<u>Answer:</u>
The correct answer option is P (S∩LC) = 0.16.
<u>Step-by-step explanation:</u>
It is known that the probability if someone is a smoker is P(S)=0.29 and the probability that someone has lung cancer, given that they are also smoker is P(LC|S)=0.552.
So using the above information, we are to find the probability hat a random person is a smoker and has lung cancer P(S∩LC).
P (LC|S) = P (S∩LC) / P (S)
Substituting the given values to get:
0.552 = P(S∩LC) / 0.29
P (S∩LC) = 0.552 × 0.29 = 0.16
Part A. You have the correct first and second derivative.
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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.
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Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out
To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0.
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well.
Answer:
11 months
Step-by-step explanation:
the math that i did was i kept adding 400 to 2500 and counted every four hundred that i added witch was 11 until i got to 7000 so the answer is 11 months. You will have some left over which you will have 300 left over.
5/4
(15/16)(4/3) The 4/3 is the reciprocal of 3/4
=5/4

As we know, slope (m) is equal to tan θ for a straight line :


So, we can say that it's a positive constant slope