Answer :
In 1976, the average blood lead level was about [tex]\( 16 \)[/tex] [tex]$\mu g / dL$[/tex].
To find the amount of lead used in a 6-month period in 1978, we need to express it in terms of [tex]\( 10^3 \)[/tex] tons. The given value is [tex]\( 12300.0 \)[/tex] tons.
Now, let’s express [tex]\( 12300.0 \)[/tex] tons as a multiple of [tex]\( 10^3 \)[/tex] (i.e., thousands of tons):
[tex]\[ 12300.0 \, \text{tons} = 12.3 \times 10^3 \text{ tons} \][/tex]
Thus, the amount of lead used in a 6-month period in 1978 was about [tex]\( 12.3 \times 10^3 \)[/tex] tons.
So, the answers are:
- In 1976, the average blood lead level was about [tex]\( 16 \)[/tex] [tex]$\mu g / dL$[/tex].
- The amount of lead used in a 6-month period in 1978 was about [tex]\( 12.3 \times 10^3 \)[/tex] tons.
To find the amount of lead used in a 6-month period in 1978, we need to express it in terms of [tex]\( 10^3 \)[/tex] tons. The given value is [tex]\( 12300.0 \)[/tex] tons.
Now, let’s express [tex]\( 12300.0 \)[/tex] tons as a multiple of [tex]\( 10^3 \)[/tex] (i.e., thousands of tons):
[tex]\[ 12300.0 \, \text{tons} = 12.3 \times 10^3 \text{ tons} \][/tex]
Thus, the amount of lead used in a 6-month period in 1978 was about [tex]\( 12.3 \times 10^3 \)[/tex] tons.
So, the answers are:
- In 1976, the average blood lead level was about [tex]\( 16 \)[/tex] [tex]$\mu g / dL$[/tex].
- The amount of lead used in a 6-month period in 1978 was about [tex]\( 12.3 \times 10^3 \)[/tex] tons.
