The inside diameter of a randomly selected piston ring is a random variable with mean value 16 cm and standard deviation 0.04 cm.
(a) if x is the sample mean diameter for a random sample of n = 16 rings, where is the sampling distribution of x centered and what is the standard deviation of the x distribution? (enter your standard deviation to five decimal places.) center cm standard deviation cm
(b) answer the questions posed in part (a) for a sample size of n = 64 rings. (enter your standard deviation to five decimal places.) center cm standard deviation cm

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

answer:     true

step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

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