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nikitadnepr [17]
3 years ago
5

20 CHILDREN WENT TO A CAFE FOR LUNCH.14 CHILDREN ATE FISH BURGERS. 10 CHILDREN ATE CHICKEN BURGERS. HOW MANY CHILDREN ATE BOTH F

ISH AND CHICKEN BURGERS?
Mathematics
1 answer:
Ber [7]3 years ago
6 0

If there are 20 children that went to eat lunch, and 14 of them ate fish burgers, and 10 ate chicken burgers, you could make a line graph with 20 points and color the kids who ate fish burgers yellow and the one who ate chicken burgers blue. The green where they overlapped, and 10 of them over lapped, meaning that <u>10 children ate both fish and chicken burgers.  </u>Hope I helped!

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A= 0
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natta225 [31]

Answers:

  • Mean = 69
  • Median = 69
  • Mode = {68, 70}

===============================================

Explanations:

To get the mean, you add up all of the values and then you divide by N, where N is the number of values in the list. In this case, you'll divide by N = 8.

Adding up all the values gets you 70+68+70+72+68+66+65+73 = 552. Dividing that over 8 then leads to the arithmetic mean of 552/8 = 69

-------

To find the median, you'll first sort the values from smallest to largest. You should have this set {65, 66, 68, 68, 70, 70, 72, 73} after sorting. Next, cross off the smallest and largest values to end up with this smaller set {66, 68, 68, 70, 70, 72}. We do this so we can locate the middle-most item.

Repeat the process of crossing off the smallest and largest items to get {68, 68, 70, 70}. Do so one more time and we get {68,70}.

The midpoint of those two values is (68+70)/2 = 138/2 = 69. The median is 69. The mean and median may not always be the same value.

--------

The mode is the most frequent value. In this case, we have 68 and 70 show up exactly twice for each unique value. This is the most compared to the other unique values that only show up exactly once.

It is possible to have more than one mode, as long those values are more frequent than the rest. We denote the mode as a set, so the mode is {68, 70}

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Step-by-step explanation:

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