![aaa813](/avatars/18630.jpg)
aaa813
28.02.2020 •
Mathematics
Where will the different sections of the box plot be
located?
The box will go from
yto 27.
Aline dividing the box will go at
The left whisker will go from 11 to..
.
The right whisker will go from
7 to 33.
Solved
Show answers
More tips
- H Health and Medicine Impeccable Memory: How to Improve It...
- L Leisure and Entertainment How Many Seasons are There in the TV Show Interns?...
- S Sport When will the Biathlon World Championships 2011 take place in Khanty-Mansiysk? Answers to frequently asked questions...
- H Health and Medicine Trading Semen for Money: Where Can You Sell and Why Would You Want to?...
- F Food and Cooking Homemade French Fries: The Ultimate Guide...
- H Health and Medicine How to Increase Blood Pressure without Medication?...
- S Style and Beauty Choosing a Hair Straightener: Specific Criteria to Consider...
- F Food and Cooking How to Make Polendwitsa at Home?...
- S Science and Technology When do we change our clocks?...
- L Leisure and Entertainment What to Give a Girl on March 8?...
Answers on questions: Mathematics
- M Mathematics Let f(x) = x − 3 and g(x) = x + 11. Find f(x) ⋅ g(x). x2 − 8x − 33 x2 − 8x + 14 x2 + 8x + 14 x2 + 8x − 33...
- M Mathematics 1. Convert the following:1.1 40 miles to km...
- M Mathematics HEY CAN ANYONE PLS ANSWER DIS MATH PROBLEM I RLLY NEED IT!!...
- H History The era between reconstruction and the progressive period represents the...
Ответ:
since each ball has a different number and if no two pairs have the same value there is going to be 14∗14 different sums. Looking at the numbers 1 through 100 the highest sum is 199 and lowest is 3, giving 197 possible sums
For the 14 case, we show that there exist at least one number from set {3,4,5,...,17} is not obtainable and at least one number from set {199,198,...,185} is not obtainable.
So we are left with 197 - 195 options
14 x 14 = 196
196 > 195
so there are two pairs consisting of one red and one green ball that have the same value
As to the comment, I constructed a counter-example list for the 13 case as follows. The idea of constructing this list is similar to the proof for the 14 case.
Red: (1,9,16,23,30,37,44,51,58,65,72,79,86)
Green: (2,3,4,5,6,7,8;94,95,96,97,98,99,100)
Note that 86+8=94 and 1+94=95 so there are no duplicated sum
Step-by-step explanation:
For the 14 case, we show that there exist at least one number from set {3,4,5,...,17} is not obtainable and at least one number from set {199,198,...,185} is not obtainable.
First consider the set {3,4,5,...,17}.
Suppose all numbers in this set are obtainable.
Then since 3 is obtainable, 1 and 2 are of different color. Then since 4 is obtainable, 1 and 3 are of different color. Now suppose 1 is of one color and 2,3,...,n−1 where n−1<17 are of the same color that is different from 1's color, then if n<17 in order for n+1 to be obtainable n and 1 must be of different color so 2,3,...,n are of same color. Hence by induction for all n<17, 2,3,...,n must be of same color. However this means there are 16−2+1=15 balls of the color contradiction.
Hence there exist at least one number in the set not obtainable.
We can use a similar argument to show if all elements in {199,198,...,185} are obtainable then 99,98,...,85 must all be of the same color which means there are 15 balls of the color contradiction so there are at least one number not obtainable as well.
Now we have only 195 choices left and 196>195 so identical sum must appear
A similar argument can be held for the case of 13 red balls and 14 green balls