Scrap tire management is primarily regulated at the state level.
Answer:
189.15cy
Explanation:
To understand this problem we need to understand as well the form.
It is clear that there is four wall, two short and two long.
The two long are 
The two long are 
The two shors are 
The height and the thickness are 14ft and 0.83ft respectively.
So we only calculate the Quantity of concrete,
![Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3](https://tex.z-dn.net/?f=Q_c%20%3D%20%5B%282%2A122.08%29%2B%282%2A86-375%29%5D%2A14%2A0.833%5C%5CQ_c%3D4864.02ft%5E3)
That in cubic yards is equal to 
Hence, we need order 5% plus that represent with the quantity

$2.
Both tickets cost $1.50
$1.50 x 2 = $3
$5 - $3 = $2
Answer: The net force in every bolt is 44.9 kip
Explanation:
Given that;
External load applied = 245 kip
number of bolts n = 10
External Load shared by each bolt (P_E) = 245/10 = 24.5 kip
spring constant of the bolt Kb = 0.4 Mlb/in
spring constant of members Kc = 1.6 Mlb/in
combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6) = 0.4 / 2 = 0.2 Mlb/in
Initial pre load Pi = 40 kip
now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them
External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip
So Total net Force on each bolt Fb = P_Eb + Pi
Fb = 4.9 kip + 40 kip
Fb = 44.9 kip
Therefore the net force in every bolt is 44.9 kip
Answer:
(a) 0.12924
(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.
Explanation:
(a)
n=200 for fair coin getting head, p= 0.5
Expectation = np =200*0.5=100
Variance = np(1 - p) = 100(1-0.5)=100*0.5=50
Standard deviation,
Z value for 108,
P( x ≥108) = P( z >1.13)= 0.12924
(b)
Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.