Which expression is equivalent to 3sqrtx^5y?

Mathematics

1 answer:

slamgirl [31]1 year ago

3 0

Answer:

The correct option is B.

Step-by-step explanation:

The given expression is

$\sqrt[3]{x^5y}$

$\sqrt[3]{x^5y}=(x^5y)^{\frac{1}{3}}$ $[\because \sqrt[n]{x}=x^{\frac{1}{n}}]$

$\sqrt[3]{x^5y}=(x^5)^{\frac{1}{3}}y^{\frac{1}{3}}$ $[\because (mn)^x=m^xn^x]$

$\sqrt[3]{x^5y}=x^{\frac{5}{3}}y^{\frac{1}{3}}$ $[\because (x^m)^n=x^{mn}]$

Therefore option B is correct.

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Jordan wants to play a basketball game at a carnival. The game costs the player $5 dollar sign, 5 to play, and the player gets t

Alinara [238K]

Answer:

The expected value of Jordan gains is -1 dollar.

Step-by-step explanation:

Consider the following random variables. X := #of shots that Jordan makes. Then, we can can define the random variable Y of the earnings of Jordan in a game as follows

Y = 5 if X=2 (since he gets 10, but invested 5), Y=0 if X=1(since he gets the 5 back) and Y=-5 if X=0(since he doesn't get the money back). Then, in this case, we can define the probability as follows.

P(Y=5) = P(X=2), P(Y=0) = P(X=1), P(Y=-5)= P(X=0).

By definition, the expected value of Y is given by

$E[Y] = 5\cdot P(Y=5)+0\cdot P(Y=0)-5 P(Y=-5)$ . By the previous analysis, we have that

$E[Y] = 5\cdot P(X=2)-5P(X=0)$

We only need to calculate the probabilities for X. In this case, we can consider each shot independt from each other. Then, we can consider X to be distributed as a binomial random variable with n=2 trials and p=0.4 of success (since he has a 40% chance of winning).

Then, by definition

$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k} = \binom{2}{k}0.4^{k}0.6^{2-k}$