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jeka57 [31]
3 years ago
5

Assume the defect length of corrosion defect in a pressurized steel pipe is normally distributed with mean value 30 mm and stand

ard deviation 7.8 mm.
a. What is the probability that defect length is at most 20 mm? Less than 20 mm?
b. What is the 75th percentile of the defect length distribution?
c. What is the 15th percentile of the defect length distribution?
d. What values separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10%?
Mathematics
1 answer:
RSB [31]3 years ago
5 0

Answer:

a) 0.1003

b) 35.26

c) 21.92

d) Middle 80% smallest 10% and largest 10% = 20.02 and 39.98

Step-by-step explanation:

Given that: mean ц = 30 , standard deviation δ = 7.8

a) P(x ≤ 20)

   = P [ (x - ц ) / δ   ≤  ( 20 -30) / 7.8 ]

   = P(z ≤ -1.28)

   = 0.1003

probability = 0.1003

b) 75th percentile

   P(Z < z) = 0.75

   z = 0.674

Using z-score formula

   x = z * δ + ц

   x = 0.674 * 7.8 + 30

   x = 35.26

c) 15th percentile

   P(Z < z) = 0.15

   z = -1.036

Using z-score formula

   x = z * δ + ц

   x = (-1.036) * 7.8 + 30

   x = 21.92

d) Middle 80%

   P(-z ≤ Z ≤ z) = 0.80

   P(Z ≤ z) - P(Z ≤ -z) = 0.80

   2P(Z ≤ z) - 1 = 0.80

   2P(Z ≤ z) = 1 + 0.80 = 1.80

   P(Z ≤ z) = 1.80 / 2 = 0.90

   P(Z ≤ ± 1.28) = 0.90

   z =  -1.28, + 1.28

   x = z * δ + ц

   x = (-1.28) * 7.8 + 30

   x = 20.02

   x = z * δ + ц

   x = 1.28 * 7.8 + 30

   x = 39.98

Middle 80% two values = 20.02 and 39.98

smallest 10%

P(Z < z) = 0.10

z = -1.28

Using z-score formula

x = z * δ + ц

x = ( -1.28) * 7.8 + 30

x = 20.02

largest 10%

P(Z > z ) = 0.10

1 - P(z < z) = 0.10

P(z < z) = 1 - 0.10 = 0.90

z = 1.28

Using z-score formula

x = z * δ + ц

x = (1.28) * 7.8 + 30

x = 39.98

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I need help fast please!!
kaheart [24]

Solution set is: (7,\frac{13}{3})

Option B is correct

Step-by-step explanation:

We need to find the solution to system of equations

y=\frac{1}{3}x+2\\y=\frac{4}{3}x-5

Let:

y=\frac{1}{3}x+2\,\,\,eq(1)\\y=\frac{4}{3}x-5\,\,\,eq(2)

Putting value of y from eq(2) into eq(1)

\frac{4}{3}x-5=\frac{1}{3}x+2\\Combining\,\,like\,\,terms:\\\frac{4}{3}x-\frac{1}{3}x=2+5\\Taking\,\,LCM\\\frac{4x-1x}{3}=7\\\frac{3x}{3}=7\\x=7

So, value of x=7

Putting value of x into equation 1 to find the value of y:

y=\frac{1}{3}x+2\\y=\frac{1}{3}(7)+2\\y=\frac{7}{3}+2\\Taking\,\,LCM\\y=\frac{7+2*3}{3}\\y=\frac{7+6}{3}\\y=\frac{13}{3}

So, value of y = \frac{13}{3}

Solution set is: (7,\frac{13}{3})

Option B is correct

Keywords: System of equations

Learn more about system of equations at:

  • brainly.com/question/9045597
  • brainly.com/question/3739260
  • brainly.com/question/13168205

#learnwithBrainly

3 0
3 years ago
F(x)=x^2. What is g(x)?
Contact [7]

Answer:

Do you have options? The answer would be

g(x)=(x/3)^2

x = 1 btw

Because x = 1, if you have options then it'll most likely be g(x)=(1/3]^2

Step-by-step explanation:

Hope this helps

4 0
2 years ago
Mr. Engen rented a car for 35.00 a day and 0.45 a mile . He had the car for 2 days and paid 166.75 . How many miles did he drove
liubo4ka [24]

d = days rented for

m = miles driven for

so the cost of the car is 35 bucks a day and 45 cents or $0.45 per mile driven, let's check it out for say a few days

1st day, 3 miles......................35(1) + 0.45(3)

2nd day, 2 miles..................35(2) + 0.45(2)

3rd day, say 5 miles............35(3) + 0.45(5)

dth day, and m miles..........35d + 0.45m

so the cost function could be written as C = 35d+0.45m.

now, we know for 2 days, d = 2, it cost 166.75 bucks, so

\bf C=35d+0.45m\implies \stackrel{C}{166.75}=35(\stackrel{d}{2})+0.45m\\\\\\166.75=70+0.45m\implies 96.75=0.45m\\\\\\\cfrac{96.75}{0.45}=m\implies 215=m

7 0
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