Please help my brain is mush i dont understand how this works nd i still have 11 assignments after this-
1 answer:
Answer:
See step by step
Step-by-step explanation:
1. First question, do they represent functions? The one o. the left is a function because if you do the vertical line test. It only passes through a point once so it is a function. while on the right if you do the vertical line test, it passes through a point twice so it not a function. For 2.. n other words, since the domain of this equation is all real numbers or negative infinity to positives infinity
let say x=1 and x=-1 and we use the equation
when we plug both those in we get
this means that we can have two different x-values to equal out to the same y value. This is the definition of a function. So the one on the right is a function.
However the one on the left is a relation. TO Prove it, the domain of that function is all real numbers equal to or greater than zero, so let use 4.
Since there are 2 possible answer choices as the y value, this isn't a function. It a relation. It maps a group of ordered sets to one x value. Which is the opposite of the function. So the one of the left is a relation
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Answer:
A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first quartile.
Correct we satisfy that 1.2 <1.7 the lower limit
Step-by-step explanation:
Assuming this complete question: A violin student records the number of hours she spends practicing during each of nine consecutive weeks:
6.2 5.0 4.3 7.4 5.8 7.2 8.4 1.2 6.3
For this case we need to sort the data first on increasing way and we got:
1.2, 4.3, 5.0, 5.8, 6.2, 6.3, 7.2, 7.4, 8.4
For this case we have 9 values the median would be on the 5 position:
The first quartile would be 5 since we analyze 1.2, 4.3, 5.0, 5.8, 6.2 and the middle point is 5.0
The third quartile would be 7.2 since we analyze 6.2, 6.3, 7.2, 7.4, 8.4 and the middle point is 7.2
Then the interquartile rnage would be:
And
The lower limit on this case would be:
And our value is 1.2 is lower than the lower limit 1.7
Considering the smallest data value (1.2) and using the 1.5 × IQR rule, we would:
A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first quartile.
Correct we satisfy that 1.2 <1.7 the lower limit
B) not classify the value 1.2 as an outlier, because it is not more than 1.5 × IQR below the first quartile.
False 1.2 is more than 1.5 IQR below the first quartile
C) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the median.
False, we analyze if is an outlier comparing with the Q1 or Q3 not with the median
D) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the mean.
False we analyze if is an outlier comparing with the Q1 or Q3 not with the mean