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4 years ago

Solve by factoring. x2 + 6x = - 9. Hint: This is a double root.

2 answers:

6 0

Subtract -9 from both sides.

x2+6x−(−9)=−9−(−9)

x2+6x+9=0

Factor left side of equation.

(x+3)(x+3)=0

Set factors equal to 0.

x+3=0 or x+3=0

x=−3

Answer:

x=−3

Doss [256]4 years ago

3 0

1
Move all terms to one side

x^2+6x+9=0

2
Rewrite x^2+6x+9 in the form a^2+2ab+b^2 , where a=x and b=3

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

3
Use Square of Sum (a+b)^2=a^2+2ab+b^2

(x+3)^2=0

4
Take the square root of both sides

x+3=0

5
Subtract 3 from both sides

x=−3

So, in short, your answer is x = - 3

I hope this helps, and have a great day. :)

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3 years ago

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3 years ago

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $5.15-1.96 \times {\frac{1.68}{\sqrt{4400} } }$ , $5.15+1.96 \times {\frac{1.68}{\sqrt{4400} } }$ ]

= [5.10 , 5.20]

Therefore, 95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

8 0

3 years ago

What is the discriminante of 9x^2+10x

Leni [432]

The discriminant of y=ax^2+bx+c is b^2-4ac

given
9x^2+10x+0
a=9
b=10
c=0

discriminant=10^2-4(9)(0)=100-0=100

the discriminant is 100

6 0

4 years ago