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vlabodo [156]
2 years ago
14

This is from Khan academy I have to attach a PNG if you can help me solve it! Thank you!

Mathematics
1 answer:
Andre45 [30]2 years ago
4 0

Step-by-step explanation:

29 {}^{ \frac{x}{2} }

((29) {}^{ \frac{1}{2} } ) {}^{x}

The answer is A

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mass is 58kg

Step-by-step explanation:

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3 years ago
The ages of a sample of fans at a rock concert are listed. 24, 27, 19, 21, 18, 23, 21, 20, 19, 33, 30, 29, 21, 18, 24, 26, 38, 1
Rudik [331]

Answer:

1. {18, 18 , 19 , 19 , 19 , 20 , 21 , 21 , 21 , 21 , 23 , 24 , 24 , 26 , 27 , 27 , 29 , 30 , 30 , 30 , 33 , 33, 34 , 35 , 38 }

2a) md= 24 b) Q1=20.5 c) Q3= 30 3) Q3-Q1 =9.5 b) 19/48

Step-by-step explanation:

To answer this question the 1st and the 2nd we need to order the data entries. So from ordering from the lowest to the highest value:

1. {18, 18 , 19 , 19 , 19 , 20 , 21 , 21 , 21 , 21 , 23 , 24 , 24 , 26 , 27 , 27 , 29 , 30 , 30 , 30 , 33 , 33, 34 , 35 , 38 }

2. There are 25 entries.

{18, 18 , 19 , 19 , 19 , 20 , 21 , 21 , 21 , 21 , 23 , 24 , 24 , 26 , 27 , 27 , 29 , 30 , 30 , 30 , 33 , 33, 34 , 35 , 38 }

In odd quantities of observations, the Median equally separates it two parts.

md=24

b) To find out the 1st quartile, we can use this way:

Q_{1}=\frac{i}{4}(n+1)\\ Q_{1}=\frac{1}{4}(25+1)\\ Q_{1}=\frac{1}{4}(26)=6.5

Then 6.5 is between the 6th and 7th position. Let's find the mean of them, now:  

Q_1=\frac{20+21}{2}= 20.5

c) Similarly toThe Third Quartile or Upper Quartile

Q_{3}=\frac{i}{4}(n+1)\\ Q_{3}=\frac{3}{4}(25+1)\\ Q_{1}=\frac{3}{4}(26)=19.5

The 19th position and 20th position average is:\frac{30+30}{2} =30

3)

a) To find the Interquartile Range, we just need to find out the difference of the upper quartile and the lower one:[tex](Q_3-Q_1)Q_3-Q_1[/tex]

(30-20.5)=9.5

b) Interquartile Ratio is given by the quotient of the Interquartile Range over the Median

\frac{IQR}{md}=\frac{9.5}{24}=\frac{19}{48}

4) Since the  Relative  frequency Histogram asked is a one with 7 classes. Let's calculate how many  values.

k=1+3.32logn

7=1+3.32logn

6=3.32logn

n≈66

Each class must have an interval of 10 ages, for (91-18)/7≈ 10. Notice the orange line intercepts the midpoint of each interval.

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What's bigger 3.05kg or 3.56
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Step-by-step explanation:

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