Here, we cannot apply the rate as is, since it is given for a year. We just have to divide the rate by the number of months in a year, since we are looking for a monthly rate. Therefore, the applicable rate is \(\displaystyle \frac{0.09}{12}\), now we can apply the \(\displaystyle Pnr\) formula and we get: \(\displaystyle 12,000\cdot\frac{0.09}{12}\cdot7\) or \(\displaystyle 630\).

Here we need our simple interest formula. There is no compounding, so all we need is:

\(\displaystyle \small I=P*r*n\)

\(\displaystyle I\) = Dollar amount of interest

\(\displaystyle P\) = Amount borrowed/invested

\(\displaystyle r\) = Annual interest rate

\(\displaystyle n\) = Number of years borrowed/loaned

Plug in what we know and solve for \(\displaystyle I\):

\(\displaystyle I= \p \$1500*0.023*0.5=\$17.25\)

So, Jasmine pays \(\displaystyle \$17.25\) in interest. 

This is a simple interest problem, therefore, the amount of interests received is given by the formula: \(\displaystyle Pnr\), where \(\displaystyle P\) is the principal,  \(\displaystyle n\) is the number of periods and \(\displaystyle r\) is the rate over that period. In this problem \(\displaystyle P\) is 140,000, \(\displaystyle n\) is 10 and \(\displaystyle r\) is 10% or 0.1. Therefore we get \(\displaystyle 140,000\cdot0.1\cdot10\) or 140,000. We add the interest received to the principal, and we get the final answer of \(\displaystyle 280,000\).

To solve this, we can create an equation for the value based on time. So if we let t be the nmbers of years that have passed, we can create a function f(t) for the value in the savings account. 

We note that f(0) =5000. (We invest 5000 at time 0.) Next year, he will have 5% more than that. To find our total value at the end of the year, we multiply 5,000 * 1.05 = 5,250. f(1) = 5000(1.05)=5,250. At the end of year 2, we will have a 5% growth rate. In other words, f(2) = (1.05)* f(1). We can rewrite this as \(\displaystyle f(2) = 5000(1.05)(1.05) = 5000 (1.05)^2\) . We can begin to see the proper equation is \(\displaystyle f(t)= 5000 (1.05)^t\). If we plug in t = 15, we will have our account balance at the end of 15 years. So, our answer is \(\displaystyle f(15)= 5000 (1.05)^{15}= 10,394.64\).

The monthly rate is \(\displaystyle \frac{6\ percent}{12}=0.5\ percent=0.005\)

3 years = 36 months

According to the compound interest formula

\(\displaystyle P=C\left ( 1+r \right )t\)

and here \(\displaystyle C=m\), \(\displaystyle r=0.005\), \(\displaystyle t=36\), so we can plug into the formula and get the value

\(\displaystyle m( 1+0.005)^{36} = (1.005)^{36} * m\)

Calculate the total amount from each bank using the following formula:

\(\displaystyle A=P\left(1+\frac{r}{n}\right)^{nt}\)

Bank A:

\(\displaystyle A=1000\left(1+\frac{0.03}{365}\right)^{(1)(365)}=1030.45\)

Bank B:

\(\displaystyle A=1000\left(1+\frac{.035}{12}\right)^{(1)(12)}=1035.57\)

\(\displaystyle 1035.57-1030.45=5.12\)

Let \(\displaystyle C\) be the amount Bryan invested in the certificate of deposit. Then he deposited \(\displaystyle \$ 8,000-C\) in a savings account. 8% of the amount in the certificate of deposit is \(\displaystyle 0.08C\), and 3% of the amount in the savings account is \(\displaystyle 0.03 \left ( 8,000-C \right )\); add these interest amounts to get $365.00.  Therefore, we can set up and solve the equation:

\(\displaystyle 0.08C + 0.03 \left ( 8,000-C \right ) = 365\)

\(\displaystyle 0.08C + 0.03 \cdot 8,000 -0.03 \cdot C = 365\)

\(\displaystyle 0.05C + 240 = 365\)

\(\displaystyle 0.05C + 240-240 = 365 -240\)

\(\displaystyle 0.05C = 125\)

\(\displaystyle 0.05C \div 0.05 = 125 \div 0.05\)

\(\displaystyle C = 2,500\)

Use the compound interest formula

\(\displaystyle A =P \left ( 1+ \frac{r}{n}\right )^{nt}\)

substituting \(\displaystyle P = 9,000\) (principal, or amount invested), \(\displaystyle r = 0.08\) (decimal equivalent of the 8% interest rate), \(\displaystyle n = 4\) (four quarters per year), \(\displaystyle t = 1\) (one year).

\(\displaystyle A =9,000 \left ( 1+ \frac{0.08}{4}\right )^{4 \cdot 1}\)

\(\displaystyle =9,000 \left ( 1.02 \right )^{4 }\)

\(\displaystyle \approx 9,741.89\)

Subtract 9,000 from this figure - the interest earned is $741.89

Since in each case the interest is compounded quarterly, the annual interest rate of 4% is divided by 4 to get 1%, the effective quarterly interest rate. 

The $10,000 remains in the savings account six months, or two quarters, so 1% is added twice - equivalently, the $10,000 is multiplied by 1.01 twice:

\(\displaystyle \$ 10,000.00 \times 1.01 = \$ 10,100.00\)

\(\displaystyle \$ 10,100.00 \times 1.01 = \$ 10,201.00\)

$5,000 is withdrawn from the savings account, leaving \(\displaystyle \$ 10,201.00 - \$ 5,000.00 = \$ 5,201.00\)

This money is untouched for six months, or two quarters, so again, we multiply by 1.01 twice:

\(\displaystyle \$ 5,201.00 \times 1.01 = \$ 5,253.01\)

\(\displaystyle \$ 5,253.01 \times 1.01 =\$ 5,305.54\)

Subtract $5,000 to get the interest:

\(\displaystyle \$ 5,305.54 - \$5,000.00 = \$305.54\)

12% annual interest compounded quarterly is, effectively, 3% interest per quarter.

Over the course of one quarter, Gary pays off \(\displaystyle \$800 \times 3 = \$2,400\), and the remainder of the loan accruses 3% interest. This happens four times, so we will subtract $2,400 and subsequently multiply by 1.03 (adding 3% interest) four times. 

First quarter:

\(\displaystyle \$10,000 - \$2,400 = \$7,600\)

\(\displaystyle \$7,600 \times 1.03 = \$7,828\)

Second quarter:

\(\displaystyle \$7,828 - \$2,400 = \$5,428\)

\(\displaystyle \$ 5,428 \times 1.03 = \$5,590.84\)

Third quarter:

\(\displaystyle \$5,590.84 - \$2,400 =\$ 3,190.84\)

\(\displaystyle \$ 3,190.84 \times 1.03 = \$3,286.57\)

Fourth quarter:

\(\displaystyle \$3,286.57 - \$2,400 =\$ 886.57\)

\(\displaystyle \$ 886.57 \times 1.03 = \$913.16\)

The thriteenth payment, with which Gary will pay off the loan, will be $913.16.