(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>
7(72) + 0.10(2574) = 504 + 257.40 = 761.40 <== this is how much she will earn. Your gonna have to pick the answer choice that is the closest since no answer choices were given.
Answer:
no they dont form a Pythagorean triple
Step-by-step explanation:
a2 + b2 = c2
The information from the first equation gives you the information needed for the second. To solve the first equation you must rearrange the equation to isolate X. In order to do that you can first move the 3 to the other side of the equation by subtracting it from both sides (5x + 3 - 3 = 4 - 3) and then simplify that to (5x = 4 - 3) and further to (5x = 1). Then to move the 5 you must divide both sides by 5 so you get (5x/5 = 1/5) which can be simplified to (x = 1/5)
From this you can use the X value and input it into the second equation
Y = -3(1/5) and then solve for Y.
Hope this helps!
Answer:
I feel that is would be D
Step-by-step explanation:
This is the most reasonable awnser.
