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A straight line is fit to a data set (ln x, y). This line intercepts the abscissa at ln x = 0.1 and with...
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A straight line is fit to a data set $(ln x, y)$. This line intercepts the abscissa at $ln x = 0.1$ and has a slope of $−0.02$. What is the value of $y$ at $x = 5$ from the fit?
I can not understand what is asking in this problem .
Can anyone please help me to understand?
numerical-methods analytic-geometry
asked Feb 1 at 6:58
sanisani
216211
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add a comment |
$begingroup$
A straight line is fit to a data set $(ln x, y)$. This line intercepts the abscissa at $ln x = 0.1$ and has a slope of $−0.02$. What is the value of $y$ at $x = 5$ from the fit?
I can not understand what is asking in this problem .
Can anyone please help me to understand?
numerical-methods analytic-geometry
asked Feb 1 at 6:58
sanisani
216211
$endgroup$
$begingroup$
For clarity, let's set $t=ln{x}$. So we assume $y=y(t)$ and that it's a linear relationship, $y=at+b$. We also know that it crosses the abscissa at $t=0.1$, in other words $y(t=0.1) = 0$ or $atimes 0.1 + b = 0$. So we have one equation for the coefficients $a$ and $b$. How would you form the second equation?
$endgroup$
– Matti P.
Feb 1 at 7:02
add a comment |
$begingroup$
A straight line is fit to a data set $(ln x, y)$. This line intercepts the abscissa at $ln x = 0.1$ and has a slope of $−0.02$. What is the value of $y$ at $x = 5$ from the fit?
I can not understand what is asking in this problem .
Can anyone please help me to understand?
numerical-methods analytic-geometry
asked Feb 1 at 6:58
sanisani
216211
$endgroup$
A straight line is fit to a data set $(ln x, y)$. This line intercepts the abscissa at $ln x = 0.1$ and has a slope of $−0.02$. What is the value of $y$ at $x = 5$ from the fit?
I can not understand what is asking in this problem .
Can anyone please help me to understand?
numerical-methods analytic-geometry
numerical-methods analytic-geometry
asked Feb 1 at 6:58
sanisani
216211
asked Feb 1 at 6:58
sanisani
216211
asked Feb 1 at 6:58
sanisani
216211
asked Feb 1 at 6:58
sanisani
216211
asked Feb 1 at 6:58
sanisani
216211
216211
$begingroup$
For clarity, let's set $t=ln{x}$. So we assume $y=y(t)$ and that it's a linear relationship, $y=at+b$. We also know that it crosses the abscissa at $t=0.1$, in other words $y(t=0.1) = 0$ or $atimes 0.1 + b = 0$. So we have one equation for the coefficients $a$ and $b$. How would you form the second equation?
$endgroup$
– Matti P.
Feb 1 at 7:02
add a comment |
$begingroup$
For clarity, let's set $t=ln{x}$. So we assume $y=y(t)$ and that it's a linear relationship, $y=at+b$. We also know that it crosses the abscissa at $t=0.1$, in other words $y(t=0.1) = 0$ or $atimes 0.1 + b = 0$. So we have one equation for the coefficients $a$ and $b$. How would you form the second equation?
$endgroup$
– Matti P.
Feb 1 at 7:02
$begingroup$
For clarity, let's set $t=ln{x}$. So we assume $y=y(t)$ and that it's a linear relationship, $y=at+b$. We also know that it crosses the abscissa at $t=0.1$, in other words $y(t=0.1) = 0$ or $atimes 0.1 + b = 0$. So we have one equation for the coefficients $a$ and $b$. How would you form the second equation?
$endgroup$
– Matti P.
Feb 1 at 7:02
$begingroup$
For clarity, let's set $t=ln{x}$. So we assume $y=y(t)$ and that it's a linear relationship, $y=at+b$. We also know that it crosses the abscissa at $t=0.1$, in other words $y(t=0.1) = 0$ or $atimes 0.1 + b = 0$. So we have one equation for the coefficients $a$ and $b$. How would you form the second equation?
$endgroup$
– Matti P.
Feb 1 at 7:02
add a comment |
1 Answer
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y and lnx have a linear relationship.
Let lnx=t then y=y(t).Therefore y=mt+c———>1
Given that the line has an x intercept at lnx=0.1(y=0).
Using 1,
$$0=m(0.1)+c$$
Implies,
$$c=-0.1m——->2$$.
Given that the slope is -0.02 which means that m=-0.02.
From 2,
$$c=-0.1(-0.02)$$
$$c=0.002$$
Therefore the line equation has to be
$$y=-0.02t+0.002$$
i.e,
$$y=-0.02lnx+0.002$$
Therefore when x=5
$$y=-0.02ln5+0.002$$
$$yapprox-0.0301$$
answered Feb 1 at 8:59
user631874user631874
84
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
y and lnx have a linear relationship.
Let lnx=t then y=y(t).Therefore y=mt+c———>1
Given that the line has an x intercept at lnx=0.1(y=0).
Using 1,
$$0=m(0.1)+c$$
Implies,
$$c=-0.1m——->2$$.
Given that the slope is -0.02 which means that m=-0.02.
From 2,
$$c=-0.1(-0.02)$$
$$c=0.002$$
Therefore the line equation has to be
$$y=-0.02t+0.002$$
i.e,
$$y=-0.02lnx+0.002$$
Therefore when x=5
$$y=-0.02ln5+0.002$$
$$yapprox-0.0301$$
answered Feb 1 at 8:59
user631874user631874
84
$endgroup$
add a comment |
$begingroup$
y and lnx have a linear relationship.
Let lnx=t then y=y(t).Therefore y=mt+c———>1
Given that the line has an x intercept at lnx=0.1(y=0).
Using 1,
$$0=m(0.1)+c$$
Implies,
$$c=-0.1m——->2$$.
Given that the slope is -0.02 which means that m=-0.02.
From 2,
$$c=-0.1(-0.02)$$
$$c=0.002$$
Therefore the line equation has to be
$$y=-0.02t+0.002$$
i.e,
$$y=-0.02lnx+0.002$$
Therefore when x=5
$$y=-0.02ln5+0.002$$
$$yapprox-0.0301$$
answered Feb 1 at 8:59
user631874user631874
84
$endgroup$
add a comment |
$begingroup$
y and lnx have a linear relationship.
Let lnx=t then y=y(t).Therefore y=mt+c———>1
Given that the line has an x intercept at lnx=0.1(y=0).
Using 1,
$$0=m(0.1)+c$$
Implies,
$$c=-0.1m——->2$$.
Given that the slope is -0.02 which means that m=-0.02.
From 2,
$$c=-0.1(-0.02)$$
$$c=0.002$$
Therefore the line equation has to be
$$y=-0.02t+0.002$$
i.e,
$$y=-0.02lnx+0.002$$
Therefore when x=5
$$y=-0.02ln5+0.002$$
$$yapprox-0.0301$$
answered Feb 1 at 8:59
user631874user631874
84
$endgroup$
y and lnx have a linear relationship.
Let lnx=t then y=y(t).Therefore y=mt+c———>1
Given that the line has an x intercept at lnx=0.1(y=0).
Using 1,
$$0=m(0.1)+c$$
Implies,
$$c=-0.1m——->2$$.
Given that the slope is -0.02 which means that m=-0.02.
From 2,
$$c=-0.1(-0.02)$$
$$c=0.002$$
Therefore the line equation has to be
$$y=-0.02t+0.002$$
i.e,
$$y=-0.02lnx+0.002$$
Therefore when x=5
$$y=-0.02ln5+0.002$$
$$yapprox-0.0301$$
answered Feb 1 at 8:59
user631874user631874
84
answered Feb 1 at 8:59
user631874user631874
84
answered Feb 1 at 8:59
user631874user631874
84
answered Feb 1 at 8:59
user631874user631874
84
84
add a comment |
add a comment |
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$begingroup$
For clarity, let's set $t=ln{x}$. So we assume $y=y(t)$ and that it's a linear relationship, $y=at+b$. We also know that it crosses the abscissa at $t=0.1$, in other words $y(t=0.1) = 0$ or $atimes 0.1 + b = 0$. So we have one equation for the coefficients $a$ and $b$. How would you form the second equation?
$endgroup$
– Matti P.
Feb 1 at 7:02