Answer:
The algorithm is as follows:
1. Declare Arr1 and Arr2
2. Get Input for Arr1 and Arr2
3. Initialize count to 0
4. For i in Arr2
4.1 For j in Arr1:
4.1.1 If i > j Then
4.1.1.1 count = count + 1
4.2 End j loop
4.3 Print count
4.4 count = 0
4.5 End i loop
5. End
Explanation:
This declares both arrays
1. Declare Arr1 and Arr2
This gets input for both arrays
2. Get Input for Arr1 and Arr2
This initializes count to 0
3. Initialize count to 0
This iterates through Arr2
4. For i in Arr2
This iterates through Arr1 (An inner loop)
4.1 For j in Arr1:
This checks if current element is greater than current element in Arr1
4.1.1 If i > j Then
If yes, count is incremented by 1
4.1.1.1 count = count + 1
This ends the inner loop
4.2 End j loop
Print count and set count to 0
<em>4.3 Print count</em>
<em>4.4 count = 0</em>
End the outer loop
4.5 End i loop
End the algorithm
5. End
Answer:
C. Exist
Hope it helps!
Answer:
a) 
, b) 
, c) 
Explanation:
a) The deceleration experimented by the commuter train in the first 2.5 miles is:
![a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5B%2815%5C%2C%5Cfrac%7Bmi%7D%7Bh%7D%20%29%5Ccdot%20%28%5Cfrac%7B5280%5C%2Cft%7D%7B1%5C%2Cmi%7D%20%29%5Ccdot%20%28%5Cfrac%7B1%5C%2Ch%7D%7B3600%5C%2Cs%7D%20%29%5D%5E%7B2%7D-%5B%2850%5C%2C%5Cfrac%7Bmi%7D%7Bh%7D%20%29%5Ccdot%20%28%5Cfrac%7B5280%5C%2Cft%7D%7B1%5C%2Cmi%7D%20%29%5Ccdot%20%28%5Cfrac%7B1%5C%2Ch%7D%7B3600%5C%2Cs%7D%20%29%5D%5E%7B2%7D%7D%7B2%5Ccdot%20%282.5%5C%2Cmi%29%5Ccdot%20%28%5Cfrac%7B5280%5C%2Cft%7D%7B1%5C%2Cmi%7D%20%29%7D)
The time required to travel is:
b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be .
c) The final constant deceleration is:
Answer:
The correct/closest option is b
Explanation:
Restriction enzymes are enzymes (endonucleases) that cut short DNA strands at specific sites. Hence, each restriction enzyme has it's own specific site (between two bases) it cuts at. There are two types of end that can be produced by this cut; the blunt end and the sticky end.
A restriction enzyme recognizes (palindromic sequence) and cut in it's own specific end.
For example, if a restriction enzyme cuts between a guanine (G) and an adenine (A), and it cuts a palindromic double stranded DNA in the manner below, it produces a sticky end.
G║AATTC
CTTAA║G
And if a restriction enzyme cuts between guanine (G) and cytosine (C) in the manner below, it produces a blunt end.
GGG║CCC
CCC║GGG
Hence, from the question, restriction enzymes (although chosen by the scientist based on desired sequence to be cut) recognize the sticky or blunt ends itself.
