T(t) = Ts + (T₀ - Ts)exp(-kt), substituting the values:
60 = 50 + (80 - 50)exp(-25k)
10/30 = exp(-25k)
k = 0.0439 °C/min
Answer:
1.734
Step-by-step explanation:
Given that:
A local trucking company fitted a regression to relate the travel time (days) of its shipments as a function of the distance traveled (miles).
The fitted regression is Time = −7.126 + .0214 Distance
Based on a sample size n = 20
And an Estimated standard error of the slope = 0.0053
the critical value for a right-tailed test to see if the slope is positive, using ∝ = 0.05 can be computed as follows:
Let's determine the degree of freedom df = n - 1
the degree of freedom df = 20 - 2
the degree of freedom df = 18
At the level of significance ∝ = 0.05 and degree of freedom df = 18
For a right tailed test t, the critical value from the t table is :
1.734
The answer:
<span>the upper and lower control limits (uclim and lclim) for mean formula is
for the mean chart
uclim= x+A2xR
where x = sum(of the value) / number of each value
and for
lclim=</span>x+A2xR
<span>
R is the range such that R= Xmax - Xmin
in the case of the sample 1: S1
the data are:
79.2 78.8 80.0 78.4 81.0
the mean is x1 = (</span>79.2 + 78.8 + 80.0 + 78.4 + 81.0) / 5= 79.48
<span>its range is R 1= 81.0 -78.4 = 2.6
we can do the same method for finding the mean chart and range for all samples
</span>S2: x2=<span> 80.14 , R2=2.3
</span>S3: x3= 80.14 , R3=1.2
S4: x4= 79.60 , R4=1.7
S5: x5= 80.02 , R5=2.0
S6: x6=80.38 , R6=1.4
<span>
therefore the average value is X= sum( x1+x2+...+x6) / 6 = 79.96
and R=sum(R1+R2+...+R6)/6=1.87
finally
range chart uclim =D4xR=3.95 and lclim is always equal to 0, because D3=0
we can say that the process is not in control.
</span>
We are given
v = 18 m/hr west
θ1 = 285°
θ2 = 340°
After 1 hours, the distance traveled by the ship is
dv = 18 mi
The distance between the ship and the lighthouse is
d = 18 / cos 340
Solve for d<span />
