Answer:1.69*10^12 J
Step-by-step explanation:
From figure above, using triangle ratio
485/755.5=y/l. Cross multiplying 485l=755.5y Divide via 485) hence l= 755.5y/485
Consider a slice volume Vslice= (755.5y/485)^2∆y; recall density =150lb/ft^3
Force slice = 150*755.5^2.y^2.∆y/485^2
From figure 2 in the attachment work done for elementary sclice
Wslice= 150.755.5^2.y^2.∆y.(485-y)/485^2
= (150*755.5^2*y^2)(485-y)∆y/485
To calculate the total work we integrate from y=0 to y= 485
Ie W=[ integral of 150*755.5^2 *y^2(485-y)dy/485] at y=0 and y= 485
Integrating the above
W= 150*755.5^2/485[485*y^3/3-y^4/4] at y= 0 and y=485
W= 150*755.5^2/485(485*485^3/3-484^4/4)-(485.0^3/3-0^4/4)
Work done 1.69*10^12joules
<h2>
Answer with explanation:</h2>
The confidence interval for population mean is given by :-
(1)
, where 
= sample mean
z* = critical value.
SE = standard error
and 
, 
= population standard deviation.
n= sample size.
As per given , we have
n= 15
It is known that ring diameter is normally distributed.
By z-table ,
The critical value for 95% confidence = z*= 1.96
A 99% two-sided confidence interval on the true mean piston diameter :
(using (1))
[Rounded to three decimal places]
∴ A 99% two-sided confidence interval on the true mean piston diameter = (74.020, 74.022)
By z-table ,
The critical value for 95% confidence = z*= 1.96
A 95% lower confidence bound on the true mean piston diameter:
(using (1))
[Rounded to three decimal places]
∴ A 95% lower confidence bound on the true mean piston diameter= 74.020
11 minutes has passed just subtract
Answer:
hb=4√3
Step-by-step explanation:
