<p>Hi I am stuck on these questions and I do not know the answers. HELP!</p>
<p>The digits 1, 2, 3, 4 are to be arranged randomly to make a positive four digit integer. What is the probability that the digits 1, 2, and 3 will be directly next to each other, in that order, from left to right?
(A) 1/12
(B) 1/8
(C) 1/6
(D) 1/4
(E) 1/3</p>
<p>The lengths of the sides of an isosceles triangle are 20, n, and n. If n is an integer, what is the smallest possible perimeter of the triangle?
(A) 40
(B) 41
(C) 42
(D) 44
(E) 60</p>
<p>A colony of bacteria grow so that t days after the start of an experiment, the number of bacteria is n2^(1/2), where n is the number of bacteria at the start of the experiment. If there are 10000 bacteria 6 day after the experiment's start, what is the value of n?
(A) 1250
(B) 1667
(C) 2500
(D) 3333
(E) 5000</p>
<p>First question: How many arrangements have 123 right next to each other, in that order, in a 4-digit number that has only the digits 1, 2, 3, and 4? There are two possibilities, 1234 or 4123. Anything else will break up 123. Now, how many possible 4-digit numbers are there, containing 1, 2, 3, and 4 with no repeats? (The statement that the digits are supposed to be arranged randomly suggests that they are not allowing 1111, or any other numbers with repeat digits.) There are 4 choices for the first number, times three choices for the second, times two choices for the third, and then there is only one final digit left. So there are 24 possibilities overall. The probability is the number of arrangements with 123 together divided by the total number of arrangements, so the answer is A) 1/12.</p>
<p>Second question: The perimeter of the triangle is 20 + 2n where n is an integer (from the information given). You want the smallest possible value of the perimeter. What you need to keep in mind is that sides of lengths (20, n, n) have to form a triangle.</p>
<p>So could the perimeter be 40? If you solve 20 + 2n = 40 for n, this gives n = 10. But could sides of that length form a triangle? Not really, because the two sides of length 10 have to be laid out in a line to stretch the distance from one endpoint of the side of length 20 to the other. I suppose that you could call this a degenerate triangle, but it would really just be three line segments, with two of them on top of the third.</p>
<p>Could the perimeter be 41? Solving 41 = 20 + 2n for n yields n = 21/2. But n is supposed to be an integer. So that’s out.</p>
<p>Could the perimeter be 42? Solving 42 = 20 + 2n for n yields n = 11. You can make an isosceles triangle with sides of (20, 11, 11). The angle opposite the side of length 20 is obtuse and quite large, but it’s possible. So this is the answer, C.</p>
<p>Third question: I think you are missing a t in the equation for the number of bacteria t days after the start of the experiment. As it stands, the question can’t be answered.</p>
<p>Thank you so much! I get them now! Also I misread the third question, it was supposed to be 2^(t/2) instead of 2^(1/2)…Thanks!</p>
<p>1) 4p4 = 24
and the possibilities are 1234 , 4321 2/24 =1/12 (A)
2) no one side of a triangle can be longer than or equal the other two sides when they are added<br>
40 = 20 + 10 +10 but it’s not possible because 10+10 =20<br>
41 = 20+ 11 +11 which is possible because ( 11 +11 = 22 ) which is bigger than 20<br>
and the question asks for the least possible perimeter (B)
3) x * 2^6/2 = 10000
x = 1250 (A)</p>