Question: 10 of 15

Two balls are chosen randomly from an urn containing 8 white 4 black, and orange balls. Suppose that we win $ 2 for each black b

Scorpion4ik [409]

(-2,-10,-1,-2,-3,-4)

8 0

3 years ago

three string are attached to a small metal ring, two of the strings make and angle of 35° with the vertical and each is pulled w

Julli [10]

No clue sorry man I would help but I need help too

6 0

3 years ago

A stream leaving a sewage pond (containing 80 mg/L of sewage) moves as a plug with a velocity of 40 m/hr. A concentration of 50

Leno4ka [110]

Answer:Decay rate constant,k = 0.00376/hr

Explanation:

IsT Order Rate of reaction is given as

In At/ Ao = -Kt

where [A]t is the final concentration at time t and [A]o is the inital concentration at time 0, and k is the first-order rate constant.

Initial concentration = 80 mg/L

Final concentration = 50 mg/L

Velocity = 40 m/hr

Distance= 5000 m

Time taken = Distance / Time

5000m / 40m/hr = 125 hr

In At/ Ao = -Kt

In 50/80 = -Kt

-0.47 = -kt

- K= -0.47 / 125

k = 0.00376

Decay rate constant,k = 0.00376/hr

8 0

3 years ago

I. The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have

drek231 [11]

Answer:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(c) 3.915 × 10⁻¹²

(d) 2.63 × 10⁻⁵

Explanation:

(a) We put time to fail = t

∴ For an exponential distribution, we have f(t) = $\lambda e^{-\lambda t}$

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

$P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1$

e^(1000·λ) - 0.1·e^(1000·λ) = 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

b) Here we have to integrate from 5000 to ∞ as follows;

$p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [ -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905$

(c) The Poisson distribution is presented as follows;

$P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!} = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!} = 3.915\times 10^{-12}$

p(x = 3) = 3.915 × 10⁻¹²

d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours

The Cumulative Distribution Function is given as follows;

p( t ≤ 1/4) $CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}$ .

4 0

3 years ago

Water flows through a multisection pipe placed horizontally on the ground. The velocity is 3.0 m/s at the entrance and 2.1 m/s a

Alex_Xolod [135]

Answer:

b. 2.3 kPa.

Explanation:

This situation can be modelled by Bernoulli's Principle, as there are no energy interaction throughout the multisection pipe and current lines exists between both ends. Likewise, this system have no significant change in gravitational potential energy since it is placed horizontally on the ground and is described by the following model:

$P_{1} + \rho \cdot \frac{v_{1}^{2}}{2} = P_{2} + \rho \cdot \frac{v_{2}^{2}}{2}$

Where:

$P_{1}$ , $P_{2}$ - Pressures at the beginning and at the end of the current line, measured in kilopascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Fluid velocity at the beginning and at the end of the current line, measured in meters per second.

Now, the pressure difference between these two points is:

$P_{1} - P_{2} = \rho \cdot \frac{v_{2}^{2}-v_{1}^{2}}{2}$

If $\rho = 1000\,\frac{kg}{m^{3}}$ , $v_{1} = 3\,\frac{m}{s}$ and $v_{2} = 2.1\,\frac{m}{s}$ , then:

$P_{1} - P_{2} = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \frac{\left(2.1\,\frac{m}{s} \right)^{2}-\left(3\,\frac{m}{s} \right)^{2}}{2}$

$P_{1} - P_{2} = -2295\,Pa$

$P_{1} - P_{2} = -2.295\,kPa$ (1 kPa is equivalent to 1000 Pa)

Hence, the right answer is B.

7 0

3 years ago