Answer:
(b) 4,445
Step-by-step explanation:
If the researcher would like to be 95% sure that the obtained sample proportion would be within 1.5% of p (the proportion in the entire population of U.S. adults), what sample size should be used?
Given a=0.05, |Z(0.025)|=1.96 (check standard normal table)
So n=(Z/E)^2*p*(1-p)
=(1.96/0.015)^2*0.5*0.5
=4268.444
Take n=4269
Answer:(b) 4,445
Answer:
First you need to set up an equation y = mx + b. m would be the monthly charge and b would be the one time fee. x represents the number of months.
To solve for the number of months with a total price of 240 we substitute 240 in for y and solve for x.
240 = 25x + 40
subtract 40 from each side
200= 25x
divide by 25
8 = x
8 months
10 + 7r = 45
7r = 35
r = 5
5 milesStep-by-step explanation:
Answer:
We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .
Step-by-step explanation:
Given -
The sample size is large then we can use central limit theorem
n = 50 ,
Standard deviation
= 7.1
Mean 
= 110
1 - confidence interval = 1 - .98 = .02
= 2.33
98% confidence interval for the mean caffeine content for cups dispensed by the machine = 
= 
= 
First we take + sign
= 112.34
now we take - sign
= 107.66
We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 .
3(0) = 2x + 6
( 0 = 2x + 6 ) - 6
( -6 = 2x ) /2
x = -3
The answer is B. (-3, 0).
Do you can posting a drawn please ?
