We will use the following formula

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Var(X) = \frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}

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GarryVolchara [31]

Answer:

D:) (2,2,) is the Answer

Step-by-step explanation:

Solve the following system:


{X - 2 Y = -2 | (equation 1)


{3 X - 2 Y = 2 | (equation 2)



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{3 X - 2 Y = 2 | (equation 1)


{X - 2 Y = -2 | (equation 2)



Subtract 1/3 × (equation 1) from equation 2:


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{0 X - (4 Y)/3 = (-8)/3 | (equation 2)



Multiply equation 2 by -3/4:


{3 X - 2 Y = 2 | (equation 1)


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{0 X+Y = 2 | (equation 2)



Divide equation 1 by 3:


{X+0 Y = 2 | (equation 1)


{0 X+Y = 2 | (equation 2)



Collect results:


Answer:  {X = 2 , Y = 2

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