Answer:
Number of fish in 2006 = 845 fishes (Approx.)
Step-by-step explanation:
Given;
Number of fish in 2000 = 1,150 fishes
Decreasing rate = 5% per year
Find:
Number of fish in 2006
Computation:
Exponential decay function:
F = P[1-r]ⁿ
Number of year = 2006 - 2000
Number of year = 6 year
Number of fish in 2006 = 1,150[1-5%]⁶
Number of fish in 2006 = 1,150[1-0.05]⁶
Number of fish in 2006 = 1,150[0.95]⁶
Number of fish in 2006 = 1,150[0.735]
Number of fish in 2006 = 845.25
Number of fish in 2006 = 845 fishes (Approx.)
Answer:
12 and 14
Step-by-step explanation:
Let the even integers be x and x+2
three times the larger number is expressed as 3(x+2)
30 more than the smaller one is x + 30
Equating both expressions
3(x+2) = x +30
Find x
3x+6 = x+30
3x-x = 30-6
2x = 24
x = 12
The second integer is 12+2 = 14
Hence the required integers are 12 and 14
