Answer:
Percent of Robbery
![=2 \%](https://img.qammunity.org/2020/formulas/mathematics/middle-school/uqbfzwi4xjsf65ei86iw5oolbokm8idqes.png)
Percent of Robbery
![=9 \%](https://img.qammunity.org/2020/formulas/mathematics/middle-school/u760x3ih1iiqyjsw0e4m6wc4w41ziw87zu.png)
Explanation:
Given that total number of crimes = 21,790,654.
total number of Robbery = 408, 016 and
total number of Burglary = 2,050,992
Now we need to find about what are their percent in crimes rounded to the nearest percent for each two crimes.
So apply the percent formula
![Percent=(Part)/(Whole)* 100 \%](https://img.qammunity.org/2020/formulas/mathematics/middle-school/o418q40qh8d26sicgah3td5ml5o3392q23.png)
Then percent of Robbery
![=(408016)/(21790654)* 100 \%=0.01872* 100 \%=2 \%](https://img.qammunity.org/2020/formulas/mathematics/middle-school/q500466719cacc9dquda6cbf2hqz0cdat8.png)
Then percent of Robbery
![=(2050992)/(21790654)* 100 \%=0.094122* 100 \%=9 \%](https://img.qammunity.org/2020/formulas/mathematics/middle-school/a8berfarbjm99zkw7xaztbsnwap0ihdvfe.png)