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statuscvo [17]
3 years ago
9

Two automobiles left simultaneously from cities A and B heading towards each other and met in 5 hours. The speed of the automobi

le that left city A was 10 km/hour less than the speed of the other automobile. If the first automobile had left city A 4 1/2 hours earlier than the other automobile left city B, then the two would have met 150 km away from B. Find the distance between A and B.
Mathematics
2 answers:
Klio2033 [76]3 years ago
6 0

Answer:

450km

Step-by-step explanation:

Take it that each automobile travels at 30 km an hour, for 150 km, meaning it will be 450 km apart.

quester [9]3 years ago
5 0

Answer:

  450 km

Step-by-step explanation:

<u>Equations</u>

We can define 3 variables: a, b, d. Let "a" and "b" represent the speeds of the cars leaving cities A and B, respectively. Let "d" represent the distance between the two cities. We can write three equations in these three variables:

1. The relation between "a" and "b":

  a = b -10 . . . . . . . the speed of car A is 10 kph less than that of car B

2. The relation between speed and distance when the cars leave at the same time:

  d = (a +b)·5 . . . . . . distance = speed × time

3. Note that the time it takes car B to travel 150 km to the meeting point is (150/b). (time = distance/speed) The total distance covered is ...

  distance covered by car A in 4 1/2 hours + distance covered by both cars (after car B leaves) = total distance

  4.5a + (150/b)(a +b) = d

__

<u>Solution</u>

Substituting for d, we have ...

  4.5a + 150/b(a +b) = 5(a +b)

  4.5ab +150a +150b = 5ab +5b^2 . . . . . . multiply by b, eliminate parentheses

  5b^2 +0.5ab -150(a +b) = 0 . . . . . . . . . . subtract the left side

Now, we can substitute for "a" and solve for b.

  5b^2 + 0.5b(b-10) -150(b -10 +b) = 0

  5.5b^2 -5b -300b +1500 = 0 . . . . . . . . eliminate parentheses

  11b^2 -610b +3000 = 0 . . . . . . . . . . . . . multiply by 2

  (11b -60)(b -50) = 0 . . . . . . . . . . . . . . . . factor

The solutions to this equation are ...

  b = 60/11 = 5 5/11 . . . and . . . b = 50

Since b must be greater than 10, the first solution is extraneous, and the values of the variables are ...

  • b = 50
  • a = b-10 = 40
  • d = 5(a+b) = 5(90) = 450

The distance between A and B is 450 km.

_____

<u>Check</u>

<em>When the cars leave at the same time</em>, their speed of closure is the sum of their speeds. They will cover 450 km in ...

  (450 km)/(40 km/h +50 km/h) = 450/90 h = 5 h

__

<em>When car A leaves 4 1/2 hours early</em>, it covers a distance of ...

  (4.5 h)(40 km/h) = 180 km

before car B leaves. The distance remaining to be covered is ...

  450 km - 180 km = 270 km

When car B leaves, the two cars are closing at (40 +50) km/h = 90 km/h, so will cover that 270 km in ...

  (270 km)/(90 km/h) = 3 h

In that time, car B has traveled (3 h)(50 km/h) = 150 km away from city B, as required.

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1. No answer ( may be due to incorrect question)

2. x = -3

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3 years ago
A radio manufacturer believes that the length of life of WPPX model radio is normal with µ=12 years and σ=2.5 years. (a) What pe
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Answer:

(a) 5.48%

(b) 17.82 years

(c) 21.19%

(d) 15.21 years

Step-by-step explanation:

z = (x-μ)/σ,

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

µ=12 years and σ=2.5 years.

(a) What percent of the radios will function for more than 16 years?

For x = 16

z = 16 - 12/2.5

z = 1.6

Probability value from Z-Table:

P(x<16) = 0.9452

P(x>16) = 1 - P(x<16)

1 - 0.9452

= 0.054799

Converting to percentage = 0.054799 × 100

= 5.4799%

Approximately = 5.48%

(b) Suppose the company decides to replace 1% of the radios. Find the length of the guarantee period. i.e. find X.

find the z score of the 99th percentile = 2.326

Hence:z = (x-μ)/σ

2.326 = x - 12/2.5

Cross Multiply

2.326 × 2.5 = x - 12

5.815 = x - 12

x = 12 + 5.815

x = 17.815

Approximately = 17.82 years

(c) What percent of the radios that will fail to satisfy the guarantee period of 10 years? , i.e. less than 10 years?

When x < 10

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Probability value from Z-Table:

P(x<10) = 0.21186

Converting to percentage

0.21186 × 100

= 21.186%

Approximately = 21.19%

(d) If 10% of the radios will function for more than X years, find X.

The z score would be : 100 - 10%

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We find the z score of the 90th percentile = 1.282

Hence:z = (x-μ)/σ

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Cross Multiply

1.282 × 2.5 = x - 12

3.205 = x - 12

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