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The normal distribution curve for the problem is shown below
We need to standardise the value X=405.5 by using the formula


We now need to find the probability of z=0.32 by reading the z-table
Note that z-table would give the reading to the left of z-score, so if your aim is to work out the area to the right of a z-score, then you'd need to do:

from the z-table, the reading

gives 0.6255
hence,

The probability that the mean weight for a sample of 40 trout exceeds 405.5 gram is 0.3475 = 34.75%
X-6y=6 slope: 1/6 y-intercept (0,1)
X= 0,6 y= -1, 0
X+3y+12=0 slope: 1/3
Y-intercept (0,4) x= -12, 0 Y= 0,4
8a-9b= 9/8 slope (0 ,7/8) x= -1,1 Y= -1/4,2
3a+b=7 1/3 (0,7/3) x= 4,7 Y=1,0
The correct answer would be-11