In a history class, the probability of earning an A is .7 times the probability of earning a B, and the probability of earning a C is 1.4 times the probability of earning a B. Assuming that all grades are A, B, or C, how many B's will there be in a history class of 31 students?
Let $x$ be the number of students who earn a B. Then we know that the number of students who get an A is $.7x$ and the number of students who earn a C is $1.4x$. Since every student in the class gets an A, B, or C and there are 31 students, this gives us the equation $.7x + x + 1.4x = 31 \Rightarrow 3.1x = 31 \Rightarrow x =\boxed{10}$.
10
Level 5
Prealgebra
Problem: In a history class, the probability of earning an A is .7 times the probability of earning a B, and the probability of earning a C is 1.4 times the probability of earning a B. Assuming that all grades are A, B, or C, how many B's will there be in a history class of 31 students?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>Let the probability of earning a B be denoted as P(B). According to the problem, the probability of earning an A is 0.7 times P(B), so P(A) = 0.7 * P(B). The probability of earning a C is 1.4 times P(B), thus P(C) = 1.4 * P(B). 

Since the only possible grades are A, B, and C, the sum of their probabilities must equal 1. Therefore, we can set up the equation:

P(A) + P(B) + P(C) = 1.

Substituting the expressions for P(A) and P(C) in terms of P(B) gives:

0.7 * P(B) + P(B) + 1.4 * P(B) = 1.

Combining the terms results in:

(0.7 + 1 + 1.4) * P(B) = 1,
3.1 * P(B) = 1.

To find P(B), we divide both sides by 3.1:

P(B) = 1 / 3.1.

Next, we calculate the number of students expected to earn a B in a class of 31 students. The expected number of B's is given by:

Number of B's = P(B) * Total Students = (1 / 3.1) * 31.

Calculating this gives:

Number of B's = 31 / 3.1 = 10.

Thus, the final answer is 10.<|END_OF_LATENT|> The final answer is $10$.