Answer:
The third month
Step-by-step explanation:
Function Modeling
We use mathematical models to approximate to the reality through numbers and symbols. It allows scientist to be able to compare, predict and take decisions based on the models.
We have two situations described as Aril and Dita donating Norwegian kroners in a monthly base.
Aril donated 150 kr the first month, and the cumulative number of kr donated increases by a factor of 2.5 each month. Being n the number of months, the function to model the donation is
[tex]A=150\cdot 2.5^n[/tex]
Note that if n=0, A is the initial donation of 150 kr
Dita donated 300 kr the first month, and the cumulative kr donated increases by 400 kro each month. It's represented as a linear function
[tex]D=300+400\cdot n[/tex]
We must find a value of n such that
[tex]A\geq D[/tex]
Or equivalently
[tex]150\cdot 2.5^n \geq 300+400\cdot n[/tex]
This is an inequality that cannot be solved for n. We'll give some values until the condition is met
[tex]n=1 \ \rightarrow \ 150\cdot 2.5^1 \geq 300+400\cdot 1 \ \rightarrow 150\geq 700[/tex]
The inequality is not true
[tex]n=2 \ \rightarrow \ 150\cdot 2.5^2 \geq 300+400\cdot 2 \ \rightarrow 937.5\geq 1100[/tex]
It's not true either
[tex]n=3 \ \rightarrow \ 150\cdot 2.5^3 \geq 300+400\cdot 3 \ \rightarrow 2343.75\geq 1500[/tex]
Now it's true, so the month where Aril's cumulative donation first exceed Dita's cumulative donation is the third
