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28_ TThe number Of deer in & park reserve was counted to be 27 on January 2000. Six months later . the number of deer was counted t0 be 38. Determine the 6-mon...

Question

28_ TThe number Of deer in & park reserve was counted to be 27 on January 2000. Six months later . the number of deer was counted t0 be 38. Determine the 6-month growth factor and 6-month percent change for the situation Determine the I-month growth factor and ~month percent change for the situation: Define function that relates the number deer in the park reserve since January 1 , 2000 in terms of the number of months that have elapsed since January 1 , 2000. (Assuming the number 0f deer c

28_ TThe number Of deer in & park reserve was counted to be 27 on January 2000. Six months later . the number of deer was counted t0 be 38. Determine the 6-month growth factor and 6-month percent change for the situation Determine the I-month growth factor and ~month percent change for the situation: Define function that relates the number deer in the park reserve since January 1 , 2000 in terms of the number of months that have elapsed since January 1 , 2000. (Assuming the number 0f deer continues t0 increase by the same percent each month.) Define function that relates the number deer in the park reserve since January , 2000 in terms of the number of 6-month intervals that have elapsed since January | , 2000. (Assuming the number of deer continues to increase by the same percent each month.) Use one of your functions to determine when the number of deer will reach approximately 125.



Answers

The exponential growth of the deer population in Massachusetts can be approximated using the model $$f(x)=50,000(1+0.06)^{x}$$ where $50,000$ is the initial deer population and 0.06 is the rate of growth. $f(x)$ is the total population after $x$ years have passed. Find each value to the nearest thousand. (a) Predict the total population after 4 yr. (b) If the initial population was $30,000$ and the growth rate was $0.12,$ how many deer would be present after 3 yr? (c) How many additional deer can we expect in 5 yr if the initial population is $45,000$ and the current growth rate is $0.08 ?$

So for this question, we asked a number things. So for part A were given a position if the vexed 0 to 50,000 times one plus 0.6 races X power where X is number years have passed for part. They were told that we are us to predict the population after four years. So this will get us approximately 6312 dear Report be. We are told that after Russian population is 30,000 and this over here should be four. So our initial population 30,000 Argo. Great! It's 12%. We're also for year olds. The mother dear. After three years, this will get us approximately 42,140 h dear, It is so for Percy. We're told that after five years Grisha population is 45,000. Growth growth rate is airplanes or eight. What is there to your population? So this will get us 66,119 points. So six we need to find are additional dear. So intake 66,119 points on six of minus 45,000. Yes, approximately 23,000. 120 dear. So these are answers

So we're talking about a deer population during different years. So for part A. How many deer are there at the beginning of a 10 year period? So the very beginning is right here. That's about 1.5 times 100. So that's 100 and 50 dear Number are part B. During which, to your period, Does the deer population appear to be increasing the most rapidly? So we're looking for the steepest slope. The steepest increase on the steepest increase appears toe happen. Mm. Look closely on the six. Looks like it's right here, right here, right here. So that's two years. So that's 1996. That's the whole year of 96 the whole year of 97. So you could say 1996 2 1998 Or, you could say years 1996 and 1997 se as the time frame. Or you could list each year and then part C. As we approach the end of the 10 year period, it seems to be leveling off. That's called the carrying capacity. So what is it in this case? And the highest number is it seems to be leveling off around eight, so that's eight times 100. So that's 800 dear

In this problem of exponential functions, we have given that the exponential growth of deer population in Massachusetts can be calculated using the formula that age T equals to 50,000. This is 50,000 multiplied with one plus points, little six to the power. And we're five where this is a 50,000 in the initial population. Oh, they did, and 0.60 So this is the rate that is 0.6 is the rate of growth and the the population after n years and where t is the population after n years and here and in the number of years, a number of years. This is the we are modeling formula and now we have to predict the population after four years. So that means and it's equals to four and now put the value so they should be teaching the total population. After four years, that is 50,000 multiplied with one plus 10.6 to the power food and now evaluated. So this will be 50,000 multiplied is 1.6 to the powerful is equal, So this is equal to 63,000. So this is equal to 63,123 0.123 points, 84 eight years. Or we can say this is equals two 63.1 to 4 years. So this is the answer. Or we can say this is approximately. This is 63,000. This So this is the answer for part A and now for the part way, in a part way, if the initial population was 30,000 and growth rate was so, if the initial population was 30,000 and the growth rate was 0.12 so the rate was 0.12 That is a developed that approximately how many days would have been present after three years? So now we have to find the value after three years. So that means energy equals to three. Now, disarray. There are 30,000, and this was one plus 0.12 to the power three. And now when we solve it, we get This is approximately equals two, 42,000 days. So this is the answer for Part B? Mm. And now for the party in the party? Yeah. How many additional there can be expected in five years if the initial population is 45,000 and the current growth rate is 0.8 So if the their initial population was 45,000 and yes, growth rate is RS equals to 0.8 and the term that the duration age of five years. So now we have to find the first number of years after five years, and then we have to subject it. So the total number of days will be here 45,000 one plus rate that is 0.8 and a number of years. That is five. And now when we evaluate it, it is equal to this is approximately equal to 66 1000 years. Yeah, and now we have to find that number of additional layers so original layer would be a listener. Years would either here present minus old. That would be here 66,000, minus 45 1000 which is equals to 21,000 additional years. So this is the answer for party that is 21,000 additional additional layers would we expect in five years

The solution to the question number 108 His here we are, given that affects is equals 2608 four into E. To the power 0.1 to zero X. So where X is a number for three years since 2000 So now for the part when X is 10 years from 2000 to 2000 and 10, we have effects is close to 6084 E. To the power 0.1 to zero in two weeks. So putting 10 in place of X. You will get effects. His calls to 6860 Yeah. So the approximate number of population given by the exponential function is six 860 million. While the actual population was 6853 Yeah, million in here. 2010. The estimated population is seven million greater than the actual population. Now, moving towards part B. When access 20 years, that is from 2000 to 2020 14. 20 in place of X. You will get your Yeah, 20 years 7734 And when X becomes 30 so half of the party will be 4 to 87 to zero. So by 2020 the estimated number of population is 7734 million. Right, Bye 2030. It is estimated to be 8720 million. Thank you.


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