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So the car is moving at 651 revolutions per minute, with wheels of a radius of 18inches
so, one revolution, is just one go-around a circle, and thus 2π, 651 revolutions is just 2π * 651, or 1302π, the wheels are moving at that "angular velocity"
now, what's the linear velocity, namely, the arc covered per minute
well
now, how much is that in miles/hrs? well
let's keep in mind that, there are 12inches in 1foot, and 5280ft in 1mile, whilst 60mins in 1hr
thus
notice, after all the units cancellations, you're only left with mi/hrs
so, one revolution, is just one go-around a circle, and thus 2π, 651 revolutions is just 2π * 651, or 1302π, the wheels are moving at that "angular velocity"
now, what's the linear velocity, namely, the arc covered per minute
well
now, how much is that in miles/hrs? well
let's keep in mind that, there are 12inches in 1foot, and 5280ft in 1mile, whilst 60mins in 1hr
thus
notice, after all the units cancellations, you're only left with mi/hrs
To find the z-score for a weight of 196 oz., use
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.