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r-ruslan [8.4K]
2 years ago
5

Write a function in any form that would match the graph shown below.

Mathematics
1 answer:
Pani-rosa [81]2 years ago
5 0

Answer:

(x-6)(x+5)^2 - 50

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Which of the following has a unit rate of $13.50 per shirt?
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B 6 shirts for 81

Step-by-step explanation:

81/6 = 13.5

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Write the expression using only positive exponents. Assume no denominator equals zero.
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The answer is

\frac{4}{ {y}^{2} }

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There is a 20% chance of thunderstorms tomorrow. Describe the likelihood of the event.
Pachacha [2.7K]

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Step-by-step explanation:

20% = 1/5 or 0.2

There is a 1/5 chance of a thunderstorm.

There really is no clear answer. I would try 0.2

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Which of the following is a solution of x2 + 5x = −2? (2 points) 5 plus or minus the square root of 33 divided by two. 5 plus or
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The easiest way it to solve it.
{x}^{2}  + 5x + 2 = 0
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5 0
3 years ago
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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.

5 0
2 years ago
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