A bag contains 40 cards numbered 1 through 40 that are either red or blue. A card is drawn at random and placed back in the bag. This is done four times. Two red cards are drawn, numbered 31 and 19, and two blue cards are drawn, numbered 22 and 7. Your friend concludes that red cards and even numbers must be mutually exclusive. Is your friend correct? Explain.

Step-by-step explanation:

Given that a bag contains 40 cards numbered 1 through 40 that are either red or blue. A card is drawn at random and placed back in the bag.

This is done four times. Two red cards are drawn, numbered 31 and 19, and two blue cards are drawn, numbered 22 and 7.

From the above we cannot conclude that red cards and even numbers are mutually exclusive

Just drawing two red cards and because the two happen to be odd we cannot generalize the red cards have odd numbers.

This might have occurred due to simple chance from a comparatively large number of 40 cards.

Suppose say we have red cards 20, and 19 red 1 blue.

Then drawing 2 from 19 red cards have more probability and this can occur by chance.

So friend's conclusion is wrong.

answer: haha (: i’ll go see what it is

step-by-step explanation:

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet