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3 years ago

14

Scores on a dental anxiety scale range from 0 (no anxiety) to 20 (extreme anxiety). The scores are normally distributed with a

mean of 1111 and a standard deviation of 44. Find the z-score for the given score on this dental anxiety scale:9

1 answer:

EastWind [94]3 years ago

5 0

<span>If scores on a dental anxiety scale range from 0 (no anxiety) to 20 (extreme anxiety), and the scores are normally distributed with a mean of 1111 and a standard deviation of 44, then the z-score for the given score on this dental anxiety scale:9 is .08.
</span>

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Solve using quadratic formula <br><br> 1) y=3x^2-x-6<br><br> 2) y=-5x^2+7x+3

Paul [167]

1) Solving $3x^2-x-6$ we get $x=\frac{1+\sqrt{73}}{6}\,\,and\,\,x=\frac{1-\sqrt{73}}{6}$

2) solving $-5x^2+7x+3$ we get $x=\frac{-7+\sqrt{109}}{-10}\,\,and\,\,x=\frac{-7-\sqrt{109}}{-10}$

Step-by-step explanation:

We need to solve the equations using quadratic formula.

The first equation is:

1) $y=3x^2-x-6$

The quadratic formula is:

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

where a =3, b = -1 and c= -6

Putting values:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-1)\pm\sqrt{(-1)^2-4(3)(-6)}}{2(3)}\\x=\frac{1\pm\sqrt{1+72}}{6}\\x=\frac{1\pm\sqrt{73}}{6}\\so, x=\frac{1+\sqrt{73}}{6}\,\,and\,\,x=\frac{1-\sqrt{73}}{6}$

The second equation is

2) $y=-5x^2+7x+3$

The quadratic formula is:

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

where a =-5, b = 7 and c= 3

Putting values:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(7)\pm\sqrt{(7)^2-4(-5)(3)}}{2(-5)}\\x=\frac{-7\pm\sqrt{49+60}}{-10}\\x=\frac{-7\pm\sqrt{109}}{-10}\\So, x=\frac{-7+\sqrt{109}}{-10}\,\,and\,\,x=\frac{-7-\sqrt{109}}{-10}$

Keywords: Solve Using Quadratic Formula

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Step-by-step explanation

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Romashka [77]

Answer: The value of the expression is $\frac{1+\sqrt{3}} {4}$ .

Explanation:

The given expression is,

$\sin (\frac{3\pi}{4})\cos ( \frac{\pi}{12} )$

Step 1: Break the angles.

$\sin (\pi-\frac{\pi}{4})\cos (\frac{\pi}{4}-\frac{\pi}{3} )$

Step 2: Use quadrant concept to find the value of $\sin (\pi-\frac{\pi}{4})$

$\sin (\frac{\pi}{4})\cos (\frac{\pi}{4}-\frac{\pi}{3} )$

Step 3:Use $\cos (A-B)=\cos A\cos B+\sin A\sin B$

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Step 4: Put these values by using trigonometric table.

$(\frac{1}{\sqrt 2})[(\frac{1}{\sqrt 2})(\frac{1}{2})+(\frac{1}{\sqrt 2})(\frac{\sqrt3}{2})]$

$(\frac{1}{\sqrt 2})[(\frac{1}{2\sqrt 2})+(\frac{\sqrt3}{2\sqrt 2})]$

$(\frac{1}{\sqrt 2})(\frac{1+\sqrt3}{2\sqrt 2})$

$\frac{1+\sqrt3}{4}$

Therefore, the value of the expression is $\frac{1+\sqrt{3}} {4}$ .

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Step-by-step explanation:

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