Gavin needs $80 to buy a fish tank. He has saved $8 and plans to work as a babysitter to earn $9 per hour. Which inequality shows the minimum number of hours, n, that Gavin should work as a babysitter to earn enough to buy the fish tank? 8 + 9n ≥ 80, so n ≥ 8 8 + 9n ≤ 80, so n ≤ 8 9n ≥ 80 + 8, so n ≥ 9.8 9n ≤ 80 + 8, so n ≤ 9.8
It's simple arithmetic. The Answer is 604661760000000000000000000000000000000000000000000000000000000000000000000000000000 and in scientific notation
60466176 x 10⁸³
Step-by-step explanation:
(a) dP/dt = kP (1 − P/L)
L is the carrying capacity (20 billion = 20,000 million).
Since P₀ is small compared to L, we can approximate the initial rate as:
(dP/dt)₀ ≈ kP₀
Using the maximum birth rate and death rate, the initial growth rate is 40 mil/year − 20 mil/year = 20 mil/year.
20 = k (6,100)
k = 1/305
dP/dt = 1/305 P (1 − (P/20,000))
(b) P(t) = 20,000 / (1 + Ce^(-t/305))
6,100 = 20,000 / (1 + C)
C = 2.279
P(t) = 20,000 / (1 + 2.279e^(-t/305))
P(10) = 20,000 / (1 + 2.279e^(-10/305))
P(10) = 6240 million
P(10) = 6.24 billion
This is less than the actual population of 6.9 billion.
(c) P(100) = 20,000 / (1 + 2.279e^(-100/305))
P(100) = 7570 million = 7.57 billion
P(600) = 20,000 / (1 + 2.279e^(-600/305))
P(600) = 15170 million = 15.17 billion
Answer:
The expected value of lateness 
hours.
Step-by-step explanation:
The probability distribution of lateness is as follows:
Lateness P (Lateness)
On Time 4/5
1 Hour Late 1/10
2 Hours Late 1/20
3 Hours Late 1/20
The formula of expected value of a random variable is:
Compute the expected value of lateness as follows:
Thus, the expected value of lateness hours.
