We need to solve two problems. (a) Simplify the expression $3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}$. (b) Find the probability that the sum of two numbers is greater than 3 and less than 7, where one number is selected randomly from the set $\{2, 3, 4\}$ and the other is selected randomly from the set $\{1, 3, 5\}$.

1. Problem Description

We need to solve two problems.

(a) Simplify the expression $3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}$.

(b) Find the probability that the sum of two numbers is greater than 3 and less than 7, where one number is selected randomly from the set $\{2, 3, 4\}$ and the other is selected randomly from the set $\{1, 3, 5\}$.

2. Solution Steps

(a) Simplify the expression $3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}$.

First convert mixed numbers to improper fractions:

$3\frac{4}{9} = \frac{3 \times 9 + 4}{9} = \frac{27+4}{9} = \frac{31}{9}$

$5\frac{1}{3} = \frac{5 \times 3 + 1}{3} = \frac{15+1}{3} = \frac{16}{3}$

$2\frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{8+3}{4} = \frac{11}{4}$

$5\frac{9}{10} = \frac{5 \times 10 + 9}{10} = \frac{50+9}{10} = \frac{59}{10}$

Now, we can rewrite the expression as:

$\frac{31}{9} \div (\frac{16}{3} - \frac{11}{4}) + \frac{59}{10}$

Find a common denominator for $\frac{16}{3}$ and $\frac{11}{4}$. The least common multiple of 3 and 4 is

1

2. $\frac{16}{3} = \frac{16 \times 4}{3 \times 4} = \frac{64}{12}$

$\frac{11}{4} = \frac{11 \times 3}{4 \times 3} = \frac{33}{12}$

So,

$\frac{16}{3} - \frac{11}{4} = \frac{64}{12} - \frac{33}{12} = \frac{64 - 33}{12} = \frac{31}{12}$

Then,

$\frac{31}{9} \div \frac{31}{12} + \frac{59}{10} = \frac{31}{9} \times \frac{12}{31} + \frac{59}{10} = \frac{12}{9} + \frac{59}{10} = \frac{4}{3} + \frac{59}{10}$

Find a common denominator for $\frac{4}{3}$ and $\frac{59}{10}$. The least common multiple of 3 and 10 is

3

0. $\frac{4}{3} = \frac{4 \times 10}{3 \times 10} = \frac{40}{30}$

$\frac{59}{10} = \frac{59 \times 3}{10 \times 3} = \frac{177}{30}$

$\frac{4}{3} + \frac{59}{10} = \frac{40}{30} + \frac{177}{30} = \frac{40 + 177}{30} = \frac{217}{30}$

$\frac{217}{30} = 7\frac{7}{30}$

(b) Find the probability that the sum of two numbers is greater than 3 and less than

7. The possible sums are obtained from the following pairs:

$\{2, 3, 4\}$ and $\{1, 3, 5\}$.

Possible pairs are:

$(2, 1) = 3$

$(2, 3) = 5$

$(2, 5) = 7$

$(3, 1) = 4$

$(3, 3) = 6$

$(3, 5) = 8$

$(4, 1) = 5$

$(4, 3) = 7$

$(4, 5) = 9$

The sums are: $\{3, 5, 7, 4, 6, 8, 5, 7, 9\}$

The sums that are greater than 3 and less than 7 are: $\{5, 4, 6, 5\}$

There are 4 such sums.

The total number of possible sums is $3 \times 3 = 9$.

The probability is $\frac{\text{number of sums greater than 3 and less than 7}}{\text{total number of sums}} = \frac{4}{9}$.

3. Final Answer

(a) $7\frac{7}{30}$

(b) $\frac{4}{9}$