Solve t/sqrt{frac{400{36}}}geq1.96

$$\frac { t } { \sqrt { \frac { 400 } { 36 } } } \geq 1.96$$

Solve for t

$$t\geq \frac{98}{15}$$

Solution Steps

Reduce the fraction \frac{400}{36} to lowest terms by extracting and canceling out 4.

$$\frac{t}{\sqrt{\frac{100}{9}}}\geq 1.96$$

Rewrite the square root of the division \frac{100}{9} as the division of square roots \frac{\sqrt{100}}{\sqrt{9}}. Take the square root of both numerator and denominator.

$$\frac{t}{\frac{10}{3}}\geq 1.96$$

Multiply both sides by \frac{10}{3}. Since \frac{10}{3} is positive, the inequality direction remains the same.

$$t\geq 1.96\times \frac{10}{3}$$

Convert decimal number 1.96 to fraction \frac{196}{100}. Reduce the fraction \frac{196}{100} to lowest terms by extracting and canceling out 4.

$$t\geq \frac{49}{25}\times \frac{10}{3}$$

Multiply \frac{49}{25} times \frac{10}{3} by multiplying numerator times numerator and denominator times denominator.

$$t\geq \frac{49\times 10}{25\times 3}$$

Do the multiplications in the fraction \frac{49\times 10}{25\times 3}.

$$t\geq \frac{490}{75}$$

Reduce the fraction \frac{490}{75} to lowest terms by extracting and canceling out 5.

$$t\geq \frac{98}{15}$$