Solve 2^-3+(2+1)^-1*3*40^-1=

$$2 ^ { - 3 } + ( 2 + 1 ) ^ { - 1 } \cdot 3 \cdot 40 ^ { - 1 } =$$

Evaluate

$$\frac{3}{20}=0.15$$

Solution Steps

Calculate 2 to the power of -3 and get \frac{1}{8}.

$$\frac{1}{8}+\left(2+1\right)^{-1}\times 3\times 40^{-1}$$

Add 2 and 1 to get 3.

$$\frac{1}{8}+3^{-1}\times 3\times 40^{-1}$$

Calculate 3 to the power of -1 and get \frac{1}{3}.

$$\frac{1}{8}+\frac{1}{3}\times 3\times 40^{-1}$$

Multiply \frac{1}{3} and 3 to get 1.

$$\frac{1}{8}+40^{-1}$$

Calculate 40 to the power of -1 and get \frac{1}{40}.

$$\frac{1}{8}+\frac{1}{40}$$

Add \frac{1}{8} and \frac{1}{40} to get \frac{3}{20}.

$$\frac{3}{20}$$

Factor

$$\frac{3}{2 ^ {2} \cdot 5} = 0.15$$