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s2008m [1.1K]
3 years ago
12

How many times will it tack when you fill 1 - liter of jar with water from the pond and uses 100 - milliliter cup to scoop water

out of the pond and pour it into the jar
Computers and Technology
1 answer:
Ne4ueva [31]3 years ago
7 0

Answer:

10 times

Explanation:

Volume of a jar = 1 liter

Use 1 liter = 1000 milliliters to convert unit of volume of a jar to milliliter.

Therefore,

Volume of a jar = 1000 milliliters

Volume of cup = 100 milliliter

A 100 - milliliter cup is used to scoop water out of the pond and pour it into the 1 - liter of jar.

Number of times = \frac{1000}{100}=10 times

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Purpose of this project is to increase your understanding of data, address, memory contents, and strings. You will be expected t
STALIN [3.7K]

Answer:

See explaination for the details

Explanation:

#Starting point for code/programm

main:

la $a0,newLine #Print a new line

li $v0,4

syscall

# Find the number of occurence of a string in the given sentence

la $a0,prompt1 # Prompt the user to enter the first string.

li $v0,4

syscall

li $v0, 8 # Service 8 = read input string

la $a0, fword

li $a1, 9

syscall

la $a0,prompt2 # Prompt the user to enter the second string.

li $v0,4

syscall

li $v0, 8 # Service 8 = read input string

la $a0, sword

li $a1, 9

syscall

# process first word

li $t4,0 # Intialize the couter to 0

la $t0,sstatement # Store the statement into $t0

nstart1: la $t1,fword # Store the search word into $t1

loop1: # loop1 finds the number of occurences

# of input word in the given statment

lb $t2,($t0) # Load the starting address(character) of

# sstatement into $t2

lb $t3,($t1) # Load the starting address of input word

# into $t3

beq $t3,'\n',inc_counter1

beqz $t3,inc_counter1 # If $t3 is null , exit loop and print output

beqz $t2,print_output1 # If $t2 is null , exit loop and print output

move $a0,$t2 # Convert $t2 to lower, if it is upper case

jal convert2lower

move $t2,$v0 # Store the return($v0) value into $t2

move $a0,$t3 # Convert $t3 to lower, if it is upper case

jal convert2lower

move $t3,$v0 # Store the return($v0) value into $t3

bne $t2,$t3,next_char1 # If both characters are not matched current

# character in the string, go to next character

addiu $t0,$t0,1 # otherwise, increment both indexes

addiu $t1,$t1,1

j loop1 # go to starting of the loop

next_char1:

la $t5,fword

bne $t5,$t1,nstart1

la $t1,fword # Store the input word into $t1

addiu $t0,$t0,1 # Increment the index to goto next character

j loop1 # go to starting of the loop

inc_counter1:

addi $t4,$t4,1 # Increment the frequency counter by 1

la $t1,fword # Store input word into $t1

j loop1 # go to starting of the loop

print_output1:

la $t0,fword

L1:

lb $a0,($t0)

beq $a0,'\n',exL1

jal convert2upper

move $a0,$v0

li $v0,11

syscall

addiu $t0,$t0,1

j L1

exL1:

la $a0,colon

li $v0,4

syscall

la $a0, dash

li $v0, 4

syscall

move $a0,$t4

li $v0,1

syscall # print new line

la $a0,newLine

li $v0,4

syscall

# process second word

li $t4,0 # Intialize the couter to 0

la $t0,sstatement # Store the statement into $t0

nstart2: la $t1,sword # Store the search word into $t1

loop2: # loop1 finds the number of occurences

# of input word in the given statment

lb $t2,($t0) # Load the starting address(character) of

# sstatement into $t2

lb $t3,($t1) # Load the starting address of input word

# into $t3

beq $t3,'\n',inc_counter2

beqz $t3,inc_counter2 # If $t3 is null , exit loop and print output

beqz $t2,print_output2 # If $t2 is null , exit loop and print output

move $a0,$t2 # Convert $t2 to lower, if it is upper case

jal convert2lower

move $t2,$v0 # Store the return($v0) value into $t2

move $a0,$t3 # Convert $t3 to lower, if it is upper case

jal convert2lower

move $t3,$v0 # Store the return($v0) value into $t3

bne $t2,$t3,next_char2 # If both characters are not matched current

# character in the string, go to next character

addiu $t0,$t0,1 # otherwise, increment both indexes

addiu $t1,$t1,1

j loop2 # go to starting of the loop

next_char2:

la $t5,sword

bne $t5,$t1,nstart2

la $t1,sword # Store the input word into $t1

addiu $t0,$t0,1 # Increment the index to goto next character

j loop2 # go to starting of the loop

inc_counter2:

addi $t4,$t4,1 # Increment the frequency counter by 1

la $t1,sword # Store input word into $t1

j loop2 # go to starting of the loop

print_output2:

la $t0,sword

L2:

lb $a0,($t0)

beq $a0,'\n',exL2

jal convert2upper

move $a0,$v0

li $v0,11

syscall

addiu $t0,$t0,1

j L2

exL2:

la $a0,colon

li $v0,4

syscall

la $a0, dash2

li $v0, 4

syscall

move $a0,$t4

li $v0,1

syscall

exit:

# Otherwise, end the program

li $v0, 10 # Service 10 = exit or end program

syscall

############################ subroutine - convert2lower #################################

convert2lower: # Converts a character(stored in $a0) to

# its lower case, if it is upper case

# and store the result(lower case) in $v0

move $v0,$a0

blt $a0,'A',return

bgt $a0,'Z',return

subi $v0,$a0,-32

return: jr $ra # Return the converted(lower case) character

############################## subroutine - convert2upper ##################################

convert2upper: # Converts a character(stored in $a0) to

# its upper case, if it is lower case

# and store the result(upper case) in $v0

move $v0,$a0

blt $a0,'a',return2

bgt $a0,'z',return2

addiu $v0,$a0,-32

return2: jr $ra # Return the converted(lower case) character

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3 years ago
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<span><span>Computer architecture was essentially a contract with software stating unambiguously what the hardware does. The architecture was essentially a set of statements of the form "If you execute <span>this </span>instruction (or get an interrupt, etc.), then that is what happens." <More sophisticated readers: I get to the term "ISA" later.>
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</span></span>Computer Architecture and Computer Organization Examples

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<span>
nVidia and Qualcomm on the other hand make GPUs (graphics processing unit as opposed to a CPU central processing unit). These GPUs are based on the ARM (Advanced RISC Machines) architecture. ARM is an example on a RISC architecture (RISC stands for Reduced Instruction Set Computer). Instructions in an ARM architecture are relatively simple and typically execute in one clock cycle. Similarly, ARM here is the computer architecture while both nVidia and Qualcomm develop their own flavor of computer organization (i.e architecture implementation)</span></span>
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<span>Print small </span>
<span>Print large </span>
<span>end</span>
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