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weqwewe [10]
3 years ago
14

A set of five distinct positive integers has a mean of $1000$ and a median of $100$. What is the largest possible integer that c

ould be included in the set?
Mathematics
1 answer:
mixer [17]3 years ago
7 0

Answer:

e = 4796

Step-by-step explanation:

given,

mean of five distinct positive number = 1000

median of the number = 100

100 is median means two number will be less than 100 and two number will be greater than 100.

let five number be

a , b, c, d, e

'e' should be the largest number

As 100 is median so 'c' = 100.

'a' and 'b' should be as small as possible and d should be the number nearest to 100.

As all the number are distinct so the least number be equal to 1 and 2

now d will be equal to 101 (nearest to 100)

now,

sum of the five number = 5 x 1000 = 5000

a + b + c + d + e = 5000

1 + 2 + 100 + 101 + e = 5000

e = 5000 - 204

e = 4796

hence, the largest number will be equal to e = 4796

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MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

Substitute the value of the variable into the expression and simplify.

Exact Form:  a = 6, 2 3

Decimal Form: a = 6 , 0. ¯ 6

6 0
3 years ago
Find the value of n.<br><br> n+27<br><br> 11n<br><br> The value of n is ? units.
Kipish [7]

Answer:2.7=n

Step-by-step explanation:

n+27=11n

Move the variable to one side so subtract n

27=10n

Divide by 10

2.7=n

I'm not sure if this what you were asking for

6 0
3 years ago
PLEASE HELP!!!!!!!!!!
Oksanka [162]

Answer:

B

Step-by-step explanation:

Data set D does not contain the value 128, which is the median value.

Data set C does not contain the outlier value 91.

Data set A contains value 168, which does not show up on the plot.

The only remaining choice is B.

_____

In order, the data values of set B are ...

... 91, 114, 120, 126, 128, 128 134, 136, 139, 142, 152

The median value of these 11 is the 6th one: 128. The median values of the remaining two sets of 5 are 120 and 139, making these values the quartiles at the ends of the box. The value 91 is more than 1.5 times the IQR (19) below the 1st quartile, so is considered an outlier. (The cutoff is 120-1.5·19=91.5.)

6 0
3 years ago
Read 2 more answers
How to simplify this
algol [13]
To simplify this, you would have to turn b^-2 into a positive exponent.
To do this, we have to flip b^-2, which would get rid of the negate from the exponent: -2

3a^4 b^-2 c^3 / b^-2

Then we get the answer:

3a^4 c^3
------------
   b^-2

I have a picture to clarify! 
I hope this helped, let me know if you don't understand! ^.^

6 0
3 years ago
B(n)=1 (-2)^n-1<br> What is the 4th term in the sequence
zzz [600]

Answer:

The fourth term of the sequence B(4) =-8

The sequence is 1,-2,4,-8...

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given that the sequence

     B(n) = 1(-2)ⁿ⁻¹

Put n=1

    B(1) = 1(-2)¹⁻¹ = 2⁰ =1

Put n=2

   B(2) = 1(-2)²⁻¹ = (-2)¹ =-2

Put n=3

  B(3) =1(-2)³⁻¹ = (-2)² =4

Put n=4

B(4) = 1(-2)⁴⁻¹ =(-2)³ =-8

The sequence

1,-2,4,-8...

5 0
3 years ago
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