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Likurg_2 [28]
3 years ago
5

Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Is this clai

m correct? a. Yes b. No
Engineering
1 answer:
Alika [10]3 years ago
5 0

Answer:

Yes this claim is correct.

Explanation:

The shear stress at any point is proportional to the velocity gradient at any that point. Since the fluid that is in contact with the pipe wall shall have zero velocity due to no flow boundary condition and if we move small distance away from the wall the velocity will have a non zero value thus a maximum gradient will exist at the surface of the pipe hence correspondingly the shear stresses will also be maximum.

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Suppose that a wireless link layer using a CSMA-like protocol backs off 1ms on average. A packet’s link and physical layer heade
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Tell me why i got this question got it right and now won’t remember but i’ll get back at you when i remember
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3 years ago
The voltage across a device and the current through it are:
butalik [34]

Answer:

attached below

Explanation:

3 0
3 years ago
Read 2 more answers
La probabilidad de que un nuevo producto tenga éxito es de 0.85. Si se eligen 10 personas al azar y se les pregunta si compraría
liq [111]

Answer:

La probabilidad pedida es 0.820196

Explanation:

Sabemos que la probabilidad de que un nuevo producto tenga éxito es de 0.85. Sabemos también que se eligen 10 personas al azar y se les pregunta si comprarían el nuevo producto. Para responder a la pregunta, primero definiremos la siguiente variable aleatoria :

X: '' Número de personas que adquirirán el nuevo producto de 10 personas a las que se les preguntó ''

Ahora bien, si suponemos que la probabilidad de que el nuevo producto tenga éxito se mantiene constante (p=0.85) y además suponemos que hay independencia entre cada una de las personas al azar a las que se les preguntó ⇒ Podemos modelar a X como una variable aleatoria Binomial. Esto se escribe :

X ~ Bi(n,p) en donde ''n'' es el número de personas entrevistadas y ''p'' es la probabilidad de éxito (una persona adquiriendo el producto) en cada caso.

Utilizando los datos ⇒ X ~ Bi(10,0.85)

La función de probabilidad de la variable aleatoria binomial es :

p_{X}(x)=P(X=x)=\left(\begin{array}{c}n&x\end{array}\right)p^{x}(1-p)^{n-x}    con x=0,1,2,...,n

Si reemplazamos los datos de la pregunta en la función de probabilidad obtenemos :

P(X=x)=\left(\begin{array}{c}10&x\end{array}\right)(0.85)^{x}(0.15)^{10-x} con x=0,1,2,...,10

Nos piden la probabilidad de que por lo menos 8 personas adquieran el nuevo producto, esto es :

P(X\geq 8)=P(X=8)+P(X=9)+P(X=10)

Calculando P(X=8), P(X=9) y P(X=10) por separado y sumando, obtenemos que P(X\geq 8)=0.820196

7 0
3 years ago
A large class with 1,000 students took a quiz consisting of ten questions. To get an A, students needed to get 9 or 10 questions
VMariaS [17]

Answer:

a. 0.11

b. 110 students

c. 50 students

d. 0.46

e. 460 students

f. 540 students

g. 0.96

Explanation:

(See attachment below)

a. Probability that a student got an A

To get an A, the student needs to get 9 or 10 questions right.

That means we want P(X≥9);

P(X>9) = P(9)+P(10)

= 0.06+0.05=0.11

b. How many students got an A on the quiz

Total students = 1000

Probability of getting A = 0.11 ---- Calculated from (a)

Number of students = 0.11 * 1000

Number of students = 110 students

So,the number of students that got A is 110

c. How many students did not miss a single question

For a student not to miss a single question, then that student scores a total of 10 out of possible 10

P(10) = 0.05

Total Students = 1000

Number of Students = 0.05 * 1000

Number of Students = 50 students

We see that 5

d. Probability that a student pass the quiz

To pass, a student needed to get at least 6 questions right.

So we want P(X>=6);

P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)

=0.08+0.12+0.15+0.06+0.05=0.46

So, the probability of a student passing the quiz is 0.46

e. Number of students that pass the quiz

Total students = 1000

Probability of passing the quiz = 0.46 ----- Calculated from (d)

Number of students = 0.46 * 1000

Number of students = 460 students

So,the number of students that passed the test is 460

f. Number of students that failed the quiz

Total students = 1000

Total students that passed = 460 ----- Calculated from (e)

Number of students that failed = 1000 - 460

Number of students that failed = 540

So,the number of students that failed is 540

g. Probability that a student got at least one question right

This means that we want to solve for P(X>=1)

Using the complement rule,

P(X>=1) = 1 - P(X<1)

P(X>=1) = 1 - P(X=0)

P(X>=1) = 1 - 0.04

P(X>=1) = 0.96

7 0
3 years ago
The domain of discourse is the members of a chess club. The predicate B(x, y) means that person x has beaten person y at some po
satela [25.4K]

Answer:A. No one has ever beat Nancy.

Explanation:

The dormain of discourse in a simple language is the set of entities upon which our discussions are based when discussing about something.

The dormain of discourse is also known simply as universe, can also be said to be a set of entities o

upon which certain variables of interest in some formal treatment may range.

The dormain of discourse is generally attributed to Augustus De Morgan, it was also extensively used by George Boole in his Laws of Thought.

THE LOGICAL UNDERSTANDING OF THE THE QUESTION IS THAT NO ONE HAS EVER BEAT NANCY.

8 0
3 years ago
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