Answer:
Step-by-step explanation:
42 ÷ 7 = 6
The number of hits on a website doubles every month. About 10,600 people visited the website during May. The expression 10,600 ·
1 answer:
Answer:
The correct option is;
B. 10,600; 2,650; About 10,600 people visited the website in May. About 2,650 people visited the website in March
Step-by-step explanation:
The given parameters are;
The number of people that visit the website every month = 2 × The number of people that visited the website the previous month
The number of people that visit the website during May ≈ 10,600
The number of visitors of the website are given by the expression 10,600 ×
For t = 0, we have;
The number of visitors of the website = 10,600 × 2⁰ = 10,600
For t = -2, we have;
The number of visitors of the website = 10,600 × 2⁻² = 2,650
Therefore, the website had about 10,650 visitors in May and about 2,650 visitors in 2 months earlier in March.
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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