Answer: 
Step-by-step explanation:
The confidence interval estimate for the population mean is given by :-
, where 
is the sample mean and ME is the margin of error.
Given : Sample mean: 
The margin of error for a 98% confidence interval estimate for the population mean using the Student's t-distribution : 
Now, the confidence interval estimate for the population mean will be :-
Hence, the 98% confidence interval estimate for the population mean using the Student's t-distribution =
X = y + 11
2x + y = -2
2(y + 11) + y = -2
2y + 22 + y = -2
3y + 22 = -2
3y = -24
y = -8
x = y + 11
x = -8 + 11
x = 3
Solution set {3, -8} (C)
Answer:
Step-by-step explanation:
Assuming the number of tickets sales from Mondays is normally distributed. the formula for normal distribution would be applied. It is expressed as
z = (x - u)/s
Where
x = ticket sales from monday
u = mean amount of ticket
s = standard deviation
From the information given,
u = 500 tickets
s = 50 tickets
We want to find the probability that the mean will be greater than 510. It is expressed as
P(x greater than 510) = 1 - P(x lesser than or equal to 510)
For x = 510
z = (510 - 500)/50 = 0.2
Looking at the normal distribution table, the probability corresponding to the z score is 0.9773
P(x greater than 510) = 1 - 0.9773 = 0.0227
Answer:
x= -1/3
Step-by-step explanation:
First, we can combine like terms on each side:
-4-18x=3x+3
Then, we move all of the variables onto one side and the constant onto the other:
-18x-3x=3+4
-21x=7
x=-7/21
x= -1/3
Hope this helps!
