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raikespeare5080
22.03.2021 •
Mathematics
Photography. In exercises 18 and 19, use the following information. Developing times of photographic prints are normally distributed with a mean of 15.4 seconds and a standard deviation of 0.48 seconds.
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Ответ:
18.The answer is 47 . 5 %
19.The answer is 83 . 85 %
Step-by-step explanation:
18 . This is a good example of normal Distribution
Z = ( Given score - Mean ) / Standard deviation
Z = (16. 36 - 15 .4 ) / 0.48
Z= 0. 96 / 0.48
Z = 2
From the table, the value of z on the right side of the normal curve
= 0.34 + 0 . 135 = 0.475
The percentage is 0. 475 x 100 = 47 . 5 %
19. We are asked to get the percentage between 13. 96 and 15. 88
Z 1 = ( 13 .96 - 15. 4 ) / 0.48
Z 1 = - 1. 44 / 0.48
Z 1 = -3
Z 2 = ( 15 . 88 - 15 . 4 ) / 0. 48
= 0. 48 / 0.48
Z 2 = 1
From the table , the value of z from - 3 to 1 is 0. 34 + 0. 135 + 0. 0235 + 0.34
= 0 .8385
The percentage is 0 . 8385 x 100 = 83 . 85 %
Ответ:
bottom right
Step-by-step explanation:
for the side that says total cost u go up to 4 and for the one that says number of puzzles u go two over