C. 0<p<3 is the answer (at most)
Hope dis helped :)
Answer:
0.1569 = 15.69%
Step-by-step explanation:
If eight calls were placed, and we need to know the probability of exactly two calls were occupied, we need to calculate a combination of 8 choose 2 (all the combinations of 2 occupied calls in the 8 total calls), and multiply by the probability of each case in the 8 calls (2 cases occupied and 6 cases not occupied):
P(8,2) = C(8,2) * p(occupied)^2 * p(not_occupied)^6
P(8,2) = (8*7/2) * (0.45)^2 * (0.55)^6
P(8,2) = 28 * 0.2025 * 0.02768 = 0.1569 = 15.69%
Let named variables as you said g- guinea pigs and c- chinchillas
System of equation is:
First is: c+ g = 60,005
Second is: c= 2g+5
When we replace second in first we get:
2g+5+g = 60,005 => 3g+5= 60,005 => 3g= 60,000
g=60.000/3 = 20,000 => g=20,000
When this result we replace in the second equation we get
c= 2 * 20,000 +5= 40,000+5 = 40,005
c= 40,005
Good luck!!!!
You just have to add or subtract the percent of a number from the original
A) 5percent of 40=2 ( 0.05*40)
Since it’s increase add 2 to 40=42
F) 40percent of 25= 10 (0.4* 25)
25-10=15
