Answer:
a) 33.33%)
b) 135 minutes
c) 8.66 min
d) 50%
Step-by-step explanation:
a) the probability for a uniform distribution is
P(b<X<a) = (a-b)/(c-d) , where c and d are the maximum and minimum values
therefore the probability that the flight is more than 140 minutes ( and less than 150 since it is the maximum value)
P(140<X<150) = (a-b)/(c-d) = (150-140)/(150-120) = 10/30 = 1/3 (33.33%)
b) the mean (expected value) for a uniform probability distribution is
E(X) = (c+d)/2 = (120+150)/2 = 135 minutes
c) the standard deviation for a uniform probability distribution is
σ²(X)= (c-d)²/12 = (150-120)²/12 = 75 min²
σ = √75 min² = 8.66 min
b) following the same procedure as in a)
P(120<X<135) = (a-b)/(c-d) = (135-120)/(150-120) = 15/30 = 1/2 (50%)
The fraction form is 26/3 or 8 2/3
Answer:
Step-by-step explanation:
11.04 = 10(1.02)^n
1.104 = 1.02^n
ln 1.104 = ln 1.02^n
ln 1.104 = n ln 1.02
n = ln 1.104/ ln 1.02
n = 4.99630409516
4.99 can be rounded to 5.
So a reasonable domain would be 0 ≤ x < 5
PART B)
f(0) = 10(1.02)^0
f(0) = 10(1)
f(0) = 10
The y-intercept represents the height of the plant when they began the experiment.
f(1) = 10(1.02)^1
f(1) = 10(1.02)
f(1) = 10.2
(1, 10.2)
f(5) = 10(1.02)^5
f(5) = 10(1.1040808)
f(5) = 11.040808
f(1)=10(1.02)^1
f(1)=10.2
Average rate= (fn2-fn1)/(n2-n1)
=11.04-10.2/(5-1)
=0.22
the average rate of change of the function f(n) from n = 1 to n = 5 is 0.22.
