Answer:
762 days
Step-by-step explanation:
Given

Let the rate be R.
So the rate change with time is represented as:

So:

To get the number of insects between day 0 and day 3, we need to integrate dR and set the bounds to 0 and 3
i.e.
becomes


Integrate
![R = 200t + \frac{10t^2}{2} + \frac{13t^3}{3} [3,0]](https://tex.z-dn.net/?f=R%20%3D%20200t%20%2B%20%5Cfrac%7B10t%5E2%7D%7B2%7D%20%2B%20%5Cfrac%7B13t%5E3%7D%7B3%7D%20%5B3%2C0%5D)
Solve for R by substituting 0 and 3 for t





<em>The population of insect between the required interval is 762</em>
Answer:
$2.45*4/7=$1.4
So felipe gets $1.4
Step-by-step explanation:
$2.45*4/7=$1.4
So felipe gets $1.4
A) 4a+3w
b) 2b+h
c) 3(w+h)
d) (4a+3w)(2b+h) - 3(w+h)
just expand brackets for d and simplify
Answer:
2 points
Step-by-step explanation:
10-14=-4
-4+6=2 or
10+6=16
16-14=2
Answer:
The standard deviation would have to be 0.05.
Step-by-step explanation:
We can solve this problem using the 68-95-99.7 rule for normal distributions:
The rule states that:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviations of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
What would the value of σ have to be to ensure that 95% of all readings are within 0.1° of μ?
This means that 2 standard deviations would need to be within 0.1 of the mean. So



The standard deviation would have to be 0.05.