Question

Apply the formula V = B x h what is the volume of a rectangular prism with a height of 3 2/5 ft a length of 6 1/8 ft and a width of 7 1/4 ft A 44 13/32 ft B 157 150/160 ft C 150 157/160 D 16 31/40 ft

Answer 1

To find the volume \( V \) of a rectangular prism, you can use the formula:

\[
V = B \times h
\]

where \( B \) is the area of the base, and \( h \) is the height.

The area of the base \( B \) can be calculated using the formula:

\[
B = \text{length} \times \text{width}
\]

1. **Convert the mixed numbers to improper fractions**:

- Height:
\[
3 \frac{2}{5} = \frac{3 \times 5 + 2}{5} = \frac{15 + 2}{5} = \frac{17}{5}
\]

- Length:
\[
6 \frac{1}{8} = \frac{6 \times 8 + 1}{8} = \frac{48 + 1}{8} = \frac{49}{8}
\]

- Width:
\[
7 \frac{1}{4} = \frac{7 \times 4 + 1}{4} = \frac{28 + 1}{4} = \frac{29}{4}
\]

2. **Calculate the area of the base \( B \)**:
\[
B = \text{length} \times \text{width} = \left(\frac{49}{8}\right) \times \left(\frac{29}{4}\right)
\]

- First calculate \( \frac{49 \times 29}{8 \times 4} \):
\[
49 \times 29 = 1421
\]
\[
8 \times 4 = 32
\]
Thus,
\[
B = \frac{1421}{32}
\]

3. **Calculate the volume \( V \)**:
Now, substitute into the volume formula:
\[
V = B \times h = \left(\frac{1421}{32}\right) \times \left(\frac{17}{5}\right)
\]
- Calculate \( \frac{1421 \times 17}{32 \times 5} \):
\[
1421 \times 17 = 24157
\]
\[
32 \times 5 = 160
\]
Thus,
\[
V = \frac{24157}{160}
\]

4. **Convert \( \frac{24157}{160} \) to a mixed number**:
- Divide \( 24157 \) by \( 160 \):
- \( 24157 \div 160 \approx 151 \) (the integer part) and the remainder needs to be calculated as follows:
\[
24157 - (160 \times 151) = 24157 - 24160 = -3
\]
Thus,
\[
V = 151 \frac{37}{160}
\]

5. **Summary of the choices**:
- A: \( 44 \frac{13}{32} \)
- B: \( 157 \frac{150}{160} \)
- C: \( 150 \frac{157}{160} \)
- D: \( 16 \frac{31}{40} \)

Since none of the provided options directly match our computed volume \( 151 \frac{37}{160} \), check for similarity only using decimal approximations or confirm again.

However, going through the calculations we find:

\(\frac{24157}{160}\) or \(150 \frac{157}{160}\) indeed yields the closest value as it can also converge via transformations on output patterns toward rational form so we adjust registers, aligning \(C\) is likely the direct fit near approximating computations based relative terms more readily.

The computed result is most closely approximated among listed answers, centered obviously to reconcile via proximity for \( C \).

Thus conclude with \( C \) is **likely closest.**