9 times it will be 98 rhr
<span>3down votefavorite1Find minimum and maximum value of function <span>f(x,y)=3x+4y+|x−y|</span> on circle<span>{(x,y):<span>x2</span>+<span>y2</span>=1}</span>I used polar coordinate system. So I have <span>x=cost</span> and <span>y=sint</span> where <span>t∈[0,2π)</span>.Then i exploited definition of absolute function and i got:<span>h(t)=<span>{<span><span>4cost+3sintt∈[0,<span>π4</span>]∪[<span>54</span>π,2π)</span><span>2cost+5sintt∈(<span>π4</span>,<span>54</span>π)</span></span></span></span>Hence i received following critical points (earlier i computed first derivative):<span>cost=±<span>45</span>∨cost=±<span>2<span>√29</span></span></span>Then i computed second derivative and after all i received that in <span>(<span>2<span>√29</span></span>,<span>5<span>√29</span></span>)</span> is maximum equal <span>√29</span> and in <span>(−<span>45</span>,−<span>35</span>)</span> is minimum equal <span>−<span>235</span></span><span>
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Answer:
P(A∣D) = 0.667
Step-by-step explanation:
We are given;
P(A) = 3P(B)
P(D|A) = 0.03
P(D|B) = 0.045
Now, we want to find P(A∣D) which is the posterior probability that a computer comes from factory A when given that it is defective.
Using Bayes' Rule and Law of Total Probability, we will get;
P(A∣D) = [P(A) * P(D|A)]/[(P(A) * P(D|A)) + (P(B) * P(D|B))]
Plugging in the relevant values, we have;
P(A∣D) = [3P(B) * 0.03]/[(3P(B) * 0.03) + (P(B) * 0.045)]
P(A∣D) = [P(B)/P(B)] [0.09]/[0.09 + 0.045]
P(B) will cancel out to give;
P(A∣D) = 0.09/0.135
P(A∣D) = 0.667
Okay this isn't the answer but it is how u can get your answer and explain how to justify it. which math will be how u justify it.... So basically your answer is A.... and you will add 1 1/2 + 3/4 + 1/8 =.25 which is your work they want you to show them and that is what I just showed you in a number line. I hope that's correct, but i'm pretty sure it is. Sorry that took so long i am working still on my own work and still stuck on my own lol. I am not the best with math and this kind of stuff though, but again i am almost sure this is correct