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N76 [4]
4 years ago
5

I dont know what a complementary angle is, can someone help me with this quistion please?

Mathematics
1 answer:
Allisa [31]4 years ago
8 0

Answer:

A, complementary angles sum up to 90°

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What is the normal square root of 20
eduard

Answer:

\displaystyle 2\sqrt{5}

Step-by-step explanation:

\displaystyle 2\sqrt{5} = \sqrt{[5][4]} \\ \\ 4,472135955 ≈ 2\sqrt{5}

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7 0
3 years ago
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
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Ana' location is further away from the entrance
3 0
3 years ago
An oil storage tank ruptures at time t = 0 and oil leaks from the tank at a rate of
wel
\displaystyle\int_0^{60}r(t)\,\mathrm dt=\int_0^{60}80e^{-0.01t}\,\mathrm dt

Let u=-0.01t so that -100\mathrm du=\mathrm dt. Then the integral becomes

\displaystyle-8000\int_0^{-0.6}e^u\,\mathrm du=8000\int_{-0.6}^0e^u\,\mathrm du=8000(e^0-e^{-0.6})\approx3610\text{ L}
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4 years ago
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