(3/5)^-2 Possible answers: 9/25 -9/25 -25/9 25/9

When converted to a household measurement, 9 kilograms is approximately equal to a

kondaur [170]

Answer:

9 kilograms is = 9000 grams

9 kilograms=317.466 ounces

Step-by-step explanation:

4 0

2 years ago

The average weight of the entire batch of the boxes of cereal filled today was 20.5 ounces. A random sample of four boxes was se

zhenek [66]

Answer:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s= 0.286$

And then the estimator for the standard error is given by:

$SE= \frac{0.286}{\sqrt{4}}= 0.143$

Step-by-step explanation:

For this case we have the following dataset given:

20.05, 20.56, 20.72, and 20.43

We can assume that the distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for this case would be:

$SE= \frac{\sigma}{\sqrt{n}}$

And we can estimate the deviation with the sample deviation:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s= 0.286$

And then the estimator for the standard error is given by:

$SE= \frac{0.286}{\sqrt{4}}= 0.143$

4 0

3 years ago

2b^2+4b–2=0 what is the answer?

s2008m [1.1K]

$2b^2+4b-2=0\\
2(b^2+2b-1)=0\\
2(b^2+2b+1-2)=0\\
2((b+1)^2-2)=0\\
2(b+1)^2-4=0\\\
2(b+1)^2=4\\\
(b+1)^2=2\\
b+1=-\sqrt2 \vee b+1=\sqrt2\\
b=-1-\sqrt2 \vee b=-1+\sqrt2

$

5 0

3 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

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:

$\alpha = \frac{1-0.95}{2} = 0.025$

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 $1-0.025 = 0.975$ , so $z = 1.96$

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.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

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

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

6 0

3 years ago

Plz help worth 40 points

denis23 [38]

Answer:

Answer is B

6 0

3 years ago

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