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:
Step-by-step explanation:
hope this helps
Answer:
.25 in per hour
Step-by-step explanation:
you just divide 28/7 and you get the per hour and you can double check your work by doing .25 * any hour until you get 28
1. . D . Identity , when you odd both properties it will always to be true on both sides
Complete question is;
An online store receives customer satisfaction ratings between 0 and 100, inclusive. In the first 10 ratings the store received, the average (arithmetic mean) of the ratings was 75. What is the least value the store can receive for the 11th rating and still be able to have an average of at least 85 for the first 20 ratings?
Answer:
50
Step-by-step explanation:
We are told that In the first 10 ratings the store received arithmetic mean of the ratings = 75.
Thus;
Sum of the first 10 ratings = 75 × 10 = 750
Now, for the mean of the first 20 ratings to be at least 85, it means that the sum of the first 20 ratings would be; 85 × 20 = 1700
Thus, the sum of the next 10 ratings would be; 1700 − 750 = 950.
If maximum rating = 100, then the maximum possible value of the sum of the 12th to 20th ratings is given by;
9 × 100 = 900.
Now, in order to make the store have an average of at least 85 for the first 20 ratings, the least value for the 11th rating is;
950 − 900 = 50
