Can All mixed numbers can be renamed as decimals?
2 answers:
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Answer:
Step-by-step explanation:
We can eliminate options C and D quite quickly because factoring either one would have one equation be y = x and the other either y = 72x - 96 or y = 72x + 108. Neither of these added to x will give either option A or B
so that leaves E.
72x² – 204x + 144 = 0
we can find the zeros
x = (204 ± √(204² - 4(72)(144))) / (2(72))
x = (204 ± 12) / 144
x = 1.5
x = 4/3
so the equation has factors
(x - 1.5)(x - 4/3)
and therefore also a factor in their product
x² - ()x + 2
which is 1/72 of the original quadratic
so all factors of E are
72(x - 1.5)(x - 4/3)
now we need to distribute factors of 72 among the other two factors so that when we add them together the x¹ terms are either 16 or 17 and the x⁰ terms sum to -12. let "a" and "b" be the factors of 72.
ab = 72
a = 72/b
-1.5a - 4b/3 = -12
1.5a + 4b/3 = 12
1.5(72/b) + 4b/3 = 12
108/b + 4b/3 = 12
324/b + 4b = 36
324 + 4b² = 36b
81 + b² = 9b
b² - 9b + 81 = 0
b = (9 ±√(9² - 4(1)(81))) / 2(1))
b = (9 ± √-243) / 2
as both of these roots are imaginary numbers
there is no valid solution to this problem as posed
IF we allow a slight edit to answer B, we can factor 72 into 9•8
y = 8(x - 1.5) y = 9(x - 4/3)
y = 8x - 12 y = 9x - 12
so the sum of the two would be 17x - 24
Answer:
a) 0.32 = 32% probability that your bid will be accepted
b) 0.72 = 72% probability that your bid will be accepted
c) An amount in excess of $15,400.
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability that we find a value X lower than x is given by the following formula.
Assume that the competitor's bid x is a random variable that is uniformly distributed between $10,400 and $15,400.
This means that
a. Suppose you bid $12,000. What is the probability that your bid will be accepted (to 2 decimals)?
You will win if the competitor bids less than 12000. So
0.32 = 32% probability that your bid will be accepted
b. Suppose you bid $14,000. What is the probability that your bid will be accepted?
You will win if the competitor bids less than 14000. So
0.72 = 72% probability that your bid will be accepted
c. What amount should you bid to maximize the probability that you get the property (in dollars)?
His bid is uniformly distributed between $10,400 and $15,400.
So, to maximize the probability that you get the property, you should bid an amount in excess of $15,400.