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algol13
1 year ago
13

University graduates have a mean job search time of 38.1 weeks, with a standard deviation of 10.1 weeks. The distribution of job

search times is not assumed to be symmetric. Between what two search times does Chebyshev's Theorem guarantee that we will find at least 89% of the graduates
Mathematics
1 answer:
Lera25 [3.4K]1 year ago
5 0

Answer:

7.1 weeks to 68.4 weeks

Step-by-step explanation:

Chebyshev's Theorem states that:

75% of the measures are within 2 standard deviations of the mean.

89% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 38.1

Standard deviation = 10.1

Between what two search times does Chebyshev's Theorem guarantee that we will find at least 89% of the graduates

Between 3 standard deviations of the mean.

So from 38.1 - 3*10.1 = 7.8 weeks to 38.1 + 3*10.1 = 68.4 weeks

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Step-by-step explanation:

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Step-by-step explanation:

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

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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,

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\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).

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