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.
If he spends 1/4 reading the news, then he spends 3/4 reading the other stuff
3/4 * 8/9 = 24/36 = 2/3 of an hr ...time spent reading sports and comics
(2/3) / 2 =
2/3 * 1/2 = 2/6 reduces to 1/3....so he spends 1/3 of an hour (or 20 minutes) reading the comics
Answer:
Step-by-step explanation:
0.04
ill help..............................................
16, because 2•x would also be 2•3 which is 6, and 5•y is rewritten as 5•2 which equals 10. 10+6=16 :)