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kifflom [539]
3 years ago
15

There is a line through the origin that divides the region bounded by the parabola y=2x-4x^2 and the x-axis into two regions wit

h equal area. What is the slope of that line? ...?
Mathematics
1 answer:
shtirl [24]3 years ago
7 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

y = 7x - 4x² 

<span>7x - 4x² = 0 </span>

<span>x(7 - 4x) = 0 </span>

<span>x = 0, 7/4 </span>

<span>Find the area of the bounded region... </span>

<span>A = ∫ 7x - 4x² dx |(0 to 7/4) </span>

<span>A = 7/2 x² - 4/3 x³ |(0 to 7/4) </span>

<span>A = 7/2(7/4)² - 4/3(7/4)³ - 0 = 3.573 </span>

<span>Half of this area is 1.786, now set up an integral that is equal to this area but bounded by the parabola and the line going through the origin... </span>

<span>y = mx + c </span>

<span>c = 0 since it goes through the origin </span>

<span>The point where the line intersects the parabola we shall call (a, b) </span>

<span>y = mx ===> b = m(a) </span>

<span>Slope = m = b/a </span>

<span>Now we need to integrate from 0 to a to find the area bounded by the parabola and the line... </span>

<span>1.786 = ∫ 7x - 4x² - (b/a)x dx |(0 to a) </span>

<span>1.786 = (7/2)x² - (4/3)x³ - (b/2a)x² |(0 to a) </span>

<span>1.786 = (7/2)a² - (4/3)a³ - (b/2a)a² - 0 </span>

<span>1.786 = (7/2)a² - (4/3)a³ - b(a/2) </span>

<span>Remember that (a, b) is also a point on the parabola so y = 7x - 4x² ==> b = 7a - 4a² </span>
<span>Substitute... </span>

<span>1.786 = (7/2)a² - (4/3)a³ - (7a - 4a²)(a/2) </span>

<span>1.786 = (7/2)a² - (4/3)a³ - (7/2)a² + 2a³ </span>

<span>(2/3)a³ = 1.786 </span>

<span>a = ∛[(3/2)(1.786)] </span>

<span>a = 1.39 </span>

<span>b = 7(1.39) - 4(1.39)² = 2.00 </span>

<span>Slope = m = b/a = 2.00 / 1.39 = 1.44</span>

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Answer:

t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539    

p_v =2*P(t_{(9)}>1.539)=0.158  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis.  

Step-by-step explanation:

Data given and notation  

Data: 7.65, 7.60, 7.65, 7.7, 7.55, 7.55, 7.4, 7.4, 7.5, 7.5

We can begin calculating the sample mean given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And the sample deviation given by:

s=\sart{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And we got:

\bar X=7.55 represent the sample mean

s=0.103 represent the sample standard deviation

n=10 sample size  

\mu_o =7.5 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 7.5 or no, the system of hypothesis would be:  

Null hypothesis:\mu = 7.5  

Alternative hypothesis:\mu \neq 7.5  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{7.55-7.5}{\frac{0.103}{\sqrt{10}}}=1.539    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=10-1=9  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(9)}>1.539)=0.158  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis.  

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Step-by-step explanation:

We are given that x is normally distributed with a known standard deviation of 12.6.

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                               P.Q.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample average mean response = 9 hours

            \sigma  = population standard deviation = 12.6

            n = sample of legal professionals = 250

            \mu = mean number of hours a legal professional works

<em>Here for constructing a 95% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

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P(-1.96 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < -1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                                       = [ 9-1.96 \times {\frac{12.6}{\sqrt{250} } } , 9+1.96 \times {\frac{12.6}{\sqrt{250} } } ]

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Therefore, a 95% confidence interval estimate of the mean number of hours a legal professional works on a typical workday is [7.44 hours, 10.56 hours].

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