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Mashcka [7]
3 years ago
12

A) -2x + 5 + 10x - 9 B) 3(x + 7) + 2(-x + 4) + 5x Simply A, and B.

Mathematics
2 answers:
sleet_krkn [62]3 years ago
7 0

Answer:

A - 8x=4

B - 6x+29

Step-by-step explanation:

I hope this helps

nlexa [21]3 years ago
7 0
Answer:

A) 8x - 4
B) 6x + 29

Step-by-Step Explanation:

A) -2x + 5 + 10x - 9
= 8x + 5 - 9
=> 8x - 4

B) 3(x + 7) + 2(-x + 4) + 5x
= 3x + 21 + 2(-x + 4) + 5x
= 3x + 21 - 2x + 8 + 5x
= 6x + 21 + 8
=> 6x + 29
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2 number cubes are rolled what is the probability that the sum if the numbers rolled is either a 4 or a 10
goldenfox [79]
With 2 cubes, there are 3 ways to roll a sum of 4... 1 and 3, 3 and 1, 2 and 2.
with 2 cubes, there are 3 ways to roll a sum of 10....6 and 4, 4 and 6, 5 and 5.

total possible outcomes = (6 * 6 ) = 36

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4 0
4 years ago
Anyone know how to do this ????
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5 0
3 years ago
Suppose a box of Cracker Jacks contains one of 5 toy prizes: a small rubber ball, a whistle, a Captain America decoder ring, a r
lakkis [162]

Answer:

11.42 boxes

Step-by-step explanation:

For the first box bought, there is a 100% chance of getting a unique toy (since you still don't have any). E₁ = 1.

After that, there is a 4 in 5 chance of getting a unique toy from the next box, the expected number of boxes required is:

E_2 = (\frac{4}{5})^{-1} = 1.25

For the next unique toy, there is now a 3 in 5 chance of getting it:

E_3 = (\frac{3}{5})^{-1} = 1.67

Following that logic, there is a 2 in 5 chance of getting the 4th unique toy:

E_4 = (\frac{2}{5})^{-1} = 2.5

Finally, there is a 1 in 5 chance to get the last unique toy:

E_5 = (\frac{1}{5})^{-1} = 5

The expected number of boxes to obtain a full set is:

E=E_1+E_2+E_3+E_4+E_5\\E=1+1.25+1.67+2.5+5\\E=11.42\ boxes

5 0
3 years ago
Use the graph to write an equation<br> y = acos[b(x – h)] + k to model the situation.
inna [77]

Answer:The amplitude:30

so a:-30

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so b: pi/6

Step-by-step explanation:

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3 years ago
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