Answer:
J (22hrs)
Step-by-step explanation:
812 divided by 14= 58
1,276 divided by 58= 22 hrs
Answer:
b³
Step-by-step explanation:
The lowest common multiple of the expression is b². This because, taking out a factor of b, b ( b² + 6b ), taking out a factor of b², b² ( b + 6 ),
taking out a factor of ( b + 6 ), ( b + 6 ) b². As a result b² is only the option which is not a factor.
Considering that each student has only one birthday, each input will be related to only one output, hence this relation is a function.
<h3>When does a relation represent a function?</h3>
A relation represents a function when each value of the input is mapped to only one value of the output.
For this problem, we have that:
- The input is the student's name.
- The output is the student's birthday.
Each student has only one birthday, hence each input will be related to only one output, hence this relation is a function.
More can be learned about relations and functions at brainly.com/question/12463448
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The correct answer is D) 4x + 3y = 90.
x = pairs of socks
y = pairs of shoes
90 = total cost of the sum of socks and shoes
Combining these with the given quantities for each variable:
4x + 3y = 90
Answer:
The students should request an examination with 5 examiners.
Step-by-step explanation:
Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the
denote the event that he passes the examination. Then,
The events (
) follows a Binomial distribution with probability of success 0.80 and the events (
) follows a Binomial distribution with probability of success 0.40.
It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,
Then,
⇒
Then,
Compute the probability that the students passes if request an examination with 3 examiners as follows:
![=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}](https://tex.z-dn.net/?f=%3D%5B%5Csum%5Climits%5E%7B3%7D_%7Bx%3D2%7D%7B%7B3%5Cchoose%20x%7D%280.80%29%5E%7Bx%7D%281-0.80%29%5E%7B3-x%7D%7D%5D%5Ctimes%5Cfrac%7B2%7D%7B3%7D%2B%5B%5Csum%5Climits%5E%7B3%7D_%7Bx%3D2%7D%7B%7B3%5Cchoose%20x%7D%280.40%29%5E%7B3%7D%281-0.40%29%5E%7B3-x%7D%7D%5D%5Ctimes%5Cfrac%7B1%7D%7B3%7D)
The probability that the students passes if request an examination with 3 examiners is 0.715.
Compute the probability that the students passes if request an examination with 5 examiners as follows:
![=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}](https://tex.z-dn.net/?f=%3D%5B%5Csum%5Climits%5E%7B5%7D_%7Bx%3D3%7D%7B%7B5%5Cchoose%20x%7D%280.80%29%5E%7Bx%7D%281-0.80%29%5E%7B5-x%7D%7D%5D%5Ctimes%5Cfrac%7B2%7D%7B3%7D%2B%5B%5Csum%5Climits%5E%7B5%7D_%7Bx%3D3%7D%7B%7B5%5Cchoose%20x%7D%280.40%29%5E%7Bx%7D%281-0.40%29%5E%7B5-x%7D%7D%5D%5Ctimes%5Cfrac%7B1%7D%7B3%7D)
The probability that the students passes if request an examination with 5 examiners is 0.734.
As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.
