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Hi, everyone.

My name is Ms. Ku and I'm really happy and excited to be learning with you today.

It's going to be a fun and interesting lesson and some of our Oak pupils will be here to help as well.

It's gonna be tricky in part, but don't worry, we'll all be here to help and I really do hope that you'll enjoy the lesson and find it challenging too.

Really excited to be working with you.

So let's make a start.

Hi, everyone and welcome to this lesson on using inequality notation for errors in calculations.

It's under the unit estimation and rounding, and by the end of the lesson, you'll be able to use inequality notation a is less then or equal to x, which is less than b to express errors in calculations.

Let's have a look at some keywords.

Remember, an error interval for a number x shows the range of possible values of x and it's written as an inequality where A is less than or equal to x less than b and today's lesson will be broken into two parts.

The first is where we'll be looking at error intervals using addition and multiplication.

And the second is where we'll be looking at error intervals in calculations.

So let's make a start.

Remember, when a number has been rounded to a degree of accuracy, we're able to identify the range of values the unrounded numbers could be, and we represent this using an error interval.

For example, 43 has been rounded to the nearest whole number.

What is the error interval? Well, a number line really does help us out here.

So drawing a number line and 43 goes in the middle.

Now, given the fact that the degree of accuracy is to the nearest whole number, we plus one and minus one from 43, giving us 42 and 44.

Then from here, we identify those middle values.

So you can see we have 42.

5 and 43.

Now from here, we need to identify the range of values which are in between 42.

5 and 43.

But remember, we're not including that 43.

So this means the error interval is 42.

5, less than or equal to x.

That means it can equal 42.

5 and x is less than 43.

5, but it can't equal our 43.

And what we've successfully done here is we found the error interval for 43 when it's been rounded to the nearest whole number.

Now, we use error intervals to help us calculate the largest and smallest values of a calculation.

And the error interval has a lower and a upper limit, a and b as you can see written here where a represents the lower limit of the error interval and b represents the upper limit of the error interval.

And it's so important to know which limits from our error intervals to use in order to calculate the largest and smallest values.

So let's have a look at an example.

Here we have a road and each length has been measured to two significant figures.

What I'd like you to do is have a look if you can identify the error interval of each length and then think, well, how are you going to use this error interval to work out the error interval of the overall length? See if you can give it a go and press pause if you need more time.

So let's have a look at road x first.

Well, we know the error interval will be between 31.

5, including that 31.

5 and 32.

5, not including our 32.

So that means our error interval is 31.

5, less than or equal to x, less than 32.

5, and our error interval for y would be 83.

5 less than or equal to y, less than 84.

Now let's see if we can work out the error interval for the overall distance.

Well, the smallest possible distance would be adding those lower limits.

31.

5, add our 83.

5 gives us 115 metres.

The largest distance can be found by summing those upper limits, 32.

5 add 84.

5, giving us 117 metres.

So that means our error interval is 115 metres less than or equal to, I'm calling it d, less than 117 metres.

Notice how we don't include the 117 and this is because we use the 32.

5 and we use that 84.

5 which are not included in the error interval.

So summarising, 115 is the sum of the lower limits of the error intervals and 117 is the sum of the upper limits of the error intervals.

So using error intervals and calculations allows us to work out the largest and smallest possible values and this is important in real life, as well as industries such as engineering, construction, medicine, so on and so forth.

So let's have a look at another example.

Here we have Alex and he's painted a wall.

He's confused.

He says, "I'm confused as I've run out of paint but I've measured everything correctly and read the instructions correctly." Lucas says, "Let's have a look." So here's the paint and the paint tin says it covers 45 metres squared, rounding to the nearest integer.

And here's the wall.

Alex measured the wall to be nine metres by five metres whereby he's rounded to the nearest integer, and he doesn't quite understand why has he run out of paint? Well, let's have a look at our paint tin.

Well, the paint tin says it covers 45 metres squared rounded to the nearest integer.

So that means our error interval must be 44.

5 less than or equal to the area covered, less than 45.

So we've got our error interval for the area covered by the paint.

Alex explains, "It means the tin of paint would cover an area between 44.

5 metre squared and 45.

5 metre squared.

It does include 44.

5 but it doesn't include 45.

5." Now let's have a look at the wall.

So Lucas says, "Let's identify the error interval of the wall." Now, given the fact that he rounded five metres to the nearest integer, he's recognised the error interval for the five-meter length wall is 4.

5 metres less than or equal to the width, which is less than 5.

5 metres.

Now let's have a look at the error interval of the length.

Well, given the fact that Alex measured the length to be nine metres to the nearest integer, that means he has an error interval of 8.

5 metres less than or equal to the length, less than or equal to 9.

5 metres.

So now let's see if we can work out the smallest area of the wall given the error intervals of the width and the length.

Well, to work it out, we're going to use those smallest possible lengths.

5 multiplied by 8.

So that means the lower limit of the area of the wall is 38.

25 metres squared.

Now let's have a look at the largest possible area.

Well, the largest possible area could be found using those upper limits.

5 multiplied by our 9.

5, giving us an upper limit of 52.

25.

Notice how it's a less than and not less than or equal to because we don't include the upper limits of the width and the length.

So now the area of the wall could be between 38.

25 metres squared and 52.

25 metres squared, including 38.

25 and not including 52.

25.

Now, given all the information that we have on the screen, Alex can now see why he doesn't have enough paint.

Do you think you can work it out as well? So hopefully you can spot the upper limit of the area covered by the paint is 45.

5 metres squared, but the upper area of the wall is 52.

25 metres squared.

So that's why there is a possible area of the wall that can be greater than 45.

5 metres squared, which is the reason why the tin of paint didn't cover the wall.

Well done if you got this.

So let's summarise what we did.

Well, to find the lower limit of the area, it was the product of the lower limits of the error intervals and the product of the upper limits of the error intervals give us the upper limit for the area of the wall.

Once again, notice how we do not include 52.

25, given the fact that we don't include the upper limits of our length and width.

So now let's move on to a check.

Here we have to work out the error interval for the perimeters of each shape and the degree of accuracy is given.

So we have a triangle.

A is given as five centimetres to one significant figure.

B is given as four centimetres to one significant figure.

And c is given us three centimetres to one significant figure.

We also have a rectangle where the width is two centimetres to one significant figure and the length is 6.

4 centimetres to one decimal place.

See if you can work out the error interval for the perimeters of each shape.

Great work, well done.

Let's see how you got on.

Well, let's identify these error intervals for a, b and c first.

Here's a, b and c.

Now, to work out the error interval for the perimeter, we're simply summing the smallest possible lengths.

5 add 3.

5 add 2.

5, thus giving us 10.

5 centimetres.

The largest perimeter will be formed by summing these upper intervals.

So 5.

5 add our 4.

5 add our 3.

5, which is 13.

So that means the error interval for our perimeter is 10.

5, less than or equal to p, less than 13.

Well done if you got this one right.

Next, let's have a look at that rectangle.

Well, here are our error intervals for the width and the length.

Now let's sum to find the perimeter.

So the smallest possible perimeter is two lots of the width and the length.

So two lots of our 1.

5, add our 6.

35 gives us 15.

7 and the largest perimeter is two lots of our width and length.

So two multiplied by our 2.

5 and our 6.

45 gives us 17.

Thus we have the error interval for our perimeter being 15.

7, less than or equal to p, less than 17.

Well done if you've got this one right.

Now let's have a look at another check, but here we're working out the error interval for the area.

See if you can give it a go and press pause if you need more time.

Great work.

So let's see how you got on.

Well, let's have a look at the triangle first.

Identifying our error interval, we have 4.

5 as our lower limit for c and 5.

5 as our upper limit for c.

5 is our lower limit for b and 4.

5 is our upper limit for b.

To work out the area, remember, the formula states 1/2 times the base times the height.

So we're multiplying those lower limits, 0.

5 times 4.

5 times 3.

5 gives us the lowest area to be 7.

875 centimetres squared.

The largest area would be 1/2 times the base times the height, so it's 1/2 times our 5.

5 multiplied by our 4.

5 giving us 12.

375 centimetres squared.

So that means we have the error interval of our area.

It's going to be 7.

875 is less than or equal to our area, which is less than 12.

375.

Great work if you got this one right.

Let's have a look at that rectangle.

Well, let's work out our error intervals and remember to work out the smallest area, we multiply our length by our width.

So it's the lower limits multiplied together gives us 9.

525 centimetres squared and the upper limit would be 2.

5 multiplied by 6.

45.

Thus giving us the error interval of our area to be 9.

525 less than or equal to our area less than 16.

125.

Great work if you got this one.

Well done, everybody.

So let's move on to your practise questions.

Here you need to work out the error interval for the perimeter of these shapes.

See if give it a go and press pause if you need more time.

Well done.

Let's move on to the next question.

Question two, I want you to work out the error interval for the area of these shapes.

See if you can give it a go and press pause if you need more time.

Great work.

Let's move on to question three.

Question three says given the error interval of the area, can you work out the rounded lengths and the degree of accuracy? See if you can give it a go and press pause if you need more time.

Well done.

Let's move on to question 3b.

Can you work out the rounded length and the degree of accuracy given the fact that we have our error interval and we have one length? See if you can give it a go.

Press pause if you need more time.

Great work, everybody.

So let's go through our answers.

For A, you should have got the error interval for the length is 2.

5 is less than or equal to l, less than 3.

5, so therefore, the perimeter would simply be multiplying the upper and lower limits by four, giving us 10 is less than or equal to p, which is less than 14.

For B, we should have the error interval for the width is given here.

The error interval for the length is given here.

Thus the perimeter is two lots of the width add the length.

So we're adding two lots of the lower limits and two lots of the upper limits, thus giving us an error interval of 20.

2, less than or equal to p, less than 20.

Great work if you got this one right.

Let's identify those error intervals for our lengths.

Here we have a, b, c and to work out the error interval for the perimeter, we're simply summing up all those lower limits and summing up all those upper limits.

Great work if you've got this and you've used the correct notation.

Moving on to question two, we had to work out the area.

So we have this error interval for our length.

Thus squaring those gives us an error interval for the area to be 18.

0625 is less than or equal to a, which is less than 18.

9225.

B, well, remember to work out the area of a right angle triangle, it's 1/2 times the base times the height, therefore we don't need c.

So we're just looking at a and b.

And lower limit would be found by 0.

5 times our 4.

5 times the 11.

5 and the upper limit is 4.

5 times 5.

5 times 12.

Thus giving us the error interval for the area to be 25.

875 is less than or equal to the area which is less than 34.

375.

Great work again if you got any of these error intervals.

Give yourself a huge well done, especially if you've used the right notation.

Question three was a great question.

It's giving you the error interval and what you have to do is you have to work out the rounded length and the degree of accuracy.

Well, hopefully you spot you need to square root both the lower limit and the upper limit as this would give you the error interval of the lengths, 8.

25 and 8.

35.

Thus, you'll be able to spot that we have lengths of 8.

3 centimetres to one decimal place.

That was a great question.

Question 3b was particularly hard and it's important to remember the formula for the area of the triangle.

It was 1/2 times the base times the height.

Now, we were given the length a, so it's important to recognise the error interval.

So using this, we can form an equation for the lower area.

5 times 2.

5 times b is equal to our lower limit of our area, 6.

3125.

Thus we can work out b to be 5.

05.

We can also write an equation to work out the upper area, 0.

5 times 3.

5 times b is 9.

0125.

So we can work out b to be 5.

15.

So that means we have the error interval for b.

Thus b must be 5.

1 centimetres to one decimal place.

Great work if you got that one right.

Great work, everybody.

So let's move on to the second part of our less than error intervals in calculations.

Now, when using subtraction, it's important to know when the largest values and when the smallest values can be calculated.

For example, here we have a jug and it was full to 100 millilitres to the nearest integer.

Laura drank 30 millilitres to the nearest millilitre.

Now, Izzy says that means the lowest amount of water that's left in the jug is 100.

5 take away 30.

Laura says no, the lowest amount of water left in the jug is 99.

5 take away 29.

And Sofia says the lowest amount of water that's left is 99.

5 subtract 30.

Who do you think is correct? Well, let's have a look at each calculation.

Well, Izzy says this, let's work it out.

That would mean Izzy's calculation would give 70 millilitres.

Laura said this, so that means Laura's calculation would give 70 millilitres.

Now, Sofia's calculation would give 69 millilitres.

So that means Sofia's right.

Izzy asks, "How did you work that out, Sofia?" And Sofia says, "Good question! What is the error interval for the amount of water that was in the jug?" Well, we know it's got to be 99.

5 is less than or equal to the water that was in the jug, which is less than 100.

So that means the water that was in the jug is given as this error interval.

Now Sofia says, "What is the error interval for the amount of water drunk by Laura?" And Izzy answers, "Well, the amount of water drank by Laura has to be 29.

5 less than or equal to d, representing the amount drank, less than 30.

5." So to work out the smallest remaining, it is the smallest amount of water that was in the jug originally subtract the largest amount of water drank by Laura, thus giving us the smallest remaining amount of water.

But how do you think we can calculate the largest amount of water left in the jug? Well, this can be worked out by identifying the largest amount of water in the jug subtract the smallest amount of water drank by Laura.

From here, that means we can identify the error interval for the amount of water that's left in the jug.

It's 69, which is the subtraction of 99.

5, the smallest amount of water, subtract the biggest amount drank, giving us 69 is less than j, the water in the jug, and the largest has to be the largest amount of water in the jug subtract the smallest amount drank.

So that means this is our error interval.

So now let's summarise this.

Well, summarising, 71 is the difference between the upper limits of w and the lower limits of d.

69 is the difference between the lower limit of w and the upper limit of d.

What's really important is that you recognise the error interval of what's remaining does not use less than or equal to.

This is because we use the upper limit of w and we use the upper limit of d and because of that, we can't use those values in our error interval.

So let's have a look at a check question.

Here you need to work out the error interval for the remaining length of rope.

A rope was originally 3.

2 metres rounded to one decimal place and a piece of 1.

6 metres was cut from the rope.

Now, what is the error interval for the remaining length? See if you can work it out and press pause if you need more time.

Well done.

Let's see how you got on.

Well, the original length has this error interval, 3.

15 less than or equal to l, less than 3.

25.

The amount cut has this error interval, 1.

55 less than or equal to c, less than 1.

65.

To work out the upper limit remaining, it is the upper limit of the length subtract the lower limit of what's cut, thus giving us 1.

7 and to work out the lower limit remaining, it's the lower limit of the length subtract the upper limit of what's cut, giving us 1.

This means the remaining length is 1.

5 less than c, less than 1.

Notice again how we don't use that less than or equal to because we used upper limits in both calculations.

Well done if you got this one right.

It was really hard.

Well done, everybody.

So let's move on to your task.

Here we have two questions.

Question one has a rope, which was originally 5.

3 centimetres to one decimal place and a piece of 3.

8 centimetres to one decimal place was cut from the rope.

You need to identify the error interval for the remaining length.

For question two, a coffee had 330 millilitres to the nearest integer and Lucas drank 32 millilitres rounded to two significant figures.

You need to identify what is the error interval for the remaining coffee? See if you can give it a go and press pause if you need more time.

Well done.

Let's move on to question three.

Question three shows a screw and it's broken off from a wall.

Originally the screw was 4.

6 millimetres to one decimal place.

Now, the broken off piece measures 3.

4 millimetres to one decimal place.

Can you work out the error interval of the screw which is lodged in the wall? Question four says a box of sweets and a chocolate weighs 43.

26 grammes correct to two decimal places.

Now, if the chocolate alone weighs 21.

14 grammes to two decimal places, what is the error interval for the mass of the box of sweets? See if you can give it a go and press pause if you need more time.

Great work, everybody.

So let's go through these answers.

Well, for question one, let's work out the error interval for the original length.

Then work out the error interval for the cut length.

Thus to work out the upper limit of the remaining, we simply subtract the upper limit of the original length, subtract the lower limit of the cut length, giving us 1.

To work out the lower limit remaining, you start with the lower limit of the original length subtract the upper limit of the cut length, giving us 1.

That means the remaining length is represented as 1.

4 less than what's remaining less than 1.

Well done if you got any of that right.

Question two, let's identify those error intervals.

We know the original full coffee has this error interval and the amount drank has this error interval.

Thus, to work out what's remaining, we do the upper limit of the original amount of coffee, subtract the lower limit of what is drank, giving us 299.

To work out the lower limit of what's remaining, we use the lower limit of what was originally in the coffee, subtract the biggest possible amount drank, thus giving us this error interval for the remaining coffee.

For question three, let's identify those error intervals again.

Well, we have our original full screw has this error interval and the broken off piece has this error interval.

So to work out the upper limit lodged in the wall, we simply do 4.

65 subtract 3.

35 giving us 1.

3 and to work out the lower limit lodged in the wall, 4.

55 take away 3.

45, which is 1.

Thus we have this error interval for the amount lodged in the wall.

1 is less than to equal to l, which is less than 1.

Amazing work if you got this one.

For question four, let's work out those error intervals.

We know this is our error interval for the total mass and this is our error interval for our chocolate.

Thus working out the upper and lowest limit of our sweets, we now have our error interval for the mass off sweets in the box to be 22.

13 is less than s, which is less than 22.

11.

Great work if you got that one right.

Amazing work, everybody.

It was a tough lesson today.

In summary, the error intervals have a lower and upper limit where a represents the lower limit of an error interval and b represents the upper limit of an error interval.

It's so important to know which limits from our error intervals to use in order to calculate the largest and smallest values.

Fantastic work, everybody.

It was great learning with you.

I've finished the video