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AleksAgata [21]
2 years ago
13

There are 3 alternative routes by which you may drive to work: Alabaster Street, Brillantine Street, and Clancy Street. It is th

e beginning of rush hour, and by experience Alabaster street will be closed by a car crash on average in 25 minutes, Brillantine street in 50 minutes, and Clancy Street in 40 minutes. Accident times are completely unpredictable. If you leave for work in one hour (60 minutes), what is the probability that (at the moment you leave) a route to work will still be open?
Mathematics
1 answer:
Solnce55 [7]2 years ago
3 0

Answer:

25%

Step-by-step explanation:

sorry i did it in my head

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Verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to
nikklg [1K]

Answer:

Part 1)

See Below.

Part 2)

\displaystyle (-0.179, -0.178) \cup (-0.010, 0.012)

Step-by-step explanation:

Part 1)

The linear approximation <em>L</em> for a function <em>f</em> at the point <em>x</em> = <em>a</em> is given by:

\displaystyle L \approx f'(a)(x-a) + f(a)

We want to verify that the expression:

1-36x

Is the linear approximation for the function:

\displaystyle f(x) = \frac{1}{(1+9x)^4}

At <em>x</em> = 0.

So, find f'(x). We can use the chain rule:

\displaystyle f'(x) = -4(1+9x)^{-4-1}\cdot (9)

Simplify. Hence:

\displaystyle f'(x) = -\frac{36}{(1+9x)^{5}}

Then the slope of the linear approximation at <em>x</em> = 0 will be:

\displaystyle f'(1) = -\frac{36}{(1+9(0))^5} = -36

And the value of the function at <em>x</em> = 0 is:

\displaystyle f(0) = \frac{1}{(1+9(0))^4} = 1

Thus, the linear approximation will be:

\displaystyle L = (-36)(x-(0)) + 1 = 1 - 36x

Hence verified.

Part B)

We want to determine the values of <em>x</em> for which the linear approximation <em>L</em> is accurate to within 0.1.

In other words:

\displaystyle \left| f(x) - L(x) \right | \leq 0.1

By definition:

\displaystyle -0.1\leq f(x) - L(x) \leq 0.1

Therefore:

\displaystyle -0.1 \leq \left(\frac{1}{(1+9x)^4} \right) - (1-36x) \leq 0.1

We can solve this by using a graphing calculator. Please refer to the graph shown below.

We can see that the inequality is true (i.e. the graph is between <em>y</em> = 0.1 and <em>y</em> = -0.1) for <em>x</em> values between -0.179 and -0.178 as well as -0.010 and 0.012.

In interval notation:

\displaystyle (-0.179, -0.178) \cup (-0.010, 0.012)

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