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Komok [63]
3 years ago
14

Use separation of variables to solve dy dx − tan x = y2 tan x with y(0) = √3. Find the value of c in radians, not degrees

Mathematics
1 answer:
a_sh-v [17]3 years ago
3 0

Answer:

y(x)=tan(-log(cos(x))+\frac{\pi }{3} )

Step-by-step explanation:

Rewrite the equation as:

\frac{dy(x)}{dx}-tan(x)=y(x)^{2} *tan(x)

Isolating \frac{dy}{dx}

\frac{dy}{dx} =tan(x)+tan(x)*y^{2}

Factor:

\frac{dy}{dx} =tan(x)*(1+y^{2} )

Dividing both sides by (1+y^{2} ) and multiplying them by dx

\frac{dy}{1+y^{2} } =tan(x)dx

Integrate both sides:

\int\ \frac{dy}{1+y^{2} } = \int\ tan(x)  dx

Evaluate the integrals:

arctan(y)=-log(cos(x))+C_1

Solving for y:

y(x)=tan(-log(cos(x))+C_1)

Evaluating the initial condition:

y(0)=\sqrt{3} =tan(-log(cos(0))+C_1)=tan(-log(1)+C_1)=tan(0+C_1)

\sqrt{3} =tan(C_1)\\arctan(\sqrt{3} )=C_1\\60=C_1

Converting 60 degrees to radians:

60degrees*\frac{\pi }{180degrees} =\frac{\pi }{3}

Replacing C_1 in the diferential equation solution:

y(x)=tan(-log(cos(x))+\frac{\pi }{3} )

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The number of events is 29​, the number of trials is 298​, the claimed population proportion is​ 0.10, and the significance leve
Nina [5.8K]

Answer:

z=\frac{0.0973 -0.1}{\sqrt{\frac{0.1(1-0.1)}{298}}}=-0.155  

p_v =2*P(Z  

And we can use excel to find the p value like this: "=2*NORM.DIST(-0.155;0;1;TRUE)"

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly different from 0.1 .  

Step-by-step explanation:

1) Data given and notation

n=298 represent the random sample taken

X=29 represent the events claimed

\hat p=\frac{29}{298}=0.0973 estimated proportion

p_o=0.1 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is 0.1 or no.:  

Null hypothesis:p=0.1  

Alternative hypothesis:p \neq 0.1  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.0973 -0.1}{\sqrt{\frac{0.1(1-0.1)}{298}}}=-0.155  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(Z  

And we can use excel to find the p value like this: "=2*NORM.DIST(-0.155;0;1;TRUE)"

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly different from 0.1 .  

We can do the test also in R with the following code:

> prop.test(29,298,p=0.1,alternative = c("two.sided"),conf.level = 1-0.05,correct = FALSE)

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