Ashipment of racquetballs with a mean diameter of 60 mm and a standard deviation of 0.9 mm is normally distributed. by how many standard deviations does a racquetball with a diameter of 58.2 mm differ from the mean?

A shipment of racquetballs with a mean diameter of 60 mm and a standard deviation of 0.9 mm is normally distributed. racquetball with a diameter of 58.2 mm differ from the mean by 1.8 mm. so the 58.2 mm racquetball differ by 2 standard deviation by the mean.

x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.