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Hello, my name is Ms. Parnham.

In this lesson, we will be learning about limits of accuracy.

Let's look at an example.

The length of a line is seven centimetres to the nearest centimetre.

What is the shortest possible length of the line? So let's think about a diagram.

So here we have seven centimetres if it's been round to the nearest centimetre, then we will mark on six centimetres below it and eight centimetres above it.

And think about the point when a number would be rounded to seven, rather than six or eight.

So the shortest possible length of the line would be halfway between six and seven, 6.

5 centimetres.

Because everything that is 6.

5 centimetres and above, will round up to seven.

What is the longest possible length of the line? Now, this is a bit trickier to think about.

We know that halfway between seven and eight, we have 7.

5.

And some people say, well, the longest possible length would be something like 7.

4 followed by a few nines, but however many nines you follow 7.

4 with, that would never be the maximum possible length of the line, 'cause you could always add another decimal place onto the decimal.

So it's actually possible to say that the longest possible length of the line is 7.

5.

Yes, this would round up to eight, but this is just exactly the same distance from seven, a 6.

5 is.

And for calculation purposes, we would say anything up to, but not quite including 7.

5 is the longest possible length of the line.

The proper name for these two numbers are the lower bound, 6.

5 centimetres, and upper bound, for 7.

5 centimetres.

Let's look at another example.

So the length of a line is 70 centimetres to the nearest 10 centimetres.

Now the shortest possible length on this line, we would use the same process of drawing a diagram.

We have 70 in the middle and this is to the nearest 10.

So let's put 60 centimetres below and 80 centimetres above.

And, thinking about the shortest possible length, let's look at the midpoint of 60 and 70.

So that's 65 centimetres.

It's the same principles as before.

So 65 centimetres is the shortest possible.

Now the longest possible, just like before, we find the midpoint of 70 and 80.

75 itself would definitely round up to 80, because that is the mathematical convention.

But just like before, my reasoning behind giving a decimal of 74.

99999, however many nines you put on that decimal, you could always add an extra decimal place and that number would be bigger.

So for calculation purposes, the longest possible length of the line, we put a 75 centimetres.

And just like before, the proper language is lower bound, 65 centimetres, and upper bound, 75 centimetres.

Here are some questions for you to try.

Pause the video to complete the task and restart the video when you're finished.

Here are the answers.

part C and E gave us rounded answers in one unit of measurement and accuracy in another.

So remembering that there are 60 seconds in a minute and 1,000 grammes in a kilogramme would help us get to the correct solutions for these.

Let's look at an example where we would be asked to calculate with upper bounds or lower bounds.

So the length and width of the rectangle have been rounded to one decimal place.

Work out the greatest possible perimeter.

We've been asked for the greatest possible perimeter.

That means it is the maximum possible, so each side is the maximum possible.

So when we calculate, we will be calculating with upper bounds.

The perimeter of a rectangle is the distance all the way around it.

So that's two lengths added to two widths.

So thinking about each dimension, the length upper bound is 8.

35 centimetres.

And don't forget, we've got two of them.

And then the length lower bound is 8.

25 centimetres.

We don't need that in this instance, but we would, if we had been asked for the least possible perimeter.

The width upper bound is 2.

95 centimetres, and the width lower bound is 2.

85 centimetre.

Again, not needed for this example because we're only working out the greatest possible perimeter.

But well-worth finding.

So we take all four sides of this rectangle, all the upper bounds, add them together, and this produces an onset of 22.

6.

Here are some questions for you to try.

Pause the video, to complete the task and restart the video when you're finished.

Here are the answers.

Using the upper bounds of 18.

05 and 12.

45, we would add them together and multiply by two for the upper bound of the perimeter.

And the lower bound for the area is the result of multiplying the two lower bounds of 17.

95 and 12.

35 together.

Here's another question for you to try.

Pause the video to complete the task and restart the video, when you're finished.

Here are the answers.

If we require a maximum cost, then we need to use the upper bound of 135.

Now that's been rounded to the nearest five.

So think about that halfway point between 135 and 140.

So that's where the 137.

5 comes from.

Here's a further question for you to try.

Pause the video to complete the task.

And restart the video when you're finished.

Here are the answers.

If we want the maximum value for a range, then we need to think about the lowest, the smallest value it can possibly be, and the highest, the largest value it can possibly be.

And then find the difference between them.

So the lower bound of the smallest value is subtracted from the upper bound of the largest value to give the answer.

Then to find the lowest possible value for a mean, we would add up all the lower bounds of those numbers and divide by five.

That's all for this lesson.

Thank you for watching.