Question

Write binary equivalent of the following octal numbers.
(i) 2306 (ii) 5610 (iii) 742 (iv) 65.203

Answer 1

Answer:

(i) 2306

= 10011000110

(ii) 5610

= 101110001000

(iii) 742

= 111100010

(iv) 65.203

= 1011

Explanation:

hope it helps u

Answer 2

Answer:

65.2038 = 6∙81+5∙80+2∙8-1+0∙8-2+3∙8-3 = 48+5+0.25+0+0.005859375 = 53.25585937510

Happened: 53.25585937510

Converting 53.25585937510 in Binary system here so:

Whole part of a number is obtained by dividing on the basis new

53 2

-52 26 2

1 -26 13 2

0 -12 6 2

1 -6 3 2

0 -2 1

1

Translation of numbers from one system to another

Happened:5310 = 1101012

The fractional part of number is found by multiplying on the basis new

Translation of numbers from one system to another

0 .255859375

. 2

0 51172

2

1 02344

2

0 04688

2

0 09375

2

0 1875

2

0 375

2

0 75

2

1 5

2

1 0

Happened:0.25585937510 = 0.0100000112

Add up together whole and fractional part here so:

1101012 + 0.0100000112 = 110101.0100000112

Result of converting:

65.2038 = 110101.0100000112