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Olin [163]
3 years ago
7

Which shows the equation below written in the form ax2 + bx + c = 0?

Mathematics
2 answers:
GalinKa [24]3 years ago
6 0
X + 9 = 2 ( x - 1 )²
x + 9 = 2 ( x² - 2 x + 1 )
x + 9 = 2 x² - 4 x + 2
- 2 x² + 4 x + x - 2 + 9 = 0
- 2 x² + 5 x + 7 = 0   / * ( -1 )
Answer:
A ) 2 x² - 5 x - 7 
nikitadnepr [17]3 years ago
6 0

Answer:

Option A is correct

2x^2-5x-7=0

Step-by-step explanation:

Given the equation:

x+9=2(x-1)^2

Using identity:

(a-b)^2=a^2-2ab+b^2

then;

x+9 = 2(x^2-2x+1)

Using distributive property: a \cdot (b+c) = a\cdot b+ a\cdot c

x+9 = 2x^2-4x+2

Subtract x from both sides we have;

9= 2x^2-5x+2

Subtract 9 from both sides we have;

0=2x^2-5x-7

or

2x^2-5x-7=0

Therefore, the equation 2x^2-5x-7=0 is written in the form of ax^2+bx+c=0

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Jada’s teacher ask her to write a three-digit number that uses the digit 4 once and the digit 7 twice. The value of one 7 needs
viva [34]
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The value of the first 7 from the left is 70.

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6 0
3 years ago
Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving
lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

P(x=0) = 0.778

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ^5C_0 (0.4)^0 * (0.6)^5

= 1 - 0.0778

= 0.9222

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

= 1 - 0.6826

= 0.3174

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0
3 years ago
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