Question

1. 0 / 1 points

as part of a training program for the boston marathon, a runner wants to build endurance by running at a rate of 9 mph for 20 min. how far will the runner travel in that time period?
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marcella leaves home at 9:00 a.m. and drives to school, arriving at 9:45 a.m. if the distance between home and school is 21 mi, what is marcellas average rate of speed?
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1. - / 1 points

a life insurance policy costs $12.07 for every $1,000 of insurance. at this rate, what is the cost of $70,000 of insurance?

Explanation:

Step1: Convert time to hours

Since \(1\) hour \( = 60\) minutes, \(20\) minutes \(=\frac{20}{60}=\frac{1}{3}\) hours.

Step2: Use the formula \(d = rt\) (distance \(d\), rate \(r\), time \(t\))

Given \(r = 9\) mph and \(t=\frac{1}{3}\) hours. Then \(d=9\times\frac{1}{3}\).

$$d = 3$$

Step3: For Marcella's speed

Time taken \(t = 45\) minutes. Convert to hours: \(t=\frac{45}{60}=\frac{3}{4}\) hours, distance \(d = 21\) miles. Use the formula \(r=\frac{d}{t}\).

$$r=\frac{21}{\frac{3}{4}}=21\times\frac{4}{3}=28$$

Step4: For insurance cost

If the cost is \(12.07\) per \(1000\) of insurance. For \(70000\) of insurance, set up a proportion. Let \(x\) be the cost. \(\frac{x}{70000}=\frac{12.07}{1000}\). Then \(x=\frac{12.07\times70000}{1000}\).

$$x = 12.07\times70=844.9$$

Answer:

1. The runner travels \(3\) miles.
2. Marcella's average rate of speed is \(28\) mph.
3. The cost of \(70000\) of insurance is \(\$844.9\).