# Which of the following equations have the same solution set? Give reasons for your answers that do not depend on solving the equations. l.x-5=3x+7 ll.3x-6=7x+8 lll.15x-9=6x+24 lV.6x-16=14x+12 V.9x+21=3x-15 Vl.-0.05+frac{x}{100}=3frac{x}{100}+0.07

Question
Equations
Which of the following equations have the same solution set? Give reasons for your answers that do not depend on solving the equations.
l.$$x-5=3x+7$$
ll.$$3x-6=7x+8$$
lll.$$15x-9=6x+24$$
lV.$$6x-16=14x+12$$
V.$$9x+21=3x-15$$
Vl.$$-0.05+\frac{x}{100}=3\frac{x}{100}+0.07$$

2020-10-28

Equations l,V and Vl all have the same solution set.
We can obtain V from l by multiplying both sides of l by 3 then applying symmetric property(switching sides):
$$(x-5)(3)=(3x+7)(3)$$
$$3x-15=9x+21$$
$$9x+21=3x-15$$
We can obtain Vl from l by multiplying both sides of l by 100 then applying commutative property on the left side:
$$\frac{x-5}{100}=\frac{3x+7}{100}$$
$$\frac{x}{100-0.05}=\frac{3x}{100+0.07}$$
$$\frac{-0.05+x}{100}=\frac{3x}{100}+0.07$$
Equations ll and lV have the same solution set.
We can obtain lV from ll by multiplying both sides of ll by 2 then substracting 4 from both sides:
$$(3x-6)(2)=(7x+8)(2)$$
$$6x-12=14x+16$$
$$6x-12-4=14x+16-4$$
$$6x-16=14x+12$$
Equations lll does not have the same solution set as the other equations since it cannot be transformed from l or ll.
Results:l,V, and Vl all have the same solution set.
ll and lV have the same solution set.
lll does not have the same solution set as the other equations.

### Relevant Questions

$$\begin{array}{cc}\hline & \text{Afraid to walk at night?} \\ \hline & \text{Yes} & \text{No} & \text{Total} \\ \hline \text{Male} & 173 & 598 & 771 \\ \hline \text{Female} & 393 & 540 & 933 \\ \hline \text{Total} & 566 & 1138 & 1704 \\ \hline \text{Source:}2014\ GSS \end{array}$$
If the chi-square $$(\chi 2)$$ test statistic $$=73.7$$ what is the p-value you would report? Use Table C.Remember to calculate the df first.
Group of answer choices
$$P>0.250$$
$$P=0.01$$
$$P<0.001$$
$$P<0.002$$

$$\begin{array}{|c|cc|}\hline & \text{Right-Tail Probability} \\ \hline df & 0.250 & 0.100 & 0.050 & 0.025 & 0.010 & 0.005 & 0.001 \\ \hline 1 & 1.32 & 2.71 & 3.84 & 5.02 & 6.63 & 7.88 & 10.83 \\ 2 & 2.77 & 4.61 & 5.99 & 7.38 & 9.21 & 10.60 & 13.82 \\ 3 & 4.11 & 6.25 & 7.81 & 9.35 & 11.34 & 12.84 & 16.27 \\ 4 & 5.39 & 7.78 & 9.49 & 11.14 & 13.28 & 14.86 & 18.47 \\ 5 & 6.63 & 9.24 & 11.07 & 12.83 & 15.09 & 16.75 & 20.52 \\ 6&7.84&10.64&12.59&14.45&16.81&18.55&22.46 \\ 7&9.04&12.02&14.07&16.01&18.48&20.28&24.32\\ 8&10.22&13.36&15.51&17.53&20.09&21.96&26.12 \\ 9&11.39&14.68&16.92&19.02&21.67&23.59&27.88 \\ 10&12.55&15.99&18.31&20.48&23.21&25.19&29.59 \\ 11&13.70&17.28&19.68&21.92&24.72&26.76&31.26 \\ 12&14.85&18.55&21.03&23.34&26.22&28.30&32.91 \\ 13&15.98&19.81&22.36 & 24.74 & 27.69 & 29.82 & 34.53 \\ 14 & 17.12 & 21.06 & 23.68 & 26.12 & 29.14 & 31.32 & 36.12 \\15 & 18.25 & 22.31 & 25.00 & 27.49 & 30.58 & 32.80 & 37.70 \\ 16 & 19.37 & 32.54 & 26.30 & 28.85 & 32.00 & 34.27 & 39.25 \\ 17 & 20.49 & 24.77 & 27.59 & 30.19 & 33.41 & 35.72 & 40.79 \\ 18 & 21.60 & 25.99 & 28.87 & 31.53 & 34.81 & 37.16 & 42.31 \\ 19 & 22.72 & 27.20 & 30.14 & 32.85 & 36.19 & 38.58 & 43.82 \\ 20 & 23.83 & 28.41 & 31.41 & 34.17 & 37.57 & 40.00 & 45.32 \\ \hline \end{array}$$
Find the discriminant of each equation and determine whether the equation has (1) two nonreal complex solutions, (2) one real solution with a multiplicity of 2, or (3) two real solutions. Do not solve the equations. $$7x^{2} - 2x - 14 = 0$$
Consider the accompanying data on flexural strength (MPa) for concrete beams of a certain type.
$$\begin{array}{|c|c|}\hline 11.8 & 7.7 & 6.5 & 6 .8& 9.7 & 6.8 & 7.3 \\ \hline 7.9 & 9.7 & 8.7 & 8.1 & 8.5 & 6.3 & 7.0 \\ \hline 7.3 & 7.4 & 5.3 & 9.0 & 8.1 & 11.3 & 6.3 \\ \hline 7.2 & 7.7 & 7.8 & 11.6 & 10.7 & 7.0 \\ \hline \end{array}$$
a) Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion. $$[Hint.\ ?x_{j}=219.5.]$$ (Round your answer to three decimal places.)
MPa
State which estimator you used.
$$x$$
$$p?$$
$$\frac{s}{x}$$
$$s$$
$$\tilde{\chi}$$
b) Calculate a point estimate of the strength value that separates the weakest $$50\%$$ of all such beams from the strongest $$50\%$$.
MPa
State which estimator you used.
$$s$$
$$x$$
$$p?$$
$$\tilde{\chi}$$
$$\frac{s}{x}$$
c) Calculate a point estimate of the population standard deviation ?. $$[Hint:\ ?x_{i}2 = 1859.53.]$$ (Round your answer to three decimal places.)
MPa
Interpret this point estimate.
This estimate describes the linearity of the data.
This estimate describes the bias of the data.
This estimate describes the spread of the data.
This estimate describes the center of the data.
Which estimator did you use?
$$\tilde{\chi}$$
$$x$$
$$s$$
$$\frac{s}{x}$$
$$p?$$
d) Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa. [Hint: Think of an observation as a "success" if it exceeds 10.] (Round your answer to three decimal places.)
e) Calculate a point estimate of the population coefficient of variation $$\frac{?}{?}$$. (Round your answer to four decimal places.)
State which estimator you used.
$$p?$$
$$\tilde{\chi}$$
$$s$$
$$\frac{s}{x}$$
$$x$$
Solve the given set of equations for value of x:
x-3z=-5
2x-y+2z=16
7x-3y-5z=19
Use the following Normal Distribution table to calculate the area under the Normal Curve (Shaded area in the Figure) when $$Z=1.3$$ and $$H=0.05$$;
Assume that you do not have vales of the area beyond $$z=1.2$$ in the table; i.e. you may need to use the extrapolation.
Check your calculated value and compare with the values in the table $$[for\ z=1.3\ and\ H=0.05]$$.
Calculate your percentage of error in the estimation.
How do I solve this problem using extrapolation?
$$\begin{array}{|c|c|}\hline Z+H & Prob. & Extrapolation \\ \hline 1.20000 & 0.38490 & Differences \\ \hline 1.21000 & 0.38690 & 0.00200 \\ \hline 1.22000 & 0.38880 & 0.00190 \\ \hline 1.23000 & 0.39070 & 0.00190 \\ \hline 1.24000 & 0.39250 & 0.00180 \\ \hline 1.25000 & 0.39440 & 0.00190 \\ \hline 1.26000 & 0.39620 & 0.00180 \\ \hline 1.27000 & 0.39800 & 0.00180 \\ \hline 1.28000 & 0.39970 & 0.00170 \\ \hline 1.29000 & 0.40150 & 0.00180 \\ \hline 1.30000 & 0.40320 & 0.00170 \\ \hline 1.31000 & 0.40490 & 0.00170 \\ \hline 1.32000 & 0.40660 & 0.00170 \\ \hline 1.33000 & 0.40830 & 0.00170 \\ \hline 1.34000 & 0.41010 & 0.00180 \\ \hline 1.35000 & 0.41190 & 0.00180 \\ \hline \end{array}$$
Solve the system. If the system does not have one unique solution, also state whether the system is onconsistent or whether the equations are dependent.
2x-y+z=-3
x-3y=2
x+2y+z=-7
Consider the curves in the first quadrant that have equationsy=Aexp(7x), where A is a positive constant. Different valuesof A give different curves. The curves form a family,F. Let P=(6,6). Let C be the number of the family Fthat goes through P.
A. Let y=f(x) be the equation of C. Find f(x).
B. Find the slope at P of the tangent to C.
C. A curve D is a perpendicular to C at P. What is the slope of thetangent to D at the point P?
D. Give a formula g(y) for the slope at (x,y) of the member of Fthat goes through (x,y). The formula should not involve A orx.
E. A curve which at each of its points is perpendicular to themember of the family F that goes through that point is called anorthogonal trajectory of F. Each orthogonal trajectory to Fsatisfies the differential equation dy/dx = -1/g(y), where g(y) isthe answer to part D.
Find a function of h(y) such that x=h(y) is the equation of theorthogonal trajectory to F that passes through the point P.
The pmf of the amount of memory X (GB) in a purchased flash drive is given as the following.
$$\begin{array}{|c|c|}\hline x & 1 & 2 & 4 & 8 & 16 \\ \hline p(x) & 0.05 & 0.10 & 0.30 & 0.45 & 0.10 \\ \hline \end{array}$$
a) Compute E(X). (Enter your answer to two decimal places.) GB
b) Compute V(X) directly from the definition. (Enter your answer to four decimal places.) $$GB^{2}$$
c) Compute the standard deviation of X. (Round your answer to three decimal places.) GB
d) Compute V(X) using the shortcut formula. (Enter your answer to four decimal places.) $$GB^{2}$$
The unstable nucleus uranium-236 can be regarded as auniformly charged sphere of charge Q=+92e and radius $$\displaystyle{R}={7.4}\times{10}^{{-{15}}}$$ m. In nuclear fission, this can divide into twosmaller nuclei, each of 1/2 the charge and 1/2 the voume of theoriginal uranium-236 nucleus. This is one of the reactionsthat occurred n the nuclear weapon that exploded over Hiroshima, Japan in August 1945.
C. In this model the sum of the kinetic energies of the two"daughter" nuclei is the energy released by the fission of oneuranium-236 nucleus. Calculate the energy released by thefission of 10.0 kg of uranium-236. The atomic mass ofuranium-236 is 236 u, where 1 u = 1 atomic mass unit $$\displaystyle={1.66}\times{10}^{{-{27}}}$$ kg. Express your answer both in joules and in kilotonsof TNT (1 kiloton of TNT releases 4.18 x 10^12 J when itexplodes).