What is the quotient of (6x2 – 19x + 15) and (2x – 3)?

Mathematics

1 answer:

Sonbull [250]3 years ago

6 0

Answer:

(3x - 5)------------------->

Step-by-step explanation:

look at the work

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Suppose there is a bag containing 5 red marbles (Upper R), 5 pink marbles (Upper P), 5 green marbles (Upper G), and 5 blac

Radda [10]

Answer:

In the Explanation

Step-by-step explanation:

(a)From the attached probability tree, the possible outcomes are:

RR,RP,RG,RB,PR,PP,PG,PB,GR,GP,GB,GG,BR,BP,BG,BB

(b)Probability Distribution of Drawing Pink Marbles

$\left|\begin{array}{c|c}Result&Probability\\----&----\\0&\frac{9}{16}\\----&---\\1& \frac{3}{8}\\----&---\\2&\frac{1}{16}\end{array}\right|$

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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 $6\cdot 0.2 = 1.2$