If x = -5 and y = x + 4, what ordered pair represents the solution? A. (5, -1) B. (-5, -1) C. (5, 1) D. (4, 4)

You and a friend are playing a game of chance. Every time you roll a 1 or 2 you are successful, and your friend will pay you $1.

LUCKY_DIMON [66]

At the end of the game you owned 13 cents if 71% of your rolls are successful and 29% of your rolls are unsuccessful option first is correct.

<h3>What is probability?</h3>

It is defined as the ratio of the number of favourable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

We have:

Every time you roll a 1 or 2 you are successful, and your friend will pay you $1. Every time you roll a 3, 4, 5 or 6, you must pay your friend $2.

71% of your rolls are successful:

= 1×0.71

= $0.71 (won)

29% of your rolls are unsuccessful

= 2×2.9

= $0.58 (lose)

The amount owned = 0.71 - 0.58 = $0.13 = 13 cents

Thus, at the end of the game you owned 13 cents if 71% of your rolls are successful and 29% of your rolls are unsuccessful option first is correct.

Learn more about the probability here:

brainly.com/question/11234923

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6 0

2 years ago

For the function below, find a formula for the upper sum obtained by dividing the interval [a comma b ][a,b] into n equal subin

Vlad [161]

Answer:

See below

Step-by-step explanation:

We start by dividing the interval [0,4] into n sub-intervals of length 4/n

$[0,\displaystyle\frac{4}{n}],[\displaystyle\frac{4}{n},\displaystyle\frac{2*4}{n}],[\displaystyle\frac{2*4}{n},\displaystyle\frac{3*4}{n}],...,[\displaystyle\frac{(n-1)*4}{n},4]$

Since f is increasing in the interval [0,4], the upper sum is obtained by evaluating f at the right end of each sub-interval multiplied by 4/n.

Geometrically, these are the areas of the rectangles whose height is f evaluated at the right end of the interval and base 4/n (see picture)

$\displaystyle\frac{4}{n}f(\displaystyle\frac{1*4}{n})+\displaystyle\frac{4}{n}f(\displaystyle\frac{2*4}{n})+...+\displaystyle\frac{4}{n}f(\displaystyle\frac{n*4}{n})=\\\\=\displaystyle\frac{4}{n}((\displaystyle\frac{1*4}{n})^2+3+(\displaystyle\frac{2*4}{n})^2+3+...+(\displaystyle\frac{n*4}{n})^2+3)=\\\\\displaystyle\frac{4}{n}((1^2+2^2+...+n^2)\displaystyle\frac{4^2}{n^2}+3n)=\\\\\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12$

but

$1^2+2^2+...+n^2=\displaystyle\frac{n(n+1)(2n+1)}{6}$

so the upper sum equals

$\displaystyle\frac{4^3}{n^3}(1^2+2^2+...+n^2)+12=\displaystyle\frac{4^3}{n^3}\displaystyle\frac{n(n+1)(2n+1)}{6}+12=\\\\\displaystyle\frac{4^3}{6}(2+\displaystyle\frac{3}{n}+\displaystyle\frac{1}{n^2})+12$