Answer:
a) 2.188
b) John's rent is not an outlier
c) His rent has to be higher than $1,235 to be an outlier
Step-by-step explanation:
The mean monthly rent of students at Oxnard University is $780 with a standard deviation of $208.
(a) John’s rent is $1,235. What is his standardized z-score?
We solve using z score formula
z = (x-μ)/σ, where
x is the raw score = $1235
μ is the population mean = $780
σ is the population standard deviation = $208
Hence,
z = 1235 - 780/208
z = 2.1875
Approximately to 3 decimal place = 2.188
(b) Is John’s rent an outlier?
No it isn't
(c) How high would the rent have to be to qualify as an outlier?
John’s rent would have to be higher than $1235
Answer:
PY = 8
Step-by-step explanation:
Lets run down the other three to show why they're wrong
<em>MY=8</em>
By P.T, that's wrong (PERPENDICULAR)
<em>PM=MY </em>
We can't prove that with the given
XY=42
Again, by P.T., this is false
BTW: You can prove PY=8 by proving XMP congruent to YMP by SAS
Answer:
Step-by-step explanation:
Given
------------------------------Has a brother | Does not have a brother
Has a sister
--------------------------- 6 ----------------- 8
Does not have a sister -----------2 ------------------ 13
Required
Probability of having a sister and a brother
First, the total number of students has to be calculated;
Total = 6 + 2+ 8 + 13
Total = 29
Number of students that have a sister and a brother is represented with data in row 1 and column 1 i.e. 6
At this point, the probability can then be calculated;
Probability = Number of students that have a sister and a brother divided by Total number of students
Answer:
Domain: All real numbers
Range: y ≥ 0
Step-by-step explanation: