Step-by-step explanation:
Given - In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen concrete has a low value of thermal conductivity in order to minimize subsequent damage due to changing temperatures. Suppose two types of concrete, a graded aggregate and a no-fines aggregate, are being considered for a certain road. The table below summarizes data on thermal conductivity from an experiment carried out to compare the two types of concrete.
Type ni xi si
Graded 42 0.486 0.187
No-fines 42 0.359 0.158
To find - a. Formulate the above in terms of a hypothesis testing problem.
b. Give the test statistic and its reference distribution (under the null hypothesis).
c. Report the p-value of the test statistic and use it to assess the evidence that this sample provides on the scientific question of difference in mean conductivity of the two materials at the 5% level of significance.
Proof -
a.)
Hypothesis testing problem :
H0 : There is significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.
H1 : There is no significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.
b)
Test statistic :




⇒Z(cal) = 3.3687
Z(tab) = 1.96
As Z (cal) > Z(tab)
So, we reject H0 at 5% Level of significance
p-value = 0.99962
Hence
There is significant difference in mean conductivity at the two materials.
Answer:
468.4 meters
Step-by-step explanation:
Find the width of the rectangle.
The area of a rectangle is A=length*width. Substitute the values of length and area and solve for width.
A=length*width
8,400=140*width
60= width
Use the width of the rectangle to find the circumference of the semicircles. Since each semicircle is half of a circle, the perimeter of the two semicircles is equal to the circumference of one circle.
The circumference of a circle is equal to pid, where d is the diameter. The diameter of the semicircle is the same as the width of the rectangle.
So, the diameter is 60 meters. Substitute the diameter into the formula for the circumference and simplify using 3.14 for pi.
≈188.4
(2 times 140)+ 188.4
So, the perimeter of the track is 468.4 meters.
We know that
if the club includes one set of identical
triplets wearing matching clothes
then
the number of different arrangements that are
possible is
10! / 3! = (10*9*8*7*6*5*4*3!)/3!
=604,800
The answer is
<span>604,800</span>
Answer:
51.0791%
Step-by-step explanation:
Calculate percentage change
from V1 = 41700 to V2 = 63000
(V2−V1)|V1|×100
=(63000−41700)|41700|×100
=2130041700×100
=0.510791×100
=51.0791%change
=51.0791%increase