# ITEC 1000 Task 2 Solutions

п»їAP/ITEC1000 Section M " Introduction to Details Technologies" Winter season 2014В

Job # two

Answers and Solutions

1)В

a) Convert 124 to their binary manifestation: 12410 =В 011111002. Great numbers are represented without any assistance. В В В В В В В В В В В В The answer: 01111100

b) Convert -124 to its binary representation: -12410 =В -011111002. Using the second method for negative numbers, change and then add 1 . В В В В В В В В В В В В The answer: В 10000100В В В В

c) Convert 53 to it is binary representation: 5310 =В 001101012. В Positive figures are always displayed by themselves. В В В В В В В В В В В В В В The response: В 00110101

2) Yes. The four least significant numbers of the ASCII code meet the 4 bits of BCD. 3) Among -7999 and +7999. It is because 16 bits provide several four-bit locations. The lowest (right most) and two central 4 portions can every single store several from zero to 9. In the maximum (left most) 4 portions, one bit is used for the sign, leaving several bits intended for digits. These kinds of 3 parts can allow for numbers between 0 and 7. 4)В В В В В В

a) 15675 is known as a positive amount and is displayed by itself В В В В В В В В В В В В В 99999 -- 8741 sama dengan 91258

b)В В 15675

+91258

106933 put the carry into the result В В В В В В В В В В В В В В В +В В В В В В 1

6934

В В В В В В В В Вc) There was zero overflow condition because the figures have contrary signs and overflow are not able to occur. В В В В В В В В В В В В В В There was a carry state because in the result, the quantity of digits surpass the specified quantity (i. elizabeth., five) and the carry was added to the effect. В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В d) 15675 is a great number and is represented independently В В В В В В В В В В В В В В 100000 - 8741 = 91259

e)В В В 15675

+91259

В В В В В В В В В В В В В В106934В В Ignore the bring and the effect is: 6934 В В В В В В В В В f) Diminished radix complement arithmetic requires an additional step of adding the bring bit for the result (in 9's compliment). В В В В В В В В В В В В Radix complement arithmetic does not require this step, for the reason that carry is merely ignored. The latter is easier. 5)В В Replace each decimal digit simply by its 4-bit binary representation and add the corresponding sign style in the low-order bits. В В В В a)В В В В -75 = 011101011101В В В В В В В В В В В В В В В b) 7657 sama dengan 01110110010101111100В В В В В В c)В В В В -4327 sama dengan 01000011001001111101В В 6)В В В В В В

a) The leading digit is between 0 and 4 and, therefore , presents a positive amount. Positive amounts are represented by themselves. В В В В В В В В The answer: В 3888В В

b)В The leading number is among 5 and 9 and, therefore , presents a negative amount. To acquire its degree, subtract the complement from your base: В В В В В В В В В 999

-717

282В В В В В В В В The response: -282

7)В

First, convert each item in the total to the floating stage sign-and-magnitude and excess-50 representation: В В В В 97. 685 sama dengan 0. 97685 x 102 = 05297685

0. 97685 sama dengan 0. 97685 x 90 = 05097685

Align the exponents by switching the lower mantissa in the second item two places correct and add both mantissas: В В В В 05297685

0520097685

05298661(85)

Cut the mantissa to five digits and present the result in floating stage format: В В В 0. 98661 * 102 = В 98. 661 (can be 98. 662 with rounding)В В В The absolute precise end result would be 98. 66185 as the floating level result is definitely 98. 661. This big difference is caused by the rotating due to the limited space for mantissa. В 8)В В В В В a)В В В В В

В В В В В В В В ВFirst, convert to the exponential explication: 11001. 0101 = 11001. 0101 times 20. В В В В В В В В В Adjust exponent to get the quantity in the normalized form simply by shifting the radix level four pieces to the left: В В В 11001. 0101 times 20 = 1 . 10010101 x twenty four. В В В В В В В В В Represent exponent in the excess-127 note: 127 + 4 sama dengan 131 = 100000112 В В В В В В В В В В В Drop initial " 1" in the mantissa, put sign little as the left the majority of bit and pad with zeros to get sole precision IEEE 754 file format: В В В В В В В В В В 01000001110010101000000000000000

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