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3 years ago

11 less than the product of a number and -4

Mathematics

1 answer:

denpristay [2]3 years ago

7 0

Answer:

-4n - 11

Step-by-step explanation:

Step 1: Convert words into an expression

11 less than the product of a number and -4

(n * -4) - 11

<em />

Answer: -4n - 11

<u>If needed solve</u>

-4n - 11 + 11 = 0 + 11

-4n / -4 = 11 / -4

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What % of 1 meter is 70 meter

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well, let's say 1 meter is the 100%, now, if we know that 1 is 100%, what percentage is 70?

$\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 1&100\\ 70&x \end{array}\implies \cfrac{1}{70}=\cfrac{100}{x}\implies x =7000$

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Two random samples are taken, with each group asked if they support a particular candidate. A summary of the sample sizes and pr

Minchanka [31]

Answer:

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98[/tex[ or 98%Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". The margin of error is the range of values below and above the sample statistic in a confidence interval. Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". [tex]p_1$ represent the real population proportion for 1

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$\hat p_2 =0.646$ represent the estimated proportion for 2

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$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

For this case we have the confidence interval given by: (-0.0313,0.2753). From this we can find the margin of erro on this way:

$ME= \frac{0.2753-(-0.0313)}{2}=0.1533$

And we know that the margin of erro is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

We have all the values except the value for $z_{\alpha/2}$

So we can find it like this:

$0.1533=z_{\alpha/2} \sqrt{\frac{0.768(1-0.768)}{92} +\frac{0.646 (1-0.646)}{95}}$

And solving for $z_{\alpha/2}$ we got:

$z_{\alpha/2}=2.326$

And we can find the value for $\alpha/2$ with the following excel code:

"=1-NORM.DIST(2.326,0,1,TRUE)"

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