Question

here are the temperatures of a glass of warm water after its put in a freezer:
t (time) c(t) (temperature)
0 80
2 64.3
4 52.7
6 42.6
8 34.4
10 27.5
which model best fits the data?
choose 1 answer:
a ( c(t) = 80 - (7.6) cdot t )
b ( c(t) = 80 cdot (0.81)^t )
c ( c(t) = 80 cdot (0.9)^t )
d ( c(t) = 80 - (15.7) cdot t )

Explanation:

Step1: Test linear model A

Calculate values for $t=2,4,6,8,10$:
$C(2)=80-7.6\times2=80-15.2=64.8$
$C(4)=80-7.6\times4=80-30.4=49.6$
$C(6)=80-7.6\times6=80-45.6=34.4$
$C(8)=80-7.6\times8=80-60.8=19.2$
$C(10)=80-7.6\times10=80-76=4$

Step2: Test exponential model B

Calculate values for $t=2,4,6,8,10$:
$C(2)=80\times(0.81)^2=80\times0.6561=52.488$
$C(4)=80\times(0.81)^4=80\times0.43046721=34.437$
$C(6)=80\times(0.81)^6=80\times0.282429536=22.594$
$C(8)=80\times(0.81)^8=80\times0.185302019=14.824$
$C(10)=80\times(0.81)^{10}=80\times0.121576654=9.726$

Step3: Test exponential model C

Calculate values for $t=2,4,6,8,10$:
$C(2)=80\times(0.9)^2=80\times0.81=64.8$
$C(4)=80\times(0.9)^4=80\times0.6561=52.488$
$C(6)=80\times(0.9)^6=80\times0.531441=42.515$
$C(8)=80\times(0.9)^8=80\times0.43046721=34.437$
$C(10)=80\times(0.9)^{10}=80\times0.34867844=27.894$

Step4: Test linear model D

Calculate values for $t=2,4,6,8,10$:
$C(2)=80-15.7\times2=80-31.4=48.6$
$C(4)=80-15.7\times4=80-62.8=17.2$
$C(6)=80-15.7\times6=80-94.2=-14.2$
$C(8)=80-15.7\times8=80-125.6=-45.6$
$C(10)=80-15.7\times10=80-157=-77$

Step5: Compare to actual data

Compare calculated values to given $C(t)$: 64.3,52.7,42.6,34.4,27.5. Model C's values are the closest.

Answer:

C. $C(t) = 80 \cdot (0.9)^t$