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Sergeeva-Olga [200]
3 years ago
9

1. The diameter of one side of a 10-foot log is approximately 13 inches. The diameter of the other side of the log is approximat

ely 11 inches. Which is the best way to estimate the volume (in cubic feet) of the log?
Mathematics
1 answer:
LuckyWell [14K]3 years ago
7 0
To best estimate the volume of the log, we assume that the diameter is the average of the diameters at both ends giving us 12 inches as the diameter. Thus, the radius is equal to 0.5 ft. The volume of the log is calculated through the equation,
                      V = πr²h = π(0.5 ft)²(10 ft) = 7.85 ft³
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Answer:

Step-by-step explanation:

League A                   League B

151.12                            163.25

148                                157

26.83                             24.93

29                                  136

136                                145

167                                178

207                               256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{26.83+29 +136+ 148+ 151.12+ 167+207}{7}\\\\Mean =123.564

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}==\sqrt{\frac{(26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2}{7}}=63.98

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

(82.5-1.5\times 76.56,159.06+1.5\times 76.56)

(-32.34,273.9)

So, There is no outlier

Maximum = 207

2)

League B in ascending order :

24.93,136,145,157,163.25,178,256

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{24.93+136+145+157+163.25+178+256}{7}\\\\Mean =151.45

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}==\sqrt{\frac{(24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2}{7}}=68.42

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Q1 is the mid value of lower quartile

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n = 4

Median = \frac{\frac{n}{2} \text{th term}+(\frac{n}{2}+1) \text{th term}}{2}\\Median = \frac{\frac{4}{2} \text{th term}+(\frac{4}{2}+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = \frac{136+145}{2}=140.5

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Q3 is the mid value of upper quartile

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Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

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