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oee [108]
4 years ago
7

The design of a digital box camera maximizes the volume while keeping the sum of the dimensions at 6 inches. If the length must

be 1.5 times the height, what should each dimension be?
Hint: Let x represent one of the dimensions, then define the other dimensions in terms of x.
Mathematics
1 answer:
love history [14]4 years ago
4 0
H = x, L = 1.5 x;  
L + W + H = 6
x + 1.5 x + W = 6
W = 6 - 2.5 x
Dimensions of the camera in terms of x:
x,  1.5 x,  6 - 2.5 x.
V = L x W x H
V = x * 1.5 x * ( 6 - 2.5 x ) = 9 x² - 3.75 x³
V ` = 18 x - 11.25 x² ( V max is when V` = 0 )
18 x - 11.25 x² = 0
x ( 18 - 11.25 x ) = 0
11.25 x = 18
x = 18 : 11.25
x = 1.6
The dimensions are:
L = 2.4
W = 2.0
H = 1.6
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Use the data set to determine which statements are correct. Check all that apply. 115, 120, 118, 104, 109, 148, 135, 141, 139

a)The median is 120.

b)The median is 109.

c)There is an outlier.

d) The lower quartile is 118.

e)The lower quartile is 112.

f)The upper quartile is 140.

g)The upper quartile is 141.

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a)The median is 120.

e)The lower quartile is 112

f)The upper quartile is 140

h)The interquartile range is 28

Step-by-step explanation:

We are given the above Data sets:

115, 120, 118, 104, 109, 148, 135, 141, 139

To confirm the correct options, we have to find the median, lower and upper quartile and the interquartile range

115, 120, 118, 104, 109, 148, 135, 141, 139

We have to find rearrange this numbers from lowest to highest

Hence we have:

104, 109, 115, 118, 120, 135, 139, 141, 148

1) Median

104, 109, 115, 118, 120, 135, 139, 141, 148

The formula for median =

1/2(n + 1)th value

n = number of terms = 9

=1/2(9 + 1)th value

= 1/2(10)th value

= 5th value

The median is the 5th value = 120

Option a) is correct

2) Lower Quartile

The formula for lower quartile=

1/4(n + 1)th value

n = 9

= 1/4(9 + 1)th value

= 1/4(10)th value

= 2.5th value

This means the value is between the 2nd and 3rd values

109= 2nd value

115= 3rd value

First Quartile = 109+ 115/2

= 224/2

= 112

Therefore, Option e)The lower quartile is 112.

3)Upper Quartile

The formula for upper quartile=

3/4(n + 1)th value

n = 9

= 3/4(9 + 1)th value

= 3/4(10)th value

= 7.5th value

This means the value is between the 7th and 8th values

139= 7th value

141 = 8th value

First Quartile = 139 + 141/2

= 280/2

= 140

Therefore, option f)The upper quartile is 140 is correct.

4) The formula for Interquartile range is

Upper quartile - Lower quartile

= 140 - 112

= 28

Therefore, the option h)The interquartile range is 28 is correct.

Looking at the data set below, their is no outlier present.

104, 109, 115, 118, 120, 135, 139, 141, 148

Hence, c)There is an outlier is not correct.

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