The sale bin in a clothing store contains an assortment of t-shirts in different sizes. There are 7 small, 8 medium, and 4 large shirts. Alan is looking for a large shirt. He starts grabbing shirts one at a time and checking the size. After he checks each shirt, he leaves it outside the bin. What is the probability that at least one of the first four shirts he checks is a large?

P(at least 1 large) = 0.648

P(at least 1 large) = 64.8%

Step-by-step explanation:

We have 7 small shirts, 8 medium shirts and 4 large shirts

Total number of shirts = 7 + 8 + 4 = 19 shirts

The probability that at least one of the first four shirts he checks is a large is given by

P(at least 1 large) = 1 - P(no large)

So first we need to find the probability that the none of the first four shirts he checks are large.

For the first check, there are 15 small and medium shirts and total 19 shirts so,

15/19

For the second check, there are 14 small and medium shirts and total 18 shirts left so,

14/18

For the third check, there are 13 small and medium shirts and total 17 shirts left so,

13/17

For the forth check, there are 12 small and medium shirts and total 16 shirts left so,

12/16

the probability of not finding the large shirt is,

P(no large) = 15/19*14/18*13/17*12/16

P(no large) = 0.352

Therefore, the probability of finding at least one large shirt is,

P(at least 1 large) = 1 - P(no large)

P(at least 1 large) = 1 - 0.352

P(at least 1 large) = 0.648

P(at least 1 large) = 64.8%

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.

the area of the rectangle is 35units^2.

step-by-step explanation:

graph it and find the distance.

it is a 5*7 rectangle.

therefore the area is 35units^2.