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Taya2010 [7]
3 years ago
7

Which represents the solution(s) of the graphed system of equations, y = –x2 + x + 2 and y = –x + 3?

Mathematics
2 answers:
lukranit [14]3 years ago
6 0

Answer:

option a or (1,2) on edge

Step-by-step explanation:

cupoosta [38]3 years ago
3 0

For this case we have the following system of equations:

y = -x ^ 2 + x + 2\\y = -x + 3

We observe that we have a quadratic equation and therefore the function is a parabola.

We have a linear equation.

Therefore, the solution to the system of equations will be the points of intersection of both functions.

When graphing both functions we have that the solution is given by:

x = 1\\y = 2

That is, the line cuts the quadratic function in the following ordered pair:

(x, y) = (1, 2)

Answer:

the solution (s) of the graphed system of equations are:

(x, y) = (1, 2)

See attached image.

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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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