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2 years ago

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Which statements are correct interpretations of this graph? Select each correct answer. A.3 pages are edited every 5 min B. 12/3

of a page is edited per minute C.61/0 of a page is edited per minute D. 5 pages are edited every 3 min there are multiple answers I NEED HELP NOW

2 answers:

Setler [38]2 years ago

5 0

Among the choices, the <span>statements that are correct interpretations of this graph are:

</span><span>A.3 pages are edited every 5 min
</span><span>C.6/10 of a page is edited per minute

The statement that "</span>3 pages are edited every 5 min" is shown by the first blue point in the graph. While the statement "6/10 of a page is edited per minute" is shown by the second blue point.

natita [175]2 years ago

5 0

Answer:

A.3 pages are edited every 5 min

C.6/10 of a page is edited per minute

Step-by-step explanation:

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The Chewy Candy Company is test marketing a new candy bar in Seoul and Manila. In Seoul, the company sold 40 candy bars the firs

mart [117]

Looking at the image attached, we can see the assumed sales of the Chewy Candy Company for its new candy bar in Manila and Seoul. Setting up the given condition for separate branches, we clearly see that during the 6th month, the total sales in Manila first exceeded the cumulative total sales in Seoul.

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2 years ago

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ACME Manufacturing claims that its cell phone batteries last more than 32 hours on average in a certain type of cell phone. Test

son4ous [18]

Answer:

We can conclude that the battery life is greater than the 32 hour claim.

Step-by-step explanation:

The null hypothesis is:

$H_{0} = 32$

The alternate hypotesis is:

$H_{1} > 32$

Our test statistic is:

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}$

In which X is the statistic, $\mu$ is the mean, $\sigma$ is the standard deviation and n is the size of the sample.

In this problem, we have that:

$X = 37.8, \mu = 32, \sigma = 10, n = 18$

So

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$t = \frac{37.8 - 32}{\frac{10}{\sqrt{18}}}$

$t = 2.46$

We need to find the probability of finding a mean time greater than 37.8. If it is 5% of smaller(alpha = 0.05.), we can conclude that the battery life is greater than the 32 hour claim.

Probability of finding a mean time greater than 37.8

1 subtracted by the pvalue of z = t = 2.46.

z = 2.46 has a pvalue of 0.9931

1 - 0.9931 = 0.0069 < 0.05

So we can conclude that the battery life is greater than the 32 hour claim.

5 0

2 years ago

Which of the following functions best represents the graph?

pav-90 [236]

Hello there!
the answer would be A

Hope this helps! :)
~Zain

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2 years ago

Read 2 more answers

Marketing companies have collected data implying that teenage girls use more ring tones on their cell phones than teenage boys d

ASHA 777 [7]

Answer:

The p-value here is 0.0061, which is very small and we have evidence that the girls' mean is higher than the boys' mean.

Step-by-step explanation:

We suppose that the two samples are independent and normally distributed with equal variances. Let $\mu_{1}$ be the mean number of ring tones for girls, and $\mu_{2}$ the mean number of ring tones for boys.

We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} > 0$ (upper-tail alternative).

The test statistic is

T = $\frac{\bar{X}_{1}-\bar{X}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$

where

$S_{p} = \sqrt{\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{n_{1}+n_{2}-2}}$ .

For this case, $n_{1}=n_{2}=20$ , $\bar{x}_{1}=3.2$ , $s_{1}=1.5$ , $\bar{x}_{2}=2.2$ , $s_{2}=0.8$ .

$s_{p} = \sqrt{\frac{(19)(1.5)^{2}+(19)(0.8)^{2}}{38}} = 1.2021$ and the observed value is

t = $\frac{3.2-2.2}{1.2021\sqrt{1/20+1/20}} = \frac{1}{0.3801} = 2.6309$ .

We can compute the p-value as P(T > 2.6309) where T has a t distribution with 20 + 20 - 2 = 38 degrees of freedom, so, the p-value is 0.0061. Because the p-value is very small, we can reject the null hypothesis for instance, at the significance level of 0.05.

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2 years ago

An Article in the Journal of Sports Science (1987, Vol. 5, pp. 261-271) presents the results of an investigation of the hemoglob

ValentinkaMS [17]

Answer:

The 95% confidence interval for the population variance is $\left[0.219, \hspace{0.1cm} 0.807\right]\\\\$

The 95% confidence interval for the population mean is $\left [15.112, \hspace{0.3cm}15.688\right]$

Step-by-step explanation:

To solve this problem, a confidence interval of $(1-\alpha) \times 100%$ for the population variance will be calculated.

$$$Sample variance: $S^2=(0.6152)^2$\\Sample size $n=20$\\Confidence level $(1-\alpha)\times100\%=95\%$\\$\alpha: \alpha=0.05$\\$\chi^2$ values (for a 95\% confidence and n-1 degree of freedom)\\$\chi^2_{\left (1-\frac{\alpha}{2};n-1\right )}=\chi^2_{(0.975;19)}=8.907\\$\chi^2_{\left (\frac{\alpha}{2};n-1\right )}=\chi^2_{(0.025;19)}=32.852\\\\$

Then, the $(1-\alpha) \times 100%$ confidence interval for the population variance is given by:

$\left [\frac{(n-1)S^2}{\chi^2_{\left (\frac{\alpha}{2};n-1\right )}}, \hspace{0.3cm}\frac{(n-1)S^2}{\chi^2_{\left (1-\frac{\alpha}{2};n-1\right )}} \right ]\\\\$ Thus, the 95% confidence interval for the population variance is: $\\\\\left [\frac{(19-1)(0.6152)^2}{32.852}, \hspace{0.1cm}\frac{(19-1)(0.6152)^2}{8.907} \right ]=\left[0.219, \hspace{0.1cm} 0.807\right]\\\\$

On other hand,

A confidence interval of $(1-\alpha) \times 100%$ for the population mean will be calculated

$$$Sample mean: $\bar X=15.40$\\Sample variance: $S^2=(0.6152)^2$\\Sample size $n=20$\\Confidence level $(1-\alpha)\times100\%=95\%$\\$\alpha: \alpha=0.05$\\T values (for a 95\% confidence and n-1 degree of freedom) T_{(\alpha/2;n-1)}=T_{(0.025;19)}=2.093\\\\$Then, the (1-\alpha) \times 100$\% confidence interval for the population mean is given by:\\\\$

\ $\left[ \bar X - T_{(\alpha/2;n-1}\sqrt{\frac{\S^2}{n}}, \hspace{0.3cm}\bar X + T_{(\alpha/2;n-1}\sqrt{\frac{\S^2}{n}} \right ]\\\\$ Thus, the 95\% confidence interval for the population mean is: $\\\\\left [15.40 - 2.093\sqrt{\frac{(0.6152)^2}{19}}, \hspace{0.3cm}15.40 + 2.093\sqrt{\frac{(0.6152)^2}{19}} \right ]=\left [15.112, \hspace{0.3cm}15.688\right] \\\\$

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2 years ago