Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).
<h3>What is a confidence interval of proportions?</h3>
A confidence interval of proportions is given by:
In which:
is the sample proportion.
In this problem, we have a 95% confidence level, hence
, z is the value of Z that has a p-value of 
, so the critical value is z = 1.96.
We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:
Hence the bounds of the interval are found as follows:
The 95% confidence interval is (0.2316, 0.3112).
More can be learned about the z-distribution at brainly.com/question/25890103
Answer: The solution is k = -171.92.
We will have the following:
* If Kristin does not decrease the price of her cakes, her projected weekly revenue from cake sales will be $2500.
*If Kristin decreases the price of her cakes, her projected weekly revalue will be $2520.
*Kristin will obtain the same revenue if she sells the cakes for $24 or $21.
The 16 oz bottle costs less. One cent less, to be exact. I came up with this by dividing the total cost (2.40) by the size of the bottle (16). I came up with .15. I did the same with the other cost and size, coming up with .16.
Answer:
t=5x=3049
Step-by-step explanation:
