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the result of (3.8621 × 1.5630) − 5.98 is properly written as a) 0.06 b…

Question

the result of (3.8621 × 1.5630) − 5.98 is properly written as a) 0.06 b) 0.056 c) 0.0565 d) 0.05646 e) 0.056462

Explanation:

Step1: Calculate the product first

$3.8621 \times 1.5630 = 5.9364623$

Step2: Subtract 5.98 from the product

$5.9364623 - 5.98 = -0.0435377$
Wait, correction: Recalculate product accurately
$3.8621 \times 1.5630 = 3.8621\times(1+0.5+0.06+0.003) = 3.8621 + 1.93105 + 0.231726 + 0.0115863 = 5.9364623$

Step3: Perform subtraction correctly

$5.9364623 - 5.98 = -0.0435377$
Wait, correction: Recognize significant figure rules. For multiplication, result has 4 sig figs (1.5630 has 5, 3.8621 has 5; least is 5? No, 3.8621 is 5 sig figs, 1.5630 is 5 sig figs, so product is 5 sig figs: $5.9365$. Then subtraction: 5.98 has 3 decimal places? No, 5.98 has 2 decimal places, 5.9365 has 4. Subtraction uses decimal places: $5.9365 - 5.98 = -0.0435$, absolute value 0.0435. Wait, no, original calculation error: $3.8621 \times 1.5630 = 5.9364623$, $5.9364623 - 5.98 = -0.0435377$. But this does not match options. Recheck multiplication:
$3.8621 \times 1.5630$:
$3.8621 * 1.563 = 3.8621*(1 + 0.5 + 0.06 + 0.003) = 3.8621 + 1.93105 + 0.231726 + 0.0115863 = 3.8621+1.93105=5.79315; 5.79315+0.231726=6.024876; 6.024876+0.0115863=6.0364623$. Oh! Critical error earlier. Correct product is $6.0364623$.

Step1 (Corrected): Calculate product correctly

$3.8621 \times 1.5630 = 6.0364623$

Step2: Subtract 5.98 from the product

$6.0364623 - 5.98 = 0.0564623$

Step3: Apply significant figure rules

For subtraction, the result has the same number of decimal places as the number with the least decimal places. 5.98 has 2 decimal places, so we round 0.0564623 to 2 decimal places in the context of the subtraction's precision? No: 6.0364623 has 6 decimal places, 5.98 has 2. The result should be rounded to 2 decimal places in the hundredths place? Wait no: 6.0364623 - 5.98 = 0.0564623. 5.98 has uncertainty in the hundredths place, so the result's hundredths place is the last certain digit? No, 5.98 is precise to 0.01, so 6.0364623 is precise to 0.0000001. The subtraction result's precision is limited by 5.98, so we look at the digits: 0.0564623. The 0.01 place is the 5 (0.05), the next digit is 6, so we round 0.05 to 0.06? No, wait, no: 6.0364623 - 5.98 = 0.0564623. 5.98 is 5.9800000, so subtraction is 6.0364623 - 5.9800000 = 0.0564623. Now, for significant figures in subtraction, we go by the number of decimal places. 5.98 has 2 decimal places, so the result should be rounded to 2 decimal places? No, 6.0364623 has 6 decimal places, 5.98 has 2. The result should have 2 decimal places? 0.0564623 rounded to 2 decimal places is 0.06? But that ignores the digits. Wait, no: the rule is that in subtraction, the result has the same number of decimal places as the measurement with the fewest decimal places. 5.98 has 2 decimal places, so we round 0.0564623 to 2 decimal places: 0.06. But wait, another approach: 3.8621 (5 sig figs) * 1.5630 (5 sig figs) = 6.0365 (5 sig figs). Then 6.0365 - 5.98 (3 sig figs, 2 decimal places). 5.98 is 5.98, so 6.0365 - 5.98 = 0.0565. Wait, no, 6.0365 - 5.98 = 0.0565, which is 0.06 when rounded to 2 decimal places, or 0.056 when rounded to 3 decimal places? Wait, no, let's use exact calculation first:
$3.8621 \times 1.5630 = 6.0364623$
$6.0364623 - 5.98 = 0.0564623$
Now, considering significant figures: 5.98 has 3 significant figures, the product has 5. When subtracting, the number of decimal places matters: 5.98 has 2 decimal places, so the result should be rounded to 2 decimal places? No, 0.0564623 to 2 decimal places is 0.06. But if we use significant figures for the final result, 0.0564623 has 5 significant figures, but the limiting term is 5.9…

Answer:

a) 0.06