Chapter 1.1, Problem 64E

To determine

To calculate: To complete the statement and round to the nearest hundredth (if required)

Answer to Problem 64E

  $\frac{7gal}{\min }\approx \frac{0.467qt}{\sec }$

Explanation of Solution

Given: Statement is $\frac{7gal}{\min }\approx \frac{\_\_\_qt}{\sec }$

Formula Used:

Conversions are

  $\begin{array}{l}1gal=4qt\\ 1\min =60\sec \end{array}$

Calculation:

Given that $\frac{7gal}{\min }\approx \frac{\_\_\_qt}{\sec }$

Converting km/min to miles/hr:

  $\begin{array}{l}\frac{7gal}{\min }\approx \frac{7\times 4qt}{60\sec }\\ \frac{7gal}{\min }\approx \frac{28qt}{60\sec }\\ \frac{7gal}{\min }\approx \frac{0.467qt}{\sec }\end{array}$

Conclusion:

Hence, $\frac{7gal}{\min }\approx \frac{0.467qt}{\sec }$