) Elise has put 5 cans (all of the same size) on her kitchen counter; 2 cans of vegetables, 2 cans of soup , and 1 can of peaches. Her son, Ryan, takes the labels off the cans and throws them away. Elise then chooses 2 cans (without replacement) at random to open. Find the probability that she will open at least 1 can of soup.

0.7 = 70%.

Step-by-step explanation:

There are 5 cans, and she will pick 2, so the number of possibilities that she can pick 2 cans is a combination of 5 choose 2:

C(5,2) = 5!(3!*2!) = 5*4/2 = 10

To find how many possibilities there are with at least 1 can of soup, we can find the number of groups that include no cans of soup, and then see how many possibilities complete the total 10:

There are 3 "no-soup" cans, so the number of possibilities is a combination of 3 choose 2:

C(3,2) = 3!/2! = 3

So, there are 3 possibilities that have no cans of soup, so the number of possibilities that have at least 1 can of soup is 10 - 3 = 7

Then, the probability is 7 / 10 = 0.7 = 70%.

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !

answer:   expression= p+p*d%

price reduction + sale price = $625

sale price: $p+$p*d%

where d% is the percentage of discount

step-by-step explanation:

original price = p

price went down by $150 price reduction $p*d%

sale price= $ 475

therefore the price before sale is: $625

and the expression that shows the price before sale is: p+p*d%