$***$Answer part d please, very bottom$*****$You are looking at a roadbed that was placed using wh it the owner is calling "State Spec Material". This road is having some "rutting issues" an I won't dry out. So you get a bu :ket, and bring it back to the garage and find:

Bucket test:

Volume $of$ soil $in$ field $=0\mathrm{.}55f{t}^3$

Weight $of$ wet soil $=65\mathrm{.}4lb.$

Weight $of$ dry soil $=56\mathrm{.}3lb.$

Other laboratory testing reveals:

$)=(\frac{{}_{\text{m}\text{u}\text{l}}}{{}_{\text{w}\text{a}\text{t}\text{e}\text{r}}}$

Fines:

$LL=20$

$PL=12\frac{?}{P}$

I $=8$

a$)$ What is the wet unit weight of the material, as it was in the field, in pounds per cubic fcot $($pcf$)?$

$y_{\text{w}\text{e}\text{t}}^{(5pts)}=\frac{W_{\text{w}\text{e}\text{t}}}{V},y_{\text{w}\text{e}\text{t}}=\frac{60\mathrm{.}416}{0\mathrm{.}55f{t}^3}$~~

b$)$ Is this believable? $($why$?$ Cite reference$($s$))5$ pts

A wet unit weight of $118\mathrm{.}91$ pet might be plansable for certein types of Soil, especially if it cantains a significant amount of fines $($sitt or clay$)$ and has undergene compaction.

reference $=$ WSDOT Errate c$)$ What is the moisture content $($wc$\%)$ of this $W_c{c}^{\text{'}\text{'}}=\frac{W_{\text{w}\text{e}\text{t}}-W_{\text{d}\text{r}\text{y}}}{W_{\text{d}\text{r}\text{y}}}*100$

$W_{c^{\text{'}}}=\frac{65\mathrm{.}4-56\mathrm{.}3}{56\mathrm{.}3}*100$

$w_c$:$=\frac{9\mathrm{.}1}{56\mathrm{.}3}*100$

$4$

d$)$ Calculate the degree of saturatio

n$,S=$ Vwater $/$ Vvoids? $(10$ pts$)$