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Softa [21]
2 years ago
9

Will Identify the outlier in the data set. 77, 73, 78, 78, 71, 75, 120, 75, 77, 72, 75, 70 The outlier is​

Mathematics
1 answer:
o-na [289]2 years ago
7 0

Answer:

120

Step-by-step explanation:

120

every other number is in the 70s

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Explain how you would simplify the expression 3x + 7y - 2x + 3y + 7?
kipiarov [429]

Answer:

Step-by-step explanation:

3x+7y-2x+3y+7

group like terms

3x-2x   +7y+3y   +7

x+10y+7

8 0
3 years ago
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L1 : (3, 5) and (2,5) L2: (2, 4) and (0, 4) A. Perpendicular B. Parallel C. Neither
Tresset [83]

answer for this question is (b)parallel

7 0
3 years ago
Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.
nignag [31]

Answer:

we need to prove : for every integer n>1, the number n^{5}-n is a multiple of 5.

1) check divisibility for n=1, f(1)=(1)^{5}-1=0  (divisible)

2) Assume that f(k) is divisible by 5, f(k)=(k)^{5}-k

3) Induction,

f(k+1)=(k+1)^{5}-(k+1)

=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1

=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k

Now, f(k+1)-f(k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k

f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k

Take out the common factor,

f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)      (divisible by 5)

add both the sides by f(k)

f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)

We have proved that difference between f(k+1) and f(k) is divisible by 5.

so, our assumption in step 2 is correct.

Since f(k) is divisible by 5, then f(k+1) must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number n^{5}-n is a multiple of 5.

3 0
3 years ago
ANSWER FREE BRINLIEST!! ANSWER THE TWOO CIRCLED ONES!!
BlackZzzverrR [31]

Answer:

alright for the first one you set them equal o each other while for the second you do the same thing

first one:8

second one:5

Step-by-step explanation:

24/3 is x and x=8

for the second one

18+12 is 30 and 30/6 is 5so the second one is 5'

if you want me to clarify more please comment

5 0
3 years ago
Read 2 more answers
The dot plots below show the scores for a group of students for two rounds of a quiz: Two dot plots are shown one below the othe
miskamm [114]

Answer:

The inference that can be made using the dot plot is:

    The range of round 1 is greater than the round 2 range.

Step-by-step explanation:

<u>Round 1:</u>

Score                Frequency

  1                          0

  2                          2

  3                          3

  4                          2  

  5                          1

Hence, the minimum score of Round 1 is: 2

maximum score is: 5

Hence, Range=Maximum value-Minimum score

                      =5-2

                       =3

Similarly, <u>Round-2</u>

Score                Frequency

  1                          0

  2                         0

  3                          0

  4                          4  

  5                          4

Hence, the minimum score of Round 1 is: 4

maximum score is: 5

Hence, Range=Maximum value-Minimum score

                      =5-4

                       =1

The scores of round 2 are higher than round-1.

Since round 2 have a higher frequency for higher scores as compared to round-1.

Hence, Range of round 1 is greater than the range of Round-2.

 

6 0
3 years ago
Read 2 more answers
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