The average human heart beats about 3.5 × 10⁷ times per year. The current population of Massachusetts is about 7 × 10⁶ people. About how many heartbeats occurs in Massachusetts per year? Write the answer in scientific notation. Express the answer to the exact decimal place.
To find the total number of heartbeats in Massachusetts per year, we need to multiply the average number of heartbeats per year per person (3.5 × 10⁷) by the current population of Massachusetts (7 × 10⁶).
Multiplying 3.5 × 10⁷ by 7 × 10⁶, we add the exponents and multiply the coefficients:
(3.5 × 7) × (10⁷ × 10⁶) = 24.5 × 10¹³
To express this in scientific notation with one decimal place, we divide 24.5 by 10, and keep the exponent 10¹³:
2.45 × 10¹⁴ heartbeats
Therefore, about 2.45 × 10¹⁴ heartbeats occur in Massachusetts per year.
To find the number of heartbeats that occur in Massachusetts per year, we need to multiply the average number of heartbeats per person per year (3.5 × 10⁷) by the current population of Massachusetts (7 × 10⁶).
3.5 × 10⁷ heartbeats/year × 7 × 10⁶ people = (3.5 × 7) × (10⁷ × 10⁶) heartbeats/year
= 24.5 × 10¹³ heartbeats/year
Since the result is already in scientific notation, we don't need to convert it. The answer is 24.5 × 10¹³ heartbeats/year.
To calculate the number of heartbeats that occur in Massachusetts per year, we need to multiply the average number of heartbeats per person by the current population of Massachusetts.
The average number of heartbeats per person per year is given as 3.5 × 10⁷.
The current population of Massachusetts is given as 7 × 10⁶ people.
To find the total number of heartbeats in Massachusetts per year, we multiply these two values:
(3.5 × 10⁷) × (7 × 10⁶)
Multiplying the values inside the parentheses:
3.5 × 7 = 24.5
Multiplying the powers of 10:
10⁷ × 10⁶ = 10¹³
Putting it all together, we have:
(3.5 × 10⁷) × (7 × 10⁶) = 24.5 × 10¹³
To express the answer in scientific notation, we combine the numeric part (24.5) with the exponent part (10¹³):
Answer: 2.45 × 10¹⁴