Answer:
- Andre subtracted 3x from both sides
- Diego subtracted 2x from both sides
Step-by-step explanation:
<u>Andre</u>
Comparing the result of Andre's work with the original, we see that the "3x" term on the right is missing, and the x-term on the left is 3x less than it was. It is clear that Andre subtracted 3x from both sides of the equation.
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<u>Diego</u>
Comparing the result of Diego's work with the original, we see that the "2x" term on the left is missing, and the x-term on the right is 2x less than it was. It is clear that Diego subtracted 2x from both sides of the equation.
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<em>Comment on their work</em>
IMO, Diego has the right idea, as his result leaves the x-term with a positive coefficient. He can add 8 and he's finished, having found that x=14.
Andre can subtract 6 to isolate the variable term, and that will give him -x=-14. This requires another step to get to x=14. Sometimes minus signs get lost, so this would not be my preferred sequence of steps.
As a rule, I like to add the opposite of the variable term with the least (most negative) coefficient. This results in the variable having a positive coefficient, making errors easier to avoid.
Answer:
Step-by-step explanation:
step 1
Find the value of x
we know that
r is the midpoint of qs
so
QR=RS
QS=QR+RS------> QS=2RS -----> equation A
RT=RS+ST ----> equation B
see the attached figure to better understand the problem
Substitute the given values in the equation B and solve for x
step 2
Find the value of RS
substitute the value of x
step 3
Find the value of QS
Remember equation A
QS=2RS
so
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
