Solve {l}{.5(1-.5)}{(.05div1.96)+[.5(1-.5)div500]}

$$\left. \begin{array} { l } { .5 ( 1 - .5 ) } \\ { ( .05 \div 1.96 ) + [ .5 ( 1 - .5 ) \div 500 ] } \end{array} \right.$$

Sort

$$\frac{2549}{98000},0.25$$

Solution Steps

Subtract 0.5 from 1 to get 0.5.

$$sort(0.5\times 0.5,\frac{0.05}{1.96}+\frac{0.5\left(1-0.5\right)}{500})$$

Multiply 0.5 and 0.5 to get 0.25.

$$sort(0.25,\frac{0.05}{1.96}+\frac{0.5\left(1-0.5\right)}{500})$$

Expand \frac{0.05}{1.96} by multiplying both numerator and the denominator by 100.

$$sort(0.25,\frac{5}{196}+\frac{0.5\left(1-0.5\right)}{500})$$

Subtract 0.5 from 1 to get 0.5.

$$sort(0.25,\frac{5}{196}+\frac{0.5\times 0.5}{500})$$

Multiply 0.5 and 0.5 to get 0.25.

$$sort(0.25,\frac{5}{196}+\frac{0.25}{500})$$

Expand \frac{0.25}{500} by multiplying both numerator and the denominator by 100.

$$sort(0.25,\frac{5}{196}+\frac{25}{50000})$$

Reduce the fraction \frac{25}{50000} to lowest terms by extracting and canceling out 25.

$$sort(0.25,\frac{5}{196}+\frac{1}{2000})$$

Least common multiple of 196 and 2000 is 98000. Convert \frac{5}{196} and \frac{1}{2000} to fractions with denominator 98000.

$$sort(0.25,\frac{2500}{98000}+\frac{49}{98000})$$

Since \frac{2500}{98000} and \frac{49}{98000} have the same denominator, add them by adding their numerators.

$$sort(0.25,\frac{2500+49}{98000})$$

Add 2500 and 49 to get 2549.

$$sort(0.25,\frac{2549}{98000})$$

Convert decimal numbers in the list 0.25,\frac{2549}{98000} to fractions.

$$\frac{1}{4},\frac{2549}{98000}$$

Least common denominator of the numbers in the list \frac{1}{4},\frac{2549}{98000} is 98000. Convert numbers in the list to fractions with denominator 98000.

$$\frac{24500}{98000},\frac{2549}{98000}$$

To sort the list, start from a single element \frac{24500}{98000}.

$$\frac{24500}{98000}$$

Insert \frac{2549}{98000} to the appropriate location in the new list.

$$\frac{2549}{98000},\frac{24500}{98000}$$

Replace the obtained fractions with the initial values.

$$\frac{2549}{98000},0.25$$

Evaluate

$$0.25,\ \frac{2549}{98000}$$