The gross weight of a truck containing 30 pigs is 8 tons 1,123 pounds and 6 ounces. The average weight of one pig is 95 pounds 1
0 ounces. Find the following: a.The net weight of the pigs in the truck. b. The weight of the empty truck.
1 answer:
Answer:
- a) 2868 lbs 12 oz
- b) 7 tons 254 lbs 10 oz
Step-by-step explanation:
a) There are 16 ounces in a pound, so 10 ounces is 10/16 = 5/8 lb. The total weight of the pigs is their number multiplied by their average weight:
... weight of pigs = 30 × 95 5/8 lb = ((30·95) + (30·5/8)) lb
... = (2850 + 150/8) lb = (2850 +18 3/4) lb = 2868 lb 12 oz
b) The truck tare weight is the gross weight less the weight of the pigs.
... truck tare = truck gross - truck load
... = (8×2000 lb) + (1123 lb) + (6/16 lb) - (2868 3/4 lb)
... = 17123 3/8 lb - 2868 3/4 lb = (17123 -2868) lb +(3/8 -3/4) lb
... = 14255 lb - 3/8 lb
... = 14254 5/8 lb
... = 14000 lb + 254 lb + 5/8 lb
... truck tare weight = 7 tons 254 lbs 10 oz
_____
Many graphing calculators will handle calculations with fractions very nicely.
You might be interested in
Answer:
a)0.7
b) 10.03
c) 0.0801
Step-by-step explanation:
Rate of return Probability
9.5 0.1
9.8 0.2
10 0.3
10.2 0.3
10.6 0.1
a.
P(Rate of return is at least 10%)=P(R=10)+P(R=10.2)+P(R=10.6)
P(Rate of return is at least 10%)=0.3+0.3+0.1
P(Rate of return is at least 10%)=0.7
b)
Expected rate of return=E(x)=sum(x*p(x))
Rate of return(x) Probability(p(x)) x*p(x)
9.5 0.1 0.95
9.8 0.2 1.96
10 0.3 3
10.2 0.3 3.06
10.6 0.1 1.06
Expected rate of return=E(x)=sum(x*p(x))
Expected rate of return=0.95+1.96+3+3.06+1.06=10.03
c)
variance of the rate of return=V(x)=
Rate of return(x) Probability(p(x)) x*p(x) x²*p(x)
9.5 0.1 0.95 9.025
9.8 0.2 1.96 19.208
10 0.3 3 30
10.2 0.3 3.06 31.212
10.6 0.1 1.06 11.236
sum[x²*p(x)]=9.025+19.208+30+31.212+11.236=100.681
variance of the rate of return=V(x)=sum(x²*p(x))-[sum(x*p(x))]²
variance of the rate of return=V(x)=100.681-(10.03)²
variance of the rate of return=V(x)=100.681-100.6009
variance of the rate of return=V(x)=0.0801