Suppose 7 quarters, 4 dimes, 4 nickels, and 3 pennies are in a box. One coin is selected at random. What is the expected value of the money drawn from the box?

First, let's get the total amount of coins inside the box.

7 quarters + 4 dimes + 4 nickels + 3 pennies = 18 total coins.

(7 quarters / 18 coins) * 100 = 38.89% chance of picking a quarter

(4 dimes / 18 coins) * 100 = 22.22% chance of picking a dime

(4 nickels / 18 coins) * 100 = 22.22% chance of picking a nickel

(3 pennies / 18 coins) * 100 = 16.67% chance of picking a dime

Since the quarters have a higher percentage of being picked, we can expect to draw $0.25 from the box

too many. (jk)

31 v bucks

step-by-step explanation:

since you bought 25 on friday and 6 on tuesday, you can just add them up.

25+6=31

check your work.

31-6=25

i hope this !

-yand3r3

assuming this is from a math graphing problem.

7 would be on the x-axis which is the horizontal line on the graph

84 would be on the y-axis which is the vertical line on the graph