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Advocard [28]
2 years ago
9

While driving home for the holidays, you can’t seem to get Little’s Law out of your mind. You note that your average speed of tr

avel is about 60 miles per hour. Moreover, the traffic report from the WXPN traffic chopper states that there is an average of 24 cars going in your direction on a one-quarter mile part of the highway. What is the flow rate of the highway (going in your direction) in cars per hour?
Mathematics
1 answer:
suter [353]2 years ago
7 0

Answer:

1,200 cars per hour

Step-by-step explanation:

Let's suppose the car flow is constant and based on the observations stated in the problem.

If we set and imaginary car counter where your car is, then there are 25 cars in ¼ of a mile (including yours) going at your same speed.

Now compute how long it will take to travel ¼ mile. As your speed is constant 60 mph, we can cross multiply  

60 miles -----------> 1 hour

¼ of a mile --------> x hours

and

60/ (¼) = 1/x ---------> x = 1/240 hours.

This means that in 1/240 hours the 25 cars will be passing the counter.

Now compute how many cars will pass through our counter in 1 hour. Again, cross multiplying

1/240 hours -------> 25 cars

1 hour --------------> x cars

and

1/240 = 25/x -------> x = 25*240 = 1,200 cars.

So, the flow rate of the highway (going in your direction) in cars per hour is 1,200 cars per hour.

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A metal alloy is a metal made by blending 2 or more types of metal. A jeweler has supplies of two metal alloys. On alloy is 30%
tensa zangetsu [6.8K]

Answer:

He should combine 1 kg of 30% alloy and 3 kg of 10% alloy.

Step-by-step explanation:

Let x = mass of 30% ally.

Let y = mass of 10% alloy.

He needs 4 kg of alloy, so the first equation, dealing with masses, is

x + y = 4

Now we deal with the amount of gold.

He uses x mass of 30% gold alloy. The amount of gold in that mass is 0.3x.

He uses y mass of 10% gold alloy. The amount of gold in that mass is 0.1y.

The total mass of gold is 0.3x + 0.1y. We are told the told the end alloy is 4 kg of 15% gold alloy. That contains 0.15(4) kg = 0.6 kg of gold.

The second equation is

0.3x + 0.1y = 0.6

Now we put together the two equations as a system of equations.

x + y = 4

0.3x + 0.1y = 0.6

Multiply both sides of the second equation by -10.

-3x - y = -6

Add this equation to the first original equation.

         x + y = 4

+     -3x - y = -6

---------------------

        -2x    = -2

            x    = 1

Now substitute x = 1 in the first original equation and solve for y.

x + y = 4

x + y = 4

y = 3

Answer: He should combine 1 kg of 30% alloy and 3 kg of 10% alloy.

6 0
2 years ago
Asha runs 5 meters every 2 seconds. How far does asha run after 3 seconds? After 5 seconds?
Brilliant_brown [7]

Answer:

2 1/2

Step-by-step explanation:

7 0
1 year ago
Read 2 more answers
I've been stuck on this for so long and I have an exam soon, anybody who can help me :'( ?
san4es73 [151]
(i)  speed = distance / time
so time =  distance / speed
here we have

time t = 1080/x  hours

(ii) return flight  time  = 1080 / (x + 30)  hours

(a)  1080/x - 1080/(x + 30) = 1/2

Multiplying  through by the LCD 2x(x + 30) we get:-

1080*2(x + 30) - 2x*1080 = x(x+30)
2160x + 64800 - 2160x = x^2 + 30x
x^2 + 30x - 64800  = 0

(b)  factoring;  -64800 = 270 * -240  ans 270-240 = 30 so we have

(x + 270)(x - 240) = 0   so x = 240  ( we ignore the negative -270)

So the speed for outward journey is 240 km/hr

(c) time ffor outward flight = 1080 / 240 =  4 1/2  hours

(d) average speed for whole flight = distance / time
   Time for outward journey = 4.5 hours and time for  return journey = d / v
= 1080 / (240+30) =  4 hours
 Therefore the average speed for whole journey =  2160 / 8.5 = 254.1 km/hr
8 0
2 years ago
A car was purchased for $25,000. Research shows that the car has an average yearly depreciation rate of 18.5%. Create a function
vitfil [10]

Answer:

V(t) = 25000 * (0.815)^t

The depreciation from year 3 to year 4 was $2503.71

Step-by-step explanation:

We can model V(t) as an exponencial function:

V(t) = Vo * (1+r)^t

Where Vo is the inicial value of the car, r is the depreciation rate and t is the amount of years.

We have that Vo = 25000, r = -18.5% = -0.185, so:

V(t) = 25000 * (1-0.185)^t

V(t) = 25000 * (0.815)^t

In year 3, we have:

V(3) = 25000 * (0.815)^3 = 13533.58

In year 4, we have:

V(4) = 25000 * (0.815)^4 = 11029.87

The depreciation from year 3 to year 4 was:

V(3) - V(4) = 13533.58 - 11029.87 = $2503.71

7 0
2 years ago
Angelo's Pizza is having a ticket raffle to raise money for new soccer uniforms. A $1 ticket gives an even chance to win $25 gif
Bingel [31]

Answer:

1-25-15-5=44

Step-by-step explanation:

7 0
2 years ago
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