Question

5. A biology class helped measure the squirrel population in San Pedro Park. First park rangers tagged 48 squirrels in the park. a week later, the class went to the same park and counted a total of 300 squirrels of which 9 were tagged. Set up a proportion to estimate the number of squirrels in San Pedro Park.

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10. The prison population in 1990 was approximately 740,000. By the year 2006 the population had reached 1,320,000. Assume the growth in prison population is linear and write a linear equation (y=mx+b) that models this prison growth, where x = 0 when the year is 1990.

According to the model in question 10, what was the prison population in 2010?

For the function f ( x ) = x 2 + 4 x, graph the function. You may use the coordinate plane in "

27. If your investment of $1000 is compounded continuously, how long will it take to double your investment at a 15% rate of interest?

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Answer 1

Solution 5 o were No. tagged 12 و۱ out of 300 squissels tagged 9 No. tagged squirrels found out of 300 Squirals (300) sporen Hence 48 No of tagged squirrel Found 900) x out of X 48 1600 Answer: No. of sqwrrel in the park - 1600) Ang Solution (10 year 1990 740000 Population population year 2006 1320000 y= = mntb x=0 for 1990 then for 2006 x=16 population diffrence slope = m = 1320000-740000 Year difference 580000 16 36250 y 36250x + b 740000 = 36250xO+b - b: 740000 Year 2010 ly = 36250 at 740000 Population = 1465000 An for year 20to 2010- 1990 = 20 11 465000 y = 36280X20 +740000

Graph will upward Paruboil fo x²+42 learling were yo f'n) - anty zo -/ = -2 TE2) = (212+ 4L-2) = 48 = my 02 +420 X(nty) --Y when a=0, fenico fal=0 parabols 2-) +-2 -3 -1 (-2,-4) Solerian ( 27) P = Investment = $1000 nale = 15t = 0.15 Continuous compounding Amont doupled = P(t) = 2x1000 IP(E)=Po est To find. time (t) Pet = 2000 Po= investment &a interest rate P(t) = Po ert oulst 2000 = to time 1000 e 2 = edist Tany log Both sicle 0.5ta ena t= ln(2) 0.15 4.62 ta 4.62 yey