Answer:
c
because oil is preventing corrosion and rust
The number of hectares of each crop he should plant are; 250 hectares of Corn, 500 hectares of Wheat and 450 hectares of soybeans
<h3>How to solve algebra word problem?</h3>
He grows corn, wheat and soya beans on the farm of 1200 hectares. Thus;
C + W + S = 12 ----(1)
It costs $45 per hectare to grow corn, $60 to grow wheat, and $50 to grow soybeans. Thus;
45C + 60W + 50S = 63750 -----(2)
He will grow twice as many hectares of wheat as corn. Thus;
W = 2C ------(3)
Put 2C for W in eq 1 and eq 2 to get;
C + 2C + S = 1200
3C + S = 1200 -----(4)
45C + 60(2C) + 50S = 63750
45C + 120C + 50S = 63750
165C + 50S = 63750 ------(5)
Solving eq 4 and 5 simultaneosly gives;
C = 250 and W = 500
Thus; S = 1200 - 3(250)
S = 450
Read more about algebra word problems at; brainly.com/question/13818690
Answer:
Communication is simply the act of transferring information from one place, person or group to another. Every communication involves (at least) one sender, a message and a recipient.
Explanation:
Answer:
938.7 milliseconds
Explanation:
Since the transmission rate is in bits, we will need to convert the packet size to Bits.
1 bytes = 8 bits
1 MiB = 2^20 bytes = 8 × 2^20 bits
5 MiB = 5 × 8 × 2^20 bits.
The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R
where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.
Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9
queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9
queueing delay for 48th packet = 0.938725181 seconds
queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds
