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timofeeve [1]
2 years ago
11

Select the expressions that are equivalent to 9q+7q.

Mathematics
2 answers:
VARVARA [1.3K]2 years ago
7 0

Hey there!


9q + 7q


All you have to do is: COMBINE the LIKE TERMS and SIMPLIFY IT!


(9q + 7q)

= 9q + 7q
= 16q


Therefore, your answer is. 16q


Good luck on your assignment & enjoy your day!



~Amphitrite1040:)

VARVARA [1.3K]2 years ago
3 0

Answer:

16q

Step-by-step explanation:

Example

Question:

Write the equivalent expression for the given expression: 3x+9

Solution:

Given expression: 3x+9

Take 3 outside from the expression, we get,

= 3(x+3), which is called the equivalent expression

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Add the numbers on the left side, which gives you -13.

-13 = -13x

Lastly, divide both sides by -13x, which will give you 1.

So x = 1
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What is the equation of the line in slope-intercept form?
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Answer:

The answer would be C because b=+16 which means that it intercepts line y at 16, which is shown in the graph.

Step-by-step explanation:

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2 years ago
Vinny has decided to order high tops in black and white at a 5:8 ratio. He believes he will need a total of 169 pairs. How many
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Write an equation of the line with a slope of 3 and a y-intercept of 4.5
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Read 2 more answers
A player of a video game is confronted with a series of four opponents and an 80% probability of defeating each opponent. Assume
mixer [17]

Answer:

(a) 0.4096

(b) 0.64

(c) 0.7942

Step-by-step explanation:

The probability that the player wins is,

P(W)=0.80

Then the probability that the player losses is,

P(L)=1-P(W)=1-0.80=0.20

The player is playing the video game with 4 different opponents.

It is provided that when the player is defeated by an opponent the game ends.

All the possible ways the player can win is: {L, WL, WWL, WWWL and WWWW)

(a)

The results from all the 4 opponents are independent, i.e. the result of a game played with one opponent is unaffected by the result of the game played with another opponent.

The probability that the player defeats all four opponents in a game is,

P (Player defeats all 4 opponents) = P(W)\times P(W)\times P(W)\times P(W)=[P(W)]^{4} =(0.80)^{4}=0.4096

Thus, the probability that the player defeats all four opponents in a game is 0.4096.

(b)

The probability that the player defeats at least two opponents in a game is,

P (Player defeats at least 2) = 1 - P (Player losses the 1st game) - P (Player losses the 2nd game) = 1-P(L)-P(WL)

                                    =1-(0.20)-(0.80\times0.20)\\=1-0.20-0.16\\=0.64

Thus, the probability that the player defeats at least two opponents in a game is 0.64.

(c)

Let <em>X</em> = number of times the player defeats all 4 opponents.

The probability that the player defeats all four opponents in a game is,

P(WWWW) = 0.4096.

Then the random variable X\sim Bin(n=3, p=0.4096)

The probability distribution of binomial is:

P(X=x)={n\choose x}p^{x} (1-p)^{n-x}

The probability that the player defeats all the 4 opponents at least once is,

P (<em>X</em> ≥ 1) = 1 - P (<em>X</em> < 1)

             = 1 - P (<em>X</em> = 0)

             =1-[{3\choose 0}(0.4096)^{0} (1-0.4096)^{3-0}]\\=1-[1\times1\times (0.5904)^{3}\\=1-0.2058\\=0.7942

Thus, the probability that the player defeats all the 4 opponents at least once is 0.7942.

3 0
3 years ago
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