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Sholpan [36]
3 years ago
9

7. A spinner has 8 equal-sized sections. To win the game, the pointer

Mathematics
1 answer:
notsponge [240]3 years ago
7 0
The answer is 50% chance for each side
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Use the elimination method to solve the system of equations.Choose the correct ordered pair. x+y=8 x-y=6
ratelena [41]

Answer:

Solution (7,1)

Step-by-step explanation:

Use the elimination method

x + y=8

x - y=6

--------------Add

2x = 14

 x = 7

Substitute x = 7 into x + y=8

7 + y=8

y = 8 - 7

y = 1

Solution (7,1)


3 0
4 years ago
Read 2 more answers
Ea+ec=5.40, ec*2=s,ea/2=s<br><br> What is Ec?
alexandr402 [8]
I have no idea I am only in  Middle School

3 0
3 years ago
2.50+0.50=21 help meeeeeeeeeee
Rzqust [24]

Answer:

You are missing some of the equation my dude

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Which expression is equivalent to *picture attached*
DiKsa [7]

Answer:

The correct option is;

4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3  \left (\dfrac{50(51) }{2} \right )

Step-by-step explanation:

The given expression is presented as follows;

\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3  \right )

Which can be expanded into the following form;

\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3  \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left  n^2 + 3  \times\sum\limits _{n = 1}^{50}  n

From which we have;

\sum\limits _{k = 1}^{n} \left  k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}

\sum\limits _{k = 1}^{n} \left  k = \dfrac{n \times (n+1) }{2}

Therefore, substituting the value of n = 50 we have;

\sum\limits _{n = 1}^{50} \left  k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}

\sum\limits _{k = 1}^{50} \left  k = \dfrac{50 \times (50+1) }{2}

Which gives;

4 \times \sum\limits _{n = 1}^{50} \left  n^2 =  4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}

3  \times\sum\limits _{n = 1}^{50}  n = 3  \times \dfrac{n \times (n+1) }{2} = 3  \times \dfrac{50 \times (51) }{2}

\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3  \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3  \times \dfrac{50 \times (51) }{2}

Therefore, we have;

4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3  \left (\dfrac{50(51) }{2} \right ).

4 0
3 years ago
PLEASE HELP Will mark brainliest
ss7ja [257]

For the first one, the range is the largest number minus the smallest number. The largest number in this case is 11 and the smallest is 4. 11 - 4 = 7 so the maximum range is 7 inches.

For the second one, the IQR is Q3 - Q1. This means 30 - 20 = IQR and 30 - 20 is 10 books.

And the last one asks for dinner customers, and there are 54 total. 28 of them chose cheesecake as their favorite dessert. This means that the amount of dinner customers that chose cheesecake as their favorite is 28/54 or 0.5185, whihc is 51.85%

Hope this helps!!

6 0
3 years ago
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