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Brilliant_brown [7]
2 years ago
13

The data below represents the relationship between the number of incidences of UFO sightings in late 1930s and early 1940s. Draw

a scatter plot for the data.
Year of Incidence
1936
1937
1938
1939
1940
1941
1942
1943
1944
1945
1946
Number of Incidences (percent)
0.9
0.8
0.8
1.3
1.4
1.2
1.7
1.8
1.6
1.5
1.5

a.
A graph has years of incidence on the x-axis from 1932 to 1948, and incidence on the y-axis, from 0 to 10. Points are around (1936, 0.9), (1937, 0.8), (1938, 0.8), (1939, 1.3), (1940, 1.4), (1941, 1.2), (1942, 1.7), (1943, 1.8), (1944, 1.6), (1945, 1.5), (1946, 1.5).
c.
A graph has years of incidence on the x-axis from 1932 to 1948, and incidence on the y-axis, from 0 to 10. Points are around (1936, 0.9), (1937, 0.8), (1938, 0.8), (1939, 1.3), (1940, 1.4), (1941, 1.2), (1942, 1.7), (1943, 1.8), (1944, 1.6), (1945, 1.5).
b.
A graph has years of incidence on the x-axis from 1932 to 1948, and incidence on the y-axis, from 0 to 10. Points are around (1936, 0.9), (1937, 1.7), (1938, 0.8), (1939, 1.3), (1940, 1.4), (1941, 1.2), (1942, 1.7), (1943, 1.8), (1944, 1.6), (1945, 1.5), (1946, 1.5).
d.
A graph has years of incidence on the x-axis from 1932 to 1948, and incidence on the y-axis, from 0 to 10. Points are around (1936, 0.9), (1937, 0.8), (1939, 1.3), (1940, 1.4), (1941, 1.2), (1942, 1.7), (1943, 1.8), (1944, 1.6), (1945, 1.5), (1946, 1.5).
Mathematics
1 answer:
vaieri [72.5K]2 years ago
4 0

The description of graph of the data about relationship of UFO sightings in late 1930s and early 1940s will be :A graph has years of incidence on the x-axis from 1932 to 1948, and incidence on the y-axis, from 0 to 10. Points are around (1936, 0.9), (1937, 0.8), (1938, 0.8), (1939, 1.3), (1940, 1.4), (1941, 1.2), (1942, 1.7), (1943, 1.8), (1944, 1.6), (1945, 1.5), (1946, 1.5).

Given:

Year of incidence                    percent

1936                                         0.9

1937                                         0.8

1938                                          0.8

1939                                          1.3

1940                                           1.4

1941                                            1.2

1942                                            1.7

1943                                             1.8

1944                                              1.6

1945                                              1.5

1946                                              1.5

We have been given the percentages of different years. The points will be (1936,0.9) , (1937,0.8) , (1938,0.8) , (1939,1.3) , (1940,1.4) , (1941,1.2) , (1942,1.7), (1943,1.8) , (1944,1.6) , (1945,1.5) , (1946,1.5).

We have to just plot the points on the graph to find scatter plot. It is called scatter plot as the points are not in line but they are scattered.

Learn more about graph at brainly.com/question/4025726

#SPJ10

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Answer:

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Step-by-step explanation:

The 2 angles on the straight line sum to 180° , that is

∠ LMP + ∠ NMP = 180 , substitute values

- 16x + 13 - 20x + 23 = 180 , that is

- 36x + 36 = 180 ( subtract 36 from both sides )

- 36x = 144 ( divide both sides by - 36 )

x = - 4

Then

∠ LMP = - 16x + 13 = - 16(- 4) + 13 = 64 + 13 = 77°

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8 0
3 years ago
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vovikov84 [41]

Answer:

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Step-by-step explanation:

By sine rule, the length of each side of a triangle is proportional to the sine value of the angle opposite to that side. For example, in this triangle \triangle ABC, angle \angle A is opposite to side BC, while \angle C is opposite to side AB. By sine rule, \displaystyle \frac{BC}{\sin{\angle A}} = \frac{AB}{\sin \angle C}.

It is already given that BC = 22.4\; \rm m and \angle A = 58^\circ. The catch is that the value of \angle C needs to be calculated from \angle A and \angle B.

The sum of the three internal angles of a triangle is 180^\circ. In \triangle ABC, that means \angle A + \angle B + \angle C = 180^\circ. Hence,

\begin{aligned}\angle C &= 180^\circ - \angle A - \angle B \\ &= 180^\circ - 58^\circ - 65^\circ \\ &= 57^\circ\end{aligned}.

Apply the sine rule:

\begin{aligned} & \frac{BC}{\sin{\angle A}} = \frac{AB}{\sin \angle C} \\ \implies & AB = \frac{BC}{\sin{\angle A}} \cdot \sin \angle C  \end{aligned}.

\begin{aligned}AB &= \frac{BC}{\sin{\angle A}} \cdot \sin \angle C \\ &= \frac{22.4\; \rm m}{\sin 58^\circ} \times \sin 57^\circ \\ &\approx 22.2\; \rm m\end{aligned}.

5 0
3 years ago
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Step-by-step explanation:

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b= x

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Equation

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3 0
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Answer:

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Step-by-step explanation:

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For the next one:

0. 1818

x     99

---------------

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16.3620

-------------

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5 0
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