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zhannawk [14.2K]
3 years ago
6

Evan's family drove to a theme park for vacation. They drove the same speed throughout the trip. The first day, they drove 300 m

iles in 6 hours. The second day, they drove 250 miles in 5 hours. The third day, they arrived at the park after driving for 3 hours. How many miles did they drive on the third day?
Mathematics
2 answers:
shtirl [24]3 years ago
7 0

Answer:

150 miles

Step-by-step explanation:

The relationship between speed, time and distance is such that the product of speed and time is distance.

Given that they drove the same speed throughout the trip

Speed on day one given that distance covered is 300 miles in 6 hours,

Speed = 300 miles/ 6 hours

= 50 miles per hour

Speed on day two given that distance covered is 250 miles in 5 hours

= 250 miles/ 5 hours

= 50 miles per hour

If on the third day, the speed is maintained and they drove for 3 hours,

Distance covered = 50 miles per hour × 3 hours = 150 miles

Lelu [443]3 years ago
5 0

Answer:

150 miles

Step-by-step explanation:

Find the unit rate (MPH) by dividing miles travlled by hours.

    300/6 = 50 MPH

    250/5 = 50 MPH

Multiply the hours on day 3 (3) by 50 MPH

    3*50 = 150 miles

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Answer:

a) s^2 =30^2 =900

b) \frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}

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Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in  the variability in the number of rooms occupied per day during a particular season of the  year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per  day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

s^2 =30^2 =900

Part b

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The degrees of freedom are given by:

df=n-1=20-1=19

Since the Confidence is 0.90 or 90%, the significance \alpha=0.1 and \alpha/2 =0.05, the critical values for this case are:

\chi^2_{\alpha/2}=30.144

\chi^2_{1- \alpha/2}=10.117

And replacing into the formula for the interval we got:

\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}

567.28 \leq \sigma^2 \leq 1690.224

Part c

Now we just take square root on both sides of the interval and we got:

23.818 \leq \sigma \leq 41.112

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