The number of permutations of the 25 letters taken 2 at a time (with repetitions) is:
The number of permutations of the 9 digits taken 4 at a time (with repetitions) is:
Each permutation of letters can be taken with each permutation of digits, therefore the total number of possible passwords is:
Answer:
y = 112
Step-by-step explanation:
If y varies directly as the square of x
mathematically;
y ∝ x²
y = kx² where 'k' is the constant of proportionality
let u know the value of k for this equation by making k the subject of the formular from y = kx²
divide both sides by x²
k = y / x²
given that y = 567 x = 9
k = 567/(9)²
k = 567/81
k = 7
now to get the value of y when x = 4
from the equation conneting x, y,
y = kx²
y = 7 × (4)²
y = 7 × 16
y = 112
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
V=xyz where x,y,z are the three dimensions...
V=(1 4/5)(1 4/5)1
V=(9/5)(9/5)=81/25
V=3 6/25 ft^3
He'll be making 1,800 more money per year
