Answer:
A
Step-by-step explanation:
Answer:
24
Step-by-step explanation:
just MultiplMultiply 8 times 3 and 24
Answer:
27000
Step-by-step explanation:
We want to round 27346 to 2 significant figures.
Now, the number already has 5 significant figures, so to reduce it to 2 sig figures, we will have to reduce the 3 digits from the right to zero so that we can have just 2 significant figures by approximating.
Thus, 3 is less than 5 and so 7 cannot increase. Thus, we will have 27000 as the number when reduced to 2 significant figures.
Answer:
5 out of 12
5/12
January, March, May, June, July
There are three months whose name starts with the letter 'j': January, June, July
There are two months whose name starts with the letter 'm': March, May
So altogether, there are 5 months whose name starts with the letter 'j' or 'm'. So the possible outcome is 5 and total number of sample space is 12.
Therefore the required probability is: 5/12
PLEASE GIVE BRAINLYEST :)
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 #)
