1. a store had 235 mp3 players in the month of january. every month, 30% of the mp3 players were sold and 50 new mp3 players were stocked in the store. which recursive function best represents the number of mp3 players in the store f(n) after n months? f(n) = 0.7 x f(n − 1) + 50, f(0) = 235, n > 0 f(n) = 237 − 0.7 x f(n − 1) + 50, f(0) = 235, n > 0 f(n) = 0.3 x f(n − 1) + 50, f(0) = 235, n > 0 f(n) = 237 + 0.7 x f(n − 1) + 50, f(0) = 235, n > 0 2. sophia invested some money in a bank at a fixed rate of interest compounded annually. the equation below shows the value of her investment after x years: f(x) = 500(1.05)x what was the average rate of change of the value of sophia's investment from the second year to the fourth year? 14.13 dollars per year 28.25 dollars per year 50.00 dollars per year 56.50 dollars per year

Part 1:

Given that a store had 235 MP3 players in the month of January and that every month, 30% of the MP3 players were sold and 50 new MP3 players were stocked in the store.

The number of MP3 players in the store after the previous months sale is given by 0.7f(n - 1) and the number of MP3 players in the store after new MP3 were added is 0.7f(n - 1) + 50.

Therefore, the recursive function that best represents the number of MP3 players in the store f(n) after n months is given by f(n) = 0.7 x f(n − 1) + 50, f(0) = 235, n > 0

Part 2:

The average rate of change of a function from a to b is given by



Given that the equation showing the value of her investment after x years is given by



Thus, the average rate of change of the value of Sophia's investment from the second year to the fourth year is given by



Therefore, the average rate of change of the value of Sophia's investment from the second year to the fourth year is 28.25 dollars per year.

1. f(n) = 0.7 x f(n â’ 1) + 50, f(0) = 235, n >= 0 2. 28.25 dollars per year 1. For this problem, let's construct the recursive formula ourselves. We know that f(0) will be 235 since that's the original stock and that n has to be greater than or equal to 0 (n=0 represents the starting stock). So we have f(n) = ?, f(0) = 235, n >= 0 Now each month we sell 30%, so we're left with 70% of the previous month's stock (100% - 30% = 70%). So let's take that into account for our function. f(n) = 0.7 x f(n-1), f(0) = 235, n >= 0 And each month we get another 50 MP3 players. Let's add that into our function. f(n) = 0.7 x f(n-1) + 50, f(0) = 235, n >= 0 And that's our answer. 2. Since we've been given the function f(x) = 500(1.05)^x, let's calculate the amount of money that we have after 2 years and 4 years. f(2) = 500(1.05)^2 = 551.25 f(4) = 500(1.05)^4 = 607.75 Now get the difference in money between those years. 607.75 - 551.25 = 56.50 And since we're talking about a period of 2 years and we want the average rate per year, divide by 2. 56.50 / 2 = 28.25 So the average change from year 2 to year 4 is $28.25 per year.

The answer to #44 is 10.4444, so g is the answer i divided the time by the miles. 47