Asample of 25 cereal boxes of granola crunch, a generic brand of cereal, yields a mean (x¯) weight of 1.02 pounds. the goal is to construct a 95% confidence interval for the mean weight (µ) of all cereal boxes of granola crunch. assume that the weight of cereal boxes is normally distributed with a population standard deviation (σ) of 0.03 pounds.

1.01 < µ < 1.03

Step-by-step explanation:

We want to construct a 95% confidence interval.  Our sample size is 25, so we use a t-value.  The degrees of freedom are 24 (always one less than the sample size).  

The t-value we get is:  2.064

We have:

x = 1.02

σ = 0.03

See the attached photo for the construction of the confidence interval...

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)