An elephant can run one over four mile in 36 seconds. Which of the following correctly shows this rate as miles per hour?

Mathematics

2 answers:

Vesnalui [34]1 year ago

7 0

It would be thirty six seconds over one-fourth mile= 144 miles per hour

ankoles [38]1 year ago

6 0

Answer: 1/4 mile over 36 seconds = 25 miles

Step-by-step explanation:

60 minutes x 60 seconds = 3600 ÷ 36 = 100 x 1/4 = 100 x .25= 25 miles

(1/4 mile = .25 because, .25 mile x 4 = 1 mile

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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: $\\\\\left [15.40 - 2.093\sqrt{\frac{(0.6152)^2}{19}}, \hspace{0.3cm}15.40 + 2.093\sqrt{\frac{(0.6152)^2}{19}} \right ]=\left [15.112, \hspace{0.3cm}15.688\right] \\\\$