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2 years ago

Solve for x: (3x−2)(2x+3)=0

2 answers:

8 0

Hi again!

(3x -2)(2x +3) = 0
6x² + 5x - 6 = 0
Now we have to factor the left side
(3x -2)(2x +3) = 0
Set factors equal to 0
3x -2 =0     or 2x + 3 =0
3x = 0 + 2    or 2x= 0 -3
3x = 2          or    2x = -3
x = 2/3 or x =-3/2

I hope that helps!

Zanzabum2 years ago

3 0

(3x−2)(2x+3)=0
(3x−2)=0; x = 2/3
(2x+3)=0; x = -3/2

answer
x = -3/2 and x = 2/3

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3 years ago

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P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample % of U.S. public high school students who were obese = 15%

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p = population percentage of all U.S. public high school students

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population proportion, p 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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.15-1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } }$ , $0.15+1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } }$ ]

= [0.110 , 0.190]

Therefore, 95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

6 0

3 years ago