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s344n2d4d5 [400]
3 years ago
15

What are the roots of the equation? 2x^2+x=6

Mathematics
2 answers:
fredd [130]3 years ago
4 0

Answer:

They would be 3/2 & -2.

ZanzabumX [31]3 years ago
4 0

Answer:

x = - 2, x = \frac{3}{2}

Step-by-step explanation:

Given

2x² + x = 6 ( subtract 6 from both sides )

2x² + x - 6 = 0

Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 2 × - 6 = - 12 and sum = + 1

The factors are + 4 and - 3

Use these factors to split the x- term

2x² + 4x - 3x - 6 = 0 ( factor the first/second and third/fourth terms )

2x(x + 2) - 3(x + 2) = 0 ← factor out (x + 2) from each term

(x + 2)(2x - 3) = 0

Equate each factor to zero and solve for x

x + 2 = 0 ⇒ x = - 2

2x - 3 = 0 ⇒ 2x = 3 ⇒ x = \frac{3}{2}

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What is the next term of the geometric sequence? 4/9, -4/3, 4
andreev551 [17]

Answer:

-12

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Copy dot flip

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Of a group of randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wa
Wittaler [7]

Answer:

We are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%

Step-by-step explanation:

We are given that in a group of randomly selected adults, 160 identified themselves as executives.

n = 160

Also we are given that 42 of executives preferred trucks.

So the proportion of executives who prefer trucks is given by

p = 42/160

p = 0.2625

We are asked to find the 95% confidence interval for the percent of executives who prefer trucks.

We can use normal distribution for this problem if the following conditions are satisfied.

n×p ≥ 10

160×0.2625 ≥ 10

42 ≥ 10 (satisfied)

n×(1 - p) ≥ 10

160×(1 - 0.2625) ≥ 10

118 ≥ 10 (satisfied)

The required confidence interval is given by

$ p \pm z\times \sqrt{\frac{p(1-p)}{n} } $

Where p is the proportion of executives who prefer trucks, n is the number of executives and z is the z-score corresponding to the confidence level of 95%.

Form the z-table, the z-score corresponding to the confidence level of 95% is 1.96

$ p \pm z\times \sqrt{\frac{p(1-p)}{n} } $

$ 0.2625 \pm 1.96\times \sqrt{\frac{0.2625(1-0.2625)}{160} } $

$ 0.2625 \pm 1.96\times 0.03478 $

$ 0.2625 \pm 0.06816 $

0.2625 - 0.06816, \: 0.2625 + 0.06816

(0.1943, \: 0.3306)

(19.43\%, \: 33.06\%)

Therefore, we are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%

5 0
3 years ago
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