BRAINLIST ASAP

Mathematics

1 answer:

kifflom [539]3 years ago

6 0

Answer:

25.50

Step-by-step explanation:

-5 notebooks $2 each

-4 pencils $1.50 each

-2 packs of markers $3.75 each

5(2)+4(1.50)+2(3.75)

$23.5

23.5 times 8.5/100

1.9975

plus 23.5

25.4975

rounded to 25.50

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PLZ HELP THIS IS DUE SOON!!!!!!! Divide. Round to the nearest hundredth if necessary.

ivolga24 [154]

Answer:

Step-by-step explanation:

when you divide 80.88 by 8.425 you get 9.6

and since youre rounding, you can round up 6 because its higher than 5

and your final answer should be 10.

hoped this helped! :)

4 0

3 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 $6\cdot 0.2 = 1.2$