Question:
In a certain factory, each day the expected number of accidents is related to the number of overtime hours by a linear equation. Suppose that on one day there were 1000 overtime hours logged and 8 accidents reported, and on another day there were 400 overtime hours logged and 5 accidents. What are the expected numbers of accidents when no overtime hours are logged?
In a certain factory, each day the expected number of accidents is related to the number of overtime hours by a linear equation. Suppose that on one day there were 1000 overtime hours logged and 8 accidents reported, and on another day there were 400 overtime hours logged and 5 accidents. What are the expected numbers of accidents when no overtime hours are logged?
Show Hint
Model accidents as a straight line in overtime hours, y = mx + c, and use the two given (hours, accidents) pairs to solve for c, the value at zero hours.
Updated On: Jul 13, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Set up the linear model.
The number of accidents depends on overtime hours through a straight line relation. Write it as \(y = mx + c\), where \(x\) is the overtime hours, \(y\) is the expected accidents, \(m\) is the slope and \(c\) is the value of \(y\) when \(x=0\), which is exactly what the question wants.
Step 2: Use the two given data points.
On one day \(x=1000\) gave \(y=8\), so \(1000m + c = 8\).
On another day \(x=400\) gave \(y=5\), so \(400m + c = 5\).
Step 3: Solve for the slope m.
Subtract the second equation from the first:
\[ (1000m + c) - (400m + c) = 8 - 5 \]
\[ 600m = 3 \]
\[ m = \frac{3}{600} = \frac{1}{200} \]
Step 4: Solve for the intercept c.
Put \(m=\frac{1}{200}\) back into \(400m+c=5\):
\[ 400 \times \frac{1}{200} + c = 5 \]
\[ 2 + c = 5 \]
\[ c = 3 \]
Step 5: Check the wrong options.
A straight line is fixed once we know two points on it, so only one value of \(c\) fits both data points at the same time. If you plug \(c=2\) or \(c=4\) or \(c=5\) into either equation, the slope you get no longer matches the other data point, so options 1, 3 and 4 (values 2, 4 and 5) all fail; only \(c=3\) is consistent with both days.
Final Answer:
With no overtime hours logged, the expected number of accidents is 3.
\[ \boxed{3} \]
The number of accidents depends on overtime hours through a straight line relation. Write it as \(y = mx + c\), where \(x\) is the overtime hours, \(y\) is the expected accidents, \(m\) is the slope and \(c\) is the value of \(y\) when \(x=0\), which is exactly what the question wants.
Step 2: Use the two given data points.
On one day \(x=1000\) gave \(y=8\), so \(1000m + c = 8\).
On another day \(x=400\) gave \(y=5\), so \(400m + c = 5\).
Step 3: Solve for the slope m.
Subtract the second equation from the first:
\[ (1000m + c) - (400m + c) = 8 - 5 \]
\[ 600m = 3 \]
\[ m = \frac{3}{600} = \frac{1}{200} \]
Step 4: Solve for the intercept c.
Put \(m=\frac{1}{200}\) back into \(400m+c=5\):
\[ 400 \times \frac{1}{200} + c = 5 \]
\[ 2 + c = 5 \]
\[ c = 3 \]
Step 5: Check the wrong options.
A straight line is fixed once we know two points on it, so only one value of \(c\) fits both data points at the same time. If you plug \(c=2\) or \(c=4\) or \(c=5\) into either equation, the slope you get no longer matches the other data point, so options 1, 3 and 4 (values 2, 4 and 5) all fail; only \(c=3\) is consistent with both days.
Final Answer:
With no overtime hours logged, the expected number of accidents is 3.
\[ \boxed{3} \]
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