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luda_lava [24]
3 years ago
12

In 8 months, 55.6 gallons were used. Find the unit rate.

Mathematics
2 answers:
OLga [1]3 years ago
8 0
6.95 gallons in 1 month
aliya0001 [1]3 years ago
8 0

Answer:

6.95 gallons/month

Step-by-step explanation:

We are given that In 8 months, 55.6 gallons were used

Amount used in 8 months = 55.6 gallons

We are supposed to find amount used per month i.e. unit rate

So, Amount used in 8 months = 55.6 gallons

Amount used in 1 month = \frac{55.6}{8} gallons

Amount used in 1 month = 6.95 gallons

So, 6.95 gallons were used in 1 month

Hence the unit rate is 6.95 gallons/month

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The mean finish time for a yearly amateur auto race was 186.32 minutes with a standard deviation of 0.305 minute. The winning​ c
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Complete Question

The mean finish time for a yearly amateur auto race was 186.94 minutes with a standard deviation of 0.305 minute. The winning car, driven by Roger, finished in 185.85 minutes. The previous year's race had a mean finishing time of 110.7 with a standard deviation of 0.115 minute. The winning car that year, driven by Karen, finished in 110.48 minutes.

Find their respective z-scores.

 Who had the more convincing victory?

A. Roger​ had a more convincing victory because of a higher z-score.

B. Karen a more convincing victory because of a higher z-score.

C. Roger had a more convincing victory, because of a lower z-score.

D. Karen a more convincing victory because of a lower z-score.

Answer:

The correct option is  D

Step-by-step explanation:

From the question we are told that

  The mean of the current year  is  \mu_c =  186.32 \  minutes

   The standard deviation is  \sigma_c  =  0.305 \  minutes

   The time taken by  the winning car for the current year  is  x =  185.85 \ minutes

  The mean of the previous  year  is  \mu_p =  110.7 \  minutes

   The standard deviation the previous  year  is  \sigma_p  = 0.115  \  minutes

 The time taken by  the winning car for the previous year  is  y =   110.48 \ minutes    

Generally the z-score for current year is mathematically evaluated as

     z_c  =  \frac{x- \mu_c}{\sigma_c }

=>     z_c  =  \frac{ 185.85- 186.32}{0.305 }

=> z_c  = -1.536

Generally the z-score for previous year is mathematically evaluated as

     z_p  =  \frac{y- \mu_p}{\sigma_p }

=>     z_p  =  \frac{ 110.48-110.7}{0.115  }

=> z_p  = -1.913

From the value obtained we see that  z_p  <  z_c , Hence  Karen had a more convincing victory because of the lower z -score.

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