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Fudgin [204]
3 years ago
5

A man walking on a railroad bridge is 2/5 of the way along the bridge when he notices a train at a distance approaching at the c

onstant rate of 45 mph . The man can run at a constant rate in either direction to get off the bridge just in time before the train hits him. How fast can the man run?
Mathematics
2 answers:
shusha [124]3 years ago
5 0

Answer: 9 mph

Step-by-step explanation:

Given that a man walking on a railroad bridge is 2/5 of the way along the bridge when he notices a train at a distance approaching at the constant rate of 45 mph .

If the man tend to run in the forward direction, he will cover another 2/5 before the train reaches his initial position. The distance covered by the man will be 2/5 + 2/5 = 4/5

The remaining distance = 1 - 4/5 = 1/5

If the man can run at a constant rate in either direction to get off the bridge just in time before the train hits him, the time it will take the man will be

Speed = distance/time

Time = 1/5d ÷ speed

The time it will take the train to cover the entire distance d will be

Time = d ÷ 45

Equate the two time

1/5d ÷ speed = d ÷ 45

Speed = d/5 × 45/d

Speed = 9 mph

makkiz [27]3 years ago
3 0

Answer:

The Man needs to run at 9 mph

Step-by-step explanation:

Let M stand for the man's speed in mph.  When the man  

runs toward point A, the relative speed of the train with respect  

to the man is the train's speed plus the man's speed (45 + M).  

When he runs toward point B, the relative speed of the train is the  

train's speed minus the man's speed (45 - M).

When he runs toward the train the distance he covers is 2 units.  

When he runs in the direction of the train the distance he covers  

is 3 units. We can now write that the ratio of the relative speed  

of the train when he is running toward point A to the relative speed  

of the train when he is running toward point B, is equal to the  

inverse ratio of the two distance units or

              (45 + M)          3

              -----------  =      ---

              (45 - M)          2

          90+2 M=135-3 M

⇒5 M = 45

⇒ M = 9 mph

The Man needs to run at 9 mph

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Lady_Fox [76]

Answer:

4:3

Step-by-step explanation:

He has 4 Soccer Medals, 3 Swimming Medals

So it would be 4:3

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3 years ago
Can you solve this??​
gavmur [86]

Answer:

The answer is c

Step-by-step explanation:

The middle problem is adding, the others are taking away, so C is the correct answer

3 0
2 years ago
Which unit is not a customary unit? <br> A. gallon<br> B. ounce<br> C. gram<br> D. cup
SOVA2 [1]

Answer:

D.Cup

Step-by-step explanation:

We humans do not use the term Cup anymore. We now use things like Grams,Ounces,Gallons,Or Quarts.

3 0
3 years ago
72
Rzqust [24]

Answer:

36

Step-by-step explanation:

The way to read this question would go: What is 72% of 50 because that will get you the number of correct solutions. When reading a statement like that, the word "of" means to multiply. However, before we multiply those two values we have to convert the value "72" into percent form. To do that you just divide it by 100 to get 0.72. Then following the statement, you would do 0.72*50 to get your final value of 36 correct questions

6 0
3 years ago
Santa Claus is assigning elves to work an eight-hour shift making toy trucks. Apprentice earn five candy canes per hour but can
dimaraw [331]

Answer:

  (a) 5 senior, 4 apprentice

  (b) 368 per shift

  (c) 7.5 senior, 0 apprentice

Step-by-step explanation:

The problem can be described by two inequalities. On describes the limit on the number of elves in the shop; the other describes the limit on the total payroll. Let x and y represent the number of apprentice and senior elves, respectively. Then the inequalities for the first scenario are ...

  x + y ≤ 9 . . . . . . . . . . . . total number of elves in the shop

  5x +8y ≤ 480/8 . . . . . . candy canes per hour paid to elves

These two inequalities are graphed in the first attachment. They describe a solution space with vertices at ...

  (x, y) = (0, 7.5), (4, 5), (9, 0)

__

(a) Santa wants to  maximize the output of trucks, so wants to maximize the function t = 4x +6y.

At the vertices of the solution space, the values of this function are ...

  t(0, 7.5) = 45

  t(4, 5) = 46

  t(9, 0) = 36

Output of trucks is maximized by a workforce of 4 apprentice elves and 5 senior elves.

__

(b) The above calculations show 46 trucks per hour can be made, so ...

  46×8 = 368 . . . trucks in an 8-hour shift

__

(c) The new demands change the inequalities to ...

  x + y ≤ 8 . . . . . . number of workers

  7x +8y ≤ 60 . . . total wages (per hour)

The vertices of the feasible region for these condtions are ...

  (x, y) = (0, 7.5), (4, 4), (8, 0)

From above, we know the truck output will be maximized at the vertex (x, y) = (0, 7.5). However, we know we cannot have 7.5 senior elves working in the shop. We can have 7 or 8 elves working.

If the workforce must remain constant, truck output is maximized by a workforce of 7 senior elves.

If the workforce can vary through the shift, truck output is maximized by adding one more senior elf in the shop for half a shift.

Santa should assign 7 senior elves for the entire shift, and 8 senior elves (one more) for half a shift.

_____

<em>Comment on apprentice elf wages</em>

At 5 candy canes for 4 trucks, apprentice elves produced trucks for a cost of 1.25 candy canes per truck. At 8 candy canes for 6 trucks, senior elves produced trucks for a cost of about 1.33 candy canes per truck. The reason for employing senior elves in the first scenario is that their productivity is 1.5 times that of apprentice elves while their cost per truck is about 1.07 times that of apprentice elves.

After the apprentice elves wages were increased, their cost per truck is 1.75 candy canes per truck, but their productivity hasn't changed. They have essentially priced themselves out of a job, because they are not competitive with senior elves.

5 0
3 years ago
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