Question The weight of an ant averages about 3×10^−6 kg. There are about 1×10^16 ants in the world. What is the approximate weight in kilograms of all the ants in the world? Write the answer in scientific notation.(1 point)

Answer 1

To find the total weight of all the ants in the world, we can multiply the average weight of an ant by the total number of ants.

The average weight of an ant is \(3 \times 10^{-6}\) kg, and the total number of ants is \(1 \times 10^{16}\).

So, the total weight \(W\) can be calculated as:

\[ W = \text{(weight of one ant)} \times \text{(number of ants)} \]

\[ W = (3 \times 10^{-6} , \text{kg}) \times (1 \times 10^{16}) \]

When multiplying these numbers in scientific notation, you multiply the coefficients (3) and add the exponents (-6 and 16):

\[ W = 3 \times 1 \times 10^{-6 + 16} = 3 \times 10^{10} , \text{kg} \]

Thus, the approximate weight of all the ants in the world is:

\[ \boxed{3 \times 10^{10}} \text{ kg} \]

Answer 2

write it regularly

Answer 3

The approximate weight of all the ants in the world is \(30,000,000,000\) kg.

Answer 4

wrong

Answer 5

I apologize for any confusion. Let's clarify and calculate once more.

To calculate the total weight of all the ants in the world:

• The average weight of an ant is \(3 \times 10^{-6}\) kg.
• The total number of ants is \(1 \times 10^{16}\).

Now, calculating the total weight:

\[ \text{Total Weight} = \text{(average weight of an ant)} \times \text{(number of ants)} \]

\[ \text{Total Weight} = (3 \times 10^{-6} , \text{kg}) \times (1 \times 10^{16}) = 3 \times 10^{10} , \text{kg} \]

In standard notation, \(3 \times 10^{10}\) kg is equal to:

\[ 30,000,000,000 , \text{kg} \]

So the answer in regular form is indeed \(30,000,000,000\) kg.

If you think there's something else wrong or if you're looking for a specific answer format, please let me know!

Answer 6

write it in scientific notation

Answer 7

The total weight of all the ants in the world, in scientific notation, is \(3 \times 10^{10}\) kg.

Answer 8

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 occur in Massachusetts per year? Write the answer in scientific notation. Express the answer to the exact decimal place.(1 point)

Answer 9

To calculate the total number of heartbeats in Massachusetts per year, we can multiply the average number of heartbeats per person per year by the population of Massachusetts.

Given:

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

The total number of heartbeats in Massachusetts per year \( H \) can be calculated as:

\[ H = (\text{heartbeats per person}) \times (\text{population}) \]

\[ H = (3.5 \times 10^7) \times (7 \times 10^6) \]

When multiplying in scientific notation, you multiply the coefficients (3.5 and 7) and add the exponents (7 and 6):

\[ H = (3.5 \times 7) \times 10^{7 + 6} \]

Calculating \(3.5 \times 7\):

\[ 3.5 \times 7 = 24.5 \]

Now, adding the exponents:

\[ H = 24.5 \times 10^{13} \]

To express \(24.5\) in proper scientific notation, we can write it as \(2.45 \times 10^1\). So we have:

\[ H = 2.45 \times 10^1 \times 10^{13} \]

Adding the exponents:

\[ H = 2.45 \times 10^{14} \]

Thus, the total number of heartbeats in Massachusetts per year, expressed in scientific notation, is:

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

Answer 10

The average human heart beats about 3.5×10^7 times per year. The current population of Massachusetts is about 7×10^6 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)

Answer 11

To calculate the total number of heartbeats that occur in Massachusetts per year, we multiply the average number of heartbeats per person per year by the population of Massachusetts.

Given:

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

Now, let's calculate the total number of heartbeats \( H \):

\[ H = (\text{heartbeats per person}) \times (\text{population}) \]

\[ H = (3.5 \times 10^7) \times (7 \times 10^6) \]

To compute this, we first multiply the coefficients:

\[ 3.5 \times 7 = 24.5 \]

Next, we add the exponents:

\[ 10^7 \times 10^6 = 10^{7+6} = 10^{13} \]

So, we have:

\[ H = 24.5 \times 10^{13} \]

To express \(24.5\) in proper scientific notation, we can write it as:

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

Thus, we can express the total heartbeats as:

\[ H = 2.45 \times 10^1 \times 10^{13} = 2.45 \times 10^{14} \]

Therefore, the total number of heartbeats in Massachusetts per year, expressed in scientific notation to the exact decimal place, is:

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