In 2010, Fernando paid $1,810 in federal income tax. In 2011, he paid $816. What is the percent of decrease in the amount paid

There were 324 adults surveyed. Among the participants, the mean number of hours of sleep each night was 7. 5 and the standard d

NeTakaya

The margin of error for 324 adults surveyed with 1.6 standard deviations is 0.1742.

<h3>What is the margin of error?</h3>

The margin of error can be defined as the amount of random sampling error in the results of a survey. It is given by the formula,

$\text{MOE}_{\gamma}=z_{\gamma} \times \sqrt{\frac{\sigma^{2}}{n}}$

$\text{MOE}$ = margin of error

$\gamma$ = confidence level

$z_{\gamma}$ = quantile

σ = standard deviation

n = sample size

As it is given that the sample size of the survey is 324, while the standard deviation of the survey is 1.6.

We know that the value of the z for 95% confidence interval is 1.96. Therefore, using the formula of the standard of error we can write it as,

$\text{MOE}_{\gamma}=z_{\gamma} \times \sqrt{\dfrac{\sigma^{2}}{n}}\\\\\text{MOE}_{\gamma}=1.96 \times \sqrt{\dfrac{1.6^{2}}{324}}\\\\\text{MOE}_{\gamma}=0.1742$

Hence, the margin of error for 324 adults surveyed with 1.6 standard deviations is 0.1742.

Learn more about Margin of Error:

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3 years ago

Martha spent 4/9 of her allowance on food and shopping. What fraction of her allowance had she left?

Mkey [24]

Her total allowance as a fraction could be expressed by 9/9. When you subtract the amount that she used, 4/9, you are left with 5/9, which is the solution.

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Daryl opens his shop at 8:30 AM and closes it at 7:50 PM. How long is the shop open?

slava [35]

Answer:

11 hours 20 minutes

Step-by-step explanation:

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3 years ago

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State the vertical asymptote of the rational function. f(x) =((x-9)(x+7))/(x^2-4)

Lubov Fominskaja [6]

$D:x^2-4\not=0\\
D:x^2\not=4\\
D:x\not=-2 \wedge x\not =2\\\\
\displaystyle
\lim_{x\to-2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\
\dfrac{(-2-9)(-2+7)}{(-2^-)^2-4}=\dfrac{-11\cdot5}{4^+-4}=\dfrac{-55}{0^+}=-55\cdot\infty=-\infty\\
\dfrac{(-2-9)(-2+7)}{(-2^+)^2-4}=\dfrac{-11\cdot5}{4^--4}=\dfrac{-55}{0^-}=-55\cdot(-\infty)=\infty
$

$\displaystyle
\lim_{x\to2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\
\dfrac{(2-9)(2+7)}{(2^-)^2-4}=\dfrac{-7\cdot9}{4^--4}=\dfrac{-63}{0^-}=-63\cdot(-\infty)=\infty\\
\dfrac{(2-9)(2+7)}{(2^+)^2-4}=\dfrac{-7\cdot9}{4^+-4}=\dfrac{-63}{0^+}=-63\cdot\infty=-\infty\\$

So, the vertical asymptotes are $x=\pm 2$

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4 years ago

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At the district competition for 5th grade jump-roper of the year, the following number of jumps without stopping were recorded p

deff fn [24]

First, we arrange the data from least to greatest:
90, 100, 100, 200, 203, 224, 233, 237, 245, 257, 264, 265, 265, 266, 277, 279, 285, 288, 288, 290, 290, 291, 291, 300

There are 24 data all in all. The mean is calculated by dividing by 24 the sum of all the data.
mean = (5828)/24242.83

The median of the number of jumps is the mean of the 12th and 13th term.
265

5 0

3 years ago