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Suppose each runner ran at the rate given in the table above for 3.1 miles. How much time will elapse between the first place fi

Scilla [17]

You didn't include the table but I found this table for the same statement, so I will answer you based on the next table:

Runner     distance       time

Arabella        7,299 feet        561 seconds
Bettina          3,425 yards     13 minutes, 12 seconds
Chandra       8,214 feet        0,195 hours
Divya            1,62 miles        732 seconds

To answer the question you must find the rate for each runner and then calculate the time to run 3.1 miles at each rate.

First you need to convert the data to obtain the rate in miles per second.

These are the main conversion identities:

1 mile = 5280 feet

1 mile = 1760 yards

1 hour = 3600 seconds

1 hour = 60 minutes

1 minute = 60 seconds

Arabella:

rate: 7,229 feet / 561 seconds * (1 mile / 5280 feet) =

= 0.00244 mile/second

Time to run 3.1 miles: V = d / t => t = d / V = 3.1 miles / 0.00244 mile/second = 1270 seconds

Bettina:

13 minutes + 12 seconds = 13*60 seconds +12 seconds = 792 seconds

rate = 3425 yards / 792 seconds * 1 mile / 1760 yards = 0.00246 mile/seconds

Time to run 3.1 miles = 3.1 miles / 0.00246 mile/second = 1260 seconds

Chandra:

rate = 8214 feet / 0.195 hours * 1 mile / 5280 feet * 1hour / 3600 seconds =

= 0.00222 seconds

Time = 3.1 mile / 0.00222 seconds = 0.389 hour = 1396 seconds

Divya:

rate = 1.62 miles / 732 seconds = 0.00221 seconds

Time = 3.1 mile / 0.00221 seconds = 1403 seconds

Now you can find the difference between fhe last and the first 1403 seconds - 1260 seconds = 143 seconds

That is equivalent to 2.38 seconds.

6 0

3 years ago

a food company delivers its fruit in two types of boxes, large a small. A delivery of 7 large boxes and 9 small boxes has a tota

ivann1987 [24]

Let L and S represent the weights of large and small boxes, respectively. The problem statement gives rise to two equations:
.. 7L +9S = 273
.. 5L +3S = 141

You can solve these equations various ways. Using "elimination", we can multiply the second equation by 3 and subtract the first equation.
.. 3(5L +3S) -(7L +9S) = 3(141) -(273)
.. 8L = 150
.. L = 150/8 = 18.75
Then we can substitute into either equation to find S. Let's use the second one.
.. 5*18.75 +3S = 141
.. S = (141 -93.75)/3 = 15.75

A large box weighs 18.75 kg; a small box weighs 15.75 kg.

5 0

3 years ago

a kid wants to buy a book but it is 18$ the book is diconted to 15 % off but she still has to pay 6% taxes how much dose she pay

baherus [9]

Discounted at 15% off....means the kid actually paid 85%

0.85(18) = 15.30

and then there is the 6% sales tax...
15.30 * 1.06 = 16.218 rounds to $ 16.22 <== what she pays

4 0

3 years ago

Plz help.........................

Salsk061 [2.6K]

The second one the are both perfect squares

3 0

3 years ago

In a random sample of 25 people, the mean commute time to work was 33.9 minutes and the standard deviation was 7.2 minutes. Ass

Phoenix [80]

Answer:

$33.9-1.32\frac{7.2}{\sqrt{25}}=31.999$

$33.9+1.32\frac{7.2}{\sqrt{25}}=35.801$

So on this case the 80% confidence interval would be given by (31.999;35.801)

And the margin of error is given by:

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

And replacing we got:

$ME = 1.32\frac{7.2}{\sqrt{25}} =1.9008$

Step-by-step explanation:

Previous concepts

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".

$\bar X=33.9$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=7.2 represent the sample standard deviation

n=25 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)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=25-1=24$

Since the Confidence is 0.80 or 80%, the value of $\alpha=0.2$ and $\alpha/2 =0.1$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,24)".And we see that $t_{\alpha/2}=1.32$

Now we have everything in order to replace into formula (1):

$33.9-1.32\frac{7.2}{\sqrt{25}}=31.999$

$33.9+1.32\frac{7.2}{\sqrt{25}}=35.801$

So on this case the 80% confidence interval would be given by (31.999;35.801)

And the margin of error is given by:

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

And replacing we got:

$ME = 1.32\frac{7.2}{\sqrt{25}} =1.9008$

7 0

4 years ago