A photo is printed on an 11-inch by 13-inch piece of paper. The photo covers 80 square inches and has a uniform border. What is
the width of the border?
2 answers:
Let
w--------> <span>width of the border
we know that
(11-2w)*(13-2w)=80--------> 11*13-22w-26w+4w</span>²=80
143-48w+4w²-80=0
4w²-48w +63=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solution are
w1=1.5
w2-10.5--------> is not solution
the answer is
the width of the border is w=1.5 in
w--------> <span>width of the border
we know that
(11-2w)*(13-2w)=80--------> 11*13-22w-26w+4w</span>²=80
143-48w+4w²-80=0
4w²-48w +63=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solution are
w1=1.5
w2-10.5--------> is not solution
the answer is
the width of the border is w=1.5 in
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9514 1404 393
Answer:
13 in by 51 in
Step-by-step explanation:
The area is the product of the dimensions, so is ...
663 = x(4x -1)
4x^2 -x -663 = 0 . . . . . . subtract 663 to put in standard form
Using the quadratic formula, we can find the solutions.
x = (-(-1) ±√((-1)^2 -4(4)(-663)))/(2(4))
x = (1 ± √10609)/8 = (1 ±103)/8
Only the positive solution is of any use in this problem, so ...
x = 104/8 = 13
4x-1 = 4(13)-1 = 51
The dimensions of the rectangle are 13 inches by 51 inches.
__
I find it easiest to solve these using a graphing calculator.
Answer:
We conclude that the true average percentage of organic matter in such soil is different from 3%.
Step-by-step explanation:
We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.
Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.
<em>Let </em><em> = true average percentage of organic matter in such soil</em>
SO, <u>Null Hypothesis</u>, :
= 3% {means that the true average percentage of organic matter in such soil is equal to 3%}
<u>Alternate Hypothesis</u>, :
3% {means that the true average percentage of organic matter in such soil is different than 3%}
The test statistics that will be used here is <u>One-sample t test statistics</u> because we don't know about the population standard deviation;
T.S. = ~
where, = sample mean amount of organic matter = 2.481%
s = sample standard deviation = 1.616%
n = sample of soil specimens = 30
So, <u><em>test statistics</em></u> = ~
= -1.759
<u></u>
<u>Now, P-value of the test statistics is given by;</u>
P-value = P( > -1.759) = <u>0.046</u> or 4.6%
- If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.
- If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.
<em>Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.</em>
Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.