Write and then solve for y = f(x) the differential equation for the statement: "The rate of change of y with respect to x is inv
1 answer:
The statement "<span>The rate of change of y with respect to x is inversely proportional to y^4" can be written mathematically as dy/dx = k/y^4
To solve the differential equation, we use variable saparable method.
y^4 dy = kdx
Integrating both sides gives,
y^5 / 5 = kx + A
y^5 = 5kx + 5A = Bx + C; where B = 5k and C = 5A
![y= \sqrt[5]{Bx+C}](https://tex.z-dn.net/?f=y%3D%20%5Csqrt%5B5%5D%7BBx%2BC%7D%20)
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To solve the differential equation, we use variable saparable method.
y^4 dy = kdx
Integrating both sides gives,
y^5 / 5 = kx + A
y^5 = 5kx + 5A = Bx + C; where B = 5k and C = 5A
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Answer:
We conclude that the new procedure will not decrease the population mean amount of time required to produce the part.
Step-by-step explanation:
We are given that a large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times follow a normal distribution with mean time μ = 45 hours.
A random sample of 25 parts will be selected and the average amount of time required to produce them will be determined. The sample mean amount of time is = 43.118 hours with the sample standard deviation s = 5.5 hours
<em>Let </em><em> = population mean amount of time required to produce an electrical part using new procedure</em>
SO, <u>Null Hypothesis</u>, :
45 hours {means that the new procedure will remain same or increase the population mean amount of time required to produce the part}
<u>Alternate Hypothesis,</u> :
< 45 hours {means that the new procedure will decrease the population mean amount of time required to produce the part}
The test statistics that will be used here is <u>One-sample t test statistics </u>because we don't know about the population standard deviation;
T.S. = ~
where, = sample mean amount of time = 43.118 hours
s = sample standard deviation = 5.5 hours
n = sample of parts = 25
So, <u><em>test statistics</em></u> = ~
= -1.711
<em>Now at 0.025 significance level, the t table gives critical value of -2.064 at 24 degree of freedom for left-tailed test. Since our test statistics is more than the critical value of t so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.</em>
Therefore, we conclude that the new procedure will remain same or increase the population mean amount of time required to produce the part.