if your asking for the equation, it's t + 7 < 3
Answer:
The population of deer at any given time = 200(e^0.03t) ÷ (1.5 + (e^0.03t))
Step-by-step explanation:
This is an example of logistic equation on population growth
carrying capacity, k = 200
Rate, r = 3% = 0.03
Initial Population, P1 = 80
P(t) =?
P(t) = (P1 (k)(e^rt)) ÷ (k- P1 + P1(e^rt))
P(t) = (80 (200)(e^0.03t)) ÷ (200 - 80 + 80(e^0.03t))
= (16000(e^0.03t)) ÷ (120 + 80(e^0.03t))
= 200(e^0.03t) ÷ (1.5 + (e^0.03t))
Due to length restrictions, we kindly invite to see the explanation below to know the answer with respect to each component of the question concerning linear equations.
<h3>How to determine a linear equation describing the daily distance of a runner</h3>
In this question we need to derive an expression of the <em>daily</em> distance as a function of time. Now we proceed to complete the components:
- <em>Linear</em> equations have an <em>independent</em> variable (t - time) and a dependent variable (x - daily distance).
- We notice that the daily distance increases linearly in time, then then we have the following pattern:
t 1 2 3 4 5 6
x 2 2.5 3 3.5 4 4.5 - The equation that represents the n-th term of the sequence is x(n) = 2 + 0.5 · (n - 1).
- The week when Susie will run 10 miles per day is:
10 = 2 + 0.5 · (n - 1)
8 = 0.5 · (n - 1)
n - 1 = 16
n = 17
Susie will run 10 miles per day in the 17th week. - It is not reasonable to think that pattern will continue indefinitely as it is witnessed in the difficulties experimented by <em>fastest</em> runners in the world to increase their <em>peak</em> speeds.
- A marathon has a distance of 26 miles, then we must solve the following equation:
26 = 2 + 0.5 · (n - 1)
24 = 0.5 · (n - 1)
48 = n - 1
n = 49
Susie should start her training 49 weeks before the marathon.
To learn more on linear equation: brainly.com/question/11897796
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Y+4=3(x-2)
Y+4=3x-6
Answer: Y=3x-10
Divide 45 by 5 and 5 hours by 5 so it’s 9 miles per hour. So the equation would be: m = 9/h (a fraction)
