Question

A typical person has an average heart rate of 75.0 beats/min. Calculate the following a)How many beats does she have in 7.0 years? b) How many beats in 7.000 years? Pay close attention to significant figures in this question.

Answer 1

Answer:

a) Beats in 7.0 years = $2.7 \times 10^8$

b) Beats in 7.000 yrs= $2.759 \times 10^8$

Explanation:

Step 1

Calculation of no. of min. in 7 year

Days in a year = 365

Hrs in a day = 24

Min. in a 1 hr = 60

Min. in a day = 60 × 24

Min. in a year = 60 × 24 × 365

Min. in 7 years

               = 60 × 24 × 365 × 7

               = 3679200 min.

a) Beats in 7.0 years

Heart rate = 75.0 beats/min.

Beats in 7.0 yrs

             = 75.0 × 3679200

             = 275940000 beats

No. of significant figure = 2

Answer has to be reported in 2 significant digits.

Beats in 7.0 years = $2.7 \times 10^8$

a) Beats in 7.000 years

Beats in 7.000 yrs = 275940000 beats

No. of significant figure in 7.000 = 4

Answer has to be reported in 4 significant figures

So, no. of beats in 7.000 yrs= $2.759 \times 10^8$

Answer 2

Let us first convert years to minute.

7 years * (365 days / year) * (24 hours / day) * (60 minutes / hour) = 3,679,200 minutes

 

a. beats = 75.0 beats/min * 3,679,200 minutes

beats = 275,940,000 beats

Since the original given is 7.0 years (2 significant figures), therefore:

beats = 270,000,000 beats

 

b beats = 75.0 beats/min * 3,679,200 minutes

beats = 275,940,000 beats

Since the original given is 7.000 years (4 significant figures), therefore:

beats = 275,900,000 beats