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emmainna [20.7K]
3 years ago
12

Find the value of y+1 given that 2y+8=4.

Mathematics
2 answers:
aleksklad [387]3 years ago
6 0

Answer:

2y + 8 = 4 \\ 2y = 4 - 8 \\ 2y =  - 4 \\ y =  \frac{ - 4}{2}  \\ y =  - 2 \\ y + 1 =  - 2 + 1 \\  \boxed{y + 1 =  - 1}

Therefore, the value of (y+1) is <u>-1</u>

  • <u>(y+1) = -1</u> is the right answer.
faltersainse [42]3 years ago
3 0

Answer: y = -2

Step-by-step explanation:

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Answer:

Step-by-step explanation:

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5 0
2 years ago
Bought a board game for $30. The board game is now worth $55. Calculate the percent change in the game's value.
Hatshy [7]
30 is the old value and 55 is the new value. In this case we have a positive change (increase) of 83.33333333 percent because the new value is greater than the old value. Using this tool you can find the percent increase for any value.
3 0
2 years ago
Can you help me with number 6? <br> Confused abit <br> Please
Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.


To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.


First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is


\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}


Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is


\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}


Third question:

To get exactly one six, it can either be the first, second or third roll.


In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.


In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus


\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}


And since there are three of these combinations, The answer is


3\frac{5^2}{6^3}


Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is


1 - \frac{5^3}{6^3}

4 0
3 years ago
Any one can help me with problem 4?
Travka [436]
One is 55%

Two is 190%

Three is 57
3 0
3 years ago
2x^2 - 5x +4=0 what is the solution
NeTakaya

Answer

x = 3.13745860… , − 0.63745860 …

Step-by-step explanation:

thats decimal form

7 0
3 years ago
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