Find the exponential model that fits the

2022-04-15

Find the exponential model that fits the points shown in the table. (Round the exponent to four decimal places.)

x05
y31


 

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Relevant Questions

asked 2022-05-09

Identify each of the following statements as true or false in relation to confidence intervals (CIs). 
Note: 0.5 marks will be taken away for each incorrect answer. The minimum score is 0.

 

 

 

 TrueFalse
A 95% CI is a numerical interval within which we are 95% confident that the true mean μμ lies.  
A 95% CI is a numerical interval within which we are 95% confident that the sample mean x¯¯¯x¯ lies.  
The true mean μμ is always inside the corresponding confidence interval.  
For a sample size n=29n=29, the number of degrees of freedom is n=30n=30.  
If we repeat an experiment 100 times (with 100 different samples) and construct a 95% CI each time, then approximately 5 of those 100 CIs would notnot contain the true mean 𝜇.  
asked 2022-04-13

Laplace transform involving step function
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asked 2022-05-08

The US government is interested in understanding what predicts death rates. They have a set of data that includes the number of deaths in each state, the number of deaths resulting from vehicle accidents (VEHICLE), the number of people dying from diabetes (DIABETES), the number of deaths related to the flu (FLU) and the number of homicide deaths (HOMICIDE).
Your run a regression to predict deaths and get the following output: 
At alpha=0.05, what is indicated by the significance F in this problem? 
A. The regression model does not significantly predict deaths. 
B. The regression model significantly predicts deaths. 
C. Two of the four independent variables significantly predict deaths. 
D. Three of the four independent variables significantly predict deaths. 
Multiple R0.874642613R Square0.764999351Adjusted R Square0.744564512Standart Error62.97881926Obsevations51 

ANOVA

 dfSSMSFSignificanceFRegression4593934..928148483.73237.436035146.36772F14Residual46182451.2571?3966ю331676   Total50776386.1851   

  CoefficientsStandartErrortStartPvalueLower95%Upper95%Intecept240.800228551.133929614.7092063982.31605E05137.8729666343.7274903VEHICLE3.0427829811.5823262211.9229808240.060685930.1422745066.227840468DIABETES11.242122651.6590664896.7761736651.97533E087.90259501714.58165028FLU12.323045842.0570122155.9907499572..9897E078.18249500716.46359668HOMICIDE1.3621284562.1135628170.6444702970.5224715012.8922528375.616509749

asked 2022-05-23

Use the table for z or t-distribution to determine the value of z or t to construct a confidence interval for a population mean for each of the following combinations of confidence and sample size:

 

Confidence interval 90% n=37

asked 2022-04-25

How to solve a cyclic quintic in radicals?
Galois theory tells us that
z111z1=z10+z9+z8+z7+z6+z5+z4+z3+z2+z+1 can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be ζ1,ζ2,,ζ10, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].
A0=x1+x2+x3+x4+x5A1=x1+ζx2+ζ2x3+ζ3x4+ζ4x5A2=x1+ζ2x2+ζ4x3+ζx4+ζ3x5A3=x1+ζ3x2+ζx3+ζ4x4+ζ2x5A4=x1+ζ4x2+ζ3x3+ζ2x4+ζx5
Once one has A0,,A4 one easily gets x1,,x5. It's easy to find A0. The point is that τ takes Aj to ζjAj and so takes Aj5 to Aj5. Thus Aj5 can be written down in terms of rationals (if that's your starting field) and powers of ζ. Alas, here is where the algebra becomes difficult. The coefficients of powers of ζ in A15 are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have A1 as a fifth root of a certain explicit complex number. Then one can express the other Aj in terms of A1. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.

asked 2022-04-01

How to solve a cyclic quintic in radicals?
Galois theory tells us that
z111z1=z10+z9+z8+z7+z6+z5+z4+z3+z2+z+1 can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be ζ1,ζ2,,ζ10, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].
A0=x1+x2+x3+x4+x5A1=x1+ζx2+ζ2x3+ζ3x4+ζ4x5A2=x1+ζ2x2+ζ4x3+ζx4+ζ3x5A3=x1+ζ3x2+ζx3+ζ4x4+ζ2x5A4=x1+ζ4x2+ζ3x3+ζ2x4+ζx5
Once one has A0,,A4 one easily gets x1,,x5. It's easy to find A0. The point is that τ takes Aj to ζjAj and so takes Aj5 to Aj5. Thus Aj5 can be written down in terms of rationals (if that's your starting field) and powers of ζ. Alas, here is where the algebra becomes difficult. The coefficients of powers of ζ in A15 are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have A1 as a fifth root of a certain explicit complex number. Then one can express the other Aj in terms of A1. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.

asked 2022-03-20

y^ prime prime -3y^ prime +2y=10 sin x+2 cos 2x . Find a general solution of the following