Answer:
T is at (13,-6)
Step-by-step explanation:
x-coordinate of T: from -9 to 2, there's 11 units, so you add 11 to 2 to find the x-coordinate of T, which is 13.
y-coordinate of T: from 4 to -1 there's -5 units, so you subtract 5 from -1 to find the y-coordinate of T, which is -6.
Answer:
The answer is "Option A and Option B".
Step-by-step explanation:
In question 1:
In all cases, the entire population is measured so that the actual medium discrepancy could be measured as well as an interval of trust cannot be used.
This issue would be that she calculated the ages with all representatives of both classes, such that she measured a whole population. It's not necessary.
In question 2:
When the p-value is 0.042. At 90% trust and 92% trust level 11 (p-value below 0.10 and 0.08) are not included. however the biggest confidence level of 92%. Consequently, the largest trust level where the 11 is Not included in the trust interval is 92% trust.
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
9514 1404 393
Answer:
water
Step-by-step explanation:
Comparison of numbers can be done a couple of different ways. One way is to subtract one from the other. If the difference is positive, the minuend (first number) is the larger of the two.
Another way is to divide one number by the other. If the ratio is more than 1, then the numerator is the larger number.
For numbers in scientific notation, the division is perhaps the easiest. Here, that gives you ...
This value is greater than 1, so the mass of a water molecule is greater.
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You will note that the multipliers (10^-26 kg) cancelled because they were equal. If they are unequal, the usual rules of exponents apply.