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vovangra [49]
3 years ago
15

When solving the equation such as 3/5x = 1/9 , why would you multiply both sides of the equation by the reciprocal of 3/5 ?

Mathematics
1 answer:
pav-90 [236]3 years ago
7 0
You would multiply each side of the equation by the reciprocal of 3/5 (5/3) because would cancel out 3/5 which would leave you with x and the answer.
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A rectangular prism measures 225 inches by 315 inches by 2 inch.
Bas_tet [7]

Answer:

No I don't agree with Priya

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(I'm sorry, I don't know what was your previous question, so I can't answer that part. If you comment it below, I will do it!)

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2 years ago
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2 years ago
Solve each pair of simultaneous equations by substitution method
WARRIOR [948]

Answer:

x = 1       y = -4

Step-by-step explanation:

- Plug the value of y into the other equation.

x + 3y = -11

x + 3(-4x) = -11

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3 0
2 years ago
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
How do I solve the problem 4040÷69.689 ?
kati45 [8]

0.05797

Use standard Division to get your answer but in most classes you won’t be given an answer that big and you should use s calculator

8 0
3 years ago
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