<p>1.How many integers are between 6 and 60 divisible by 6?</p>
<p>This does not include 6 and 60 because they are not between 6 and 60. Numbers divisible by 6 are 12, 18, 24, 30, 36, 42, 48, and 54. 8 integers</p>
<p>2.The sum of a set of consecutive even integers is 12.If the least is -10,how many elements are in the set?</p>
<p>Consecutive even integers are 2, 4, 6, 8, 10, 12, 14, etc. The difference between any two consecutive even integers is 2 (because you are counting by 2). If the least is -10, the consecutive ones are -8, -6, -4, -2, 0, 2, 4, 6, 8, and 10. Notice you are going from -10 to 0, and then from 0 to 10. The sum of the consecutive even integers in the set (-10, 10) is 0 because you are going from -10 to 10. Going up one more number you would arrive at 12. This would make the sum of the set 0+12=12. The set should therefore be -10 to 12, with [12-(-10)]/2 = 12 as the amount of elements in the set:</p>
<p>[-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12]</p>
<p>3.Whats the remainder when 2^400 is divided by 10?</p>
<p>2^400 is 1 multiplied by 2 400 times. Basically there are 400 2s multiplied by each other.</p>
<p>2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
…
2^12 = 4096
…
2^16 = 65536</p>
<p>If you continue you will notice that the values of 2^4, 2^8, 2^12, 2^16, etc. (numbers whose powers are multiples of 4) end with 6. The power of 2^400, which is 400, is a multiple of 6; therefore, 2^400 should follow this pattern and it should end with 6.
If you divide a number that ends with 6 by 10, the remainder will be 6.
36 = 10 + 10 + 10 + 6 = three tens and a six (36/10 is 3, with a remainder of 6)
46 = 10 + 10 + 10 + 10 + 6 = four tens and a six (46/10 is 4, with a remainder of 6)</p>
<p>4.Class A and class B have an equal number of students.If a total of 3 students move form class A to B,the ratio of students in class A to B is 5:30.How many students were originally in the class?</p>
<p>Either you typed this question wrong or this question was faulty, because I can’t answer this question.</p>
<p>5.A two sided coin is tossed three times.What is the probablity that “heads” will be the result exactly two times ? (i got the answer as either 3/8 or 3/4 …which onez right?)</p>
<p>There are 3 flips. Think of this as 3 slots.</p>
<hr>
<p>Each slot has 2 possibilities: head or tail
2 2 2
You multiply the numbers in the slots to find the amount of combinations in the entire set of 3 flips</p>
<p>2<em>2</em>2 = 8</p>
<p>There are 8 possible combinations if you flip a coin 3 times.</p>
<p>You can get 2 heads in 3 different combinations.
H = head, T = tail
HHT
HTH
THH</p>
<p>answer is 3/8</p>
<p>6.A single dice with 6 faces numbered 1 through 10 is thrown twice.If the numeral that faces as the result of row is recorded,what is the probability that the sum of the two recorded numbers is less than 10?</p>
<p>This question makes no sense. How is it numbered 1 through 10 if it only has 6 faces? If it is numbered 1 through 6, the amount of possible combinations is 6*6=36.</p>
<p>You need to get a sum less than 10. You can fail this by getting 10, 11, or 12 (12 is the max since the most you can get is 6 twice). You get 10 by getting 5+5, 6+4, or 4+6. You get 11 by getting 5+6, 6+5. You get 12 by getting 6+6. So, there are 6 combinations that do not give you a sum of less than 10. Therefore there are 36-6=30 combinations that DO give you a sum of less than 10. The answer is 30/36, or 5/6</p>
<p>7.A square is inscribed in a circle.If a point is selected at random from the interior of the circle,what’s the probability that the point lies outside the square?</p>
<p>This question takes too long to explain, and I would need a diagram. You need to use the Pythagorean theorem, a^2 + b^2 = c^2, where the hypotenuse c is both the diameter of the circle and a diagonal of the square. </p>
<p>8.On a two part exam,a student answered 40% questions in the first part correctly and 60% of the questions in the second part correctly.If the student answered a total of 70 questions correctly.How many questions were there in the second part of the exam? </p>
<p>40% of 70 is (70)<em>(0.40) = 28
60% of 70 is (70)</em>(0.60) = 42</p>