A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is $6600 dollars. How much did he invest at each rate?

Let x represent amount invested in the higher-yielding account.

We have been given that a man puts twice as much in the lower-yielding account because it is less risky. So amount invested in the lower-yielding account would be .

We are also told that his annual interest is $6600 dollars. We know that annual interest for one year will be principal amount times interest rate.

, where,

I = Amount of interest,

P = Principal amount,

r = Annual interest rate in decimal form,

t = Time in years.

We are told that interest rates are 6% and 10%.

Amount of interest earned from lower-yielding account: .

Amount of interest earned from higher-yielding account: .

Let us solve for x.

Therefore, the man invested $30,000 at 10%.

Amount invested in the lower-yielding account would be .

Therefore, the man invested $60,000 at 6%.

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.