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vlabodo [156]
3 years ago
9

What is this -5g-13g=54?

Mathematics
2 answers:
lara31 [8.8K]3 years ago
7 0
The answer is g= -3

Hope this helps
-BARSIC- [3]3 years ago
4 0
The answer is -3. Now I just got 100 points!!!
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Step-by-step explanation:

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If RV=w and SU=w–30, what is the value of w?
DIA [1.3K]

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w = 60

Step-by-step explanation:

the midsegment SU is half the measure of side RV, then

SU =  RV , so

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2 years ago
Compute the probability of randomly selecting an eight or jack
Anit [1.1K]

Answer:

2/13

Step-by-step explanation:

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6 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

5 0
3 years ago
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