5) "Viral" Video. You hear about a super-funny video that's been posted on YouTube. When you go to view it you see that 1,345 ,397 people have viewed it since it was posted 3 days and 10 hours ago. The increase in the number of viewings follows the law of uninhibited growth with k=0.1721 when time is measured in hours yassume a starting value of 1 viewing). 4) How many people had viewed this video after it had been posted 2 hours? 2 days? b) When did the number of times viewed exceed 1 million?

Question

Question

5) "Viral" Video. You hear about a super-funny video that's been posted on YouTube. When you go to view it you see that 1,345 ,397 people have viewed it since it was posted 3 days and 10 hours ago. The increase in the number of viewings follows the law of uninhibited growth with k=0.1721 when time is measured in hours yassume a starting value of 1 viewing). 4) How many people had viewed this video after it had been posted 2 hours? 2 days? b) When did the number of times viewed exceed 1 million?

Answer 1

## For 2 hours: After performing the calculation, we find that \(N(2) \approx 1.37\). Therefore, approximately 1.37 people had viewed the video after it had been posted for 2 hours. ## For 2 days: After performing the calculation, we find that \(N(48) \approx 349.6\). Therefore, approximately 349.6 people had viewed the video after it had been posted for 2 days. ## When the views exceeded 1 million: After solving the equation, we find that \(t \approx 79.6\) hours. Therefore, the number of views exceeded 1 million approximately 79.6 hours after the video was posted.

Answer 2

## Step 1: Understanding the problem The problem is about the law of uninhibited growth, which is a type of exponential growth. The general formula for this type of growth is \(N(t) = N_0 \cdot e^{kt}\), where \(N(t)\) is the final amount, \(N_0\) is the initial amount, \(k\) is the growth rate, and \(t\) is the time. In this case, the initial amount \(N_0\) is 1 (the starting view), the growth rate \(k\) is 0.1721, and the time \(t\) is in hours. ## Step 2: Calculating the number of views after 2 hours and 2 days To find the number of views after 2 hours and 2 days, we substitute the respective values of \(t\) into the formula. ### For 2 hours: \(N(2) = 1 \cdot e^{0.1721 \cdot 2}\) ### For 2 days (48 hours): \(N(48) = 1 \cdot e^{0.1721 \cdot 48}\) ## Step 3: Calculating when the views exceeded 1 million To find out when the views exceeded 1 million, we set \(N(t) = 1000000\) and solve for \(t\). ### \(1000000 = 1 \cdot e^{0.1721 \cdot t}\)

Question

Answer

Answer

Explain

Hot Questions