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Ray Of Light [21]
3 years ago
5

A motorcycle can travel 70 miles per gallon. Approximately how many gallons of fuel will the motorcycle need to travel 60 km?

Mathematics
2 answers:
xxMikexx [17]3 years ago
6 0

Answer:

B

Step-by-step explanation:

70 miles * [1.6 km/mile] = 112 km

Notice when you multiply by km/mile by miles the miles cancel out.

1 gallon will get 112 km

x gallon will get 60 km

  • 1/x = 112/60                     cross multiply
  • 1 * 60 = 112 * x                 divide by 112
  • 60/112 = x                        switch
  • x = 60/112                        Do the division
  • x = 0.54                           Which rounds to 0.5
kap26 [50]3 years ago
4 0

Answer:

We have the following given

mileage = 70 miles per gallon

distance = 60 km


Required: amount of fuel in gallons


First, we must convert the distance to miles, so

60 km (1 mi/1.6 km) = 37.5 mi


So, the amount of fuel is

37.5 mi / (70 mi/gal) = 0.54 gallon


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How to solve 4n-40=7(-2n+2)
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A sample size 25 is picked up at random from a population which is normally
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Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 100 and variance of 36.

This means that \mu = 100, \sigma = \sqrt{36} = 6

Sample of 25:

This means that n = 25, s = \frac{6}{\sqrt{25}} = 1.2

(a) P(X<99)

This is the pvalue of Z when X = 99. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{99 - 100}{1.2}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

Z = \frac{X - \mu}{s}

Z = \frac{100 - 100}{1.2}

Z = 0

Z = 0 has a pvalue of 0.5

X = 98

Z = \frac{X - \mu}{s}

Z = \frac{98 - 100}{1.2}

Z = -1.67

Z = -1.67 has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

So

P(98 < X < 100) = 0.4525

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