Answers:
1.) 78 would be the range
3.) 18.8 or letter C would be correct
4.) 9.4
- A) 2/10 + 8/100 = <u>0</u><u>.</u><u>2</u><u>8</u>
- B) 13/100 + 4/10 = <u>0</u><u>.</u><u>5</u><u>3</u>
- C) 6/10 + 39/100 = <u>0</u><u>.</u><u>9</u><u>9</u>
- D) 70/100 + 3/10 = <u>1</u>
<u></u>
<em><u>#</u></em><em><u>carry </u></em><em><u>on </u></em><em><u>learning </u></em><u><</u><u>3</u>
So add all of them and then see smallest one sorry if I'm wrong
In order to answer the above question, you should know the general rule to solve these questions.
The general rule states that there are 2ⁿ subsets of a set with n number of elements and we can use the logarithmic function to get the required number of bits.
That is:
log₂(2ⁿ) = n number of <span>bits
</span>
a). <span>What is the minimum number of bits required to store each binary string of length 50?
</span>
Answer: In this situation, we have n = 50. Therefore, 2⁵⁰ binary strings of length 50 are there and so it would require:
log₂(2⁵⁰) <span>= 50 bits.
b). </span><span>what is the minimum number of bits required to store each number with 9 base of ten digits?
</span>
Answer: In this situation, we have n = 50. Therefore, 10⁹ numbers with 9 base ten digits are there and so it would require:
log2(109)= 29.89
<span> = 30 bits. (rounded to the nearest whole #)
c). </span><span>what is the minimum number of bits required to store each length 10 fixed-density binary string with 4 ones?
</span>
Answer: There is (10,4) length 10 fixed density binary strings with 4 ones and
so it would require:
log₂(10,4)=log₂(210) = 7.7
= 8 bits. (rounded to the nearest whole #)
