Ask question

Ask question

All categories

1 year ago

6

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

owing: (a) Cumulative distribution function of ﬂange thickness (b) Proportion of ﬂanges that exceeds 1.02 millimeters (c) Thickness exceeded by 90% of the ﬂanges (d) Mean and variance of ﬂange thickness

1 answer:

Mrrafil [7]1 year ago

4 0

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

= 0.000833

You might be interested in

Fill in the blank. Given O below, you can conclude that xz is congruent to

Illusion [34]

Answer:

D

Step-by-step explanation:

I do not know the mathematical answer all i know is that they are congruent

7 0

1 year ago

Read 2 more answers

A large city in the ancient world was square-shaped. measured in miles, its area numerically exceeded its perimeter by about 132

melisa1 [442]

Area = perimeter + 132.

Let each side of the city be x miles long, then:-

x^2 = 4x + 132
x^2 - 4x - 132 = 0

x = [-(-4) +/- sqrt((-4)^2 - 4 * 1 *-132)] / 2

x = 13.66, -9.66 We ignore the negative

So the city has dimension of 13.66 * 13.66

13.7 * 13.7 to nearest 10th

7 0

1 year ago

If a cow has a mass of 9×102 kilograms, and a blue whale has a mass of 1.8×105 kilograms, which of these statements is true?

lana66690 [7]

Answer:

The mass of the Blue whale is 200 times the mass of the cow

Step-by-step explanation:

Given

Mass of Cow = 9 * 10² kg

Mass of Blue Whale = 1.8 * 10⁵ kg

Required

Determine the relationship between both weights

Represent the mass of the cow with C and the mass of the whale with B

$C = 9 * 10^2kg$

$C = 9 *100kg$

$C = 900kg$

$B = 1.8 * 10^5kg$

$B = 1.8 * 100000kg$

$B = 180000kg$

Divide the bigger weight by the smaller weight

$\frac{B}{C} = \frac{180000kg}{900kg}$

$\frac{B}{C} = \frac{180000}{900}$

$\frac{B}{C} = {200}{}$

Multiply both sides by C

$C * \frac{B}{C} = {200}{} * C$

$B = {200}{} * C$

$B = 200}C$

<em>From the expression above, it can be concluded that the mass of the Blue whale is 200 times the mass of the cow</em>

7 0

1 year ago

A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12

marissa [1.9K]

Answer:

Step-by-step explanation:

Hello!

You have two random samples obtained from two different normal populations.

Sample 1

n₁= 15

X[bar]₁= 350

S₁= 12

Sample 2

n₂= 17

X[bar]₂= 342

S₂= 15

At α: 0.05 you need to obtain the p-value for testing variances for a one tailed test.

If the statistic hypotheses are:

H₀: σ₁² ≥ σ₂²

H₁: σ₁² < σ₂²

The statistic to test the variances ratio is the Stenecor's-F test. $F_{H_0}=(\frac{S^2_1}{Sigma^2_1}) * (\frac{S^2_2}{Sigma^2_2} )$ ~ $F_{n_1-1;n_2-1}$

$F_{H_0}= \frac{(12)^2}{(15)^2} * 1= 0.64$

The p-value is:

P( $F_{14;16}$ ≤0.64)= 0.02

I hope it helps!

7 0

2 years ago

You drive 30 miles due east in a half hour. Then, you turn left and drive 30 miles north in 1 hour. What are your average speed

garri49 [273]

Original position: P1=(0,0)
<span>You drive 30 miles due east in a half hour: x=+30 miles, t1=1/2 hour=0.5 hours
</span><span>Then, you turn left and drive 30 miles north in 1 hour: y=+30 miles, t2=1 hour
Rectangular coordinates of final position: P2=(x,y)→P2=(30,30)
Total time: t=t1+t2=0.5 hours+1 hour→t=1.5 hours

Average speed: S ave=d/t
Total distance: d=x+y=30 miles+30 miles→d=60 miles
S ave = 60 miles / (1.5 hours)
S ave = 40 miles/hour

Velocity is a vector, the magnitude of this vector is the magnitude of the vector of change of position dividing by the total time t
The vector of change of position: s=P1-P2=(30,30)-(0,0)=(30-0,30-0)→
s=(30,30)
Magnitude of vector s=sqrt[30^2+30^2]=sqrt[30^2*2]=sqrt[30^2]*sqrt(2)
Magnitude of vector s=30*sqrt(2) miles

Magnitude of velocity vector = Magnitud of vector s / t
Magnitude of velocity vector = [30*sqrt(2) miles] / (1.5 hours)
Magnitude of velocity vector = 20*sqrt(2) miles / hour
Magnitude of velocity vector=20*1.4142 miles / hour
Magnitude of velocity vector=28.284 miles/hour

Polar coordinates of your position=(r, theta)
r=Magnitude of vector s=30*sqrt(2) miles
theta=tan^(-1) (y/x) = tan^(-1) [(30 miles) / (30 miles)]
theta=tan^(-1) (1)→theta=45°=Pi/4 (Pi=3.1416)
Polar coordinates of your position: ( 30*sqrt(2) miles, 45°)
Polar coordinate of your position: ( 30*sqrt(2) miles, Pi/4 )

Answers:
Average speed: 40 miles / hour
Velocity: 20*sqrt(2) miles / hour = 28.284 miles / hour
Rectangular coordinates of your position = (30,30)
Polar coordinates of your position=(30*sqrt(2) miles,45°)
Polar coordinates of your position=(30*sqrt(2) miles,Pi/4)</span>

4 0

1 year ago