Answer:
Step-by-step explanation:
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Retro Rides is a club for owners of vintage cars and motorcycles. Every year the club gets together for a ride. This year, 38 ve
2 answers:
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Answer: On the 29th day
Step-by-step explanation:
According to this problem, no lilypad dies and the lilypads always reproduce, so we can apply the following reasoning.
On the first day there is only 1 lilypad in the pond. On the second day, the lilypad from the first reproduces, so there are 2 lilypads. On day 3, the 2 lilypads from the second day reproduce, so there are 2×2=4 lilypads. Similarly, on day 4 there are 8 lilypads. Following this pattern, on day 30 there are 2×N lilypads, where N is the number of lilypads on day 29.
The pond is full on the 30th day, when there are 2×N lilypads, so it is half-full when it has N lilypads, that is, on the 29th day. Actually, there are lilypads on the 30th, and
lilypads on the 29th. This can be deduced multiplying succesively by 2.
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