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ArbitrLikvidat [17]
3 years ago
7

Low-fuel landing 25% instrument failure 15% bad weather 5% failed landing gear 15% engine failure 20% autopilot failure 15% norm

al conditions 5% At a flight-training academy, each pilot is given an emergency condition in a simulator during testing (the probabilities are shown in the table). In the current class, 25 trainees failed the test and 115 passed. How many trainees were tested in the most common scenario? 25 35 15 10 © 2015 Edmentum. All rights reserved.
Mathematics
1 answer:
docker41 [41]3 years ago
8 0
The most common scenario was the low-fuel landing. It was there in 25 % of the tests. The number of tests was 25+115=140 tests (adding up the failed and passed test. 
So to find our how many were trained in the most common scenario, we have to calculate25% of 140. 
25/100*140=(i calculate frist 25/100. I divide both by 25, we have 1/4)1/4*140 = In order to calculate this, we need to divide 140/4. This is:
  35____140:4 12   20   20   ==

so the correct answer is 35. 
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A(x) = 2 (3x - 2)(x - 5)
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Answer:

Quadratic polynomial can be factored using the transformation ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

), where x  

1

​  

 and x  

2

​  

 are the solutions of the quadratic equation ax  

2

+bx+c=0.

−x  

2

−3x+5=0

All equations of the form ax  

2

+bx+c=0 can be solved using the quadratic formula:  

2a

−b±  

b  

2

−4ac

​  

 

​  

. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=  

2(−1)

−(−3)±  

(−3)  

2

−4(−1)×5

​  

 

​  

 

Square −3.

x=  

2(−1)

−(−3)±  

9−4(−1)×5

​  

 

​  

 

Multiply −4 times −1.

x=  

2(−1)

−(−3)±  

9+4×5

​  

 

​  

 

Multiply 4 times 5.

x=  

2(−1)

−(−3)±  

9+20

​  

 

​  

 

Add 9 to 20.

x=  

2(−1)

−(−3)±  

29

​  

 

​  

 

The opposite of −3 is 3.

x=  

2(−1)

3±  

29

​  

 

​  

 

Multiply 2 times −1.

x=  

−2

3±  

29

​  

 

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is plus. Add 3 to  

29

​  

.

x=  

−2

29

​  

+3

​  

 

Divide 3+  

29

​  

 by −2.

x=  

2

−  

29

​  

−3

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is minus. Subtract  

29

​  

 from 3.

x=  

−2

3−  

29

​  

 

​  

 

Divide 3−  

29

​  

 by −2.

x=  

2

29

​  

−3

​  

 

Factor the original expression using ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

). Substitute  

2

−3−  

29

​  

 

​  

 for x  

1

​  

 and  

2

−3+  

29

​  

 

​  

 for x  

2

​  

.

−x  

2

−3x+5=−(x−  

2

−  

29

​  

−3

​  

)(x−  

2

29

​  

−3

​  

)

EVALUATE

5−3x−x  

2Quadratic polynomial can be factored using the transformation ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

), where x  

1

​  

 and x  

2

​  

 are the solutions of the quadratic equation ax  

2

+bx+c=0.

−x  

2

−3x+5=0

All equations of the form ax  

2

+bx+c=0 can be solved using the quadratic formula:  

2a

−b±  

b  

2

−4ac

​  

 

​  

. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=  

2(−1)

−(−3)±  

(−3)  

2

−4(−1)×5

​  

 

​  

 

Square −3.

x=  

2(−1)

−(−3)±  

9−4(−1)×5

​  

 

​  

 

Multiply −4 times −1.

x=  

2(−1)

−(−3)±  

9+4×5

​  

 

​  

 

Multiply 4 times 5.

x=  

2(−1)

−(−3)±  

9+20

​  

 

​  

 

Add 9 to 20.

x=  

2(−1)

−(−3)±  

29

​  

 

​  

 

The opposite of −3 is 3.

x=  

2(−1)

3±  

29

​  

 

​  

 

Multiply 2 times −1.

x=  

−2

3±  

29

​  

 

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is plus. Add 3 to  

29

​  

.

x=  

−2

29

​  

+3

​  

 

Divide 3+  

29

​  

 by −2.

x=  

2

−  

29

​  

−3

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is minus. Subtract  

29

​  

 from 3.

x=  

−2

3−  

29

​  

 

​  

 

Divide 3−  

29

​  

 by −2.

x=  

2

29

​  

−3

​  

 

Factor the original expression using ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

). Substitute  

2

−3−  

29

​  

 

​  

 for x  

1

​  

 and  

2

−3+  

29

​  

 

​  

 for x  

2

​  

.

−x  

2

−3x+5=−(x−  

2

−  

29

​  

−3

​  

)(x−  

2

29

​  

−3

​  

)

EVALUATE

5−3x−x  

2

Step-by-step explanation:

7 0
3 years ago
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