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3 years ago

Pls help me!!!! Im stuck!!!

Mathematics

2 answers:

blondinia [14]3 years ago

8 0

Answer:

infinite solutions

Step-by-step explanation:

you can solve this by the elimination method. take the rover a equation and multiply it by -7 and then combine the equations. when you do this, you can see that either side equals 0 which means that there are infinite solutions.

-7(y= 1.5x - 6) which equals the equation below

-7y= -10.5x-42 then combine the other equation

7y=10.5x+42 and that equals

0=0

exis [7]3 years ago

3 0

Answer:

has no solution.

Step-by-step explanation:

you can see that the two equations are almost identical except for the fact that one is multiplied by 7, but it isnt completely multiplied correctly so there would be no solutions.

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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 Two marbles are withdrawn randomly.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

topjm [15]

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We can assume that both marbles are withdrawn at the same time.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 The number of different ways in which we can withdrawn 2 marbles out of 10 is $\binom{10}{2}$ .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

E(X^2) = (1.10^2)*probability of winning + probability of losing = $1.10^2\cdot \frac{4}{9}+\frac{5}{9}=\frac{82}{75}$

We will use the following formula

$Var(X) = E(X^2)-E(X)^2$

Thus

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

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