5pi/6 r^2 = 15pi
R^2 = 18
R= 3sqrt(3)
Answer:
6°F
58°F
Step-by-step explanation:
From the data Given :
Temperature rise in Buffalo by noon time = 14°F
The initial temperature at Buffalo = - 8°F
Rise in temperature = 14°F
Therefore, temperature at noon = (-8 + 14)°F = 6°F
B.)
Temperature at 6am in St. Louis = 32°F
Temperature at 6am in Juneau = - 26°F
Temperature difference :
(32 - (-26))°F
32 + 26
= 58°F
Temperature in St. Louis is 58°F higher
Answer:
well this is very easy all you have to do is divide 4 by 5 and you will get 4 with a remander of 1 or 0.8 or if this was a real life scenario then you could just take that extra slice and cut it in five ways. hope this helped! can i get brainiest plzz? im only ONE away from getting up to expert!
Step-by-step explanation:
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.