Round to the nearest whole number:
a) 14 ⇒ 13.5 ; 13.9 ; 14.1 ; 14.4
b) 28 ⇒ 27.5 ; 27.9 ; 28.1 ; 28.4
c) 32 ⇒ 31.5 ; 31.9 ; 32.1 ; 32.4
d) 50 ⇒ 49.5 ; 49.9 ; 50.1 ; 50.4
e) 67 ⇒ 66.5 ; 66.9 ; 67.1 ; 67.4
f) 71 ⇒ 70.5; 70.9 ; 71.1 ; 71.4
g) 88 ⇒ 87.5; 87.9 ; 88.1 ; 88.4
h) 95 ⇒ 94.5; 94.9 ; 95.1 ; 95.4
If the following number is within the range of 1 to 4 then round down.
If the following number is within the range of 5 to 9 then round up.
Answer: 2 yards
Step-by-step explanation: there are 36 inches in a yard so divide 72 by 36 to get 2 which is the answer to the question
hope this helps mark me brainliest if you can
Answer:
$27428.57
Step-by-step explanation:
To solve this problem, we can use the formula for compound interest:
P = Po * (1 + r)^(t)
Where P is the final value, Po is the inicial value, r is the rate of interest, t is the time.
With r = 0.06, t = 4 (The rate is for every 15 days, and the time is 2 months, so we have that 2 months = 4 periods of 15 days) and P = Po + 7200, we have:
Po + 7200 = Po * (1 + 0.06)^4
Po + 7200 = Po * 1.2625
Po * 0.2625 = 7200
Po = 7200 / 0.2625 = $27428.57
Answer:
Check the explanation
Step-by-step explanation:
1) Algorithm for finding the new optimal flux: 1. Let E' be the edges eh E for which f(e)>O, and let G = (V,E). Find in Gi a path Pi from s to u and a path 
, from v to t.
2) [Special case: If , and 
have some edge e in common, then Piu[(u,v)}uPx has a directed cycle containing (u,v). In this instance, the flow along this cycle can be reduced by a single unit without any need to change the size of the overall flow. Return the resulting flow.]
3) Reduce flow by one unit along 
4) Run Ford-Fulkerson with this sterling flow.
Justification and running time: Say the original flow has see F. Lees ignore the special case (4 After step (3) Of the elgorithuk we have a legal flaw that satisfies the new capacity constraint and has see F-1. Step (4). FOrd-Fueerson, then gives us the optimal flow under the new cePacie co mint. However. we know this flow is at most F, end thus Ford-Fulkerson runs for just one iteration. Since each of the steps is linear, the total running time is linear, that is, O(lVl + lEl).
