Newton's Law of Cooling
Tf=Ts+(Ti-Ts)e^(-kt) where Tf is temp at time t, Ts is temp of surroundings, Ti is temp of object/fluid. So we need to find k first.
200=68+(210-68)e^(-10k)
132=142e^(-10k)
132/142=e^(-10k)
ln(132/142)=-10k
k=-ln(132/142)/10
k≈0.0073 so
T(t)=68+142e^(-0.0073t) so how long until it reaches 180°?
180=68+142e^(-0.0073t)
112=142e^(-0.0073t)
112/142=e^(-0.0073t)
ln(112/142)=-0.0073t
t= -ln(112/142)/(0.0073)
t≈32.51 minutes
Answer:
d. The histogram would be approximately bell-shaped, and the normal quantile plot would have data points have follow a straight-line pattern.
Step-by-step explanation:
Since the variable is normally distributed, the histogram of women's height should be approximately bell shaped (if the data was obtained form a random sample).
Again, the variable is normally distributed, therefore, the quantile plot should follow a straight line pattern (a diagonal line to be more precise).
Answer:
(3,16)
Step-by-step explanation:
The degrees of freedom of the critical value F are (k-1,n-k).
We are given that there are four sample group, so,
k=4.
Also, we are given that the each four groups contains five observations, so,
n=4*5
n=20
The critical value F has degree of freedom
(k-1,n-k)
(4-1,20-4)
(3,16).
Thus, the degrees of freedom for the critical value of F are (3,16).
Answer:
1 hour = 7/18 of the distance between two ports
Step-by-step explanation:
Given that:
In 3/7 hours distance traveled is 1/6
For finding distance in one hour, divide both sides by 7/3 so that 3/7 would be cancelled out:
3/7 * 7/3 hours = 1/6 * 7/3 distance
By simplifying:
1 hour = 7/18 of the distance between two ports
This means that in one hour 7/18 of the distance is covered between two ports.
i hope it will help you!
Answer: its A
Step-by-step explanation:
Because
