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erastovalidia [21]
3 years ago
7

The table shows the results of an experiment in which subjects taking either a drug or a placebo either reported fatigue or did

not report Teague.
Which statements are true?

Chose all answers that are correct.

- more subjects took the drug than took the placebo.

- twenty-eight percent of the subjects did not report fatigue.

- a greater percent of the subjects did not report fatigue.

-a greater percentage of subjects taking the placebo did not report fatigue.

Mathematics
2 answers:
Angelina_Jolie [31]3 years ago
3 0

Answer:

- more subjects took the drug than took the placebo.  

- a greater percent of the subjects did not report fatigue.  

- a greater percentage of subjects taking the placebo did not report fatigue.

Step-by-step explanation:

60% of the subjects took the drug and 40% took the placebo.  

56% of the subjects did not report fatigue and 44% reported fatigue.

28%  of subjects taking the placebo did not report fatigue and 12% report fatigue.

Hitman42 [59]3 years ago
3 0

Answer:

<h2>- More subjects took the drug than took the placebo.</h2><h2>- A greater percent of the subjects did not report fatigue.</h2>

Step-by-step explanation:

A cross-table is about several variables, where we can get different analysis about each aspect of each variable.

In this case, we have Drug and Placebo, each of them has certain percentage about Reported Fatigue and Did not Report Fatigue.

According to the table, 60% of the people took drug, and 40% took placebo. That means more subjects took the drug than took the placebo.

Additionally, 56% of the people didn't report fatigue, and 44% reported fatigue. Therefore, a greater percent of the subjects did not report fatigue.

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What is 1/5 times 10x?
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Step-by-step explanation:

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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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