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Amanda [17]
3 years ago
15

Jonathan has 8 stamps.some are 3 cent stamps and some are 5 cent stamps.if the total value of the stamps is 36 cents. How many o

f each type of stamp does he have ?
Mathematics
1 answer:
tatiyna3 years ago
8 0
Let x= number of 3 cent stamps 
let y=number of 5 cent stamps
x+y=8
3x+5y=36

3x+5y=36    3x-5y=36
-3(x+y=8) .  -3y-3y=-24
                       2y=12
                       y=6

x+6=8
-6 .   -6
x=2

There will be 2 3 cent stamps and 6 5 cent stamps



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The roots of  equations are as m =  \frac{-3}{2} + \frac{\sqrt{11} }{2}  And n =  \frac{-3}{2} - \frac{\sqrt{11} }{2}    

Step-by-step explanation:

The given quadratic equation is 2 x² + 6 x - 1 = 0

This equation is in form of a x² + b x + c = 0

Let the roots of the equation are ( m , n )

Now , sum of roots = \frac{ - b}{a}

And products of roots = \frac{c}{a}

So, m + n = \frac{ - 6}{2} = - 3

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Or, (m - n)² = (m + n)² - 4mn

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Or, (m - n)² = 9 + 2 = 11

I.e m - n = \sqrt{11}

Again m + n = - 3    And m - n = \sqrt{11}

Solving this two equation

(m + n) + ( m - n) = - 3 + \sqrt{11}

I.e 2 m =  - 3 + \sqrt{11}

Or, m = \frac{-3}{2} + \frac{\sqrt{11} }{2}

Similarly n =  \frac{-3}{2} - \frac{\sqrt{11} }{2}      

Hence the roots of  equations are as m =  \frac{-3}{2} + \frac{\sqrt{11} }{2}  And n =  \frac{-3}{2} - \frac{\sqrt{11} }{2}      Answer

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DerKrebs [107]
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Step-by-step explanation:

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, where \hat{p} is the sample proportion, n is the sample size , z_{\alpha/2} is the critical z-value.

Given : Significance level : \alpha:1-0.99=0.01

Sample size : n= 85

Critical value : z_{\alpha/2}=2.576

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Now, the  99% confidence level will be :

\hat{p}\pmz_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.42\pm(2.576)\sqrt{\dfrac{0.42(1-0.42)}{85}}\\\\\approx0.42\pm0.053\\\\=(0.42-0.053,\ 0.42+0.053)=(0.367,\ 0.473)

Hence, the  99% confidence interval estimate of the true proportion of families who own at least one DVD player is  (0.367,\ 0.473)

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