Answer: The correr solution is 
Step-by-step explanation:
Since the steps in Lin's solution are not shown, we are not able to find her erros. However, we can solve it step by step and find the correct result:
Firstly, in the left part of the equation we have to multiply 
by the numbers inside the parenthesis; and in the right side we have to solve first what is inside the parenthesis:
Adding similar terms in the left side and multiplying 
by the number inside the parenthesis in the right side:
Isolating 
:
This is the correct solution
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 #)
Answer:
Mean:
Median: 63, 66, 67, 68, 73, 74, 76, 78, 80, 83, 86, 87, 88, 89, 89, 89, 93, 93, 95, 99, 100
= 169/2
Mode: 100
Step-by-step explanation:
Answer:
3.33 and 1/3
Step-by-step explanation:
"Dense" here means that there are infinite irrational numbers between two rational numbers. Also, there are infinite rational numbers between two rational numbers. That's the meaning of dense. Actually, that can be apply to all real numbers, there always is gonna be a number between other two.
But, to demonstrate that irrationals are dense, we have to based on an interval with rational limits, because the theorem about dense sets is about rationals, and the dense irrational set is a deduction from it. That's why the best option is 2, because that's an interval with rational limits.
Hey! My last name's Alvarez too! Anyway, the answer came out to 180cm squared. Hope I helped! Since you can barely get more than one or two answers on this site
