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zvonat [6]
2 years ago
6

A train is leaving in 11 minutes and you are one mile from the station. Assuming you can walk at 4mph and run at 8mph, how much

time can you afford to walk before you must run in order to catch the train?
Mathematics
1 answer:
ddd [48]2 years ago
6 0

7 minutes is the time can you afford to walk

<em><u>Solution:</u></em>

Given that, train is leaving in 11 minutes and you are one mile from the station

Let "x" represent how much time can you afford to walk

Then, (11 - x) is the time you run

You can walk at 4mph and run at 8mph

<em><u>Convert to miles per minute</u></em>

1 hour = 60 minutes

4\ mph = 4 \times \frac{1}{60} \text{ miles per minute } = \frac{1}{15} \text{ miles per minute }\\\\8\ mph = 8 \times \frac{1}{60}\ \text{ miles per minute } = \frac{2}{15} \text{ miles per minute }

We know that,

Distance = speed \times time

Given that, you are one mile from the station

Therefore,

1 mile = distance walk + distance run

1 = x \times \frac{1}{15} + (11-x) \times \frac{2}{15}\\\\1 = \frac{1}{15}(x + 2(11-x))\\\\1 = \frac{1}{15}(x + 22 - 2x)\\\\x + 22 - 2x = 15\\\\x = 22 - 15\\\\x = 7

Thus, 7 minutes is the time can you afford to walk

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Vadim26 [7]

Answer:

<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>

Step-by-step explanation:

Given the function  j(x) = 11.6e^x and k(x) = ln \dfrac{x}{11.6}, to show that both equality functions are true, all we need to show is that both  j(k(x)) and k(j(x)) equal x,

For j(k(x));

j(k(x)) = j[(ln x/11.6)]

j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}

j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)

j[(ln x/11.6)] = 11.6 * x/11.6

j[(ln x/11.6)] = x

Hence j[k(x)] = x

Similarly for k[j(x)];

k[j(x)] = k[11.6e^x]

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k[11.6e^x]  = ln(e^x)

exponential function will cancel out the natural logarithm leaving x

k[11.6e^x]  = x

Hence k[j(x)] = x

From the calculations above, it can be seen that j[k(x)] =  k[j(x)]  = x, this shows that the functions j(x) = 11.6e^x and k(x) = ln \dfrac{x}{11.6} are inverse functions.

4 0
2 years ago
In a sample of 100 steel canisters, the mean wall thickness was 8.1 mm with a standard deviation of 0.5 mm. Someone says that th
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Answer:

This statement can be made with a level of confidence of 97.72%.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.1 mm

Standard Deviation, σ = 0.5 mm

Sample size, n = 100

We are given that the distribution of thickness is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling:

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P(mean thickness is less than 8.2 mm)

P(x < 8.2)

P( x < 8.2)\\\\ = P( z < \displaystyle\frac{8.2 - 8.1}{0.05})\\\\ = P(z < 2)

Calculation the value from standard normal z table, we have,  

P(x < 8.2) =0.9772 = 97.72\%

This statement can be made with a level of confidence of 97.72%.

8 0
2 years ago
The price per night for a hotel room is different at three different hotels.
Damm [24]
I think you meant it to be not repeating 3 times so
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What is the value of k such that x-5 is a factor of x3 – x2 + kx - 30?
Paraphin [41]

Answer:

k = - 14

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given that (x - 5) is a factor of the polynomial then x = 5 is a root and

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5³ - 5² + 5k - 30 = 0

125 - 25 + 5k - 30 = 0

70 + 5k = 0 ( subtract 70 from both sides )

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