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A student's calculation was found to have a 35.5% error, and his experimental measurement was 15.6 cm. What are the two possible values for the actual measurement?

User Sccs
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2 Answers

2 votes

Final answer:

The two possible values for the actual measurement are approximately 10.063 cm and 21.137 cm.

Step-by-step explanation:

To calculate the two possible values for the actual measurement, we first need to find the maximum and minimum values that the experimental measurement can be off by. Since the student's calculation has a 35.5% error, we can find the maximum value by adding 35.5% of the experimental measurement to it and find the minimum value by subtracting 35.5% of the experimental measurement from it.

Maximum value: 15.6 + (15.6 * 0.355) = 15.6 + 5.537 = 21.137 cm

Minimum value: 15.6 - (15.6 * 0.355) = 15.6 - 5.537 = 10.063 cm

Therefore, the two possible values for the actual measurement are approximately 10.063 cm and 21.137 cm.

User Imcc
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Let the two possible actual measurements be .x' and 'y', where x is a measurement less than 15.6 and y is a measurement more than 15.6

x + 35.5% of x = 15.6
y - 35.5% of y

35.5% as decimal is 0.355

x + 0.355x = 15.6, and
y - 0.355y = 15.6

Solving x + 0.355x = 15.6
1.355x = 15.6
x = 15.6÷1.355 = 11.53

Solving y - 0.355x = 15.6
0.645y = 15.6
y = 15.6 - 0.645
y = 24.19

Two possible measurements are 11.53 and 24.19
User Arthur Weborg
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6.7k points
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