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2 years ago

Use the accompanying data set to complete the following actions.

1 answer:

tatiyna2 years ago

5 0

A quartile divides data into three points a lower quartile (Q₁), median (Q₂), and upper quartile (Q₃) to form four groups of the dataset.

a. Quartiles Q₁= 57, Q₂= 59, Q₃= 62

b. Interquartile range, (IQR) = 5

c. 0 and 75 are outlines.

First arrange the given data into increasing order,

We get, 0, 55, 56, 57, 57, 57, 57, 59, 59, 59, 60, 62, 62, 64, 65, 75

Divide into Quartiles,

( 0, 55, 56, 57 ) ( 57, 57, 57, 59) ( 59, 59, 60, 62) ( 62, 64, 65, 75)

Q₁ = $\frac{57+57}{2}=57$

Q₂ = $\frac{59+59}{2}=59$

Q₃= $\frac{62 + 62}{2}=62$

Interquartile range (IQR) = Q₃ - Q₁=62 - 57 = 5

High outliner = 1.5 (IQR) + Q₃

= 1.5(5) + 62

= 69.5

low outliner = Q₁ - 1.5 (IQR)

= 57 - 1.5 (5)

= 49.5

All the values from arrange data set which are lies outside from 69.5 and 49.5 known as outliners.

Here, 0 and 75 are outliners.

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Based on a poll, 40% of adults believe in reincarnation. Assume that 77 adults are randomly selected, and find the indicated

damaskus [11]

To solve this problem, we make use of the binomial probability equation:

P = [n! / (n – r)! r!] p^r * q^(n – r)

where,

n is the total number of adult samples = 77

r is the selected number of adults who believe in reincarnation

p is the of believing in reincarnation = 40% = 0.40

q is 1 – p = 0.60

a. What is the probability that exactly 66 of the selected adults believe in reincarnation?

So we use r = 66

P = [77! / (77 – 66)! 66!] 0.40^66 * 0.60^(77 – 66)

P = 1.32 x 10^-16

b. What is the probability that all of the selected adults believe in reincarnation?

So we use r = 77

P = [77! / (77 – 77)! 77!] 0.40^77 * 0.60^(77 – 77)

P =2.28 x 10^-31

c. What is the probability that at least 66 of the selected adults believe in reincarnation?

So we use r = 66 to 77

P (r=66) = 1.32 x 10^-16

P (r=67) = [77! / (77 – 67)! 67!] 0.40^67 * 0.60^(77 – 67) = 1.44 x 10^-17

P (r=68) = [77! / (77 – 68)! 68!] 0.40^68 * 0.60^(77 – 68) = 1.42 x 10^-18

P (r=69) = [77! / (77 – 69)! 69!] 0.40^69 * 0.60^(77 – 69) = 1.23 x 10^-19

P (r=70) = [77! / (77 – 70)! 70!] 0.40^70 * 0.60^(77 – 70) = 9.38 x 10^-21

P (r=71) = [77! / (77 – 71)! 71!] 0.40^71 * 0.60^(77 – 71) = 6.17 x 10^-22

P (r=72) = [77! / (77 – 72)! 72!] 0.40^72 * 0.60^(77 – 72) = 3.43 x 10^-23

P (r=73) = [77! / (77 – 73)! 73!] 0.40^73 * 0.60^(77 – 73) = 1.56 x 10^-24

P (r=74) = [77! / (77 – 74)! 74!] 0.40^74 * 0.60^(77 – 74) = 5.64 x 10^-26

P (r=75) = [77! / (77 – 75)! 75!] 0.40^75 * 0.60^(77 – 75) = 1.50 x 10^-27

P (r=76) = [77! / (77 – 76)! 76!] 0.40^76 * 0.60^(77 – 76) = 2.64 x 10^-29

P (r=77) = 2.28 x 10^-31

The total is the sum of all:

P(total) = 1.48 x 10^-16

d. If 77 adults are randomly selected, is 66 a significantly high number who believe in reincarnation?

C.Yes, because the probability that 66 or more of the selected adults believe in reincarnation is less than 0.05

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3 years ago

How to I find a? It’s round to the nearest tenth.

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Answer:

a ≈ 21.8

Step-by-step explanation:

We require the third angle in the triangle

Subtract the sum of the 2 angles from 180

third angle = 180° - (75 + 31.8)° = 180° - 106.8° = 73.2°

Using the Sine rule, that is

$\frac{a}{sin73.2}$ = $\frac{22}{sin75}$ ( cross- multiply )

a × sin75° = 22 × sin73.2° ( divide both sides by sin75° )

a = $\frac{22sin73.2}{sin75}$ ≈ 21.8 ( to the nearest tenth )

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Answer:

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Mars Gravitational Force : 3.711m/s

So an object or person on Mars would weigh 37.83% its weight on earth. Therefore, a person would be much lighter on mars. A person is about 62.17% heavier on earth than on Mars.

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Lines k and n are perpendicular.if the slope of line k is -6,what is the slope of line n?

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Answer:

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Step-by-step explanation:

Well perpendicular lines are reciprocals of each other,

meaning if like k has a slope of -6 then line n will have a slope of positive 1/6.

Thus,

line n has a slope of 1/6.

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