Answer 1 #
Question: A distance is measured as $8 \text{ m}$ to the nearest metre.
Step 1: Identify the accuracy.
The distance is measured “to the nearest metre”, so the accuracy is $1$ metre.
Step 2: Work out half of the accuracy.
Half of the accuracy = $\frac{1}{2} \times 1 = 0.5$ metres
Step 3: Find the lower bound by subtracting the half-accuracy from the given number.
Lower bound = $8 – 0.5 = 7.5$ metres
Step 4: Find the upper bound by adding the half-accuracy to the given number.
Upper bound = $8 + 0.5 = 8.5$ metres
Step 5: Write the final answer clearly.
Upper bound = $8.5 \text{ m}$ (but not including $8.5$)
We can write this as: $7.5 \leq x < 8.5$
Answer 2 #
Question: A number is $350$ to the nearest 10.
Step 1: Identify the accuracy.
The number is rounded “to the nearest 10”, so the accuracy is $10$.
Step 2: Work out half of the accuracy.
Half of the accuracy = $\frac{1}{2} \times 10 = 5$
Step 3: Find the lower bound by subtracting the half-accuracy from the given number.
Lower bound = $350 – 5 = 345$
Step 4: Find the upper bound by adding the half-accuracy to the given number.
Upper bound = $350 + 5 = 355$
Step 5: Write the final answer clearly.
Upper bound = $355$ (but not including $355$)
We can write this as: $345 \leq x < 355$
Answer 3 #
Question: A weight is recorded as $2.7 \text{ kg}$ to 1 decimal place.
Step 1: Identify the accuracy.
The weight is rounded “to 1 decimal place”, so the accuracy is $0.1$ (one tenth).
Step 2: Work out half of the accuracy.
Half of the accuracy = $\frac{1}{2} \times 0.1 = 0.05$
Step 3: Find the lower bound by subtracting the half-accuracy from the given number.
Lower bound = $2.7 – 0.05 = 2.65$ kg
Step 4: Find the upper bound by adding the half-accuracy to the given number.
Upper bound = $2.7 + 0.05 = 2.75$ kg
Step 5: Write the final answer clearly.
Upper bound = $2.75 \text{ kg}$ (but not including $2.75$)
We can write this as: $2.65 \leq x < 2.75$
Answer 4 #
Question: A number is $4500$ to the nearest 100.
Step 1: Identify the accuracy.
The number is rounded “to the nearest 100”, so the accuracy is $100$.
Step 2: Work out half of the accuracy.
Half of the accuracy = $\frac{1}{2} \times 100 = 50$
Step 3: Find the lower bound by subtracting the half-accuracy from the given number.
Lower bound = $4500 – 50 = 4450$
Step 4: Find the upper bound by adding the half-accuracy to the given number.
Upper bound = $4500 + 50 = 4550$
Step 5: Write the final answer clearly.
Upper bound = $4550$ (but not including $4550$)
We can write this as: $4450 \leq x < 4550$
Answer 5 #
Question: A time is measured as $15.83$ seconds to 2 decimal places.
Step 1: Identify the accuracy.
The time is rounded “to 2 decimal places”, so the accuracy is $0.01$ (one hundredth).
Step 2: Work out half of the accuracy.
Half of the accuracy = $\frac{1}{2} \times 0.01 = 0.005$
Step 3: Find the lower bound by subtracting the half-accuracy from the given number.
Lower bound = $15.83 – 0.005 = 15.825$ seconds
Step 4: Find the upper bound by adding the half-accuracy to the given number.
Upper bound = $15.83 + 0.005 = 15.835$ seconds
Step 5: Write the final answer clearly.
Upper bound = $15.835 \text{ seconds}$ (but not including $15.835$)
We can write this as: $15.825 \leq x < 15.835$
Answer 6 #
Question: A number is $7200$ to 2 significant figures.
Step 1: Identify the accuracy.
The number $7200$ has 2 significant figures: $7$ and $2$. The zeros are not significant.
The last significant figure ($2$) is in the hundreds place, so the accuracy is $100$.
Step 2: Work out half of the accuracy.
Half of the accuracy = $\frac{1}{2} \times 100 = 50$
Step 3: Find the lower bound by subtracting the half-accuracy from the given number.
Lower bound = $7200 – 50 = 7150$
Step 4: Find the upper bound by adding the half-accuracy to the given number.
Upper bound = $7200 + 50 = 7250$
Step 5: Write the final answer clearly.
Upper bound = $7250$ (but not including $7250$)
We can write this as: $7150 \leq x < 7250$
Answer 7 #
Question: A length is $0.056 \text{ m}$ to 3 significant figures.
Step 1: Identify the accuracy.
The number $0.056$ has 3 significant figures: $5$, $6$, and $6$. The zeros at the start are not significant.
The last significant figure ($6$) is in the thousandths place, so the accuracy is $0.001$.
Step 2: Work out half of the accuracy.
Half of the accuracy = $\frac{1}{2} \times 0.001 = 0.0005$
Step 3: Find the lower bound by subtracting the half-accuracy from the given number.
Lower bound = $0.056 – 0.0005 = 0.0555$ m
Step 4: Find the upper bound by adding the half-accuracy to the given number.
Upper bound = $0.056 + 0.0005 = 0.0565$ m
Step 5: Write the final answer clearly.
Upper bound = $0.0565 \text{ m}$ (but not including $0.0565$)
We can write this as: $0.0555 \leq x < 0.0565$
Answer 8 #
Question: A number is $0.0048$ to 2 significant figures.
Step 1: Identify the accuracy.
The number $0.0048$ has 2 significant figures: $4$ and $8$. The zeros at the start are not significant.
The last significant figure ($8$) is in the ten-thousandths place, so the accuracy is $0.0001$.
Step 2: Work out half of the accuracy.
Half of the accuracy = $\frac{1}{2} \times 0.0001 = 0.00005$
Step 3: Find the lower bound by subtracting the half-accuracy from the given number.
Lower bound = $0.0048 – 0.00005 = 0.00475$
Step 4: Find the upper bound by adding the half-accuracy to the given number.
Upper bound = $0.0048 + 0.00005 = 0.00485$
Step 5: Write the final answer clearly.
Upper bound = $0.00485$ (but not including $0.00485$)
We can write this as: $0.00475 \leq x < 0.00485$
Answer 9 #
Question: A mass is recorded as $12000 \text{ g}$ to 3 significant figures.
Step 1: Identify the accuracy.
The number $12000$ has 3 significant figures: $1$, $2$, and the first $0$ after $2$. The other zeros are not significant.
The last significant figure is in the hundreds place, so the accuracy is $100$.
Step 2: Work out half of the accuracy.
Half of the accuracy = $\frac{1}{2} \times 100 = 50$
Step 3: Find the lower bound by subtracting the half-accuracy from the given number.
Lower bound = $12000 – 50 = 11950$ g
Step 4: Find the upper bound by adding the half-accuracy to the given number.
Upper bound = $12000 + 50 = 12050$ g
Step 5: Write the final answer clearly.
Upper bound = $12050 \text{ g}$ (but not including $12050$)
We can write this as: $11950 \leq x < 12050$
Answer 10 #
Question: A temperature is $-4.5°\text{C}$ to 1 decimal place.
Step 1: Identify the accuracy.
The temperature is rounded “to 1 decimal place”, so the accuracy is $0.1$ (one tenth).
Step 2: Work out half of the accuracy.
Half of the accuracy = $\frac{1}{2} \times 0.1 = 0.05$
Step 3: Find the lower bound by subtracting the half-accuracy from the given number.
Lower bound = $-4.5 – 0.05 = -4.55°\text{C}$
Step 4: Find the upper bound by adding the half-accuracy to the given number.
Upper bound = $-4.5 + 0.05 = -4.45°\text{C}$
Step 5: Write the final answer clearly.
Upper bound = $-4.45°\text{C}$ (but not including $-4.
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