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Anni [7]
3 years ago
11

in the past month,karen rented  7 7 videos games and  6 d v d . the rental price for each video game  was $ 2

. 8 0 the rental priece for each.What is the total amount that karen spent on video game and dvd rentals in the past month?
Mathematics
1 answer:
Savatey [412]3 years ago
3 0
232.4 hope tthis helps
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Which steps will generate an expression equivalent to 17(x+5)
Eddi Din [679]
Multiply 17x5=85,85+5=90 so aswer is 90
3 0
3 years ago
Read 2 more answers
Please help me answer question B!
skelet666 [1.2K]

Answer:

  • The program counter holds the memory address of the next instruction to be fetched from memory
  • The memory address register holds the address of memory from which data or instructions are to be fetched
  • The memory data register holds a copy of the memory contents transferred to or from the memory at the address in the memory address register
  • The accumulator holds the result of any logic or arithmetic operation

Step-by-step explanation:

The specific contents of any of these registers at any point in time <em>depends on the architecture of the computer</em>. If we make the assumption that the only interface registers connected to memory are the memory address register (MAR) and the memory data register (MDR), then <em>all memory transfers of any kind</em> will use both of these registers.

For execution of the instructions at addresses 01 through 03, the sequence of operations may go like this.

1. (Somehow) The program counter (PC) is set to 01.

2. The contents of the PC are copied to the MAR.

3. A Memory Read operation is performed, and the contents of memory at address 01 are copied to the MDR. (Contents are the LDA #11 instruction.)

4. The MDR contents are decoded (possibly after being transferred to an instruction register), and the value 11 is placed in the Accumulator.

5. The PC is incremented to 02.

6. The contents of the PC are copied to the MAR.

7. A Memory Read operation is performed, and the contents of memory at address 02 are copied to the MDR. (Contents are the SUB 05 instruction.)

8. The MDR contents are decoded and the value 05 is placed in the MAR.

9. A Memory Read operation is performed and the contents of memory at address 05 are copied to the MDR. (Contents are the value 3.)

10. The Accumulator contents are replaced by the difference of the previous contents (11) and the value in the MDR (3). The accumulator now holds the value 11 -3 = 8.

11. The PC is incremented to 03.

12. The contents of the PC are copied to the MAR.

13. A Memory Read operation is performed, and the contents of memory at address 03 are copied to the MDR. (Contents are the STO 06 instruction.)

14. The MDR contents are decoded and the value 06 is placed in the MAR.

15. The Accumulator value is placed in the MDR, and a Memory Write operation is performed. Memory address 06 now holds the value 8.

16. The PC is incremented to 04.

17. Instruction fetch and decoding continues. This program will go "off into the weeds", since there is no Halt instruction. Results are unpredictable.

_____

Note that decoding an instruction may result in several different data transfers and/or memory and/or arithmetic operations. All of this is usually completed before the next instruction is fetched.

In modern computers, memory contents may be fetched on the speculation that they will be used. Adjustments need to be made if the program makes a jump or if executing an instruction alters the data that was prefetched.

4 0
3 years ago
Solve for g. g – (–43) = 19 A. –62 B. –24 C. 24 D. 62
olasank [31]
G-(-43)=19
g+43=19
g=19-43
g=-24

B.


5 0
3 years ago
A batch consists of 12 defective coils and 88 good ones. Find the probability of getting two good coils when two coils are rando
Allushta [10]

<u>Answer:</u>

The probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made is 0.7744  

<u>Solution:</u>

Total number of coils = number of good coils + defective coils = 88 + 12 = 100

p(getting two good coils for two selection) = p( getting 2 good coils for first selection ) \times p(getting 2 good coils for second selection)

p(first selection) = p(second selection) = \frac{\text { number of good coils }}{\text { total number of coils }}

Hence, p(getting 2 good coil for two selection) = \frac{88}{100} \times \frac{88}{100} =\bold{0.7744}

5 0
3 years ago
The sum of three consecutive odd integers is 69. find the integers
Marta_Voda [28]
Let the integers be x-2, x, x+2.

Given,

x -2 + x + x + 2 = 69

3x = 69

x = 23

Hence, the integers are 21, 23 and 25.
6 0
3 years ago
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