Check the picture below.
A)
well, we start off by making a table of values for the function, with a few "t" values, as you see in the picture, then graph those points to get the graph.
B)
at t = 0, h(0) = 98, and at t = 1, h(1) = 82, so it went from 98 to 82, for a difference of 16.
C)
at t = 1, h(1) = 82, and at t = 2 h(2) = 34, so it went from 82 to 34, for a difference of 48.
does it fall the same distance on both intervals? well, they're different, so nope.
Answer:
Proportion of all bearings falls in the acceptable range = 0.9973 or 99.73% .
Step-by-step explanation:
We are given that the diameters have a normal distribution with a mean of 1.3 centimeters (cm) and a standard deviation of 0.01 cm i.e.;
Mean, = 1.3 cm and Standard deviation,
= 0.01 cm
Also, since distribution is normal;
Z = ~ N(0,1)
Let X = range of diameters
So, P(1.27 < X < 1.33) = P(X < 1.33) - P(X <=1.27)
P(X < 1.33) = P( <
) = P(Z < 3) = 0.99865
P(X <= 1.27) = P( <
) = P(Z < -3) = 1 - P(Z < 3) = 1 - 0.99865
= 0.00135
P(1.27 < X < 1.33) = 0.99865 - 0.00135 = 0.9973 .
Therefore, proportion of all bearings that falls in this acceptable range is 99.73% .
