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Inessa05 [86]
4 years ago
9

It is given that f(x) = 2x2 - 12x + 10.

Mathematics
1 answer:
Furkat [3]4 years ago
8 0

Answer:

a = 2, b = - 3, c = - 8

Step-by-step explanation:

Expand f(x) = a(x + b)² + c and compare coefficients of like terms, that is

a(x + b)² + c ← expand (x + b)² using FOIL

= a(x² + 2bx + b²) + c ← distribute parenthesis by a

= ax² + 2abx + ab² + c

Compare like terms with f(x) = 2x² - 12x + 10

Compare coefficients x² term

a = 2

Compare coefficients of x- term

2ab = - 12, substitute a = 2

2(2)b = - 12

4b = - 12 ( divide both sides by 4 )

b = - 3

Compare constant term

ab² + c = 10 , substitute a = 2, b = - 3

2(- 3)² + c = 10

18 + c = 10 ( subtract 18 from both sides )

c = - 8

Then a = 2, b = - 3, c = - 8

 

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If a fair coin is flipped 15 times, what is the probability that there are more heads than tails?
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The probability that there are more heads than tails is equal to \dfrac{1}{2}.

Step-by-step explanation:

Since the number of flips is an odd number, there can't be an equal number of heads and tails. In other words, there are either

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Let the event that there are more heads than tails be A. \lnot A (i.e., not A) denotes that there are more tails than heads. Either one of these two cases must happen. As a result, P(A) + P(\lnot A) = 1.

Additionally, since this coin is fair, the probability of getting a head is equal to the probability of getting a tail on each toss. That implies that (for example)

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Due to this symmetry,

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In other words P(A) = P(\lnot A).

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\begin{aligned}P(A) =& P(n \ge 8) \cr =& \sum \limits_{i = 8}^{15} {15 \choose i} (0.5)^{i} (0.5)^{15 - i}\cr =& \sum \limits_{i = 8}^{15} {15 \choose i} (0.5)^{15}\cr =& (0.5)^{15} \left({15 \choose 8} + {15 \choose 9} + \cdots + {15 \choose 15}\right) \cr =& (0.5)^{15} \left({15 \choose (15 - 8)} + {15 \choose (15 - 9)} + \cdots + {15 \choose (15 - 15)} \right) \cr =& (0.5)^{15} \left({15 \choose 7} + {15 \choose 6} + \cdots + {15 \choose 0}\right)\end{aligned}

\begin{aligned}\phantom{P(A)} =& \sum \limits_{i = 0}^{7} {15 \choose i} (0.5)^{15}\cr =& P(n \le 7) \cr =& P(\lnot A)\end{aligned}.

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