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3 years ago

Write the quadratic equation in general form. (x - 3)^ 2 = 0

2 answers:

5 0

The answer is: x² – 6x + 9 = 0 .
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Explanation:
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Given: (x – 3)² = 0 ; write as: general form: "ax² + bx + c = 0"; a ≠ 0 .

Note: (x – 3)² = (x – 3)(x – 3) = x² – 3x – 3x + 9 = x² – 6x + 9 ;
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Rewrite: (x – 3)² = 0 ; →
as: x² – 6x + 9 = 0 ; which is our answer.
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→ x² – 6x + 9 = 0 ; is in "general form", or "standard equation format"; that is: " ax² + bx + c = 0 "; (a ≠ 0) ;
→ in which:
a = 1 (implied coefficient, since anything multiplied by "1" is that same value);
b = -6;
c = 9
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Kisachek [45]3 years ago

5 0

Easy
expand
(x-3)^2=0
(x-3)(x-3)
foil
first: x time x=x²
outter: -3 times x=-3x
inner:x times -3=-3x
last: -3 times -3=9
add them
x²-3x-3x+9=0
x²-6x+9=0

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Convert 11.42424242 to a rational expression in the form of a/b , where b ≠ 0.

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The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth

Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = $P(Laptop) = p_{X} = \frac{4}{9}$

The probability of selecting a desktop is = $P(Desktop) = p_{Y} = \frac{5}{9}$

Then both X and Y follows Binomial distribution.

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The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select k objects from n distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

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${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$

Thus, there are 70 different ways set up 4 computers out of 8.

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It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

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