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koban [17]
4 years ago
14

Naturally, it would be nice to convert the entire output of make-change to the RLE format. Write a SCHEME function (rle-all lcoi

ns) which, when given a list of coin changes, produces a list of RLE encoded coin-changes. For instance the call (rle-all (make-change 11 '(25 10 5 1))) produces (((11 . 1)) ((6. 1) (1 . 5)) ((1 . 1) (2 . 5)) ((1 . 1) (1 . 10))) which is a list of 4 lists (since there are four ways to give change on 11 cents) where each list is an RLE encoding.
Engineering
1 answer:
Stolb23 [73]4 years ago
6 0

Answer:

Bubble sort, sometimes referred to as sinking sort, is a simple sorting algorithm that repeatedly steps through the list, compares adjacent elements and swaps them if they are in the wrong order. The pass through the list is repeated until the list is sorted

Explanation:

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One of the best methods for understanding a company’s policies and procedures is
11Alexandr11 [23.1K]

Answer:

Usually a handbook or website

7 0
3 years ago
(Tacoma Bridge Failure) Are There Relevant Ethical Issues Or Just 20-20 Hindsight?
NemiM [27]

Answer:

i have no idea

Explanation:

7 0
3 years ago
3. (a) (5 points) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queue
Bezzdna [24]

Answer:

(N-1) × (L/2R) = (N-1)/2

Explanation:

let L is length of packet

R is rate

N is number of packets

then

first packet arrived with 0 delay

Second packet arrived at = L/R

Third packet arrived at = 2L/R

Nth packet arrived at = (n-1)L/R

Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R

Now

L / R = (1000) / (10^6 ) s = 1 ms

L/2R = 0.5 ms

average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2

the average queuing delay of a packet = 0 ( put N=1)

4 0
4 years ago
A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. Calculate the number of atoms in t
uysha [10]

Answer:288 pm

Explanation:

Number of atoms(s) for face centered unit cell -

Lattice points: at corners and face centers of unit cell.

For face centered cubic (FCC), z=4.

- whereas

For an FCC lattices √2a =4r =2d

Therefore d = a/√2a = 408pm/√2a= 288pm

I think with this step by step procedure the, the answer was clearly stated.

8 0
3 years ago
A subsurface exploration report shows that the average water content of a fine-grained soil in a proposed borrow area is 22% and
Morgarella [4.7K]

Answer:

shrinkage ratio  = 1.538

Explanation:

given data

water content = 22 %

dry density γ = 82 pcf

required dry density specified γ' = 96 pcf

required to produce = 50,000 yd³  = 50000 × 27 = 1350,000 ft³

solution

we get here first volume of borrow pit that is we know that

dry density ∝  \frac{1}{volume}  

so \frac{\gamma d}{\gamma 'd}  = \frac{v'}{v}

\frac{82}{96}  = \frac{1350000}{v}

v = 1580487.8 ft³

v = 58536.58 yd³

so here

shrinkage ratio will be as

shrinkage ratio = \frac{96}{62.4}  

shrinkage ratio  = 1.538

6 0
3 years ago
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