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shutvik [7]
3 years ago
11

Lin and Andre start walking toward each other at the same time from opposite ends of the 22-mile walking trail. Lin walks at a s

peed of 2.5 miles per hour. Andre walks at a speed of 3 miles per hour
Mathematics
2 answers:
goldfiish [28.3K]3 years ago
8 0

Answer:

Step-by-step explanation:

let they meet after t hours

2.5t+3t=22

5.5 t=22

55/10 t=22

t=22×10/55=4 hours

after 4 hours they meet.

Genrish500 [490]3 years ago
3 0

Answer:

they would meet after 4 hour since the start of their journey

Step-by-step explanation:

Supposing we have to calculate in what time will they cross each other

Given:

distance between the opposite ends = 22 miles

speed of Lin= 2.5 miles/hr

speed of Andre= 3 miles/hr

they both walk towards each other at the same time

Now the relative speed of approach = 3+2.5= 5.5 miles/hr

time of meet = distance/ speed of approach

= 22/5.5= 4 hours

so, they would meet after 4 hour since the start of their journey

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Find ten rational numbers between -2/5 and 1/2
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Answer:

-1/3, -1/4, -3/10, -1/8, -1/16, 1/3, 1/4, 3/10, 1/8, 2/5.

Step-by-step explanation:

-2/5 is -0.4 and 1/2 is 0.5

Ten rational numbers in between are, -0.33333 (-1/3), -0.25 (-1/4), -0.3 (-3/10), -0.125 (-1/8), -0.0625 (-1/16), 0.33333 (1/3), 0.25 (1/4), 0.3 (3/10), 0.125 (1/8), 0.4 (2/5)

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A system consists of two components C1 and C2, each of which must be operative in order for the overall system to function. Let
hram777 [196]

Answer:

The reliability of the first system to work is 0.72 whereas the reliability of the second system to work is 0.98.As the reliability of the second system is more than the first one so the second system is more reliable.

Step-by-step explanation:

For first system as given in the attached diagram gives,

P(W_1W_2)=P(W_1) \times P(W_2)

As the systems are independent.

The given data indicates that

  • P(W_1) is given as 0.9
  • P(W_2) is given as 0.8

Now the probability of the system is given as

P(W_1W_2)=P(W_1) \times P(W_2)\\P(W_1W_2)=0.9 \times 0.8\\P(W_1W_2)=0.72

So the reliability of the first system to work is 0.72.

For the second system is given as

P(W_1W_2)=1-P(W_1'W_2')

Where

  • P(W_1'W_2') is the probability where both of the systems does not work. this is calculated as

P(W_1'W_2')=P(W_1') \times P(W_2')\\P(W_1'W_2')=(1-P(W_1)) \times (1-P(W_2))\\P(W_1'W_2')=(1-0.9) \times (1-0.8)\\P(W_1'W_2')=(0.1) \times (0.2)\\P(W_1'W_2')=0.02

So now the probability of the second system is given as

P(W_1W_2)=1-P(W_1'W_2')\\P(W_1W_2)=1-0.02\\P(W_1W_2)=0.98

So the reliability of the second system to work is 0.98.

As the reliability of the second system is more than the first one so the second system is more reliable.

8 0
2 years ago
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