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2 years ago

6

Vince weighs 160 pounds and his friend nadir weighs 140 nadir calculated that his weight on another planet would be about 56lb a

pproximately what would Vince weigh on the other planet

2 answers:

zzz [600]2 years ago

7 0

His weight would be 64lb

chubhunter [2.5K]2 years ago

3 0

Vince would weigh 64 pounds

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John is participating in a marathon that is 26.2 miles. His distance (d, in miles) depends on his time (t, in hours). Which is a

madreJ [45]

The range of a function is representative of the values on the y-axis. In this case, the graph will contain distance on the y-axis at is the dependent variable, while the independent variable is time.

We know that the minimum value of the distance will be 0, given that there can be no negative distance. Moreover, the maximum value is 26.2 miles, since the marathon will then be over. Therefore, a good range for the situation will be 0-27 miles.

6 0

2 years ago

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 Two marbles are withdrawn randomly.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We can assume that both marbles are withdrawn at the same time.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 The number of different ways in which we can withdrawn 2 marbles out of 10 is $\binom{10}{2}$ .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) = $\frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}$

7 0

2 years ago

What is the equation of <br> -2(x-3)+4x=-(-x+1)

BabaBlast [244]

Answer:

Exact form: $-5/3$

Decimal form: $-1.66$ (repeating)

Mixed number form: $-1\frac{2}{3}$

8 0

1 year ago

Read 2 more answers

Ken grew 4/5 of an inch last year. Saying grew 3/8 of an inch. Who grew more and that by how much?

Romashka-Z-Leto [24]

Ken grew more by 17/40 of an inch

4 0

1 year ago

A rectangular area is formed having a perimeter of 40 cm. Determine the length and breadth of the rectangle if it is to enclose

Vilka [71]

Let x and y be the dimensions of the rectangle. If the perimeter is 40, we have

$2(x+y)=40 \iff x+y=20$

We can expression one variable in terms of the others as

$x+y=20 \iff x=20-y$

Since the area is the product of the dimensions, we have

$xy=(20-y)y=-y^2+20y$

This is a parabola facing down, so it's vertex is the maximum:

$f(y)=-y^2+20y \implies f'(y)=-2y+20$

So, the maximum is

$f'(y)=0 \iff -2y+20=0 \iff 2y=20 \iff y=10$

And since we know that $x+y=20$ , we have $x=10$ as well.

This is actually a well known theorem: out of all the rectangles with given perimeter, the one with the greatest area is the square.

5 0

2 years ago