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:
option (c) n = 201
Step-by-step explanation:
Data provided in the question:
Standard deviation, s = 5.5 ounce
Confidence level = 99%
Length of confidence interval = 2 ounces
Therefore,
margin of error, E = (Length of confidence interval ) ÷ 2
= 2 ÷ 2
= 1 ounce
Now,
E = 
here,
z = 2.58 for 99% confidence interval
n = sample size
thus,
1 = 
or
n = (2.58 × 5.5)²
or
n = 201.3561 ≈ 201
Hence,
option (c) n = 201
Think of the entire population of 6th graders here. 3/7 are boys and 4/7 are girls. 5/8 of these boys are in Ms. Jones' class; that fraction would be (5/8)(3/7), or 15/56. This 15/56 represents the fraction of the entire sixth grade class who are in Ms. Jones' class.
