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irina1246 [14]
3 years ago
15

Femi drove 290 miles from his college to home and used 23.2 gallons of gasoline. His sister, Kehinde, drove 225 miles from her c

ollege to home and used 15 gallons of gasoline. Whose vehicle had better gas mileage? [5]
Mathematics
2 answers:
alina1380 [7]3 years ago
4 0

Find the unit cost (that is, the number of miles per gallon of gas) in each case.

Case 1:

290 mi

------------- = 12.5 mpg

23.2 gal


Case 2:

225 mi

------------ = 15 mpg

15 gal


Kehinde's car got the better mileage:  15 mpg versus 12.5 for Femi's car.

Nata [24]3 years ago
3 0

to find miles per gallon we simply do D/G

D = distance

G = gallons

290/23.2 = 12.5 mpg

225/15 = 15 mpg

Kehinde has better gas mileage.

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There are 100 students in a hostel. Food provision for them is 20 days.How long will these provision last if 25 more students jo
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Currently 100 students can eat for 20 days, so we need to find the number of days 125 students can eat.

Originally:
100 students eat for 20 days

New:
125 students eat for x days. Multiply original number of students by original number of days which will give the total amount of students per day. Then divide the total amount of students per day by the new number of students to find the new number of days.

x= (100 students * 20 days)/ (100+25)
x= (100 * 20)/ 125
x= 2000/125
x= 16 days

ANSWER: 125 students can eat for 16 days.

Hope this helps! :)
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What is the exact distance between points A and B?
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It would be the square root of 85 which is 9.2
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Determine which equation below can be solved to find the value of Angle<br> A. *
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Step-by-step explanation:

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Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a
Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

P(X

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The cumulative distribution for this function is given by:

F(X) = 1- e^{-\lambda x}, x\ geq 0

We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

8 0
3 years ago
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