From the Arithmetic Information given, Pr = 0.049375 ≠ -8.050625 ≠ 0.0325 See explanation below.
<h3>
What are the step by steps solution to the questions above?</h3>
First, lets us restate the question properly. We have
Pr = 1 - ((6*12+6)/80)² = 1 - ((9*12²)/1600) - ((3*12*6)/1600) - (6²/6400) = 1 - 0.005625 * 12² - 0.001875 * 12 * 6 - 0.00015625 * 6²
Note that there are three equals signs. So lets divide the problem according and solve for the different parts.
Lets 1 - ((6*12+6)/80)² ............A
1 - ((9*12²)/1600) - ((3*12*6)/1600) - (6²/6400) .............B; and
1 - 0.005625 * 12² - 0.001875 * 12 * 6 - 0.00015625 * 6² ........C
Solving for A we have
1 - (78/80)²
= 1 - (0.975)²
= 1 - 0.950625
A = 0.049375
Solving for B we have
1 - ((9*12²)/1600) - ((3*12*6)/1600) - (6²/6400)
= 1- (14,256/1600) - (216/1600) - (36/6400)
= 1 - 8.91 - 0.135 - 0.005625
B = -8.050625
Solving for C we have
1 - 0.005625 * 12² - 0.001875 * 12 * 6 - 0.00015625 * 6²
= 1 - 0.005625 * 144 - 0.001875 * 12 * 6 - 0.00015625 * 144
= 0.0325
In summary we can state that:
A = 0.049375
B = -8.050625
C = 0.0325
Given that there were no abstract quantities, we can state that
Pr = A ≠ B ≠C or
Pr = 0.049375 ≠ -8.050625 ≠ 0.0325
Learn more about equations with equal signs at
brainly.com/question/18924248
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Answer:
the maximum number of crates that can be stacked between the floor and ceiling
where 1 foot = 12 inches. SO the answer is that a maximum of 10 crates can be stacked from floor to ceiling.
Step-by-step explanation:
i) the maximum number of crates that can be stacked between the floor and ceiling
where 1 foot = 12 inches
Answer:
I believe the answer is 152 ft
Step-by-step explanation:
On the side, you would find the area of the square (8x5=40). Then, you would find the area of the triangular front (6x4=24 divded by 2 so it's 12). There are two sides of each, so 24+80=104. Then, you dfind the aarea of the bottom square (6x8=48) and add them all together to make 152
do parentheses first which gives you 42-4+111 then aubtract to get 38+111 which equals 149
