Answer:
Step-by-step explanation:
Given the integral equation
According to integration by part;
u = x, dv = e^7x
du/dx = 1
du = dx
Substitute the given values into the formula;
At the end of the zeroth year, the population is 200.
At the end of the first year, the population is 200(0.96)¹
At the end of the second year, the population is 200(0.96)²
We can generalise this to become at the end of the nth year as 200(0.96)ⁿ
Now, we need to know when the population will be less than 170.
So, 170 ≤ 200(0.96)ⁿ
170/200 ≤ 0.96ⁿ
17/20 ≤ 0.96ⁿ
Let 17/20 = 0.96ⁿ, first.
log_0.96(17/2) = n
n = ln(17/20)/ln(0.96)
n will be the 4th year, as after the third year, the population reaches ≈176
18.432 + 5.55 + 19.3
We can notice that the last number in the process of addition is 19.3 | with 1 decimal place value / 3 significant figures |
It is better to change all values to 1 decimal place value / 3 significant figures
18.432 → 18.4
5.55 → 5.6 ( 5 rounds up )
19.3 → 19.3
Let us add:
18.4 + 5.6 + 19.3 = 43.3
If you add the original numbers, you will get
18.432 + 5.55 + 19.3 = 43.282
43.282 ≈ 43.3
Answer:
The answer is B
Step-by-step explanation:
