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Alex_Xolod [135]
3 years ago
15

A taxi ride cost $2 for the first 1/5 of a km and 40 cents for each additional fifth of a km. Dan has $9.60. What is the longest

taxi ride he can afford, in kilometres?
Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
5 0

Answer:

4 kilometers

Step-by-step explanation:

<em>First! Let's create an equation, with our variable (x) representing the number of additional fifth kilometers Dan can afford. </em>

2 + 0.40x = 9.60

<em>Now, we have to solve for our x value by subtracting 2 from both sides.</em>

0.40x = 7.60

<em>Next, we can divide both sides by 0.40 to find out what our x value is equal too.</em>

x = 19

<em>Don't forget this x value is just for the number of </em><em>additional</em><em> fifth kilometers Dan can afford on top of his first fifth mile.</em>

<em />

<em>So! Dan can afford 20 fifth kilometers, which in turn in is 4 kilometers in total. </em>

<em />

Hope this Helps! :)

<em>Have any questions? Ask below in the comments and I will try my best to answer.</em>

-SGO

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The probability that it rains at most 2 days is 0.00005995233 and the variance is 0.516

<h3>The probability that it rains at most 2 days</h3>

The given parameters are:

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The probability that it rains at most 2 days is represented as:

P(x ≤ 2) = P(0) + P(1) + P(2)

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P(x) = ^nC_x * p^x * (1 - p)^{n - x}

So, we have:

P(0) = ^7C_0 * (92\%)^0 * (1 - 92\%)^{7 - 0} = 0.00000002097

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P(2) = ^7C_2 * (92\%)^2 * (1 - 92\%)^{7 - 2} = 0.00005824315

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P(x ≤ 2) =0.00000002097 + 0.00000168821 + 0.00005824315

P(x ≤ 2) = 0.00005995233

Hence, the probability that it rains at most 2 days is 0.00005995233

<h3>The mean</h3>

This is calculated as:

Mean = np

So, we have:

Mean = 7 * 92%

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Hence, the mean is 6.44

<h3>The standard deviation</h3>

This is calculated as:

σ = √np(1 - p)

So, we have:

σ = √7 * 92%(1 - 92%)

Evaluate

σ = 0.718

Hence, the standard deviation is 0.718

<h3>The variance</h3>

We have:

σ = 0.718

Square both sides

σ² = 0.718²

Evaluate

σ² = 0.516

This represents the variance

Hence, the variance is 0.516

Read more about normal distribution at:

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