Automobile Repairs The standard deviation of a sample of 15 automobile repairs at a local garage was $120.82. Assume the variable is normally distributed. Find the 90% confidence of the true variance and standard deviation.

i believe it is 0.1m because the measurement of 30.4m limits how precise you can be

you can only be as precise as the least precise measurement

5: 20

step-by-step explanation:

hello there!

i believe the answer is

(cot2+1)/sec2=cot2

    csc2/sec2=cot2

  (1/sin2)/(1/cos2)=cot2

  (1/sin2)(cos2/1)=cot2

  cos2/sin2=cot2

  (1/tan2)=cot2

  (cot2)=(cot2)

hope this !

hyperz ^_^