Question

A car factory checks 40 cars and 6 of them have defect that prevent them from being sold. Give the relative frequency (probability) of a sellable (non-defective) car from this group of 40. Express your answer as a fraction in simplest form.



(1 point)

Answer 1

To find the relative frequency (probability) of a sellable (non-defective) car, we first need to determine the number of sellable cars. There are 40 cars in total, and 6 of them are defective.

The number of non-defective (sellable) cars is:

\[
40 - 6 = 34
\]

Now, to find the relative frequency of a sellable car, we use the formula:

\[
\text{Relative frequency} = \frac{\text{Number of sellable cars}}{\text{Total number of cars}}
\]

Plugging in the values we have:

\[
\text{Relative frequency} = \frac{34}{40}
\]

Next, we simplify this fraction. Both the numerator and the denominator can be divided by 2:

\[
\frac{34 \div 2}{40 \div 2} = \frac{17}{20}
\]

Thus, the relative frequency (probability) of a sellable (non-defective) car from this group of 40 is:

\[
\boxed{\frac{17}{20}}
\]