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3 years ago

14

How do you factor 2x^2-9x-5?

1 answer:

maks197457 [2]3 years ago

7 0

Hello from MrBillDoesMath!

Answer: (2x + 1) (x-5)

Discussion:

To factor the polynomial let's find its roots using the quadratic formula:

x = ( -b -+\- sqrt ( b^2 - 4ac)) / 2a which in our case is:

x = ( - (-9) +\- sqrt ( (-9^2 - 4*2*(-5)) ) /(2*2)

x = ( 9 +\- sqrt (81 + 40))/4 =>

x = ( 9 +\- sqrt(121)) / 4 =>

x = ( 9 +\- 11) /4 as sqrt (121) =11

Finally then,

x = ( 9 - 11) /4 = -2/4 = -1/2 and

x = (9 + 11)/4 = 20/4 = 5

So the polynomial factors as follows:

( 2x - (-1) ) ( x - 5) or

(2x + 1) (x-5)

Thank you,

MrB

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Factor each.<br><br><br> Please help!

Naddika [18.5K]

Answer:

1) The factors of $x^2-3x+2=0$ are $\mathbf{(x-1)(x-2)=0}$

Option C is correct.

3) The factors of $x^2+2x-8=0$ are $\mathbf{(x-2)(x+4)=0}$

Option B is correct.

Step-by-step explanation:

1) Factor : $x^2-3x+2=0$

For factoring we need to break the middle term, such that there sum is equal to middle term of expression and product is equal to product of first and last term.

The middle term : -3x

We can break them as (-2x)( -x)

Solving:

$x^2-3x+2=0\\x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0$

So, factors of $x^2-3x+2=0$ are $\mathbf{(x-1)(x-2)=0}$

Option C is correct.

3) Factor: $x^2+2x-8=0$

For factoring we need to break the middle term, such that there sum is equal to middle term of expression and product is equal to product of first and last term.

The middle term : 2x

We can break them as (4x)( -2x)

Solving

$x^2+2x-8=0\\x^2+4x-2x-8=0\\x(x+4)-2(x+4)=0\\(x-2)(x+4)=0$

So, factors of $x^2+2x-8=0$ are $\mathbf{(x-2)(x+4)=0}$

Option B is correct.

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3 years ago

A 95% confidence interval for a population mean is computed from a sample of size 400. Another 95% confidence interval will be c

Anna [14]

Answer:

The interval from the sample of size 400 will be approximately <u>One -half as wide</u> as the interval from the sample of size 100

Step-by-step explanation:

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the 95% confidence interval is dependent on the value of the margin of error at a constant sample mean or sample proportion

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

Here assume that $Z_{\frac{\alpha }{2} } \ and \ \sigma \$ is constant so

$E = \frac{k}{\sqrt{n} }$

=> $E \sqrt{n} = K$

=> $E_1 \sqrt{n}_1 = E_2 \sqrt{n}_2$

So let $n_1 = 400$ and $n_2 = 100$

=> $E_1 \sqrt{400} = E_2 \sqrt{100}$

=> $E_1 = \frac{\sqrt{100} }{\sqrt{400} } E_2$

=> $E_1 = \frac{1}{2 } E_2$

So From this we see that the confidence interval for a sample size of 400 will be half that with a sample size of 100

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