Answer:
Option c.
No damping
Step-by-step explanation:
We can easily solve this question by using a graphing calculator or any plotting tool.
The function is
f(x) = (√11)*cos(3.7x)
Which can be seen in the picture below
We can notice that f(x) is a cosine with maximum amplitude of (√11). Neither this factor nor the multiplication of x by 3.7 serve as a damping factor since they are constants.
f(x) does not present any dampening
<span>$ 2037.40 </span>
<span>Interesting that there was no down payment..........The staggering monthly payment above would be enough to pay off the total mortgage in just over 5 years. Nick's parents would be much better off to save their money for a few years.</span>
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.
Answer:
40
Step-by-step explanation:
The entire line is 180°
since 140° is occupied, there is a 40° angle.
I hope isn't too late :'(