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Lana71 [14]
3 years ago
7

Su now wants to modify the text box that contains her numbered list. She accesses the Format Shape pane. In what ways can Su mod

ify the text box using the pane? Check all that apply.
A.She can rotate the text box.
B.She can add color to the text box.
C.She can add a shape to the text box.
D.She can add a border to the text box.
E.She can insert a picture in the text box.
Computers and Technology
2 answers:
Fiesta28 [93]3 years ago
7 0
I believe its E she can insert a picture in the text book
Ganezh [65]3 years ago
7 0

Answer:

Explanation:

One group of students did an experiment to study the movement of ocean water. The steps of the experiment are listed below.

Fill a rectangular baking glass dish with water.

Place a plastic bag with ice in the water near the left edge of the dish.

Place a lighted lamp near the left edge of the dish so that its light falls directly on the plastic bag.

Put a few drops of ink in the water.

The student did not observe any circulation of ink in the water as expected because the experiment had a flaw. Which of these statements best describes the flaw in the experiment? (2 points)

Not enough ink was added.

Not enough water was taken.

The dish was too small for the experiment.

The lamp and the ice bag were at the same place.

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Answer:

Explanation:

application form and any fees

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3 years ago
1. How many bits would you need to address a 2M × 32 memory if:
Dominik [7]

Answer:

  1. a) 23       b) 21
  2. a) 43        b) 42
  3. a) 0          b) 0

Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8;  L = ( 2M * 32 ) / 8 =  2M * 4

hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

<u>2) How many bits are required to address a 4M × 16 main memory</u>

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

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3 years ago
Consider two different implementations, M1 and M2, of the same instruction set. There are three classes of instructions (A, B, a
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Explanation:

A.)

we have two machines M1 and M2

cpi stands for clocks per instruction.

to get cpi for machine 1:

= we multiply frequencies with their corresponding M1 cycles and add everything up

50/100 x 1 = 0.5

20/100 x 2 = 0.4

30/100 x 3 = 0.9

CPI for M1 = 0.5 + 0.4 + 0.9 = 1.8

We find CPI for machine 2

we use the same formula we used for 1 above

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20/100 x 3 = 0.6

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B.)

CPU execution time for m1 and m2

this is calculated by using the formula;

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