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Andru [333]
3 years ago
7

List P contains m numbers; list Q contains n numbers. If the two lists are combined to produce list R, containing m + n numbers,

is the median of list R greater than the median of list P? (1) The smallest number in list Q is greater than the largest number in list P. (2) m = n
Mathematics
1 answer:
Anarel [89]3 years ago
5 0

Answer: YES

Step-by-step explanation:

We need to write out the expressions

P= {m}

Q= {n}

R= {m+n}

If 2m=n then we can say;

P= {½n} Q= {n} & R= {³/²n}

It is obvious that the smaller number in Q is greater than the largest number in P

We can make some assumptions.

Let n= (x,y,z)

Consequently,

P={½x,½y,½z} Q={x,y,z} and R= {1.5x,1.5y,1.5z}

Therefore the median will be the middle element,

Median of P= ½y

Median of Q = y

Median of R = 1.5y

And 1.5y>1.5y

Then we can agree that the median of R is greater than the median of both P and Q

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elena-s [515]

Answer:

first is 3.5, second is 1

Step-by-step explanation:

-4.5 = x + 2x - 6

-4.5 - 6 = 3x

10.5 = 3x

10.5/3 = x

3.5 = x

2(3.5) - 6

7 - 6

1

8 0
3 years ago
Solve the equation 3x + 4 = 5x - 8​
CaHeK987 [17]

Answer:

x = 6

Step-by-step explanation:

3x + 4 = 5x - 8

==> add 8 to both sides

in doing so we get 3x + 4 + 8 = 5x - 8 + 8

the -8 and the +8 cancels out and 4 + 8 = 12

we are left with 3x + 12 = 5x

==> subtract 3x from both sides

in doing so we get 3x - 3x + 12 = 5x - 3x

the 3x and -3x cancel out and 5x - 3x = 2x

we are left with 12 = 2x

==> divide both sides by 2

in doing so we get 12/2 which equals 6 and 2x/2x which leaves us with

6 = x

3 0
2 years ago
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Mr. franklin paid $33.20 for 8 gallons of gas. What is the price of 1 gallon ?
evablogger [386]
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Find how many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6 and 7 (with repetitions), if:
Goshia [24]

Answer:

case 1 = 2592

case 2 =  729

case 1 + case 2 =  2916

(this is not a direct adition, because case 1 and case 2 have some shared elements)

Step-by-step explanation:

Case 1)

6 digits numbers that can be divided by 25.

For the first four positions, we can use any of the 6 given numbers.

For the last two positions, we have that the only numbers that can be divided by 25 are numbers that end in 25, 50, 75 or 100.

The only two that we can create with the numbers given are 25 and 75.

So for the fifth position we have 2 options, 2 or 7,

and for the last position we have only one option, 5.

Then the total number of combinations is:

C = 6*6*6*6*2*1 = 2592

case 2)

The even numbers are 2,4 and 6

the odd numbers are 3, 5 and 7.

For the even positions we can only use odd numbers, we have 3 even positions and 3 odd numbers, so the combinations are:

3*3*3

For the odd positions we can only use even numbers, we have 3 even numbers, so the number of combinations is:

3*3*3

we can put those two togheter and get that the total number of combinations is:

C = 3*3*3*3*3*3 = 3^6 = 729

If we want to calculate the combinations togheter, we need to discard the cases where we use 2 in the fourth position and 5 in the sixt position (because those numbers are already counted in case 1) so we have 2 numbers for the fifth position and 2 numbers for the sixt position

Then the number of combinations is

C = 3*3*3*3*2*2 = 324

Case 1 + case 2 = 324 + 2592 = 2916

4 0
3 years ago
Is 66,369,027 divisible by 4?
larisa86 [58]

Answer: no because you get a decimal

Step-by-step explanation:

6 0
3 years ago
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