Answer:
The first year in which Clara will see that Investment B's value will exceed Investment A's value will be year 14.
Step-by-step explanation:
Since Clara made two investments, and Investment A has an initial value of $ 500 and increases by $ 45 every year, while Investment B has an initial value of $ 300 and increases by 10% every year, and Clara checks the value of her investments once to year, at the end of the year, to determine what is the first year in which Clara sees that Investment B's value has exceeded investment A's value, the following calculation must be performed:
500 + (45 x X) = A
300 x 1.1 ^ X = B
A = 500 + 45 x 5 = 500 + 225 = 725
B = 300 x 1.1 ^ 5 = 483.15
A = 500 + 45 x 10 = 950
B = 300 x 1.1 ^ 10 = 778.12
A = 500 + 45 x 15 = 1175
B = 300 x 1.1 ^ 15 = 1253.17
A = 500 + 45 x 14 = 1,130
B = 300 x 1.1 ^ 14 = 1,139.25
Therefore, the first year in which Clara will see that Investment B's value will exceed Investment A's value will be year 14.
Answer:
The population for the first year is 75.35.
Step-by-step explanation:
Solution:
Given Data;
Carrying capacity (K) = 100
Growth rate (r) = 1.8% = 0.018
Initial population (Po) = 75
time = first year
Population at any given time (pt) = ?
Using the population growth rate formula, we have
<h2>
dp/dt = rp (1-p/t)---------------------------1</h2>
Where r is the growth rate, t is the given time and p is the population growth
Equation 1 is differentiated and is obtained as;
<h2>Pt = K/ [1 + (K-Po/Po)e^-rt]------------2</h2>
Substituting into equation 2, we have
Pt = 100/[ 1+ (100-75/75) *e^-0.018*1]
Pt = 100/[1+ (0.333 * e^-0.018]
= 100/[1 +(0.333*0.982)]
= 100/[1 + 0.327]
= 100/1.327
=75.35
Answer:
Nice.
Step-by-step explanation:
The answer is D because you can switch around the two numbers with the commutative property to make it -7-8.
