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2 years ago

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A salesman made a trip of 420 miles by bus and train. He traveled 3 hours by bus and 5 hours by train. If the train averaged 12

mph more than the bus, find the rate of each.

1 answer:

asambeis [7]2 years ago

3 0

Answer:

The speed at which the buss made the trip was 45 mph and the train was 57 mph.

Step-by-step explanation:

The average speed is given by:

speed = distance / time

Therefore we can manipulate it to give us the distance:

distance = speed*time

The time distance of each stage of his trip summed must be equal to the total distance of the trip. Since he made a trip in two legs, one by bus that lasted 3 h at a speed of "x" and one that lasted 5 hours at a speed of "x + 12". We have:

420 = 3*x + 5*(x + 12)

3*x + 5*(x + 12) = 420

3*x + 5*x + 60 = 420

8*x = 420 - 60

8*x = 360

x = 45 mph

The speed at which the buss made the trip was 45 mph and the train was 57 mph.

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A project’s critical path is composed of activities A (variance .33), B (variance .67), C (variance .33), and D (variance .17).

julia-pushkina [17]

Answer:

The standard deviation on the critical path = 0.834

Step-by-step explanation:

Variance of activity A ( $V_{1}$ ) = 0.33

Variance of activity B ( $V_{2}$ ) = 0.67

Variance of activity C ( $V_{3}$ ) = 0.33

Variance of activity D ( $V_{4}$ ) = 0.17

Thus the standard deviation on the critical path ( $\alpha ^{2}$ ) = $V_{1} ^{2}$ + $V^2_{2}$ + $V^2_{3}$ + $V^2_{4}$

$\alpha ^{2}$ = $0.33^{2}$ + $0.67^{2}$ + $0.33^{2}$ + $0.17^{2}$

$\alpha ^{2}$ = 0.6956

$\alpha$ = 0.834

The standard deviation on the critical path = 0.834

4 0

2 years ago

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

2 years ago

Barbara drives between Miami, Florida, and West Palm Beach, Florida. She drives 50 mi in clear weather and then encounters a thu

NeX [460]

Distance traveled in clear weather = 50 miles

Distance traveled in thunderstorm = 15 miles

Let speed in clear weather = x

⇒ Speed in thunderstorm = x-20

Total time taken for trip = 1.5 hours

We need to determine average speed in clear weather (i.e. x) and average speed in the thunderstorm (i.e. x-20 ).

Total time taken for trip = Time taken in clear weather + Time taken in thunderstorm

⇒ Total time taken for trip = $\frac{Distance covered in clear weather}{Speed in clear weather}$ + $\frac{Distance covered in thunderstorm}{Speed in thunderstorm}$

⇒ 1.5 = $\frac{50}{x}$ + $\frac{15}{x-20}$

⇒ 1.5 = $\frac{50(x-20)+15(x)}{(x)(x-20)}$

⇒ 15*x*(x-20) = 10*[50*(x-20)+15*x]

⇒ 15x² - 300x = 500x - 10,000 + 150x

⇒ 15x² - 300x = 650x - 10,000

⇒ 15x² - 950x + 10,000 = 0

⇒ 3x² - 190x + 2,000 = 0

The above equation is in the format of ax² + bx + c = 0

To determine the roots of the equation, we will first determine 'D'

D = b² - 4ac

⇒ D = (-190)² - 4*3*2,000

⇒ D = 36,100 - 24,000

⇒ D = 12,100

Now using the D to determine the two roots of the equation

Roots are: x₁ = $\frac{-b+\sqrt{D}}{2a}$ ; x₂ = $\frac{-b-\sqrt{D}}{2a}$

⇒ x₁ = $\frac{-(-190)+\sqrt{12,100}}{2*3}$ and x₂ = $\frac{-(-190)-\sqrt{12,100}}{2*3}$

⇒ x₁ = $\frac{190+110}{6}$ and x₂ = $\frac{190-110}{6}$

⇒ x₁ = $\frac{300}{6}$ and x₂ = $\frac{80}{6}$

⇒ x₁ = 50 and x₂ = 13.33

So speed in clear weather can be 50 mph or 13.33 mph. However, we know that in thunderstorm was 20 mph less than speed in clear weather.

If speed in clear weather is 13.33 mph then speed in thunderstorm would be negative, which is not possible since speed can't be negative.

Hence, the speed in clear weather would be 50 mph, and in thunderstorm would be 20 mph less, i.e. 30 mph.

7 0

2 years ago

Tia’s tent is in the form of a triangular prism as shown below if Tia plans to waterproof the tent excluding the pace what is th

Marat540 [252]

just multiple it all together

7 0

2 years ago

A surveyor, Toby, measures the distance between two landmarks and the point where he stands. He also measured the angles between

Tju [1.3M]

The Set Up:

x² = (Side1)² + (Side2)² - 2[(Side1)(Side2)]

Solution:

cos(Toby's Angle) • x² = 55² + 65² - 2[(55)(65)] cos(110°)

x² = 3025 + 4225 -7150[cos(110°)]

x² = 7250 - 2445.44x =

√4804.56x = 69.31m

The distance, x, between two landmarks is 69.31m.

Note: The answer choices given are incorrect.

3 0

2 years ago

Read 2 more answers