(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:
90°.
Step-by-step explanation:
if AR is diameter and the angle ∠AOR (it's m∠2) is based on this diameter, then its measure is 90°. It is central angle.
So, both equations are essentially linear equations.
Linear equations are written in the format
y = mx+b, where
m represents the
slope/slope intercept and
b represents the
y-intercept.
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Part Athe slope and y-intercept
y = mx +b
m - the slope; b- y-intercept
therefore;
y = 6x - 4
m = 6; b = -4y = 5x - 3
m = 5; b = -3The coordinates of the point where the lines are crossed are the solution to the system of linear equations.
How to graph the lines:
y = 6x - 4
y-intercept (0; -4)
for x = 1 ⇒ y = 6 · 1 - 4 = 6 - 4 = 2 ⇒ (1; 2)
y = 5x - 3
y-intercept (0; -3)
for x=1 ⇒ y = 5 · 1 - 3 = 5 - 3 = 2 ⇒ (1; 2)
***look at the img for graph reference***Part B:
x = 1; y = 2
We need to find the angle (rounded to the nearest whole degree) of a circle that would represent the percent of students taking mathematics.
Since a complete turn around the circle has 360º, we have the following proportion:
<em>Number of students Angle</em>
734 360º
442 x
Then, cross multiplying the above values, we obtain:

<em>Answer</em>: 217º
You have a 50% chance of picking a green tile so your answer would be 0.5