$3\mathrm{.}9.$ By hand, write a VBA function named random$\_$exact that generates a very precise random number which rounds to a particular rounded number of your choice $($have one in mind, and write your function with that one in mind$),$ and returns the very precise random number to the Excel cell from which the function random$\_$exact is called. So$,$ for example, the VBA function might generate the very precise number $2\mathrm{.}48763123$ if the rounded number you have in mind is $2\mathrm{.}5.$ The variable low$\_$poss is to represent the lowest possible exact value that rounds to that particular rounded number. The variable $w$ is to represent the width of the range of the exact values that round to that particular rounded number. The values of low$\_$poss and $w$ are user$-$provided. You don't actually have to provide the values of low$\_$poss and $w,$ because this code is written by hand, and will not be executed.

$3\mathrm{.}10.$ Repeat #$3\mathrm{.}9,$ but write execute the VBA function in an actual module. You must provide values of low$\_$poss and $w$ as arguments when you enter the function in a spreadsheet cell, but do not use $2\mathrm{.}5$ as the value to which the very precise number is to round. You must choose your own for credit.

$3\mathrm{.}11.s,$ the length of the side of a square, is known to two significant digits to be $4\mathrm{.}5.$

a$.$ Show how the range of possible exact values for the function $A(s),$ the area of the square, is related to the range of possible exact values for the independent variable $s$ by plotting the function $A(s),$ and shading in the portions of the A and $s$ axes occupied by possible exact values.

b$.$ State the range of possible exact values for $s$ and for $A(s).$

c$.$ State the value of $A(4\mathrm{.}5)$ with the greatest warranted precision.

d$.$ How many significant digits are lost in the output $($the function$)$ relative to the input $($the independent variable$)?$

e$.$ If you feel that a calculated result expressed to the greatest warranted precision indicates an unacceptably broad range of possible exact values, what alternatives do you have?

f$.$ Suppose that, like in part $a,s$ is known to two significant digits to be $4\mathrm{.}5.$

i$.)$ In Excel, use the rand$()$ function to randomly generate $10$ possible values of $s,$ and then use relative cell addresses to generate $10$ corresponding values of $A(s),$ to at least $5$ significant digits. Then copy and paste only the values $($not the function $-$ do this by selecting the $"123"$ paste option$)$ from these latter $10$ cells into $10$ other cells, so that the values are "frozen", and then box in these frozen values as your answer.

ii$.)$ Based on these $10$ randomly generated boxed$-$in values, what would be the value of $A(4\mathrm{.}5),$ expressed with the greatest precision warranted?

iii.$)$ If you were to increase the number of random generations in ii$.,$ would you become more confident that the complete range of possible values will be adequately represented? Briefly explain your answer.

iv$.)$ If you were to write a VBA function to randomly draw a billion numbers each of which round off to $4\mathrm{.}5,$ what do you think the minimum value of those numbers would be$,$ to $3$ significant digits? Explain your answer.

g$.$ Calculate the condition number for $A(4\mathrm{.}5).$ Based on this condition number, what is the estimated number of significant digits lost in the output relative to the input? Also, based on this condition number, what is the estimated number of significant digits to which we would know A$(4\mathrm{.}5)?$ How does this compare to the number of significant digits to which we actually know A$(4\mathrm{.}5)?$

$3\mathrm{.}12.$ Convert the following binary numbers into decimal numbers.

a$.101$

b$.1011$

c$.1010101$

d$.110\mathrm{.}101$

e$.11\mathrm{.}0010$

$3\mathrm{.}13.$ Convert the following decimal numbers into binary numbers.

a$.5$

b$.35$

c$.89$

d$.5\mathrm{.}5$

e$.5\mathrm{.}75$

Please show $3\mathrm{.}10$ and $3\mathrm{.}11$ in excel if possible!