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1 year ago

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A photograph is 3 in. longer than it is wide. When a 2-in. border is placed around the photograph, the total area of the photogr

aph and the border is 108 in^2. Find the dimensions of the photograph.

1 answer:

Vsevolod [243]1 year ago

4 0

width = 7

length = 10

Step-by-step explanation:

solve for the one letter that gets talked about the most

which is w or width

width = w

length = 3 + w

2 inch border means

width = w + 2

length = 3 + w + 2

area = length times width

area = ( 3 + w + 2 ) times (w + 2 )

108 = ( 3 + w + 2 ) times (w + 2 )

108 = ( w + 5 ) times (w + 2 )

( w + 5 ) times (w + 2 ) = 108

w^2 + 7w +10 = 108

w^2 + 7w - 98 = 0

going to make w = x

x^2+7x-98=0

(x-7)(x+14)

x = 7

x = -14

w=7

w= -14

cant have a negative measurement so w=7 is used

width = w

length = 3 + w

width = 7

length = 3 + 7 = 10

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Step-by-step explanation:

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The first sample mean is $\= x _1 = 8$

The second sample mean is $\= x _2 = 10$

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The first variance is $v_2 = 0.55$

Given that the confidence level is 95% then the level of significance is 5% = 0.05

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

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=> $\sigma_1 = 0.5$

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$\sigma_2 = \sqrt{v_2}$

=> $\sigma_2 = \sqrt{0.55}$

=> $\sigma_2 = 0.742$

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Generally the second standard error is

$SE_2 = \frac{\sigma_2}{\sqrt{n} }$

$SE_2 = \frac{0.742}{\sqrt{60} }$

$SE_2 = 0.09$

Generally the standard error of the difference between their mean scores is mathematically represented as

$SE = \sqrt{SE_1^2 + SE_2^2 }$

=> $SE = \sqrt{0.06^2 +0.09^2 }$

=> $SE = 0.66}$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } * SE) < \mu_1 - \mu_2 < (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } * SE)$

=> $(8 -10) -(1.96 * 0.66) < \mu_1 - \mu_2 < (8-10) +(Z_{\frac{\alpha }{2} } * 0.66)$

=> $-3.29 < \mu_1 - \mu_2 < -0.70$

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