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

x | 0 | 5 |
---|---|---|

y | 3 | 1 |

2022-04-15

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

x | 0 | 5 |
---|---|---|

y | 3 | 1 |

You can still ask an expert for help

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.**

True | False | |
---|---|---|

A 95% CI is a numerical interval within which we are 95% confident that the true mean μ$\mu $ lies. | ||

A 95% CI is a numerical interval within which we are 95% confident that the sample mean x¯¯¯$\overline{x}$ lies. | ||

The true mean μ$\mu $ is always inside the corresponding confidence interval. | ||

For a sample size n=29$n=29$, the number of degrees of freedom is n=30$n=30$. | ||

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 not$not$ contain the true mean 𝜇. |

asked 2022-04-13

Laplace transform involving step function

$f\left(t\right)=\frac{\mathrm{sin}\left(2t\right)}{{e}^{2t}}+t\xb7u(t-4)$

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.

$\begin{array}{|cc|}\hline \text{Multiple R}& 0.874642613\\ \text{R Square}& 0.764999351\\ \text{Adjusted R Square}& 0.744564512\\ \text{Standart Error}& 62.97881926\\ \text{Obsevations}& 51\\ \hline\end{array}\phantom{\rule{0ex}{0ex}}$

$\text{ANOVA}$

$\begin{array}{|cccccc|}\hline & df& SS& MS& F& SignificanceF\\ \text{Regression}& 4& \mathrm{593934..928}& 148483.732& 37.43603514& 6.36772F-14\\ \text{Residual}& 46& 182451.2571?3966\u044e331676& \phantom{\rule{0ex}{0ex}}& \phantom{\rule{0ex}{0ex}}& \phantom{\rule{0ex}{0ex}}\\ \text{Total}& 50& 776386.1851& \phantom{\rule{0ex}{0ex}}& \phantom{\rule{0ex}{0ex}}& \phantom{\rule{0ex}{0ex}}\\ \hline\end{array}$

$\phantom{\rule{0ex}{0ex}}\begin{array}{|ccccccc|}\hline & Coefficients& StandartError& tStart& P-value& Lower95\mathrm{\%}& Upper95\mathrm{\%}\\ \text{Intecept}& 240.8002285& 51.13392961& 4.709206398& 2.31605E-05& 137.8729666& 343.7274903\\ \text{VEHICLE}& 3.042782981& 1.582326221& 1.922980824& 0.06068593& -0.142274506& 6.227840468\\ \text{DIABETES}& 11.24212265& 1.659066489& 6.776173665& 1.97533E-08& 7.902595017& 14.58165028\\ \text{FLU}& 12.32304584& 2.057012215& 5.990749957& \mathrm{2..9897}E-07& 8.182495007& 16.46359668\\ \text{HOMICIDE}& 1.362128456& 2.113562817& 0.644470297& 0.522471501& -2.892252837& 5.616509749\\ \hline\end{array}$

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

$\frac{{z}^{11}-1}{z-1}={z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z}^{6}+{z}^{5}+{z}^{4}+{z}^{3}+{z}^{2}+z+1$ 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 $\zeta}^{1},{\zeta}^{2},\dots ,{\zeta}^{10$, 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].

$\begin{array}{rl}{A}_{0}& ={x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}\\ {A}_{1}& ={x}_{1}+\zeta {x}_{2}+{\zeta}^{2}{x}_{3}+{\zeta}^{3}{x}_{4}+{\zeta}^{4}{x}_{5}\\ {A}_{2}& ={x}_{1}+{\zeta}^{2}{x}_{2}+{\zeta}^{4}{x}_{3}+\zeta {x}_{4}+{\zeta}^{3}{x}_{5}\\ {A}_{3}& ={x}_{1}+{\zeta}^{3}{x}_{2}+\zeta {x}_{3}+{\zeta}^{4}{x}_{4}+{\zeta}^{2}{x}_{5}\\ {A}_{4}& ={x}_{1}+{\zeta}^{4}{x}_{2}+{\zeta}^{3}{x}_{3}+{\zeta}^{2}{x}_{4}+\zeta {x}_{5}\end{array}$

Once one has $A}_{0},\dots ,{A}_{4$ one easily gets $x}_{1},\dots ,{x}_{5$. It's easy to find $A}_{0$. The point is that $\tau$ takes $A}_{j$ to $\zeta}^{-j}{A}_{j$ and so takes $A}_{j}^{5$ to $A}_{j}^{5$. Thus $A}_{j}^{5$ can be written down in terms of rationals (if that's your starting field) and powers of $\zeta$. Alas, here is where the algebra becomes difficult. The coefficients of powers of $\zeta$ in $A}_{1}^{5$ 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 $A}_{1$ as a fifth root of a certain explicit complex number. Then one can express the other $A}_{j$ in terms of $A}_{1$. 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

Galois theory tells us that

$\frac{{z}^{11}-1}{z-1}={z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z}^{6}+{z}^{5}+{z}^{4}+{z}^{3}+{z}^{2}+z+1$ 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 $\zeta}^{1},{\zeta}^{2},\dots ,{\zeta}^{10$, 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].

$\begin{array}{rl}{A}_{0}& ={x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}\\ {A}_{1}& ={x}_{1}+\zeta {x}_{2}+{\zeta}^{2}{x}_{3}+{\zeta}^{3}{x}_{4}+{\zeta}^{4}{x}_{5}\\ {A}_{2}& ={x}_{1}+{\zeta}^{2}{x}_{2}+{\zeta}^{4}{x}_{3}+\zeta {x}_{4}+{\zeta}^{3}{x}_{5}\\ {A}_{3}& ={x}_{1}+{\zeta}^{3}{x}_{2}+\zeta {x}_{3}+{\zeta}^{4}{x}_{4}+{\zeta}^{2}{x}_{5}\\ {A}_{4}& ={x}_{1}+{\zeta}^{4}{x}_{2}+{\zeta}^{3}{x}_{3}+{\zeta}^{2}{x}_{4}+\zeta {x}_{5}\end{array}$

Once one has $A}_{0},\dots ,{A}_{4$ one easily gets $x}_{1},\dots ,{x}_{5$. It's easy to find $A}_{0$. The point is that $\tau$ takes $A}_{j$ to $\zeta}^{-j}{A}_{j$ and so takes $A}_{j}^{5$ to $A}_{j}^{5$. Thus $A}_{j}^{5$ can be written down in terms of rationals (if that's your starting field) and powers of $\zeta$. Alas, here is where the algebra becomes difficult. The coefficients of powers of $\zeta$ in $A}_{1}^{5$ 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 $A}_{1$ as a fifth root of a certain explicit complex number. Then one can express the other $A}_{j$ in terms of $A}_{1$. 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