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
The answer is c u welcome
Answer:
$1296
Step-by-step explanation:
the markup is 700 so add that to the 500, then sales tax adds 96
Answer:
Step-by-step explanation:
- Orchestra Seats: Balcony seats + 7
- Orchestra Seats: 600
- Balcony Seats: 120
- Total Amount of Sales: 13,560
A.) 600 x 20 = 12,000
120 x 27 = 3,240
12,000 + 3,240 = 15,240
B.) 600 x 13 = 7,800
120 x 20 = 2,400
7,800 + 2,400 = 10,200
C.) 600 x 18 = 10,800
120 x 25 = 3,000
10,800 + 3,000 = 13,800
D.) 600 x 11 = 6,600
120 x 18 = 2,160
6,600 + 2,160 = 8,760
3 is not the GCF, 30 is the GCF. The answer is 30(2+3).