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Fudgin [204]
2 years ago
10

a car uses 9.5 L petrol per 100km a) determine the rate in km/L b) how far can the car travel with 47,5L petrol c) how many litr

es of petrol do they need to travel 350 km​
Mathematics
1 answer:
Lana71 [14]2 years ago
6 0

Step-by-step explanation:

a) determine the rate in km/L

==> 9.5 L/100km = 10.526316 km/L

b) how far can the car travel with 475L petrol

==> 0.210526315789473km/l

c) how many litres of petrol do they need to travel 350 km

==>Liters per km (l/km)2.86×10-3

Liters per 10 km (l/10 km)0.03

Liters per 100 km (l/100km)0.29

Kilometer per liter (km/l)350

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While researching the cost of school lunches per week across the state, you use a sample size of 45 weekly lunch prices. The sta
Drupady [299]

We assume the lunch prices we observe are drawn from a normal distribution with true mean \mu and standard deviation 0.68 in dollars.


We average n=45 samples to get \bar{x}.


The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write


\sigma = 0.68 / \sqrt{45} = 0.101


Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains \mu.


Our interval takes the form of ( \bar{x} - z \sigma, \bar{x} + z \sigma ) as \bar{x} is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".


Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.


Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.


With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.


We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is


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in other words a margin of error of


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3 0
3 years ago
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Sarah needs to buy 3 identical shirts to wear. She had $15 gift certificate. Including using the gift certificate, she can spend
Marianna [84]
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