We are given several arithmetic problems involving fractions and decimals. We need to solve each one. The problems are labeled b, c, d, e, f, g, and h.

1. Problem Description

We are given several arithmetic problems involving fractions and decimals. We need to solve each one. The problems are labeled b, c, d, e, f, g, and h.

2. Solution Steps

b) $2\frac{3}{4} + 4\frac{2}{4}$

First, we add the whole numbers: $2+4 = 6$.

Next, we add the fractions: $\frac{3}{4} + \frac{2}{4} = \frac{3+2}{4} = \frac{5}{4}$.

Since $\frac{5}{4} = 1\frac{1}{4}$, we have $6 + 1\frac{1}{4} = 7\frac{1}{4}$.

c) $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8}$

We can write this as $1 + \frac{4}{8} + \frac{2}{8} + \frac{1}{8} = 1 + \frac{4+2+1}{8} = 1 + \frac{7}{8} = 1\frac{7}{8}$.

d) $15 + \frac{1}{25} + \frac{1}{125}$

We can write this as $15 + \frac{5}{125} + \frac{1}{125} = 15 + \frac{5+1}{125} = 15 + \frac{6}{125} = 15\frac{6}{125}$.

e) $9 + \frac{3}{7}$

This is simply $9\frac{3}{7}$.

f) $\frac{4}{20} + \frac{16}{60} - \frac{11}{80}$

Simplify the fractions first:

$\frac{4}{20} = \frac{1}{5}$

$\frac{16}{60} = \frac{4}{15}$

$\frac{11}{80}$ remains the same.

We need a common denominator for $\frac{1}{5}$, $\frac{4}{15}$, and $\frac{11}{80}$. The least common multiple of 5, 15, and 80 is

2

4

0. $\frac{1}{5} = \frac{48}{240}$

$\frac{4}{15} = \frac{64}{240}$

$\frac{11}{80} = \frac{33}{240}$

Now, we have $\frac{48}{240} + \frac{64}{240} - \frac{33}{240} = \frac{48+64-33}{240} = \frac{112-33}{240} = \frac{79}{240}$.

g) $234.43 + 56.7 + 23.145$

$234.430 + 56.700 + 23.145 = 314.275$

h) $132.332 - 44.12$

$132.332 - 44.120 = 88.212$

3. Final Answer

b) $7\frac{1}{4}$

c) $1\frac{7}{8}$

d) $15\frac{6}{125}$

e) $9\frac{3}{7}$

f) $\frac{79}{240}$

g) $314.275$

h) $88.212$