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Lena [83]
1 year ago
14

Please help! This assignment is due soon! (In The photo)

Mathematics
1 answer:
eimsori [14]1 year ago
4 0

Answer

Elena must have substracted 1/2x from both sides of the equation.

Lin must have multiplied both sides of the equation by 2

Explanation

The equation given is

\frac{1}{2}x+3=\frac{7}{2}x+5

For Elena to have arrived at

3=\frac{7}{2}x-\frac{1}{2}x+5

Then Elena must have substracted 1/2x from both sides of the equation.

That is;

\begin{gathered} \frac{1}{2}x+3=\frac{7}{2}x+5 \\ \text{Substracting }\frac{1}{2}x\text{ from both sides of the equation will give Elena first step} \\ \frac{1}{2}x+3-\frac{1}{2}x=\frac{7}{2}x+5-\frac{1}{2}x \\ 3=\frac{7}{2}x-\frac{1}{2}x+5 \end{gathered}

For Lin to have arrived at

x+6=7x+10

It shows Lin must have multiplied both sides of the equation by 2

That is;

\begin{gathered} \frac{1}{2}x+3=\frac{7}{2}x+5 \\ \text{Multiply both sides of the equation by the lowest common mutiple } \\ \text{of the denominator which is 2.} \\ \frac{1}{2}x(2)+3(2)=\frac{7}{2}x(2)+5(2) \\ x+6=7x+10 \end{gathered}

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A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi
34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

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

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

 \mu_{X}=n\mu

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

\sigma_{X}=\sqrt{n}\sigma

It is provided that:

\mu=\$27\\\sigma=\$18\\n=310

As the sample size is quite large, i.e. <em>n</em> = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let <em>s</em> denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding <em>z-</em>value is, -2.33.

Compute the value of <em>s</em> as follows:

z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

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Answer:

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Step-by-step explanation:

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3 years ago
A circle has center at –6 + 2i and a point on the circle at 4 + 26i. Which of the following points is also on the circle?
Alenkasestr [34]
Answer:
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