In 2004, about 14 percent of the estimated total population of the United States was found to be Hispanic. At that time about 1/25 of the Hispanic Population was Cuban, and about 2/3 of the Cuban population lived in Florida. What fraction of the total U.S. population was Cuban and living in Florida in 2004???

The answer to this question was 7/1875.
Because my teacher gave me the answer since I got it wrong and I need to figure out how to solve it because I tried many ways and I never get 7/1875. so can someone please help me to learn how to solve his question please???

1/25 , 2/3 , and 7/1875 are fractions

anyone....

To solve this question, let's break it down step by step.

Step 1: Calculate the Hispanic population in 2004.
We know that 14 percent of the total U.S. population was Hispanic. So if we let "P" represent the total U.S. population, we can write the equation: 0.14P = Hispanic population.

Step 2: Calculate the Cuban population.
We are given that 1/25 of the Hispanic population was Cuban, so we can write the equation: (1/25)(0.14P) = Cuban population.

Step 3: Calculate the number of Cubans living in Florida.
We know that 2/3 of the Cuban population lived in Florida, so we can write the equation: (2/3)(1/25)(0.14P) = Cubans living in Florida.

Step 4: Calculate the fraction of the total U.S. population that was Cuban and living in Florida.
To do this, we need to divide the number of Cubans living in Florida by the total U.S. population: [(2/3)(1/25)(0.14P)] / P.

Now, let's simplify the equation:

Step 1: 0.14P = Hispanic population
Step 2: (1/25)(0.14P) = Cuban population
Step 3: (2/3)(1/25)(0.14P) = Cubans living in Florida
Step 4: (2/3)(1/25)(0.14P) / P = (2/3)(0.14) / 25
= (2/3)(14/100) / 25
= (2/3)(7/50) / 25
= 7/75 / 25
= 7/75 x 1/25
= 7/1875.

Therefore, the fraction of the total U.S. population that was Cuban and living in Florida in 2004 is 7/1875.