Melody

3xy-3xy   =   

x*y = y*x

so

3xy = 3yx

so

3xy-3xy= 0

You have asked this question at least twice and it has been answered at least twice!

Do not waste answerers time like this again!

\(1- 0.15z-1 - 0.1z-2\\ =1-1-2  - 0.15z - 0.1z\\ =   -2  - 0.25z \\ = -0.25(8+z) \)

or

\(1- 0.15z^{-1} - 0.1z^{-2}\\ 1- \frac{0.15}{z} -\frac{ 0.1}{z^2}\\ \frac{1}{100z^2}(100z^2- 15z -10)\\\)

sqrt(15^2+4*100*10) = 65

(15+65)/200 = 0.4

(15-65)/200 = -0.25

\(=\frac{1}{100z^2}(z-0.4)(z+0.25)\\ =\frac{1}{100z^2}(z-0.4)(z+0.25)\\\)

asinus is telling you that this is the general shape but the exact shape depends on the value of a.

There is no greatest value :/

There should be more to the question :)

Write (2√3)(5√27^3) as a single exponent

\((2\sqrt3)(5(\sqrt{27})^3)\\ =(2\sqrt3)(5(\sqrt{9*3})^3)\\ =(2\sqrt3)(5(3\sqrt{3})^3)\\ =(2\sqrt3)(5(3^3(\sqrt{3})^3)\\ =(2\sqrt3)(5(3^3(3\sqrt{3}))\\ =2*\sqrt3*5*3^3*3\sqrt{3}\\ =2*5*3^3*3\sqrt{3}*\sqrt3\\ =10*3^3*3*3\\ =10*3^5\\ \)

The following data shows the weight (in kg) of 13 crabs found in a restaurant on a particular evening: 3.4 1.2 1.7 2.4 2.4 1.1 0.9 0.8 1.2 1.3

(i) Compute the mean and median

3.4      1.2     1.7      2.4      2.4    1.1    0.9    0.8    1.2     1.3         I onlu count 10 crabs,

For the median I must put them in order.

0.8      0.9      1.1     1.2     1.2     1.3     1.7    2.4      2.4    3.4      

There are 2 middle ones so the median is    (1.2+1.3)/2= 1.25Kg

For the mean, add them all up and divide by 10 (because there are 10 crabs)

(0.8+0.9+1.1+1.2+1.2+1.3+1.7+2.4+2.4+3.4)/10 = 1.64Kg                       

(ii) Determine the shape of the distribution based on the sample data. Explain your conclusion

Is 3.5%<7.6*10^-3<3.8*10^-1 true?

lets see

3.5% = 0.035

\(7.6*10^{-3}=0.0076\)

0.035 is not less than 0.0076 so the statement is not true. 

Well here is your graph :)