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Questmn 3014Acontrac requires lease payrents of $400 at the beginning of every month for 6 yearsa What is the present value of the contract if the ease rate is 5.31...

Question

Questmn 3014Acontrac requires lease payrents of $400 at the beginning of every month for 6 yearsa What is the present value of the contract if the ease rate is 5.319 compounded annually?Rovnj the near-s: centb. What is the present value Of the contract ifthe lease rate is 5.31% compounded daily?Ravngnc Ncjrest -N

Questmn 3014 Acontrac requires lease payrents of $400 at the beginning of every month for 6 years a What is the present value of the contract if the ease rate is 5.319 compounded annually? Rovnj the near-s: cent b. What is the present value Of the contract ifthe lease rate is 5.31% compounded daily? Ravng nc Ncjrest -N



Answers

Helen deposits $\$ 100$ at the end of each month into an account that pays $6 \%$ interest per year compounded monthly. The amount of interest she has accumulated after $n$ months is given by
$$I_{n}=100\left(\frac{1.005^{n}-1}{0.005}-n\right)$$ a. Find the first six terms of the sequence. b. Find the interest she has accumulated after 5 years.

All right, So we have this sequence here which I believe is just slightly different from the textbook on Lian that I'd put X in everywhere there wasn't in and just speak to this if you didn't see the previous video, this comes from Dez Most, which is an online graphing calculator that is free for anybody to use. That's how I generated dysfunction. And this function here in my table is actually all there. It's a little bit cut off, but you can kind of see there would be basically a minus. X got cut off, and I used that to help me generate the 1st 6 terms in my Siri's, as opposed to going in and plugging in a one and then plugging in the two and so forth. I let that graphing calculator plugging the 123 all the way down the six. And so here are the 1st 6 terms right there. If we look at the next question, the next question asks. Basically, how much interest is it gonna be accumulated after six years? And that word accumulated means we do have toe add all of these things together, which is why I used as most again to set up a summation here. And so let me just speak to the summation a little bit. This summation basically says from month one all the way and I don't know if I said All right, I think it says five years all the way up to month 60. I got 60 because five times 12 is 60. So in five years, with 12 months, there's gonna be 60 months ago, but that go by. So adding up the interest, which is what the summation does for a month, one all the way to month 60. What is that? Total interest that will be accumulated. And here it is, right here. If you want a round that off, that three means around down it will be $19,377.61 that would be accumulated again. You can create this and generate this. If you're a graphing calculator, can't do it itself. Through an online program called Dez Mose, I highly encourage you to use it

Yes. At this time we're giving her sequence, which is I end, and we want to find our 1st 6 times. So let's plug in or en values from 1 to 6 into our equation here to see what we get. So I'm just gonna play this in your calculator. So this would be one point, does it all five to the power of an integral to one minus one, divided by to appoint Jos. 05 and minus. It's time here. Okay, So I made a mistake. You're So what's wrong? Let's see. So we want to divide this. Perhaps I need implication here. Okay. Okay. So now let's find our 1st 6 Terps. So this is our first six times, and now we want to also find the interest after five years. So it's not that end is in months. So for five years, that's going to be equal to five times fall once that 60 itself. That gives us 927

In this question I'm going to use on a carbon calculator. It is going to this most. Yeah, well, given the sequence and And you got your 100 hands with that? 201 and, uh, 1.5. Yeah. And, Pamela, uh, in this corner, you want to find the first six stems from the sequence. So me one too. 34 506. Immigration in March. So we see the body will be here. So the first six steps will be the first moment. 1000 and five. The next one will be the 201 the third one. No far the fifth and a six year. And then when you find a balance after the five years bye, computing the 60 times in the sequence. So when the sixties here inside. So in two minutes, it would be out here. That's gonna be the parents after and five years. And then we have the balance into the 20 years. So 240 here. And that was after all. So this will be the after the five years. This will be the on the balance after the 20 years

We're going to use our compound interest formula we're going to solve for t the number of years it takes when you start with $100 end with $450 compounding four times a year at a 6% interest rate. So let's go ahead and substitute all the numbers into that equation. 4 50 is the final amount. 100 is the starting amount, that principle. And we have one plus 0.0 6/4 raised to the power 40. We want to solve this for tea. What if we simplified one plus 10.6 divided by four just to make it easier to write, it will be 1.15 Sweet 450 equals 100 times 1.15 raised to the fourty. Okay, so this is an exponential equation, and the first step in solving it will be two to divide both sides by 100. So we get 4.5 equals 1.15 to the power 40. Now is the time to use logarithms so we can take the log of both sides. I'm going to use law based 10 but you could use natural log or any other base you like. So the log of the left equals the log of the right, and then we can use the power property of logarithms and bring the fourty out to the front. So now we have log of 4.5 equals 40 times the log of 1.15 Now, if we're trying to solve for T, what we need to do is divide both sides by four and divide both sides by log of 1.15 So that gives us T equals log of 4.5, divided by four log of one point 015 Now let's go ahead and put that in a calculator and approximate it and we can round it to the nearest well, let's maybe rounded to the nearest 10th so we have log of 4.5. I don't think the problem gives us directions about rounding. So, perhaps, and is that, But maybe we'll round to the nearest 10th. 25.3 is the number of years that would take


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