If a 0.584-inch diameter pin slips into a hole dimensioned as 9/16 ± 1/64 inches, what can be concluded?

In this scenario, you're evaluating the fit between a pin and a hole based on their dimensions. The hole measurement is given as 9/16 inches with a tolerance of ± 1/64 inches. This means the hole can range from 9/16 - 1/64 inches to 9/16 + 1/64 inches.

First, convert the dimensions into a common format for easier comparison. The exact value of 9/16 inches is 0.5625 inches. The tolerance of ± 1/64 inches translates to approximately ± 0.015625 inches. Therefore, the minimum and maximum acceptable dimensions for the hole can be calculated as follows:

• Minimum size of the hole: 0.5625 - 0.015625 = 0.546875 inches

• Maximum size of the hole: 0.5625 + 0.015625 = 0.578125 inches

Now we compare this range to the diameter of the pin, which is 0.584 inches. Since 0.584 inches is greater than the maximum dimension of the hole at 0.578125 inches, it indicates that the pin is larger than the hole can accommodate.

Given the defined range