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iogann1982 [59]
2 years ago
12

Solve for x. Round your answer to the nearest hundredth.

Mathematics
2 answers:
anastassius [24]2 years ago
7 0

Answer:

Choice C is correct answer.

Step-by-step explanation:

Given equation is :

2x²-6x+3= 0

ax²+bx+c = 0 is general quadratic equation.

x = (-b±√b²-4ac) / 2a is quadratic formula to solve general quadratic equation.

comparing given equation with general quadratic equation,we get

a = 2 , b = -6 and c = 3

putting above values in quadratic formula: we get

x = (-(-6)±√(-6)²-4(2)(3)) / 2(2)

x = (6±√36-24) / 4

x = (6±√12) / 4

x = (6±2√3) / 4

x = 2(3±√3) / 4

x = 3±√3 / 2

x = 3±1.73 / 2

x = (3+1.73) / 2 or x = (3-1.73) / 2

x = 4.73 / 2 or x = 1.27 / 2

x =  2.37  or x = .63 is solution of 2x²-6x+3 = 0.



Zarrin [17]2 years ago
5 0

Answer:

Option B is correct

Step-by-step explanation:

2x² - 6x + 3 = 0

Using Quadratic formula:

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

a = 2, b = -6, c = 3

x = \frac-(-6)+-{\sqrt{(-6)^{2}-4(2)(3)}}{2(2)}

x = \frac{-(-6)+-\sqrt{(-6)^{2}-4(2)(3)}}{2(2)}\\

=\frac{6+-\sqrt{36-24}}{4}\\\\=\frac{6+-\sqrt{12}}{4}\\\\=\frac{6+-3.46}{4}\\\\

=\frac{6+3.46}{4}

x = 2.37  


=\frac{6-3.46}{4}

 x = 0.63      

   


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Use the empirical rule to solve the problem. The annual precipitation for one city is normally distributed with a mean of 288 in
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Answer:

z=-1.99

z=1.99

And if we solve for a we got

a=288 -1.99*3.7=214.4

a=288 +1.99*3.7=295.4

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the annual precipitation of a population, and for this case we know the distribution for X is given by:

X \sim N(288,3.7)  

Where \mu=288 and \sigma=3.7

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Using this condition we can find the limits

z=-1.99

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Answer:

  • <u>5</u><u> </u>is the value which makes the equation true .

Step-by-step explanation:

In this question we have provided an equation that is <u>9</u><u> </u><u>(</u><u> </u><u>3x</u><u> </u><u>-</u><u> </u><u>1</u><u>6</u><u> </u><u>)</u><u> </u><u>+</u><u> </u><u>1</u><u>5</u><u> </u><u>=</u><u> </u><u>6</u><u>x</u><u> </u><u>-</u><u> </u><u>2</u><u>4</u><u> </u>. And we are asked to <u>write </u><u>the </u><u>steps </u><u>to </u><u>solve </u><u>the </u><u>equation </u><u>with </u><u>explanation </u> and <u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>X </u><u>.</u>

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\longmapsto \quad \: 9(3x - 16) + 15 = 6x - 24

<u>Step </u><u>1</u><u> </u><u>:</u> Solving parenthesis on left side using distributive property which means multiplying 9 with 3x as well as -16 :

\longmapsto \quad \:27x -  \bold{144 }+  \bold{15 }= 6x - 24

<u>Step </u><u>2</u><u> </u><u>:</u> Solving like terms on left side that are -144 and 15 :

\longmapsto \quad \:27x -129 = 6x - 24

<u>Step </u><u>3 </u><u>:</u> Adding 129 on both sides :

\longmapsto \quad \:27x - \cancel{129} -  \cancel{129} = 6x \bold{ - 24 } +  \bold{129}

Now on cancelling -129 with 129 on left side and solving the terms that are -24 and 129 on right side , We get :

\longmapsto \quad \:27x = 6x + 105

<u>Step </u><u>4</u><u> </u><u>:</u> Subtracting with 6x on both sides :

\longmapsto \quad \: \bold{27x} -  \bold{6x} =  \cancel{6x} +105 - \cancel{ 6x}

On calculating further, We get :

\longmapsto \quad \:21x =  105

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\longmapsto \quad \: \dfrac{ \cancel{21}x}{ \cancel{21}}  = \dfrac{105}{ 21}

Now , by cancelling 21 with 21 on left side , We get :

\longmapsto \quad \:x =  \cancel{\dfrac{105}{21}}

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  • 9 ( 15 - 16 ) + 15 = 30 - 24

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  • -9 + 15 = 6

  • 6 = 6

  • L.H.S = R.H.S

  • Hence , Verified .

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