Answer:
The 98% confidence interval for the mean amount of money spent on lunch per week for all students is between $13.3 and $16.7.
Step-by-step explanation:
A confidence interval has the following format.
![\mu_{x} \pm M](https://tex.z-dn.net/?f=%5Cmu_%7Bx%7D%20%5Cpm%20M)
In which
is the mean of the sample and M is the margin of error.
In this problem, we have that:
![\mu_{x} = 15, M = 1.7](https://tex.z-dn.net/?f=%5Cmu_%7Bx%7D%20%3D%2015%2C%20M%20%3D%201.7)
![\mu_{x} - M = 15 - 1.7 = 13.3](https://tex.z-dn.net/?f=%5Cmu_%7Bx%7D%20-%20M%20%3D%2015%20-%201.7%20%3D%2013.3)
![\mu_{x} + M = 15 + 1.7 = 16.7](https://tex.z-dn.net/?f=%5Cmu_%7Bx%7D%20%2B%20M%20%3D%2015%20%2B%201.7%20%3D%2016.7)
The 98% confidence interval for the mean amount of money spent on lunch per week for all students is between $13.3 and $16.7.
Where’s the angle? Is there a picture of it?
Answer:
8/12 + 3/12
Step-by-step explanation: multiply 2 and 3 by 4 to add these two fractions correctly.
You could do 228/6 = 38
2280/60 = 38
hope this helps you
Answer:
The best answer would be the 25% off the entire purchase. Say for example Mrs. Walsh was willing to spend 100$ that would be 25% dollars she saved. But when it comes to save the most amount of money. If she spends under 60 dollars use the 15 dollars off. If it’s over 60 use the 25% off coupon