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Ahat [919]
3 years ago
7

What is the equation of the graphed line written in standard form?

Mathematics
2 answers:
hodyreva [135]3 years ago
7 0

Answer:

C. y = -2/3x -2

Step-by-step explanation:

We know the slope is negative. You can find slope using rise/run. The y-intercept is -2 because the line meets -2 at the y-intercept.    

TEA [102]3 years ago
3 0

Answer:

The first answer (top)

Step-by-step explanation:

Standard form is ax+by=c

a and b are the coefficients

Take a point from the given line

I chose (0, -2)

Then, plug in the values of x and y

2*0+3*-2=-6

-6=-6

Since it works the standard form of the line is 2x+3y=-6

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Proof :

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I hope this makes sense and helps.

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Step-by-step explanation:

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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

8 0
3 years ago
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