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beks73 [17]
2 years ago
13

Solve for y 2x+y-3z=5

Mathematics
2 answers:
lisabon 2012 [21]2 years ago
5 0

Answer:

Option A

Step-by-step explanation:

Since the "y" variable is inside a root, we need to square both sides of the equation to open the root and solve for y.

<u>When squaring both sides, we get;</u>

\implies \sqrt{2x + y - 3z}  = 5

\implies(\sqrt{2x + y - 3z})^{2}   = 5^{2}

\implies2x + y - 3z  = 25

Now, simply isolate the y-variable to determine its value. This can be done by subtracting 2x on both sides of the equation.

\implies2x + y - 3z  = 25

\implies2x + y - 3z - 2x = 25 - 2x

\implies y - 3z = 25 - 2x

Now, add 3z both sides of the equation to further isolate the y-variable.

\implies y - 3z + 3z = 25 - 2x + 3z

\implies y = 25 - 2x + 3z

Therefore, Option A is correct.

Elza [17]2 years ago
4 0

Answer:

<h3>A. y=-2x+3z+25</h3>

Step-by-step explanation:

Isolate the term of x and y from one side of the equation.

<u>To solve:</u>

  • The value of y.

\Longrightarrow: \sf{\sqrt{2x+y-3z}=5}

<h3>2x+y-3z=25</h3>

<u>First, you have to subtract by 2x-3z from both sides.</u>

\Longrightarrow: \sf{2x+y-3z-\left(2x-3z\right)=25-\left(2x-3z\right)}}

<u>Solve.</u>

\Longrightarrow: \boxed{\sf{y=25-2x+3z}}}

  • <u>Therefore, the correct answer is "A. y=-2x+3z+25".</u>

I hope this helps, let me know if you have any questions.

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Answer:

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Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we select appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

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And the standard deviation of the distribution of the sum of values of X is given by,  

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The information provided is:

<em>μ</em> = $970

<em>σ</em> = $129

<em>n</em> = 102

Since the sample size is quite large, i.e. <em>n</em> = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

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Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

P (\sum X \leq  100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}

                              =P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

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