if `a=[(0,1),(1,0)]` then `a^4` is (a) `[(0,0),(1,1)]` (b) `[(1,1),(0,0)]` (c) `[(0,1),(1,0)]` (d) `[(1,0),(0,1)]`
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a - ii, b - iv, c - i, d - iii * If the number ends with a $0$ then there are $9$ choices for the first digit, $8$ for the second and $7$ for the third, which makes $1\times9\times8\times7= 504$ possibilities.
* If the number is even ending with something else than $0$ then there are $4$ choices for the last digit, $8$ choices for the first digit (no $0$ nor the last digit), $8$ for the second digit and $7$ for the third digit, which makes $4\times8\times8\times7 =1792$


Together, this gives $2296$ numbers with $4$ distinct digits that are even. Note that this does not allow leading $0$, as you see to want it based from the question.


Correct Answer: $B$ the sum of a pmf over its entire domain must equal 1.
so cς(x=1 to ∞)σ(y=1 to ∞)[(1/4^x)(1/3^y)] = 1.
 
ς(i=1 to ∞)[1/n^i] = 1/(n-1). so
 
ς(y=1 to ∞)[(1/3^y)] = 1/(3-1) = 1/2 and
ς(x=1 to ∞)[(1/4^x)] = 1/(4-1) =1/3.
 
so, σ(x=1 to ∞)σ(y=1 to ∞)[(1/4^x)(1/3^y)] = (1/2)*(1/3) = 1/6.
 
to make the sum = 1, then c=6. 4x3 – 12x2 + 11x – 5 is divided by 2x – 1remainder theorem states that if p(x) is any polynomial and a is any real number and if p(x) is divided by the linear polynomial (x – a) , then the remainder is p(a).let p(x) = 4x3 – 12x2 + 11x – 5 and we have (2x – 1)the zero of (2x – 1) is now using remainder theorem,p(x) = 4x3 – 12x2 + 11x – 5 is divided by 2x – 1 then, is the remainderremainder = –2 1 a c d
2 a
3 c
4 b
5 c
i got a 100% a, b and c %%
var1= [1 1 2 4 5 7 7 7 8 10 11 13 14 14 17 17 17 17 18 20];
var2= [0.5 0.3 0.1 0.6 0.2 0.4 0.8 0.9 0.3 0.2 0.1 0.4 0.8 0.7 0.5 0.9 0.2 0.8 0.3 0.7];

x=min(var1):1:max(var1);
newele1=setdiff(x,var1);
newele2=nan(size(newele1));

newvar1=[var1,newele1];
newvar2=[var2,newele2];

[newvar1,index]=sort(newvar1);
newvar2=newvar2(index) 1) p(x) = 1/x2 + 1 if x ≤ 3 it is clear that x= 3 is included, so you have to use this function.2) loo k at table d, and compare the given information with the data written inside the table. let put a(x2+4)+(bx+c)(2x+1) =17put 2x+1=0a=4by equating the coefficient of x2 and constant term,a+2b=04+2b=0b=-24a+c=174×4+c=17c=1from the equation(1), we get, >> get(groot,'defaultaxescolororder')
 ans =
 0 0.4470 0.7410
 0.8500 0.3250 0.0980
 0.9290 0.6940 0.1250
 0.4940 0.1840 0.5560
 0.4660 0.6740 0.1880
 0.3010 0.7450 0.9330
 0.6350 0.0780 0.1840
 >> 

the letter monikers for colors are the pure colors, 'b' --> [0 0 1], etc., ...