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lys-0071 [83]
2 years ago
14

John traveled 150km in 6 hours and alice traveled 180km in 4 hours. what is the ratio of the average speed of john to average sp

eed of alice?
a. 3:2
b. 2:3
c. 5:9
d. 5:6
Mathematics
1 answer:
alekssr [168]2 years ago
3 0
Answer will be d. 5:6
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Write a multiplication fact that goes with this division fact. 24 ÷ 4 = 6
Mekhanik [1.2K]
Well I wild have to say 6x4=24
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Which system of equations below has no solution? A. y = 4x + 5 and y = 4x – 5 B. y = 4x + 5 and 2y = 8x + 10 C. y = 4x + 5 and y
Stels [109]

Answer: A  y=4x -5 and y=4x+5

Step-by-step explanation:

They have no solutions because they have the same slopes but different y intercepts. That works with any equation.

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Complete the coordinate proof of the theorem. Given: A B C D is a parallelogram. Prove: The diagonals of A B C D bisect each oth
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2 years ago
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The thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters. Determine the foll
Mrrafil [7]

Answer

given,

thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters.

X = U[0.95,1.05]           0.95≤ x ≤ 1.05

the cumulative distribution function of flange

F(x) = P{X≤ x}=\dfrac{x-0.95}{1.05-0.95}

                     =\dfrac{x-0.95}{0.1}

b) P(X>1.02)= 1 - P(X≤1.02)

                   = 1- \dfrac{1.02-0.95}{0.1}

                   = 0.3

c) The thickness greater than 0.96 exceeded by 90% of the flanges.

d) mean = \dfrac{0.95+1.05}{2}

              = 1

   variance = \dfrac{(1.05-0.95)^2}{12}

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4 0
2 years ago
A quality control engineer tests the quality of produced computers. suppose that 5% of computers have defects, and defects occur
11111nata11111 [884]
(a) 0.059582148 probability of exactly 3 defective out of 20

 (b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.

  (a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So

 0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)

 = 0.05^3 * 0.95^17 * 20! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)

 = 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)

 = 0.05^3 * 0.95^17 * 20*19*3

 = 0.000125* 0.418120335 * 1140

 = 0.059582148

  (b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.

 The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)

 = 0.05*0.95^3*24/(1!3!)

 = 0.05*0.857375*24/6

 = 0.171475

 
 So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
3 0
2 years ago
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