Answer:
C
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.
The sum of the interior and exterior angle need to equal 180 degrees.
Let the exterior angle = X
You are told the interior angle = 38x +24
Now you have x + 38x +24 = 180
Combine like terms:
39x +24 = 180
Subtract 24 from each side:
39x = 156
Divide both sides by 39:
x = 156 / 39
x = 4
The number of sides is found by dividing 360 by the exterior angle:
360 / 4 = 90
There are 90 sides.