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-Dominant- [34]
2 years ago
10

margaret drove to a business appointment at 70 mph. her average speed on the return trip was 60 mph. the return trip took 1 6 hr

longer because of heavy traffic. how far did she travel to the​ appointment?
Mathematics
1 answer:
Anon25 [30]2 years ago
4 0

The distance travelled to the appointment is 6720 miles

<h3>How to find the distance she travelled for the appointment?</h3>

Since margaret drove to a business appointment at 70 mph.

So, the distance on the trip,d = vt where

  • v = speed to business appointment = 70 mph and
  • t = time to business appointment

So, d = 70t

Since her average speed on the return trip was 60 mph. the return trip took 16 hr longer because of heavy traffic.

The distance on the return trip, D = v't' where

  • v = speed on return trip = 60 mph and
  • t = time on return trip = t + 16

So, D = 60(t + 16)

Since both distances are equal, we have that

d = D

70t = 60(t + 16)

70t = 60t + 960

70t - 60t = 960

10t = 960

t = 960/10

t = 96 h

Since the distance to the appointment d = 70t

Substituting t = 96 into the equation, we have

d = 70t

d = 70 × 96

d = 6720 miles

So, the distance to the appointment is 6720 miles

Learn more about distance here:

brainly.com/question/24295081

#SPJ1

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a) The 99% confidence interval would be given by (589.588;731.038)

b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.  

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".

The data is:

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In order to calculate the mean and the sample deviation we need to have on mind the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n}  

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=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

On this case the average is \bar X= 660.313

=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

The sample standard deviation obtained was s=95.898

Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =16<30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. The degrees of freedom are given by:

df=n-1=16-1=15

We can find the critical values in excel using the following formulas:

"=T.INV(0.005,15)" for t_{\alpha/2}=-2.95

"=T.INV(1-0.005,15)" for t_{1-\alpha/2}=2.95

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[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]  

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