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3 years ago

Divide 5/9 round to the nearest tenth

Mathematics

2 answers:

almond37 [142]3 years ago

4 0

Answer:

0.6

Step-by-step explanation:

5/9=0.55555555555...

since the hundredth digit is a 5 you round up to 0.6

zhenek [66]3 years ago

3 0

Answer:

0.55

Step-by-step explanation:

5 divided by 9 = 0.55

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Answer:

B. 3.99x10^-6

Step-by-step explanation:

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3 years ago

If the simple interest on $5,000 for 9 years is $1,800, then what is the interest rate?

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$I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill & \$1800\\ P=\textit{original amount deposited}\dotfill & \$5000\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &9 \end{cases} \\\\\\ 1800=5000(\frac{r}{100})(9)\implies \cfrac{1800}{5000(9)}=\cfrac{r}{100}\implies \cfrac{1}{25}=\cfrac{r}{100} \\\\\\ 100=25r\implies \cfrac{100}{25}=r\implies \stackrel{\%}{4}=r$

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2 years ago

What are two solutions to x>29

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Step-by-step explanation:

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3 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$