The population of one type of fish, P1, and another type of fish, P2, are modeled by the equations P1 = 1000(2/3)^t and P2 = 1000(1/3)^t where t is the time in years. Which expression describes the relationship between the populations P1 and P2 in 3 years?

A. P2 = 1/2P1
B. P2 = 1/8P1
C. P2 = 2/9P1
D. P2 = 1/27P1
E. P2 = 8/729P1

Need ! give step by step solutions on how to solve number one [tex]\frac{9-2\sqrt{3} }{12+\sqrt{3} }[/tex] number two [tex]x+4=\sqrt{13x-20}[/tex] number three (domain and range) [tex]f(x)=2\sqrt[3]{x} +1[/tex]

2people are trying different phones to see which one has the better talk time. chatlong has an average of 18 hours with a standard deviation of 2 hours and talk-a-lot has an average of 20 hours with a standard deviation of 3. the two people who tested their phones both got 20 hours of use on their phones. for the z-scores i got 1 for the chatlong company and 0 for talk-a-lot. what is the percentile for each z-score?

Since opening night, attendance at play a has increased steadily, while attendance at play b first rose and then fell. equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. on what day(s) was the attendance the same at both plays? what was the attendance? play a: y = 8x + 191 play b: y = -x^2 + 26x + 126 a. the attendance was never the same at both plays. b. the attendance was the same on day 5. the attendance was 231 at both plays on that day. c. the attendance was the same on day 13. the attendance was 295 at both plays on that day. d. the attendance was the same on days 5 and 13. the attendance at both plays on those days was 231 and 295 respectively.