The average human heart beats about 3.5 x 107 times per year. The current population of Massachusetts is about 7 x 10 people. About how many heartbeats occur in Massachusetts per year? Write the answer in scientific notation. Express the answer to the exact decimal place. (1 point)

1 answer

To find the total number of heartbeats that occur in Massachusetts per year, we'll multiply the average number of heartbeats per heart per year by the population of Massachusetts.

Given:

  • Average heartbeats per year = \( 3.5 \times 10^7 \)
  • Population of Massachusetts = \( 7 \times 10^6 \)

Now we multiply these two quantities:

\[ \text{Total heartbeats} = (3.5 \times 10^7) \times (7 \times 10^6) \]

Using the properties of multiplication (specifically \(a \times b = (a \times c) \times (b \times d)\)), we can rearrange it:

  1. Multiply the coefficients: \(3.5 \times 7 = 24.5\)
  2. Add the exponents of 10 in the scientific notation: \(10^7 \times 10^6 = 10^{7+6} = 10^{13}\)

Now we combine these results:

\[ \text{Total heartbeats} = 24.5 \times 10^{13} \]

To express this in proper scientific notation, we can move the decimal point in 24.5 one place to the left:

\[ 24.5 = 2.45 \times 10^1 \]

Now, we substitute this back into our multiplication:

\[ 24.5 \times 10^{13} = (2.45 \times 10^1) \times 10^{13} = 2.45 \times 10^{1 + 13} = 2.45 \times 10^{14} \]

Thus, the total number of heartbeats that occur in Massachusetts per year is:

\[ \boxed{2.45 \times 10^{14}} \]