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16. $p = \frac{2e}{\sqrt{f}}$ $e = 6.8$ correct to 1 decimal place - Edexcel - GCSE Maths - Question 17 - 2022 - Paper 3

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16.-$p-=-\frac{2e}{\sqrt{f}}$----$e-=-6.8$-correct-to-1-decimal-place-Edexcel-GCSE Maths-Question 17-2022-Paper 3.png

16. $p = \frac{2e}{\sqrt{f}}$ $e = 6.8$ correct to 1 decimal place. $f = 0.05$ correct to 1 significant figure. Work out the upper bound for the value of $... show full transcript

Worked Solution & Example Answer:16. $p = \frac{2e}{\sqrt{f}}$ $e = 6.8$ correct to 1 decimal place - Edexcel - GCSE Maths - Question 17 - 2022 - Paper 3

Step 1

Find the upper bound for e

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Answer

The value of e is given as 6.8 correct to 1 decimal place. Therefore, the upper bound can be calculated as:

Upper bound of e = 6.8 + 0.05 = 6.85

Step 2

Find the upper bound for f

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Answer

The value of f is given as 0.05 correct to 1 significant figure. Thus, the upper bound is:

Upper bound of f = 0.05 + 0.005 = 0.055

Step 3

Substitute upper bounds into the equation for p

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Answer

Now we can substitute the upper bounds into the formula to find p:

p=2×6.850.055p = \frac{2 \times 6.85}{\sqrt{0.055}}

Calculating 0.055\sqrt{0.055}:

0.0550.2344\sqrt{0.055} \approx 0.2344

Substituting that back into the formula gives:

p=2×6.850.234458.5p = \frac{2 \times 6.85}{0.2344} \approx 58.5

Step 4

Provide the answer to 3 significant figures

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Answer

The final answer for pp, rounded to 3 significant figures, is:

p58.5p \approx 58.5

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