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algol13
3 years ago
12

Please help!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
ololo11 [35]3 years ago
8 0
1st step is distributing 5
2nd step is adding 40

s = -10

Hope this helps :)
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Question 5 (Solving Problema) lon 5 Solving Problems] Write 24 18 in slmplest form​
otez555 [7]

Answer:

4/3

Step-by-step explanation:

reduce the fraction by a factor of 6

pretty sure you mean write 24/18 in simplest form so that's the answer

Hope this helps! :)

8 0
2 years ago
Reduce the fraction 36/48 to its lowest
KonstantinChe [14]

Answer:

3/4

Step-by-step explanation:

36/48

Divide the top and bottom by 12

36/12 = 3

48/12 =4

36/48 = 3/4

4 0
3 years ago
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Factorise x2-10x-24 plz help ​
Papessa [141]

Answer:

(х-4)(х-6)

Step-by-step explanation:

х¹= 4

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3 years ago
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Are 0,2,5 closed or not under addition
kvasek [131]

Answer:

There are closed addition.

Step-by-step explanation:

Closure (mathematics) ... For example, the positive integers are closed under addition, but not under subtraction: is not a positive integer even though both 1 and 2 are positive integers. Another example is the set containing only zero, which is closed under addition, subtraction and multiplication.

So, 0,2,5 are closed addition

Please mark me brainliest.

3 0
3 years ago
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
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