D. 14
LL = 1/2 * Hsqrt3
7sqrt3 = 1/2 * ysqrt3
24.25 = ysqrt3
24.25/sqrt3 = y
y = 14
Answer:
The correct answer is 1,51%
Step-by-step explanation:
To get the probability equally likely of the first 6 examined that have a defective compressor, we use the following formula
P=# of possibilities that meet the condition / #of equally likely possibilities.
As there are multiple events , and all must be achieve, we multiply the probabilities.
P1=7 / 11
P2=6 / 10
P3=5 / 9
P4=4 / 8
P5=3 / 7
P6=2 / 6
P(x)=7 / 11*6 / 10*5 / 9*4 / 8*3 / 7*2 / 6=0,01515151515
P(x)=1,51%
The function f(x) is defined in to different ways depending on the value of X. When x is < 5, use the expression 2x - 1 to evaluate the function When x is ≥ 5, use the expression x2 - 3 for example f(1) = 2(1) - 1 = 1 f(10) = (10)2 - 3 = 97
Answer:
4 (9+4)
Step-by-step explanation:
its 4(9+4) because we are trying to make 36+16
4(8+5) is the same answer but it would be 32+20
while 4(9+4) would be 36+16
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 #)
