1/2 * 5 = 2.5.
You would need 2.5 boxes to build 5 birdhouses.
Whole
so think of it as a interger
(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>
The correct answer is b. 14(m-8).
This is because it says that the weight of the boxes drew lifted were 8 pounds less than what marco lifted. This can be represented by m-8. You then need to multiply this by the amount of boxes that drew lifted, or 14.
The first thing I'll do is solve "5y = 2x + 20" for "<span>y=</span>", so that I can find my reference slope:
y = (2/5)x + 4;
So the reference slope from the reference line is <span>m = 2/5;</span>.
Now I need to find two new slopes, and use them with the point they've given me; namely, with the point (-1, 3). They want me to find the line through (4, –1) that is parallel to 5y = 2x + 20; that is, through the given point, they want me to find a line that has the same slope as the reference line.
Since a parallel line has an identical slope, then the parallel line through (-1, 3) will have slope <span>m = 2/5</span>. Now I have a point and a slope! So I'll use the point-slope form to find the line: y - 3 = (2/5)( x + 1);
Finally, y = (2/5)x + 17/5;
