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Luba_88 [7]
3 years ago
8

¿como represento 2/3 en decimal?​

Mathematics
2 answers:
3241004551 [841]3 years ago
8 0

Answer:

0.66666666 o 0.66. :)((((((()(

son4ous [18]3 years ago
3 0

Answer: 0.66

Step-by-step explanation:

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saveliy_v [14]
This one is super difficult, I see your struggle but the answer is A.
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Evaluate the expression 4x^4 y^3 when x=1/5 and y=6
wariber [46]

Answer:

5.5296

Step-by-step explanation:

to evaluate the expression 4x^4 y^3 we would substitute the value of x and y into it and evaluate. since x = 1/5 and y = 6

4x^4 y^3

4 × x^4 × 4× y³

4 × (1/5)∧4 × 4 × 6³

4× (0.2)∧4 × 4 × 216

4 × 0.0016 × 864

0.0064 × 864

5.5296

5 0
3 years ago
Jason wanted to buy postcards for his cousin. The price was 2 cards for $1. He had some change in his pocket. What combination o
Zina [86]

Answer:

2 dimes, 1 nickel and 3 quarters.

Step-by-step explanation:

The price of 2 postcards is $1. Any combination of coins that adds to $1 will work here. Remember, a penny is worth 0.01, a nickel is worth 0.05, a dime is worth 0.10, and a quarter is worth 0.25.

Adding 2 dimes, 1 nickel, and 3 quarters would be 2(0.10)+1(0.05)+3(0.25)=1.


8 0
3 years ago
7(3w-3)=-12<br> HURRY PLEASE
Lady_Fox [76]

Answer:

3/7

Step-by-step explanation:

21w - 21 = -12

21w = -12 +21

21w = 9

w = 9/21

i.e w= 3/7

8 0
2 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
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