In this question it basically wants you to leave Y alone in a side of the equation.
In this case,
For 3y=c
For Ay=w
For Y/c=w
Y=cw
For y/a=2c
y=2ac
For a=y+p
y=a-p
For C=y-k
y=C+k
Answer:
Step-by-step explanation:
The exponential function that decrease by 15% every time x increases by 1 is given by:
We simplify the parenthesis to get:
Therefore the decrease by 15% every time x increases by 1 is
The second choice is correct.
Step-by-step explanation:
to find a common denominator, you have to find a number that "works" with every other number.
for example, say you have
2/4 and 8/12
First you need to find the common factor between 4 and 12, so list all your fours
4, 8, 12, 16, 20
Now list all your twelves
12, 24, 36, 48, 60
to find the common factor you look at both your list of numbers and find one that's the same, sometimes it takes a long list of numbers to find the common factor, but you will run into one.
So by looking at our list we see that 4 and 12 share the common factor of 12. but since 8/12 already has a denominator of 12, we are going to leave it alone.
now think about what you would multiply 4 by, to get to 12. The answer is
4 x 3 = 12
to make the numerator correct, you multiply it by the same number you did 4, so since your faction is 2/4 you should do 2 x 3 = 6
now you have your answer,
2/4 and 8/12 turns into
6/12 and 8/12
and that's how you find it, let me know if you have questions :)
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.
