It is about <span>1.37638192.
The cotangent is just one divided by the tangent. So, find the tangent of pi/5 = </span><span>0.726542528, then divide 1 by that number.</span>
Answer:
The probability that the good exam belongs to student <em>X</em> is 0.8571.
Step-by-step explanation:
It is provided that the probability that <em>X</em> did well in the exam is, P (X) = 0.90 and the probability that <em>X</em> did well in the exam is, P (Y) = 0.40,
Compute the probability that exactly one student does well in the exam as follows:
![P(Either\ X\ or\ Y\ did\ well)=P(X\cap Y^{c})+P(X^{c}\cap Y)\\=P(X)P(Y^{c})+P(X^{c})P(Y)\\=P(X)[1-P(Y)]+[1-P(X)]P(Y)\\=(0.80\times0.60)+(0.20\times0.40)\\=0.56](https://tex.z-dn.net/?f=P%28Either%5C%20X%5C%20or%5C%20Y%5C%20did%5C%20well%29%3DP%28X%5Ccap%20Y%5E%7Bc%7D%29%2BP%28X%5E%7Bc%7D%5Ccap%20Y%29%5C%5C%3DP%28X%29P%28Y%5E%7Bc%7D%29%2BP%28X%5E%7Bc%7D%29P%28Y%29%5C%5C%3DP%28X%29%5B1-P%28Y%29%5D%2B%5B1-P%28X%29%5DP%28Y%29%5C%5C%3D%280.80%5Ctimes0.60%29%2B%280.20%5Ctimes0.40%29%5C%5C%3D0.56)
Then the probability that <em>X</em> is the one who did well in the exam is:
![P(X\ did\ well\ in\ the\ exam)=\frac{P(X\cap Y^{c})}{P(X\cap Y^{c})+P(X^{c}\cap Y)}\\ =\frac{P(X)[1-P(Y)]}{P(X\cap Y^{c})+P(X^{c}\cap Y)} \\=\frac{0.80\times0.60}{0.56}\\=0.857143\\\approx0.8571](https://tex.z-dn.net/?f=P%28X%5C%20did%5C%20well%5C%20in%5C%20the%5C%20exam%29%3D%5Cfrac%7BP%28X%5Ccap%20Y%5E%7Bc%7D%29%7D%7BP%28X%5Ccap%20Y%5E%7Bc%7D%29%2BP%28X%5E%7Bc%7D%5Ccap%20Y%29%7D%5C%5C%20%3D%5Cfrac%7BP%28X%29%5B1-P%28Y%29%5D%7D%7BP%28X%5Ccap%20Y%5E%7Bc%7D%29%2BP%28X%5E%7Bc%7D%5Ccap%20Y%29%7D%20%5C%5C%3D%5Cfrac%7B0.80%5Ctimes0.60%7D%7B0.56%7D%5C%5C%3D0.857143%5C%5C%5Capprox0.8571)
Thus, the probability that the good exam belongs to student <em>X</em> is 0.8571.
The simpilest form will be A. Hope this helps :)
Answer:
60≥20x+50y
Step-by-step explanation:
Let x = notebooks and y = pens (all per week)
One solution would be x ≥ 1 and y ≥ 1 as at least one per each item per week