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3 years ago

For the data set below, which of the following are true? {88, 79, 86, 93, 76, 80, 82, 72, 95, 79}

Mathematics

1 answer:

FrozenT [24]3 years ago

6 0

D. The interquartile range is 9

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If Owen and Eliza had a combined taxable income of $68,200 and filed their federal income tax return with the Married Filing Joi

tangare [24]

Answer:

9425

Step-by-step explanation:

They are in the third tax bracket down. That means that their joint income is over 67900 but less than 137050.

The base rate is 9350. But they are 68200 - 67900 = 300 over the minimum.

So 25% of that 300 is also to be paid to the government 25/100 * 300 = 75 dollars

Now they must pay 9350 + 75 = 9425

Answer D

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3 years ago

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Simplify this expression: cos t(sec t − cos t)

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$\cos t \, (\sec t - \cos t)$
$= \cos t \, (\frac{1}{\cos t} - \cos t)$
$= 1 - \cos^2 t$
$= \bf sin^2 t$

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3 years ago

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professor190 [17]

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2 years ago

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Anna35 [415]

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3 years ago

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A population has a known standard deviation of 1.27 and a sample space contains 85 values find the margin of error needed to cre

r-ruslan [8.4K]

Answer:

The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

In this question:

$\sigma = 1.27, n = 85$

$M = z*\frac{\sigma}{\sqrt{n}}$

$M = 2.575*\frac{1.27}{\sqrt{85}}$

$M = 0.3547$

The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547

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4 years ago