Using radiclas, write equivalent expression for the expression y to the power of 2 1/3

Mathematics

1 answer:

SashulF [63]3 years ago

6 0

Answer:

$\large\boxed{y^{2\frac{1}{3}}=y^2\sqrt[3]y}$

Step-by-step explanation:

$y^{2\frac{1}{3}}\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=y^{2+\frac{1}{3}}=y^2\cdot y^\frac{1}{3}\\\\\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=y^2\sqrt[3]y$

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Answer:

a) p=0.2

b) probability of passing is 0.01696

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Step-by-step explanation:

a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.

b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6

$P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} = \binom{6}{k}0.2^{k}(0.8)^{6-k}$ .

Recall that $\binom{n}{k}= \frac{n!}{k!(n-k)!}$ In this case, the student passes if X is at least four correct questions, then

$P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=\binom{6}{4}0.2^{4}(0.8)^{6-4}+\binom{6}{5}0.2^{5}(0.8)^{6-5}+\binom{6}{6}0.2^{6}(0.8)^{6-6}= 0.01696$

c)The expected value of a binomial random variable with parameters n and p is $E[X] = np$ . IN our case, n=6 and p =0.2. Then the expected value of correct answers is $6\cdot 0.2 = 1.2$