Let:
C ( 0 ) = 0 ∘ C C ( 100 ) = 100 ∘ C Y ( 0 ) = 10 ∘ Y ( 100 ) = 130 ∘ Y ( 127 ) = 127 ∘ \begin{array}{l}
C(0) = 0{}^{\circ} \mathrm{C} \\
C(100) = 100{}^{\circ} \mathrm{C} \\
Y(0) = 10{}^{\circ} \\
Y(100) = 130{}^{\circ} \\
Y(127) = 127{}^{\circ} \\
\end{array} C ( 0 ) = 0 ∘ C C ( 100 ) = 100 ∘ C Y ( 0 ) = 10 ∘ Y ( 100 ) = 130 ∘ Y ( 127 ) = 127 ∘ C ( 127 ) = ? C(127) = ? C ( 127 ) = ?
It's possible to present dependence C from Y as equation:
C ( x ) = a ∗ Y ( x ) + b C(x) = a * Y(x) + b C ( x ) = a ∗ Y ( x ) + b
Write a system of equations:
{ 10 a + b = 0 130 a + b = 100 \left\{
\begin{array}{l}
10a + b = 0 \\
130a + b = 100
\end{array}
\right. { 10 a + b = 0 130 a + b = 100
Solve a system:
{ b = − 10 a 130 a + b = 100 b = − 10 a 130 a − 10 a = 100 \left\{
\begin{array}{l}
b = -10a \\
130a + b = 100 \\
b = -10a \\
130a - 10a = 100
\end{array}
\right. ⎩ ⎨ ⎧ b = − 10 a 130 a + b = 100 b = − 10 a 130 a − 10 a = 100 { 10 a + b = 0 a = 100 120 \left\{
\begin{array}{l}
10a + b = 0 \\
a = \frac{100}{120}
\end{array}
\right. { 10 a + b = 0 a = 120 100 a = 10 12 , b = − 100 / 12 a = \frac{10}{12}, \quad b = -100/12 a = 12 10 , b = − 100/12
Equation:
C ( x ) = 10 12 Y ( x ) − 100 / 12 C(x) = \frac{10}{12} Y(x) - 100/12 C ( x ) = 12 10 Y ( x ) − 100/12
Enter data:
C ( 127 ) = 127 10 12 − 100 12 C ( 127 ) = 97.5 ∘ C \begin{array}{l}
C(127) = 127 \frac{10}{12} - \frac{100}{12} \\
C(127) = 97.5{}^{\circ} \mathrm{C} \\
\end{array} C ( 127 ) = 127 12 10 − 12 100 C ( 127 ) = 97.5 ∘ C
Answer:
"A" - 97.5°C