Question
Write $\Big(\frac{1}{9}\Big)^{-\frac{1}{2}}\times(64)^{-\frac{1}{3}}$ as a rational number.
✓
Answer
We have to find the value of $\Big(\frac{1}{9}\Big)^{-\frac{1}{2}}\times(64)^{-\frac{1}{3}}$
So, $\Big(\frac{1}{9}\Big)^{-\frac{1}{2}}\times(64)^{-\frac{1}{3}}=\Big(\frac{1}{9}\Big)^{\frac{-1}{2}}\times(64)^{\frac{-1}{3}}$
$=\Big(\frac{1}{3^2}\Big)^{\frac{-1}{2}}\times(4^3)^{\frac{-1}{3}}$
$=\Bigg(\frac{1}{3^{2\text{x}\frac{-1}{2}}}\Bigg)\times\Big(4^{3\times\frac{-1}{3}}\Big)$
$\Big(\frac{1}{9}\Big)^{\frac{-1}{2}}\times(64)^{\frac{-1}{3}}=\frac{1}{3^{-1}}\times4^{-1}$
$=\frac{1}{\frac{1}{3}}\times\frac{1}{4}$
$=1\times\frac{3}{1}\times\frac{1}{4}$
$=\frac{3}{4}$ Hence the value of the value of $\Big(\frac{1}{9}\Big)^{\frac{-1}{2}}\times(64)^{\frac{-1}{3}}$ is $\frac{3}{4}.$
So, $\Big(\frac{1}{9}\Big)^{-\frac{1}{2}}\times(64)^{-\frac{1}{3}}=\Big(\frac{1}{9}\Big)^{\frac{-1}{2}}\times(64)^{\frac{-1}{3}}$
$=\Big(\frac{1}{3^2}\Big)^{\frac{-1}{2}}\times(4^3)^{\frac{-1}{3}}$
$=\Bigg(\frac{1}{3^{2\text{x}\frac{-1}{2}}}\Bigg)\times\Big(4^{3\times\frac{-1}{3}}\Big)$
$\Big(\frac{1}{9}\Big)^{\frac{-1}{2}}\times(64)^{\frac{-1}{3}}=\frac{1}{3^{-1}}\times4^{-1}$
$=\frac{1}{\frac{1}{3}}\times\frac{1}{4}$
$=1\times\frac{3}{1}\times\frac{1}{4}$
$=\frac{3}{4}$ Hence the value of the value of $\Big(\frac{1}{9}\Big)^{\frac{-1}{2}}\times(64)^{\frac{-1}{3}}$ is $\frac{3}{4}.$
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On a particular day, at a crossing in a city, the various types of $240$ vehicles going past during a time interval were observed, as under:
Out of these vehicles, one is chosen at random. What is the probability that the chosen vehicle is a two-wheeler?
→|
Types of vehicle
|
Two-wheelers
|
Three-wheelers
|
Four-wheelers
|
|
Frequency
|
$84$
|
$68$
|
$88$
|