In order to rationalize the denominator of each expression, we need to multiply the expression by the same radical in the denominator, this way we can remove the radical from the denominator.

$\frac{5\sqrt{4}}{\sqrt{3}}(\cdot\sqrt{3})=\frac{5\sqrt{4}\sqrt{3}}{(\sqrt{3})^2}=\frac{5\cdot2\cdot\sqrt{3}}{3}=\frac{10\sqrt{3}}{3}$

10)

$-\frac{5}{3\sqrt{2}}(\cdot\sqrt{2})=-\frac{5\sqrt{2}}{3(\sqrt{2})^2}=-\frac{5\sqrt{2}}{3\cdot2}=-\frac{5\sqrt{2}}{6}$

11)

$\frac{2\sqrt{3}}{4\sqrt{5}}(\cdot\sqrt{5})=\frac{2\sqrt{3}\sqrt{5}}{4(\sqrt{5})^2}=\frac{2\sqrt{15}}{4\cdot5}=\frac{\sqrt{15}}{2\cdot5}=\frac{\sqrt{15}}{10}$

12)

$\frac{\sqrt{5}}{\sqrt{2}}(\cdot\sqrt{2})=\frac{\sqrt{5}\sqrt{2}}{(\sqrt{2})^2}=\frac{\sqrt{10}}{2}$

3 0

1 year ago

Aja's favorite cereal is running a promotion that says 111-in-444 boxes of the cereal contain a prize. Suppose that Aja is going

Bond [772]

Using the binomial distribution, there is a 0.6328 = 63.28% probability that she wins at most 1 prize.

For each box, there are only two possible outcomes, either it has a prize, or it does not. The probability of a box having a prize is independent of any other box, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$