This is the answer to the question
Answer: The question is incomplete, here is the complete question ;
Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose that for banner-tailed kangaroo rats, X has an exponential distribution with parameter ? = 0.01327. (a) What is the probability that the distance is at most 100 m? At most 200 m? Between 100 and 200 m? (Round your answers to four decimal places.
Step-by-step explanation:
- X = denote the distance (m)
- P (X > x) = exp( - 0.01342x)
- Therefore, P (X < = 100) = 1 - P (X > 100) = e-0.01342 X 100 = 0.2613 ; the probability that the distance is at most 100 m
- Similarly, P (X < =200) = e-0.01342 X 200 = 0.0683 ; At most 200 m
- P(100 < X <200) = 0.2613 - 0.0683 = 0.193 ; Between 100 and 200 m
First factor x^2+13x+36 using the AC method. Then if any individual factor on the left side of the equation is equal to zero the whole expression will equal zero. After that set x+4 equal to zero and solve for x and do the same for 9. Lastly your answer is all the values of (x+4)(x+9)=0 true.
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