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.
Answer:
3, 4, 8
Step-by-step explanation:
Answer:
2%
Step-by-step explanation:
Let x be the first number
It increases by 70 %
The new number is
m = x+ .70x
= 1.7x
Let y be the second number
It decreases by 40 %
The new number is
n =y - .40 y
= .6y
The product of the original numbers is xy
The product of the new number is
mn = (1.7x * .6y) = 1.02xy
The new number is larger than the old number so it is an increase.
Percent increase is new - original divided by original times 100%
Percent increase = (1.02 xy - xy)
----------------- * 100%
xy
= .02 xy
------- * 100 %
xy
= .02 * 100 %
= 2%
Answer:
46.6
Step-by-step explanation:

The actual wingspan of the airplane would be 504 which isn't an answer but maybe you just typed A. in incorrectly <span />