Karla invests $300 compounded every 6 months at a rate of 10% for 3 years. At the end of three years, Karla will have $402.03 in her account. Show how to arrive at this amount.

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1 + r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $300

r = 10% = 10/100 = 0.1

n = 2 because it was compounded 2 times in a year(6 months).

t = 3 years

Therefore,

A = 300(1 + 0.1/2)^2 × 3

A = 300(1 + 0.05)^6

A = 300(1.05)^6

A = $402.03

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b