An exponential decay function, and use it to approximate the population in 2027 is P(t) = 4001e^-(0.0037)(28) = P(t) = 4001e^-0.1036
<h3>Exponential equation</h3>
The standard exponential function is expressed as:
P(t) = P0e^-rt
- P0 is the initial population = 4001
- r is the rate. = 0.37% = 0.0037
- t is the time
Substitute
P(t) = 4001e^-(0.0037)t
If t = 28 (by 2027)
An exponential decay function, and use it to approximate the population in 2027 is P(t) = 4001e^-(0.0037)(28) = P(t) = 4001e^-0.1036
Learn more on exponential function here: brainly.com/question/12940982
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nothing to say this app is so bad this is not useful at all
The distance traveled, d = 2/5 miles = 0.4 miles.
The time taken to travel 0.4miles is given as, t = 1/4 hr = 0.25 hr.
Thus, the average speed, s = distance traveled divided by time taken.
i.e.
![\begin{gathered} \text{average speed = }\frac{distance\text{ traveled}}{\text{time taken}}\text{ = }\frac{0.4}{0.25}\text{ } \\ \text{average sp}eed\text{ = 1.6 miles/hr} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7Baverage%20speed%20%3D%20%7D%5Cfrac%7Bdistance%5Ctext%7B%20traveled%7D%7D%7B%5Ctext%7Btime%20taken%7D%7D%5Ctext%7B%20%3D%20%20%7D%5Cfrac%7B0.4%7D%7B0.25%7D%5Ctext%7B%20%7D%20%5C%5C%20%5Ctext%7Baverage%20sp%7Deed%5Ctext%7B%20%3D%201.6%20miles%2Fhr%7D%20%5Cend%7Bgathered%7D)
we can now go through the options and see the correct answers:
option A is correct because unit rate per hour is the same thing as average speed
option B is wrong because average speed is greater than 1mile/hr from our calculation above
option C is wrong because 13/5 gives us 2.6 miles/hr
option D is correct because the average speed is indeed greater than 1
option E is wrong because average speed is 1.6 miles/hr not 2 1/4 miles/hr
Thus, the answers are:
Option A and Option D
Answer:
The percentage of admitted applicants who had a Math SAT of 700 or more is 48.48%.
Step-by-step explanation:
The Bayes' theorem is used to determine the conditional probability of an event <em>E</em>
, belonging to the sample space S = (E₁, E₂, E₃,...Eₙ) given that another event <em>A</em> has already occurred by the formula:
![P(E_{i}|A)=\frac{P(A|E_{i})P(E_{i})}{\sum\limits^{n}_{i=1}{P(A|E_{i})P(E_{i})}}](https://tex.z-dn.net/?f=P%28E_%7Bi%7D%7CA%29%3D%5Cfrac%7BP%28A%7CE_%7Bi%7D%29P%28E_%7Bi%7D%29%7D%7B%5Csum%5Climits%5E%7Bn%7D_%7Bi%3D1%7D%7BP%28A%7CE_%7Bi%7D%29P%28E_%7Bi%7D%29%7D%7D)
Denote the events as follows:
<em>X</em> = an student with a Math SAT of 700 or more applied for the college
<em>Y</em> = an applicant with a Math SAT of 700 or more was admitted
<em>Z</em> = an applicant with a Math SAT of less than 700 was admitted
The information provided is:
![P(Y)=0.36\\P(Z)=0.18\\P(X|Y)=0.32](https://tex.z-dn.net/?f=P%28Y%29%3D0.36%5C%5CP%28Z%29%3D0.18%5C%5CP%28X%7CY%29%3D0.32)
Compute the value of
as follows:
![P(X|Z)=1-P(X|Y)\\=1-0.32\\=0.68](https://tex.z-dn.net/?f=P%28X%7CZ%29%3D1-P%28X%7CY%29%5C%5C%3D1-0.32%5C%5C%3D0.68)
Compute the value of P (Y|X) as follows:
![P(Y|X)=\frac{P(X|Y)P(Y)}{P(X|Y)P(Y)+P(X|Z)P(Z)}](https://tex.z-dn.net/?f=P%28Y%7CX%29%3D%5Cfrac%7BP%28X%7CY%29P%28Y%29%7D%7BP%28X%7CY%29P%28Y%29%2BP%28X%7CZ%29P%28Z%29%7D)
![=\frac{(0.32\times 0.36)}{(0.32\times 0.36)+(0.68\times 0.18)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%280.32%5Ctimes%200.36%29%7D%7B%280.32%5Ctimes%200.36%29%2B%280.68%5Ctimes%200.18%29%7D)
![=0.4848](https://tex.z-dn.net/?f=%3D0.4848)
Thus, the percentage of admitted applicants who had a Math SAT of 700 or more is 48.48%.
Answer:
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Step-by-step explanation:
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