Answer:
The 90% confidence interval for the true mean natural gas bill for homes in the 19808 zip code area in February is between $292.48 and $331.32
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 21 - 1 = 20
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 20 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 1.7247
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 311.9 - 19.42 = $292.48
The upper end of the interval is the sample mean added to M. So it is 311.9 + 19.42 = $331.32
The 90% confidence interval for the true mean natural gas bill for homes in the 19808 zip code area in February is between $292.48 and $331.32
Answer:
1. y=-x
2. y=-3/2x-2
3. x=4
4. y=-2x-1
Step-by-step explanation:
slope intercept form is y=mx+b with mx being the slope and b being the y intercept
if its parallel it has to have the same slope.
perpendicular lines have the opposite slope and sign. so if its a whole number you put it over 1 and if its a fraction you switch the values in the numerator and denominator in addition to changing the sign. You keep the y intercept the same.
Answer:
the answer for this is:
subtract 7 from 4 to get -3
you then have 2^-3
remove the negative exponent by rewriting as 1/(2^3)
Step-by-step explanation:
Answer:
And the best option would be:
c. 1450 +/- 12
Step-by-step explanation:
Information provided
represent the sample mean for the SAT scores
population mean (variable of interest)
represent the sample variance given
n=25 represent the sample size
Solution
The confidence interval for the true mean is given by :
(1)
The sample deviation would be 
The degrees of freedom are given by:
The Confidence is 0.954 or 95.4%, the value of 
and 
, assuming that we can use the normal distribution in order to find the quantile the critical value would be 
The confidence interval would be
And the best option would be:
c. 1450 +/- 12
