ALGEBRA 2-Alternate Exam 4 - continued On June 1st, Hannah shoots a video and up

Question

ALGEBRA 2-Alternate Exam 4 - continued On June 1st, Hannah shoots a video and uploads it to the Internet. She shares the video with her friends and encourages them to share it with others. By the end of June 2nd, her video has been 37 times. Over each of the next 9 days, the total number of views increases by 35%. How many views is the video projected to have by the end of June 11th? 27. ??8 views 28. However, on the night of June 10th, a few bloggers link to Hannah's video. These links lead to 855 more views than expected on the 11th. After June 11th, the number of views increases by 55% each day. At this rate, on what day is the video anticipated to have 100,000 views? tog 35) G,55 29. Suppose that the increased popularity from blog traffic lasts only through the 18th. For the remainder of June, the growth rate slows to 25% per day. Find the average daily growth rate from June 2 through June 30 to the nearest tenth of a percent.

Explanation / Answer

27. The sum of the n terms of a geometric series with 1st term a and common ratio r is a(1-r)n/(1-r). Hence the no. of views from the 3rd to 11th June is 37[1-(1.15)9]/(1-1.15) = 37(1-3.517876292)/(-0.15) = 37*2.517876291/0.15 = 621 ( on rounding off to the nearest whole no.). Hence the total no. of views from the 1st June, till the3 11th June is 37+621 = 658.

28. If the no. of views on 11th June increase by 855, then the no. of views till 11th June are 621+855 = 1476. Further, the nth term of a geometric series with 1st term a and common ratio r is arn-1. Hence the no. of views, before the bloggers joined, on 11th June was expected to be 37*(1.15)9-1 = 37*3.059 =113          ( on rounding off to the nearest whole no.). After, the group of bloggers joined, the total no. of views on 11th June was 113+855 = 968. Now, let us suppose that the total no. of views reaches 100000 after in            n days after the 11th June. Then 100000 = 1476+968[1-(1.55)n]/(1-1.55) or, (100000-1476)*0.55 = (1.55)n-1 or, (1.55)n = (100000-1476)*0.55 +1 = 54189( on rounding off to the nearest whole no.). Now, on taking log of both the sides, we get n log 1.55 = log54189 so that n = log 54189/log 1.55 = 4.733911137/0.190331698 = 25 (on rounding off to the nearest whole no.).

29. The rate of daily growth in views after the 2nd June, till the 10th June is 15 %. The total no. of views on 10th June was 37*(1.15)8-1 = 98. Therefore, the rate of growth in views on 11th June is (968-98)*100/98 = 887.75 % . The daily growth rate from 12th June till 18th June is 55% and the daily rate of decline from 19th till 30th June is 25 %. Hence the average rate of daily growth in views after the 2nd June, till the 30th June is [ 8*15+ 887.75+ 7*55-12*25]/28 = (120+887.75+385-300)/28 = 1092.75/28 = 39.02678571 % or, 39 % ( on rounding off to the nearest tenth).

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