Answer: The function is M(t) = 40mg*0.992^t
Step-by-step explanation:
Half life refers to the time that a given quantity to reduce to half its initial value.
This is usually a exponential decay, so we want a function of the form:
M(t) = A*r^(*t)
Were t is time, in this case, t is the number of days
A is the initial value
r is the rate of decrease, we will find that for exponential decays r must be a number between 0 and 1. If r is closer to 0, we will have a fast exponential decay, if r is closer to 1 we will have a slower exponential decay.
We have that the initial mass is 40mg, this is at t = 0.
M(0) = A*r^(0) = A = 40mg
So now we have the value of A.
Now, we know that in t = 84, the mass of the substance will be half its initial value, so it will be:
A/2 = 40mg/2 = 20mg
this means that:
M(84) = 40mg*r^(84) = 20mg
r^(84) = 20mg/40mg = 1/2
![r = \sqrt[84]{1/2}](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B84%5D%7B1%2F2%7D)
= 0.992
Then the equation is:
M(t) = 40mg*0.992^t
Answer:
(x + 2)^2 + (y + 2)^2 = 16
Step-by-step explanation:
B they only have one pair of parallel sides. The top and bottom
The statement third, “This is voluntary response bias. The result overestimates true support for firing the coach” is correct.
<h3>What is a survey?</h3>
A survey is a means of gathering information from a sample of people using pertinent questions with the goal of understanding populations as a whole.
We have:
A local baseball team is struggling this season, and many fans of the team believe it may be time to replace the head coach.
Number of votes V(n) = 2367
After the value of V(n) 79% of those who responded felt the coach should be fired.
Based on the data given, we can say this is referred to as voluntary response bias. The outcome exaggerates the level of support for firing the coach.
Thus, the statement third “This is voluntary response bias. The result overestimates true support for firing the coach” is correct.
Learn more about the survey here:
brainly.com/question/17373064
#SPJ1
Answer:
Step-by-step explanation:
Reducing 3 from numerator and denominator,
