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Akimi4 [234]
2 years ago
8

Separate 846 into 3 parts so that the second part is twice the first part and the third part

Mathematics
2 answers:
wolverine [178]2 years ago
6 0

Answer:

it is 1 ok have fun babby and enjoy your Christmas ok baby bye bye

neonofarm [45]2 years ago
5 0

Answer:

1) 94, 2) 188, 3) 564

Step-by-step explanation:

This is pretty simple. The first thing you want to do is come up with an algebraic equation to solve this. It will be x + 2x + 6x=846 . The first x represents the smallest part, or the first part. The second part is represented by 2x because it is twice the first, and 6x is 3 times the second part which is 6 times the first part. Combine like terms to get 9x = 846. Now isolate the variable by dividing 9 from both sides. This will give you x = 94. Now just plug in the x to find the second and third part.

Hope this helped :)

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Evgesh-ka [11]

Answer:

Each combo costed $4.95

Step-by-step explanation:

To solve, simply divide 103.95 by 21, and then put a dollar sign. (Most teachers would count no dollar sign as incorrect/points off, so don't forget it!)

103.95 ÷ 21 = 4.95 → $4.95

<em>Hope this helps!</em>

5 0
3 years ago
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masya89 [10]
This is in its simplest form.
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Express 8% as a decimal
konstantin123 [22]

Answer:

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3 years ago
Read 2 more answers
A raffle offers one $8000.00 prize, one $4000.00 prize, and five $1600.00 prizes. There are 5000 tickets sold at $5 each. Find t
Harman [31]

Answer:

The expectation is  E(1 )= -\$ 1

Step-by-step explanation:

From the question we are told that  

     The first offer is  x_1 =  \$ 8000

     The second offer is  x_2 =  \$ 4000

      The third offer is  \$ 1600

      The number of tickets is  n  =  5000

      The  price of each ticket is  p= \$ 5

Generally expectation is mathematically represented as

             E(x)=\sum  x *  P(X = x )

     P(X =  x_1  ) =  \frac{1}{5000}    given that they just offer one

    P(X =  x_1  ) = 0.0002    

 Now  

     P(X =  x_2  ) =  \frac{1}{5000}    given that they just offer one

     P(X =  x_2  ) = 0.0002    

 Now  

      P(X =  x_3  ) =  \frac{5}{5000}    given that they offer five

       P(X =  x_3  ) = 0.001

Hence the  expectation is evaluated as

       E(x)=8000 *  0.0002 + 4000 *  0.0002 + 1600 * 0.001

      E(x)=\$ 4

Now given that the price for a ticket is  \$ 5

The actual expectation when price of ticket has been removed is

      E(1 )= 4- 5

      E(1 )= -\$ 1

4 0
3 years ago
Will Give You Brainliest :)
Misha Larkins [42]

178 is the answer

mark brainliest

7 0
3 years ago
Read 2 more answers
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