Given the three integers are
, we arrive at

We can merge the fractions on the left side:


Answer:
1/6 ton of fish is provided during the night feeding.
Step-by-step explanation:
The fish consumed in the morning is
tons, and the amount of fish served in the afternoon is
greater than in the morning. This means

Let’s represent the quantity of fish given at night as
, and if the total fish fed throughout the day equals
, we have

By summing the numerators on the left side, we derive:

and subtracting
from both sides allows us to isolate 

Since
, we derive

The common denominator for the fractions is 6; hence, we write the equation in the form

and simplifying the numerators results in:

which indicates the amount of tons fed during the night feeding.
a) The hypothesis states that the mean is different from 0.5025 and has been rejected. b) The p-value computed is 0. c) A confidence interval is established as 0.50456 < u < 0.50464. The hypothesis testing is based on a given standard deviation (s.d.) of 0.0001, a sample mean of x = 0.5046 from a sample size of n = 25. The two-sided alternative and significance level of 0.05 were assumed. A Z-score was calculated to show that the hypothesis is rejected since the mean differs from 0.5025.