Allie works for Canton Water Works. Her annual salary is $75,964.

se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y

alexandr1967 [171]

Answer:

It is proved that $f_x, f_y$ exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

$f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)$

To show exixtance of $f_x(0,0), f_y(0,0)$ we take,

$f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0$ exists.

And,

$f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}$ exists.

To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

$\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}$ where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

4 0

2 years ago

The floor area of a hall is 1044 m2. The length is 36 m. Give a quick estimate of width of the hall. ?

wolverine [178]

36m. Im not sure but thats what i got, hope it helps...

4 0

1 year ago

The community center has a pottery class each month. Each student pays $15 for the class and $27 for materials. This month the p

Dominik [7]

If you add $15 + $27 because that's how much it costs for each student, you get $42. Then you divide $714 by $42, you get 17 and that's how many students are in the class this month.

3 0

1 year ago

Read 2 more answers

There are 200 students in eleventh grade high school class. There are 40 students in the soccer team and 50 students in the bask

bekas [8.4K]

Answer:

P(A) = 0.2

P(B) = 0.25

P(A&B) = 0.05

P(A|B) = 0.2

P(A|B) = P(A) = 0.2

Step-by-step explanation:

P(A) is the probability that the selected student plays soccer.

Then:

$P(A)=\dfrac{40}{200}=0.2$

P(B) is the probability that the selected student plays basketball.

Then:

$P(B)=\dfrac{50}{200}=0.25$

P(A and B) is the probability that the selected student plays soccer and basketball:

$P(A\&B)=\dfrac{10}{200}=0.05$

P(A|B) is the probability that the student plays soccer given that he plays basketball. In this case, as it is given that he plays basketball only 10 out of 50 plays soccer:

$P(A|B)=\dfrac{P(A\&B)}{P(B)}=\dfrac{10}{50}=0.2$

P(A | B) is equal to P(A), because the proportion of students that play soccer is equal between the total group of students and within the group that plays basketball. We could assume that the probability of a student playing soccer is independent of the event that he plays basketball.

4 0

2 years ago

Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum

dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 --- Approximated

c.

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

=.

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

Thus it 30.671% probable that 0, 1, or 2 students received accommodation.

e.

The expected value from d) is .6

The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours

8 0

2 years ago