Answer:
Researchers are<u> 95%</u> certain that the interval ($2005.76, $2024.24) contains the mean monthly rental cost of <u>all rent-controlled apartments in Brooklyn</u>.
Step-by-step explanation:
Last part of the question and choices are missing. Full question is like this:
Researchers conducted a study to determine the monthly rental cost of rent-controlled apartments in the five boroughs of New York City in 2012. The study randomly sampled 203 apartment records from Brooklyn, obtained from a large collection of income- and expense-filing statements. The 95% confidence interval of rent-controlled apartment costs in Brooklyn was $1,025.00 ± ± $9.24. The cost of all rent-controlled apartments in Brooklyn has a standard deviation of $67.20. State the conclusion of the z z -confidence interval for the mean.
Researchers are ________certain that the interval ($2005.76, $2024.24) contains the mean monthly rental cost of _____.
all rent-controlled apartments in New York City.
all rent-controlled apartments in Brooklyn.
the 98 apartments in the sample.
all apartments in Brooklyn.
Confidence Intervals give a range of values that the statistic (mean cost of all rent-controlled apartments in Brooklyn) can assume in a <em>confidence level </em>(95%). Confidence levels give the levels we can be sure (95%) that the statistic falls within the interval.
Since the sample is about the rent-controlled apartments in Brooklyn, the confidence interval gives an estimate of the rent-controlled apartments in Brooklyn, not in New York.
Answer:
x = - 3
Step-by-step explanation:
2x - 10 + 6 = 7x - 7 - 2x + 12
2x - 4 = 5x + 5
- 5 - 4 = 5x - 2x
-9 = 3x
x = - 9/3
x = - 3
Answer:
the answer is D
Step-by-step explanation:
this is because the number is unknown so you have to put a variable
Answer:
17160
Step-by-step explanation:
first there are 13 options
then there are 13 - 1 that was chosen
then there are 13 - 2 that were chosen
then there are 13 - 3 that were chosen
so that's 13 options, then 12 options, then 11 options, and then 10 options
and in order to figure out all of the possibilties we mutiply the options so
13*12*11*10 = 17160
little further explaining of why this works:
we have a set of letters (3 in a set)
A B C
what are the possible combinations?
AB
AC
BC
BA
CA
CB
the answer is 6 which is also 3 * 2 * 1 = 6
HOPE THAT HELPS!1! ^_^
