## MATH1011 Quizzes

Quiz 4: Data scaling
Question 1 Questions
It is suspected that the set of data $\begin{array}{cccccc}\hfill t\hfill & \hfill 1.0\hfill & \hfill 2.0\hfill & \hfill 5.0\hfill & \hfill 10.0\hfill & \hfill 15.0\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 3.5\hfill & \hfill 3.9\hfill & \hfill 5.3\hfill & \hfill 8.7\hfill & \hfill 14.34\hfill \end{array}$ is exponential. Find the best approximation for the given data.
 a) $y=3.2{e}^{0.1t}$ b) $y=2×1{0}^{-4}{e}^{10t}$ c) $y=3.5{t}^{0.13}$ d) $y=3.5{t}^{7.7}$

Choice (a) is correct!
We do a semi-log transformation and to 2 decimal places we find $\begin{array}{cccccc}\hfill t\hfill & \hfill 1.0\hfill & \hfill 2.0\hfill & \hfill 5.0\hfill & \hfill 10.0\hfill & \hfill 15.0\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill Y=lny\hfill & \hfill 1.25\hfill & \hfill 1.36\hfill & \hfill 1.67\hfill & \hfill 2.16\hfill & \hfill 2.66\hfill \end{array}$ A scatterplot shows that the relationship between $t$ and $Y$ is linear. You should draw the scatterplot to verify this. So letting $Y=mt+b$ we find $m=0.09$ and $b=1.15$. Thus $y={e}^{1.15}{e}^{0.1t}=3.2{e}^{0.1t}$.
Choice (b) is incorrect
Choice (c) is incorrect
This equation does not define an exponential relationship.
Choice (d) is incorrect
This equation does not define an exponential relationship.
It is suspected that the set of data $\begin{array}{ccccccc}\hfill t\hfill & \hfill 0.25\hfill & \hfill 0.5\hfill & \hfill 0.75\hfill & \hfill 1.0\hfill & \hfill 1.25\hfill & \hfill 1.5\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 0.66\hfill & \hfill 1.09\hfill & \hfill 1.79\hfill & \hfill 2.96\hfill & \hfill 4.87\hfill & \hfill 8.03\hfill \end{array}$ is exponential. Find the best approximation for the given data.
 a) $y=0.6{e}^{0.5t}$ b) $y=0.4{e}^{2t}$ c) $y=1.7{t}^{0.7}$ d) $y=4.6{t}^{1.4}$

Choice (a) is incorrect
Choice (b) is correct!
We do a semi-log transformation and to 2 decimal places we find $\begin{array}{ccccccc}\hfill t\hfill & \hfill 0.25\hfill & \hfill 0.5\hfill & \hfill 0.75\hfill & \hfill 1.0\hfill & \hfill 1.25\hfill & \hfill 1.5\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill Y=lny\hfill & \hfill -0.42\hfill & \hfill 0.09\hfill & \hfill 0.58\hfill & \hfill 1.09\hfill & \hfill 1.58\hfill & \hfill 2.08\hfill \end{array}$ With $Y=mt+b$ we find $m=0.09$ and $b=-0.92$. Thus $y={e}^{-0.92}{e}^{2t}=0.4{e}^{2t}$.
Choice (c) is incorrect
This equation does not define an exponential relationship.
Choice (d) is incorrect
This equation does not define an exponential relationship.
It is suspected that the following set of data $\begin{array}{cccccc}\hfill x\hfill & \hfill 5.0\hfill & \hfill 5.5\hfill & \hfill 6.0\hfill & \hfill 6.5\hfill & \hfill 7.0\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 1.36\hfill & \hfill 1.50\hfill & \hfill 1.66\hfill & \hfill 1.83\hfill & \hfill 2.03\hfill \end{array}$ is either exponential or allometric. Find the best approximation for the given data.
 a) $y=2×1{0}^{-11}{e}^{5x}$ b) $y=0.5{e}^{0.2x}$ c) $y=1.1{x}^{0.16}$ d) $y=0.5{x}^{6.2}$

Choice (a) is incorrect
Choice (b) is correct!
Transforming the data by a log-log transformation and a semi-log transformation we see that the semi-log transformation is closest to linear.
Choice (c) is incorrect
Choice (d) is incorrect
The pH of a solution is defined to be $pH=-lo{g}_{10}\left[{H}^{+}\right]$, where $\left[{H}^{+}\right]$ is its hydrogen ion concentration. Hydrochloric acid has a pH of 1, while water has a pH of 7. Which of the following is correct?
 a) The number of hydrogen ions in a cup of hydrochloric acid is $1{0}^{6}$ times the number in a cup of water. b) The number of hydrogen ions in a cup of hydrochloric acid is $1{0}^{-6}$ times the number in a cup of water. c) The number of hydrogen ions in a cup of hydrochloric acid is $1{0}^{7}$ times the number in a cup of water. d) The number of hydrogen ions in a cup of hydrochloric acid is $1{0}^{-7}$ times the number in a cup of water.

Choice (a) is correct!
${\left[{H}^{+}\right]}_{HCl}=\frac{1}{10}$ whilst ${\left[{H}^{+}\right]}_{{H}_{2}O}=\frac{1}{1{0}^{7}}$
Choice (b) is incorrect
Incorrect answer.
Choice (c) is incorrect
Incorrect answer.
Choice (d) is incorrect
Incorrect answer.
Uranium-238 has a half-life of $4.46×1{0}^{9}$ years. A radioactive rock is found to contain 16 grams of Uranium-238. After how many years does it contain just 4 grams?
 a) $4.46×1{0}^{9}$ years b) $2.23×1{0}^{9}$ years c) $8.92×1{0}^{9}$ years d) $1.784×1{0}^{16}$ years

Choice (a) is incorrect
Incorrect answer.
Choice (b) is incorrect
Incorrect answer.
Choice (c) is correct!
It will take two half-lives for the sample to reach a quarter of its original size, since $\frac{1}{4}={\left(\frac{1}{2}\right)}^{2}$.
Choice (d) is incorrect
Incorrect answer.
Which of the following transformations linearises the data? $\begin{array}{cccccc}\hfill x\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 1\hfill & \hfill -5\hfill & \hfill -15\hfill & \hfill -29\hfill & \hfill -47\hfill \end{array}$
 a) $X={x}^{2},Y=y$ b) $X=\frac{1}{x},Y=y$ c) $X=\sqrt{x},Y=y$ d) $X=x,Y=y$

Choice (a) is correct!
Choice (b) is incorrect
Incorrect answer
Choice (c) is incorrect
Incorrect answer
Choice (d) is incorrect
Incorrect answer