Answer:
16.8 ÷ 8 = 2.1
( For the other ones with the " tenths " confuses because I dont know where its going gl l though )
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The median number of minutes for Jake and Sarah are equal, but the mean numbers are different.
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For this, you never said the choices, but I’ve done this before, so I’m going to use the answer choices I had, and hopefully they are right.
Our choices are -
• The median number of minutes for Jake is higher than the median number of minutes for Sarah.
• The mean number of minutes for Sarah is higher than the mean number of minutes for Jake.
• The mean number of minutes for Jake and Sarah are equal, but the median number of minutes are different.
• The median number of minutes for Jake and Sarah are equal, but the mean number of minutes are different.
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So to answer the question, we neee to find the median and mean for each data set, so -
Jack = [90 median] [89.6 mean]
Sarah = [90 median] [89.5 mean]
We can clearly see the median for both is 90, so we can eliminate all the choices that say they are unequal.
We can also see that Jack has a higher mean (89.6) compared to Sarah (89.5).
We can eliminate all the choices that don’t imply that too.
That leaves us with -
• The median number of minutes for Jake and Sarah are equal, but the mean number of minutes are different.
Answer:
Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, ![[ab]_K](https://tex.z-dn.net/?f=%20%5Bab%5D_K%20)
is equal to ![[ba]_K](https://tex.z-dn.net/?f=%20%5Bba%5D_K%20)
, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.
Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, ![[ab]_H = [ba]_H](https://tex.z-dn.net/?f=%20%5Bab%5D_H%20%3D%20%5Bba%5D_H%20)
. Since a and b were generic elements of H, then H/G is abelian.
Number of tickets= X
To find the number of tickets, divide the money Jan has with the cost of each ticket.
So,
X =37.50/ 5.25=7.14
But since she can't buy part of a ticket, Jan can buy 7 tickets
