A$20 bill two $10 bills three $5 bills and four $1 bills are placed in a bag. if a bill is chosen at random what is the expected value for the amount chosen?

Hello mate!

You have one $20 bill, two $10 bills, three $5 bills and four $1 bills in a bag.

This adds to a total of 10 bills in the bag, and you want to know the expected value if you took one at random.

The expected value is defined as ∑pₙxₙ

where pₙ is the probability for each event (in this case is the number of bills of a given price divided the number of bills in the bag) and xₙ is the value of the bill.

Then we can write the expected value as:

E = (1/10)*$20 + (2/10)*$10 + (3/10)*$5 + (4/10)*$1 = (1/10)( 20$ + 2*$10 + 3*$5 + 4*$1) = (1/10)*($59) = $5.90

The expected value is $5.90.

the correct option is b

step-by-step explanation:

answer: the resulting cross section is a rectangle. solution: a plane parallel to the base of a triangular prism will intersect a cross section that is the same shape as its bases. so the cross section is a triangle.

step-by-step explanation: