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olga nikolaevna [1]
3 years ago
5

Denise has 5 hours to spend training for an upcoming race. She completes her training by running full speed the distance of the

race and walking back the same distance to cool down. If she runs at a speed of 9mph and walks back at a speed of 3mph, how long should she plan to spend walking back?
Mathematics
1 answer:
Rainbow [258]3 years ago
6 0
<span>To solve this problem, we can use this formula d = rd (distance = rates x time) She runs at a speed of 9 mph and walks at a speed of 3 mph. Her distance running is d = 9tr where tr is the time she spends running Her distance walking is d = 3tw where tw is the time she spends walking The distances are the same so 9tr = 3tw We also know that the total time is 5 hours tr + tw = 5 tr = 5-tw Substitute this value of tr in the first equation 9tr = 3tw 9(5-tw) = 3tw 45-9tw = 3tw 45 = 12tw 3.75= tw Denise will spend 3.75 hours (3 hours, 45 minutes) walking back and 1.25 hours (1 hour, 15 minutes) running.</span>
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10 whole boards

Step-by-step explanation:

1. 36 divided by 3.75 = 9.6

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3. 9.6 rounded = 10

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A pet store employee measures the lengths of two pet snakes, Noodles and Squiggles. Noodles measures 30 inches long, and Squiggl
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4

Squiggles

Step-by-step explanation:

steps

  1. convert the length of squiggles from feet to inches
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Answer:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

Step-by-step explanation:

Notation

\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

For this case the 9% confidence interval is given by:

8.8104 \leq \mu \leq 11.1248

We can calculate the mean with the following:

\bar X = \frac{8.8104 +11.1248}{2}= 9.9676

And we can find the margin of error with:

ME= \frac{11.1248- 8.8104}{2}= 1.1572

The margin of error for this case is given by:

ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE

And we can solve for the standard error:

SE = \frac{ME}{t_{\alpha/2}}

The critical value for 95% confidence using the normal standard distribution is approximately 1.96 and replacing we got:

SE = \frac{1.1572}{1.96}= 0.5904

Now for the 98% confidence interval the significance is \alpha=1-0.98= 0.02 and \alpha/2 = 0.01 the critical value would be 2.326 and then the confidence interval would be:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

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3 years ago
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