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The weight of the Most Significant Bit (MSB) in a 5-bit binary number is 16. In binary representation, each bit position corresponds to a power of 2, starting from the right with 2^0. Therefore, the MSB, which is the leftmost bit in a 5-bit number, represents 2^4 or 16 in decimal.
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What is the MSB of a positive binary number?
The most significant byte (MSB) of a positive binary number is the decimal value of the left-most bit.For example, the binary number 10111001011 is 11 bits, meaning it's 11 digits long. Thus, the decimal value of the left-most bit, the MSB, is 1 X 210 = 1024. The reason why it's not 1 X 211 is that the decimal value of the right-most bit is represented by raising 2 to the 0th power, not the first power. In this case, the right-most bit has a decimal value of 1 X 20 = 1.
Twos complement of a given 3 or more bit binary number of non-zero magnitude is the same the original number if all bits except the?
ANSWER: MSB IS 1 In the 2's complement representation, the 2's complement of a binary number is obtained by first finding the one's complement (flipping all the bits), and then adding 1 to the result. This representation is commonly used to represent signed integers in binary form. Now, if all bits except the sign bit are the same, taking the 2's complement of the binary number will result in the negative of the original number. The sign bit (the leftmost bit) is flipped, changing the sign of the entire number. For example, let's take the 4-bit binary number 1101 The 2's complement would be obtained as follows: Find the one's complement: 0010 Add 1 to the one's complement: 0011
How do you use Multiple Comparisons of Means (Two-way ANOVA) on Surface?
H_0:μ_(1∙)=μ_(2∙)=⋯=μ_(a∙) F=MSA/MSE F_(ν_1,ν_2 ) ν_1=a-1 , ν_2=(a-1)(b-1) H_0:μ_(∙1)=μ_(∙2)=⋯=μ_(∙b) F=MSB/MSE F_(ν_1,ν_2 ) ν_1=b-1 , ν_2=(a-1)(b-1)  
What is Multiple Comparisons of Means (Two-way ANOVA) in Depth?
H_0:μ_(1∙)=μ_(2∙)=⋯=μ_(a∙) F=MSA/MSE F_(ν_1,ν_2 ) ν_1=a-1 , ν_2=ab(n-1) H_0:μ_(∙1)=μ_(∙2)=⋯=μ_(∙b) F=MSB/MSE F_(ν_1,ν_2 ) ν_1=b-1 , ν_2=ab(n-1) H_0: there is not interactions between factors F=(MS(AB))/MSE F_(ν_1,ν_2 ) ν_1=(a-1)(b-1), ν_2=N-ab Contrast H_0:c_1 μ_1+c_2 μ_2+⋯c_k μ_k=0 ∑▒c=0 t=(∑_(i=1)^k▒〖c_i x ̅_i 〗)/(s_p √(∑_(i=1)^k▒(c_i^2)/n_i )) Student t(υ) ν=n_1+⋯+n_k-k
