F = t ⇨ df = dt
dg = sec² 2t dt ⇨ g = (1/2) tan 2t
⇔
integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt
u = 2t ⇨ du = 2 dt
As integral of tan u = - ln (cos (u)), you get :
integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant
integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant
integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer
The correct option regarding the outliers of the data-set is given by:
The greatest value, 78, is the only outlier.
<h3>How to use the quartiles of a data-set to identitfy outliers?</h3>
- The median of the data-set separates the bottom half from the upper half, that is, it is the 50th percentile.
- The first quartile is the median of the first half of the data-set.
- The third quartile is the median of the second half of the data-set.
- The interquartile range is the difference of the third quartile with the first quartile.
- Measures that are more than 1.5 IQR from Q1 and Q3 are considered outliers.
The IQR for this problem is:
IQR = 37 - 13 = 24.
Hence the bounds for outliers are:
- Less than 13 - 1.5 x 24 = -23.
- Greater than 37 + 1.5 x 24 = 73,
The options are:
- Both 1 and 78 are outliers.
Hence the correct option is that only 78 is an outlier.
More can be learned about outliers of a data-set at brainly.com/question/17083142
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Answer:
Pay attention in class
Step-by-step explanation:
By paying more attention in class, surely you will begin to understand the material. If you are having a problem concentrating in class then I recommend talking to a teacher or your parent.
There are three groups. One group for girls. One group for half the boys which is equal to one group of girls and another identical sized group of boys because there are twice as many boys as girls.
48=total
3=number of groups
48/3=16
16 is equal to the number of girls because it is equal to half the number of boys
16*2=32
32+16=48
Number of boys plus number number of girls: 32+16=48
