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devlian [24]
3 years ago
12

Solve using quadratic formula.

Mathematics
2 answers:
svp [43]3 years ago
6 0

Answer:

1. 2/5,-3 2. x=\frac{-5+-\sqrt{13} }{6}

Step-by-step explanation:

i used the quadratic formula to find x also please note that 2 has 2 answers bc of the +- beofre the sqrt of 13  

marin [14]3 years ago
4 0

Step-by-step explanation:

<h2>1).</h2>

5x² + 13x - 6 = 0

Using the quadratic formula

x =  \frac{ - b± \sqrt{ {b}^{2}  - 4ac} }{2a}

a = 5 , b = 13 c = - 6

We have

x =  \frac{  - 13± \sqrt{ {13}^{2} - 4(5)( - 6) } }{2(5)}

x =  \frac{ - 13± \sqrt{169  + 120} }{10}

x =  \frac{ - 13± \sqrt{289} }{10}

x =  \frac{ - 13±17}{10}

x =  \frac{ - 13 + 17}{10}  \:  \:  \:  \:  \: or \:  \:  \:  \: x =  \frac{ - 13 - 17}{10}

<h3>x = 2/5 or x = - 3</h3>

<h2>2).</h2>

3x² + 5x + 1 = 0

a = 3 , b = 5 , c = 1

x =  \frac{ -5 ±  \sqrt{ {5}^{2}  - 4(3)(1)} }{2(3)}

x =  \frac{ - 5± \sqrt{25 - 12} }{6}

x =  \frac{ - 5± \sqrt{13} }{6}

x =  \frac{ -  5 +  \sqrt{13} }{6}  \:  \:  \:  \: or \:  \:   \: x =  \frac{ - 5 -  \sqrt{13} }{6}

Hope this helps you

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Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

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

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So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

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P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

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Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

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