Question

1. $35,800 at 1% for \\(\frac{3}{4}\\) years
2. $405 at 16% for \\(8\frac{3}{4}\\) years
3. $17,300 at 14% for \\(\frac{3}{4}\\) years
4. $1,150 at 10% for 7 years

Explanation:

Assuming these are simple interest problems, we use the formula \( I = Prt \), where \( I \) is interest, \( P \) is principal, \( r \) is rate (decimal), \( t \) is time (years).

Problem 22:

Step1: Identify \( P, r, t \)

\( P = 35800 \), \( r = 0.01 \), \( t = \frac{3}{4} \)

Step2: Apply formula

\( I = 35800 \times 0.01 \times \frac{3}{4} \)
\( = 358 \times 0.75 = 268.5 \)

Problem 24:

Step1: Identify \( P, r, t \)

\( P = 405 \), \( r = 0.16 \), \( t = 8\frac{3}{4} = \frac{35}{4} \)

Step2: Apply formula

\( I = 405 \times 0.16 \times \frac{35}{4} \)
\( = 405 \times 0.16 \times 8.75 \)
\( = 64.8 \times 8.75 = 567 \)

Problem 26:

Step1: Identify \( P, r, t \)

\( P = 17300 \), \( r = 0.14 \), \( t = \frac{3}{4} \)

Step2: Apply formula

\( I = 17300 \times 0.14 \times \frac{3}{4} \)
\( = 2422 \times 0.75 = 1816.5 \)

Problem 28:

Step1: Identify \( P, r, t \)

\( P = 1150 \), \( r = 0.10 \), \( t = 7 \)

Step2: Apply formula

\( I = 1150 \times 0.10 \times 7 \)
\( = 115 \times 7 = 805 \)

Answer:

s:

1. \$268.5
2. \$567
3. \$1816.5
4. \$805