Someone plz help me with number 13 on this document

Mathematics

1 answer:

alukav5142 [94]3 years ago

3 0

The answer will be -1
(4,-1)

You might be interested in

Translate the following word phrases to an algebraic expression and simplify: “8 times the difference of 6 times a number and 3”

Ostrovityanka [42]

SOLUTION:

Step 1:

In this question, we are meant to:

Translate the following word phrases to an algebraic expression and simplify:

“8 times the difference of 6 times a number and 3”

Step 2:

Assuming the unknown number be y, we have that:

$\begin{gathered} 8\text{ ( 6y - 3 )} \\ =\text{ 48 y - 24} \end{gathered}$

CONCLUSION:

The final answer is:

$48y\text{ - 24}$

3 0

11 months ago

Solve it I need help

rusak2 [61]

It opens upward and the vertex is (2,-6) here are some plotted points: (0,6),(1,-3),(2,-6),(3,-3),(4,6). Hope that helps. I hate graphing but if you do it enough times you'll get the hang of it!

6 0

2 years ago

Discrete R.V. Assume a student shows up for EGR 280 completely unprepared and the instructor gives a pop quiz. Assume there are

Tom [10]

Answer:

a) p=0.2

b) probability of passing is 0.01696

c) The expected value of correct questions is 1.2

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 $6\cdot 0.2 = 1.2$