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Descnbe the Confidanca Interval In Iwo lo three complete sentencel, uxplain whal A coritidence interval meens [ ganerel; IN1 yOJ WLre talkig sonxo1 Kto hu" no...

Question

Descnbe the Confidanca Interval In Iwo lo three complete sentencel, uxplain whal A coritidence interval meens [ ganerel; IN1 yOJ WLre talkig sonxo1 Kto hu" not teken sulistcs complele sentarcos; oxplain What thils confcaicu interval menjia for thla parlicular gud}: Corstruat a contkderce Iriturval for each ccnfidenc# Wvul given.ConndencelctaEBPTErOr BounoConildonceIntanani5092WatTValat tuppong Ine EDP 1 {7a contance luvot mcm_ 4" ? Doun Ina *iuth oline contioonca Interynl Ircruacn Ar

Descnbe the Confidanca Interval In Iwo lo three complete sentencel, uxplain whal A coritidence interval meens [ ganerel; IN1 yOJ WLre talkig sonxo1 Kto hu" not teken sulistcs complele sentarcos; oxplain What thils confcaicu interval menjia for thla parlicular gud}: Corstruat a contkderce Iriturval for each ccnfidenc# Wvul given. Conndencelcta EBPTErOr Bouno ConildonceIntanani 5092 WatT Valat tuppong Ine EDP 1 {7a contance luvot mcm_ 4" ? Doun Ina *iuth oline contioonca Interynl Ircruacn Ar dcciensa? Explain AIYthe Txppen 711.17 {1 0,] '888



Answers

In Problems 65-72, verify each function has a zero in the indicated interval. Then use the Intermediate Value Theorem to approximate the zero correct to three decimal places by repeatedly subdividing the interval containing the zero into 10 sub intervals. $$ f(x)=x^{3}-4 x+2 ; \text { interval: }(1,2) $$

For the given problem, we're going to be um looking closely at these functions, we want to verify that each function has a zero in the indicated interval. And then use the intermediate value theorem to approximate the correct zero to three decimal places. So if we have a full backs equals x cubed, Let's react -5. And without even looking at this, we know that over the interval since this is a polynomial, we know it's going to be continuous. FF one gives us a negative number, but FF two gives us a positive number. So we know that somewhere in between there were going to reach um zero. And if we use the intermediate value theorem to approximate the zero by dividing subdividing the interval and containing zero, We see that if we bring this down further, so 1.5 or one point 01, you see it's so negative at 1.05 Or 1.1, we see that we're getting closer to the zero so we can approximate it And we end up approximating it to being 1.154. And that's gonna be our final answer.

For this problem, we are asked to partition the the interval from 1-4 into three sub intervals of equal length. So we first want to figure out the length of our intervals, so it's going to be the end point minus. The beginning point divided by n four minus one, is three divided by three, gives us one. So our sub intervals will be uh let me figure out how to write this precisely here. The first sub interval will be just 1-2 non inclusive. second sub interval will be 2-3 non inclusive, And the third sub interval will be 3-4 inclusive.

For the given problem, we want to consider Effa Becks Equal to X to the 4th -2 x cubed. Yeah. Mhm Plus 21 x minus 23. There's going to be from the interval of 1- two. So you see that this is going to include the zero by the intermediate value theorem because this is continuous and this goes from negative to positive. So we see if we do about 1.1, we still have a negative value, we're going to make this 1.3 and it's positive. 1.2 weeks ago is positive. So I'm more concerned about 1.15 for example, and this will be 1.16. So we see 1.16 is going to be I'm an approximation. But if we get closer we see 1.157 is negative. 1.158 is positive. So based on this we see 1.157 Is our best three decimal approximation.

With a given problem, we want to consider the function F. Of X. Equalling X to the 4th minus X cubed. Uh huh. Class acts right next to. We're looking over the interval from 1-2. So f of one, I'm gonna give us negative. One effort to is going to give us eight. Um and then we're going to bring this closer, so 1.3. Um and then about 1.4. So actually that's very close because this is negative and this positive Gonna be closer though to 1.3. So if we bring this down to 1.31 At the me here, so 1.305, It's also negative. So we would say 1.306307. Um so about 1.308 or 1.309 Based on this 1.309 will be a better approximation.


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