Aresearcher is interested in the lengths of salvelinus fontinalis (brook trout), which are known to be approximately normally distributed with mean 80 centimeters and standard deviation 5 centimeters. to preserve brook trout populations, some regulatory standards need to be set limiting the size of fish that can be caught.
what proportion of fish are larger than 86 centimeters in length

a. 0.0228
b. 0.0548
c. 0.1151
d. 0.8849

c. 0.1151

Step-by-step explanation:

d) the probability that, on average, fish are larger than 86 centimeters in length.The area under part of a normal probability curve is  directly proportional to probability and the value is calculated as

z = (x₁−x) /σ

where z = propability of normal curve

x₁ = variate mean = 86cm

x = mean of 80cm

σ = standard deviation = 5cm

applying the formula,

z= (86-80)/5

z = 6/5 =1.2

Using a table of partial areas beneath the standardized normal curve (see Table of normal curve, a z-value of 1.2 corresponds to an area of 0.3849 between the mean value. but, because the standard curve has 0.5, then will minus 0.3849 from 0.5= 0.5 - 0.3849 = 0.1151

Thus the probability of a fish are larger than 86 centimeters in length is 0.1151

ok

step-by-step explanation:

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.