Answer:
1 = 0.61, 2 = 1.13
Step-by-step explanation:
The first is 25 + 25 + 10 + 1. This is 61 cents which is 0.61 of 1 dollar. The second is 100 + 10 + 1 + 1 + 1. This is equal to 113, 0r 1.13
BRAINLIEST PLEASE
Answer:
(3,10)
Step-by-step explanation:
When x is 3
y is 11, and this is invalid because it is not at accord.
For this case we must find an expression equivalent to:
We look for the factors of 10 and 30:
10: 1,2,5,10
30: 1,2,3,5,6,10,15,30
The greatest common factor of both numbers is 10.
Then, we can rewrite the expression as:
Answer:
(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:
4951 students
Step-by-step explanation:
If there was 99 students from each state, that would make the total to be:
99 * 50 = 4950
We can use the pigeon hole principle to make this the minimum number required.
The pigeon hole principle basically tells us that if you have more "pigeons" than "holes" then there must be one hole with multiple objects in it.
So, using this idea, we see that:
we need at least 1 more to ensure this
So, min number required would be
4950 + 1 = 4951
