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
Remember that when you bring ÷ to the other side of the equal sign it'll become multiply instead
Answer:
Step-by-step explanation:
Corresponding scores before and after taking the course form matched pairs.
The data for the test are the differences between the scores before and after taking the course.
μd = scores before taking the course minus scores before taking the course.
a) For the null hypothesis
H0: μd ≥ 0
For the alternative hypothesis
H1: μd < 0
b) We would assume a significance level of 0.05. The P-value from the test is 0.65. The p value is high. It increases the possibility of accepting the null hypothesis.
Since alpha, 0.05 < than the p value, 0.65, then we would fail to reject the null hypothesis. Therefore, it does not provide enough evidence that scores after the course are greater than the scores before the course.
c) The mean difference for the sample scores is greater than or equal to zero
Answer:
The mean is 19.5
Step-by-step explanation:
You have to add all of the numbers up then divide by how many numbers there are.
Ex: 5, 5, 6, 7
add them up and you get 23
Divide 23 by how many numbers you have, which in this case, is 4
The answer to this example would be 5.75
(x + 4) (x + 2)
=x^2 +2x + 4x + 8
=x^2 + 6x + 8
the other factor is (x + 2)
