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3 years ago

9

Please help with this

1 answer:

horrorfan [7]3 years ago

4 0

Answer:

Is D

Step-by-step explanation:

Because is moving four times right and four times down.

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I really need help with this please help best aswer gets brainliest

svet-max [94.6K]

Answer:

Relative frequency is 7.41% or 0.0741

Step-by-step explanation:

Given

The Attached Table

Required

Calculate the relative frequency of the class with lower limit 27

Relative Frequency is calculated by dividing individual frequency by the total frequency

Mathematically,

$Relative\ Frequency = \frac{Individual\ Frequency}{Total\ Frequency}$

The total frequency of the given data is 6+8+4+2+5+2

$Total\ Frequency = 27$

The class with lower limit 27 has a frequency of 2;

Hence;

$Relative\ Frequency = \frac{Individual\ Frequency}{Total\ Frequency}$ becomes

$Relative\ Frequency = \frac{2}{27}$

$Relative\ Frequency = 0.07407407407$

$Relative\ Frequency = 0.0741$ (Approximated)

You may also leave your answer in percentage form

$Relative\ Frequency = 0.0741 * 100\%$

$Relative\ Frequency = 7.41 \%$

Hence, the relative frequency is 7.41% or 0.0741

7 0

3 years ago

Read 2 more answers

Rewrite 14 1/3 as an improper fraction

Anastaziya [24]

Answer:

Step-by-step explanation:

((14×3) + 1 )/3 = (43/3)

5 0

2 years ago

How many solutions are there to this equation 3(x-4)+5-x=2x-7

Svetlanka [38]

3x-12+5-x=2x-7
2x-7 =2x-7

Means there are infinite solutions.

8 0

2 years ago

What is the volume, in cubic inches, of one cube with an edge length of 1/3 inch

DanielleElmas [232]

Answer:

1/27 cubic in.

Step-by-step explanation:

1/3*1/3*1/3= 1/27

3 0

2 years ago

Find the area of the shaded region. Round the nearest hundredth if necessary. YZ=14.2m

andreyandreev [35.5K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

$A_1 = 67.58 \ in^2$

2

$A_2 =415.4 \ ft^2$

3

$A_3 = 8.48 \ cm^2$

4

$A_4 = 480.38 \ m^2$

Step-by-step explanation:

Generally the area of a sector is mathematically represented as

$A = \frac{\theta}{360} * \pi r^2$

Now at $r_1 = 11 in$ and $\theta_1 = 64^o$

$A_1 = \frac{64}{360} * 3.142 * 11^2$

$A_1 = 67.58 \ in^2$

Now at $r_2 = 20 ft$ in and $\theta_2 = 119 ^o$

$A_2 = \frac{119}{360} * 3.142 * 20^2$

$A_2 =415.4 \ ft^2$

Now at $r_3 = 6.5 cm$ and $\theta_3 = 23 ^o$

$A_3 = \frac{23}{360} * 3.142 * 6.5 ^2$

$A_3 = 8.48 \ cm^2$

Now at $r_4 = 14.2 m$ and $\theta_4 = 360 -87 = 273 ^o$

$A_4 = \frac{273}{360} * 3.142 * 14.2^2$

$A_4 = 480.38 \ m^2$

6 0

2 years ago