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2 years ago

47. A car travels at a constant speed of 50 miles per hour.

Mathematics

1 answer:

SIZIF [17.4K]2 years ago

3 0

Answer:

The inverse of the given function is $t(d)=\frac{d}{50}$ . The time taken to travel 180 miles is 3.6 hours.

Step-by-step explanation:

It is given that a car travels at a constant speed of 50 miles per hour. A function for distance traveled is also given as d(t)=50t.

It is required to find the inverse of the function by expressing the time travel in terms of distance traveled and also find t(180).

To find the inverse, express time traveled in terms of distance traveled using normal algebraic methods and then substitute 180 for t and explain its results.

Step 1 of 2

The given function is d(t)=50t.

Consider d(t) as d and t as t(d) and rewrite the equation.

The rewritten function is d=50 t(d)

Divide by 50 on both sides of the equation.

$\begin{aligned}&d=50 t(d) \\&\frac{d}{50}=\frac{50 t(d)}{50} \\&t(d)=\frac{d}{50}\end{aligned}$

Step 2 of 2

Substitute 180 for $t$ in the simplified expression and interpret the results.

$\begin{aligned}&t(d)=\frac{d}{50} \\&t(d)=\frac{180}{50} \\&t(d)=3.6\end{aligned}$

The time taken to travel 180 miles is 3.6 hours.

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From the question, the given parameters are:

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Evaluate the like terms

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$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

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# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

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