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Natasha2012 [34]
3 years ago
14

Determine whether the following arguments are inductive or deductive. If an argument is inductive, determine whether it is stron

g or weak. If it is deductive, determine whether it is valid or invalid. When a cook cannot recall the ingredients in a recipe, it is appropriate that she refresh her memory by consulting the recipe book. Similarly, when a student cannot recall the answers during a final exam, it is appropriate that she refresh her memory by consulting the textbook.
Mathematics
1 answer:
Akimi4 [234]3 years ago
3 0

Answer:

inductive; weak

Step-by-step explanation:

The argument is inductive because the main point of the argument or theory was presented later at the end of the argument, after opening the argument with anecdotal evidence or observations. However, no matter how true the premises may seem to be, the argument is weak, because the conclusion is false and does not really follow from the premises. Nothing in her premises suggests consulting her textbook during final exams is acceptable. Consulting textbook during exams is certainly a prohibited act, and as such, is not the best way to refresh her memory.

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An article in Technometrics (Vol. 19, 1977, p. 425) presents the following data on the motor fuel octane ratings of several blen
Slav-nsk [51]

Answer:

Median = 90.4

Q_1 = 88.6

Q_3 = 92.2

Step-by-step explanation:

Given

The above data

Required

- A stem and leaf display

- The median

- The quartiles

First, determine the range of the data

Smallest = 83.4

Highest = 100.3

Next, group each dataset base on common whole numbers.

So, we have:

83.4

84.3\ 84.3

85.3

86.7\ 86.7\ 86.7

87.4\ 87.5\ 87.6\ 87.7\ 87.8\ 87.9

88.2\ 88.3\ 88.3\ 88.3\ 88.4\ 88.5\ 88.5\ 88.6\ 88.6\ 88.7\ 88.9

89.0\ 89.2\ 89.3\ 89.3\ 89.6\ 89.7\ 89.8\ 89.8\ 89.9\ 89.9

90.0\ 90.1\ 90.1\ 90.1\ 90.3\ 90.4\ 90.4\ 90.4\ 90.5\ 90.6\ 90.7\ 90.8\ 90.9

91.0\ 91.0\ 91.0\ 91.1\ 91.1\ 91.1\ 91.2\ 91.2\ 91.2\ \ 91.5\  91.6\ 91.6\ 91.8\ 91.8

92.2\ 92.2\ 92.2\ 92.3\ 92.6\ 92.7\ 92.7\ 92.7

93.0\ 93.2\ 93.3\ 93.3\ 93.4\ 93.7

94.2\ 94.2\ 94.4\ 94.7

96.1\ 96.5

98.8

100.3

Next, we construct the stem and leaf plot.

The whole numbers will be the stem while the decimal parts will be the leaf.

So, we have:

\begin{array}{ccc}{Stem} & {} & {Leaf} & {83} & {|} & {.4} & {84} & {|} & {.3\ .3} & {85} & {|} & {.3} & {86} & {|} & {.7\ .7\ .7} & {87} &{|} & {.4\ .5\ .6\ .7\ .8\ .9} & {88} & {|} & {.2\ .3\ .3\ .3\ .4\ .5\ .5\ .6\ .6\ .7\ .9} &{89} & {|} & {.0\ .2\ .3\ .3\ .6\ .7\ .8\ .8\ .9\ .9} & {90} & {|} &{.0\ .1\ .1\ .1\ .3\ .4\ .4\ .4\ .5\ .6\ .7\ .8\ .9} & {91} &{|}&{.0\ .0\ .0\ .1\ .1\ .1\ .2\ .2\ .2\ .5\ .6\ .6\ .8\ .8} &{92} &{|} &{.2\ .2\ .2\ .3\ .6\ .7\ .7\ .7} \ \end{array}

  \begin{array}{ccc} {93} & {|} & {.0\ .2\ .3\ .3\ .4\ .7} & {94} &{|} & {.2\ .2\ .4\ .7} &{96} & {|} & {.1\ .5} & {98} & {|} & {.8} & {100} &{|} &{.3} \ \end{array}

From the above plot,

n = 83

The median is calculated as:

Median = \frac{n+1}{2}th

Median = \frac{83+1}{2}th

Median = \frac{84}{2}th

Median = 42nd

i.e. the median is at the 42nd position.

From the above stem and leaf plot.

The 42nd position is at stem 90 and the leaf  .4

So the median is:

Median = 90.4

The lower quartile (Q) is calculated as:

Q_1 = \frac{n+1}{4}th

Q_1 = \frac{83+1}{4}th

Q_1 = \frac{84}{4}th

Q_1 = 21st

i.e. the lower quartile is at the 21st position.

From the above stem and leaf plot.

The 42nd position is at stem 88 and the leaf .6

So the lower quartile is:

Q_1 = 88.6

The upper quartile (Q3) is calculated as:

Q_3 = 3 * \frac{n+1}{4}th

Q_3 = 3 * \frac{83+1}{4}th

Q_3 = 3 * \frac{84}{4}th

Q_3 = 3 * 21th

Q_3 = 63rd

i.e. the upper quartile is at the 63rd position.

From the above stem and leaf plot.

The 63rd position is at stem 92 and the leaf .2

So the upper quartile is:

Q_3 = 92.2

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3 years ago
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Vitek1552 [10]

The length of side of triangle is 5 units and length of side of pentagon is 1.5 units.

<u>SOLUTION:</u>

Given, An equilateral triangle has sides of length (3k-2) units.  

Then, perimeter of the triangle will be 3 \times(3 k-2)=9 k-6

A regular pentagon (5 sides) has sides of (2k+0.5) units.  

Then, perimeter of the pentagon will be 5 \times(2 k+0.5)=10 k+2.5

We have to find the dimensions of each shape . Now, the perimeter of the triangle is twice the perimeter of the pentagon,  

So, \text {perimeter of triangle} = 2\times \text{perimeter of pentagon}

\begin{array}{l}{\rightarrow 9 k-6=2(10 k+2.5)} \\\\ {\rightarrow 9 k-6=20 k+5} \\\\ {\rightarrow 20 k-9 k=-6-5} \\\\ {\rightarrow 11 k=-11} \\\\ {\rightarrow k=-1}\end{array}

Then, length of side of triangle = 3(-1)-2 = - 5

Length of side of pentagon = 2(-1) + 0.5 = - 1.5

We have to neglect, negative sign as lengths can’t be negative. Even if we change the sign above all conditions are satisfied.

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