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3 years ago

The box plots below show the distribution of salaries, in thousands, among employees of two small companies.

Mathematics

2 answers:

Shtirlitz [24]3 years ago

7 0

Answer:

Mean and IQR

Step-by-step explanation:

The measure of centre gives the central or the measure which gives the best mid term of a distribution. Based in the details of the box plot, the median is the value which divides the box in the box plot.

For company A:

Range = 25 to 80 with a median value at 30 ; this means the median does not give a good centre measure of the distribution ad it is very close to the minimum value. This goes for the Company B plot too; with values ranging from (35 to 90) and the median designated at 40.

Hence, the mean will be the best measure of centre rather Than the median in this case.

For the variability, the interquartile range would best suit the distribution. With the lower quartile and upper quartile both having reasonable width to the minimum and maximum value of the distribution.

jeyben [28]3 years ago

3 0

Answer:

Step-by-step explanation:

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4 years ago

Read 2 more answers

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mars1129 [50]

Answer:

GCF is 1

Step-by-step explanation:

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3 years ago

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4 years ago

Find the length of the smooth arc 1/3(x2 + 2)3/2 + 24 from x=0 to x=3.

marishachu [46]

Answer:

L = 12

Step-by-step explanation:

$\int\limits^a_b {\sqrt{1+(f'(x))^2} } \, dx$

$f(x) = \frac{(x^2+2)^3^/^2}{3}+24\\f'(x)=\frac{3}{2}*\frac{(x^2+2)^1^/^2*2x}{3}\\ f'(x) = x\sqrt{x^2+2}\\ \\\\L = \int\limits^3_0 {\sqrt{1+(x\sqrt{x^2+2})^2 } \, dx \\\\\\L = \int\limits^3_0 {\sqrt{1+x^2(x^2+2) } \, dx\\\\\\\\L = \int\limits^3_0 {\sqrt{1+x^4+2x^2 } \, dx\\$

Let u = x^2

$\int\limits^3_0{\sqrt{u^2+2u+1} } \, dx \\\\\\\int\limits^3_0 {\sqrt{(u+1)^2} } \, dx \\\int\limits^3_0 {x^2+1} \, dx \\\\L = x^3/3+x (0to3)\\L = 27/3 + 3\\L = 9 + 3\\L = 12$

6 0

3 years ago