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Serggg [28]
3 years ago
10

There are two sets of rainfall data for two different time periods in the same location.

Mathematics
2 answers:
Cerrena [4.2K]3 years ago
7 0

Answer:

Box - and - whisker and scatter plots

Step-by-step explanation:

Monica [59]3 years ago
3 0

Answer:

two different time periods in the same location.

Period 1: 2.3, 2.1, 2.2, 2.2, 2.2, 2.1, 2.4, 2.5, 2.2, 2.0, 1.9, 1.9, 2.1, 2.2, 2.3

Period 2: 2.3, 2.1, 3.3, 1.5, 3.6, 1.6, 3.0, 1.1, 4.7, 2.1, 2.4, 1.9, 2.8, 0.5, 2.3

What type of visual display could be used to compare the two data sets?

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Let sin A= -3/5 with 270°&lt; A&lt; 360°. Find the following.<br> sin A/2
GREYUIT [131]

Answer:

1/sqrt10

Step-by-step explanation:

1) Find out cosA using formula (cosA)^2+(sinA)^2=1

The module of cosA= sqrt (1- (-3/5)^2)= sqrt 16/25=4/5

So cosA=-4/5 or cosA=4/5.

Due to the condition 270degrees< A<360 degrees, 0<cosA<1 that's why cosA=4/5.

2) Find sinA/2 using a formula cosA= 1-2sinA/2*sinA/2 where cosA=4/5.

(sinA/2)^2= 0.1

sinA= sqrt 0.1= 1/ sqrt10 or sinA= - sqrt 0.1= -1/sqrt10

But 270°< A< 360°, then 270/2°<A/2<360/2°

135°<A/2<180°, so sinA/2 must be positive and the only correct answer is

sin A/2= 1/sqrt10

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3 years ago
Solve for X: Correctly answer the question
Klio2033 [76]

Answer:

x=16

Step-by-step explanation:

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(8+24) * 8 = x^2

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256  =x^2

Take the square root of each side

sqrt(256) = sqrt(x^2)

16 = x

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