29 because first u add 5 then 8 3 times.
(e) Each license has the formABcxyz;whereC6=A; Bandx; y; zare pair-wise distinct. There are 26-2=24 possibilities forcand 10;9 and 8 possibilitiesfor each digitx; yandz;respectively, so that there are 241098 dierentlicense plates satisfying the condition of the question.3:A combination lock requires three selections of numbers, each from 1 through39:Suppose that lock is constructed in such a way that no number can be usedtwice in a row, but the same number may occur both rst and third. How manydierent combinations are possible?Solution.We can choose a combination of the formabcwherea; b; carepair-wise distinct and we get 393837 = 54834 combinations or we can choosea combination of typeabawherea6=b:There are 3938 = 1482 combinations.As two types give two disjoint sets of combinations, by addition principle, thenumber of combinations is 54834 + 1482 = 56316:4:(a) How many integers from 1 to 100;000 contain the digit 6 exactly once?(b) How many integers from 1 to 100;000 contain the digit 6 at least once?(a) How many integers from 1 to 100;000 contain two or more occurrencesof the digit 6?Solutions.(a) We identify the integers from 1 through to 100;000 by astring of length 5:(100,000 is the only string of length 6 but it does not contain6:) Also not that the rst digit could be zero but all of the digit cannot be zeroat the same time. As 6 appear exactly once, one of the following cases hold:a= 6 andb; c; d; e6= 6 and so there are 194possibilities.b= 6 anda; c; d; e6= 6;there are 194possibilities. And so on.There are 5 such possibilities and hence there are 594= 32805 such integers.(b) LetU=f1;2;;100;000g:LetAUbe the integers that DO NOTcontain 6:Every number inShas the formabcdeor 100000;where each digitcan take any value in the setf0;1;2;3;4;5;7;8;9gbut all of the digits cannot bezero since 00000 is not allowed. SojAj= 9<span>5</span>
Answer:
a =1 and a=4.
Step-by-step explanation:
The function is
If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.
Now, for a a positive real we have that 
will annulate the denominator, i.e
. But, if a = 1 we have:
so, the value 
won't annulate the denominator.
Now, for a = 4 we have:
so, the value 
won't annulate the denominator.
In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.
A. likely
B. unlikely
C. There is a greater chance in picking to purple marbles because there are less of the white and more purple.
Hallie should spin the wheel because u have a greater chance of landing on a 5 then rolling two number cubes and hoping both number cubes come to equal 5.
