3

Is the gap store an example of commercial construction? Yes or no

user100 [1]

Answer:

no

Explanation:

5 0

3 years ago

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8 What is the probability of observing

Viktor [21]

Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

When we only have the mean, we use the Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that $\mu = 0.8$

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

$P(X \leq 4) + P(X > 4) = 1$

We want P(X > 4). Then

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

$P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595$

$P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438$

$P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383$

$P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9986 = 0.0014$

0.14% probability of observing more than 4 errors in the carpet

5 0

3 years ago

The collar A, having a mass of 0.75 kg is attached to a spring having a stiffness of k = 200 N/m . When rod BC rotates about the

gladu [14]

Answer:

Speed=1.633 m/s

Force= 20 N

Explanation:

Ideally, $v^{2}=\frac {ks^{2}}{m}$ hence $v=s\sqrt {\frac {k}{m}}$ where v is the speed of collar, m is the mass of collar, k is spring constant and s is the displacement.

In this case, s=100-0=100mm=0.1m since 1 m is equivalent to 1000mm

k is given as 200 N/m and mass is 0.75 Kg

Substituting the given values

$v=0.1 m\sqrt \frac {200 N/m}{0.75 Kg}=1.632993162 m/s\approx 1.633 m/s$

Therefore, <u>the speed is 1.633 m/s</u>

The sum of vertical forces is given by mg where g is acceleration due to gravity and it's value taken as $9.81 m/s^{2}$

Therefore, $F_y=0.75\times 9.81=7.3575 N\approx 7.36 N$

The sum of forces in normal direction is given by $Ma_n=Ks$ therefore

$Ma_n=200*0.1=20 N$

Therefore, <u>normal force on the rod is 20 N</u>

5 0

3 years ago

How are scientific discoveries used in engineering design?

Citrus2011 [14]

Answer:

one is technology

6 0

3 years ago

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by

Stells [14]

Answer:

$\dfrac{\bar{h}}{h}=\dfrac{5}{4}$

Explanation:

Given that

$Nu_x=0.035Re_x^{0.8} Pr^{1/3}$

We know that

Rex=ρvx/μ

So

$Nu_x=0.035Re_x^{0.8} Pr^{1/3}$

$Nu_x=0.035\times\left(\dfrac{\rho vx}{\mu}\right)^{0.8}Pr^{1/3}$

All other quantities are constant only x is a variable in the above equation .so lets take all other quantities as a constant C

$Nu_x=C.x^{0.8}=C.x^{4/5}$

We also know that

Nux=hx/K

$C.x^{4/5}=\dfrac{hx}{k}$

m is the constant

$h=mx^{-1/5}$

This is local heat transfer coefficient

The average value of h given as

$\bar{h}=\dfrac{\int_{0}^{L}hdx}{L}$

$\bar{h}=\dfrac{5m}{4}\times\dfrac{L^{4/5}}{L}$

$\bar{h}=\dfrac{5m}{4}L^{-1/5}$ ---------1

The value of local heat transfer coefficient at x=L

$h=mx^{-1/5}$

$h=mL^{-1/5}$ -----------2

From 1 and 2 we can say that

$\dfrac{\bar{h}}{h}=\dfrac{5}{4}$

8 0

3 years ago