Question

Determine the result and the propagation of error in the following equation.

(2.8542 + 0.0070) + (2.431 + 0.100)

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(1.4915 + 0.0069) x (3.459 + 0.049)

Record your answer in the essay (next question) for the error write it as +/-, so your answer should be written like 3.4563 +/0.0034

Answer 1

Let c = $\frac{(2.8542 \pm 0.0070) + (2.431 \pm 0.100)}{(1.4915 \pm 0.0069) \times (3.459 \pm 0.049)}$

Let $n = (2.8542 \pm 0.0070) + (2.431 \pm 0.100)$ and $d =(1.4915 \pm 0.0069) \times (3.459 \pm 0.049)$

Then,

and $\Delta n = \sqrt{0.0070^2+0.100^2} = 0.1002$

$d = 1.4915 \times 3.459 = 5.1591$ and $\Delta d = d\sqrt{(\frac{0.0069}{1.4915})^2+(\frac{0.049}{3.459})^2} = 5.1591\times 0.0149 = 0.0769$

$c = \frac{5.2852}{5.1591} = 1.0244$ and $\Delta c = c\sqrt{(\frac{0.1002}{5.2852})^2+(\frac{0.0769}{5.1591})^2} = 1.0244\times 0.0241 = 0.0247$

Therefore, result is $1.0244 \pm 0.0247$