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The largest two primes that are both greater than 10 are 11 and 13. Their product is 11 x 13 = 143, which is less than 200.

We can list all possible products of two different primes greater than 10 and less than 200 by listing all possible pairs of primes and multiplying them:

11 x 13 = 143
11 x 17 = 187
11 x 19 = 209 (too large)
11 x 23 = 253 (too large)
11 x 29 = 319 (too large)
13 x 17 = 221 (too large)
13 x 19 = 247 (too large)
13 x 23 = 299 (too large)
13 x 29 = 377 (too large)
17 x 19 = 323 (too large)
17 x 23 = 391 (too large)
17 x 29 = 493 (too large)
19 x 23 = 437 (too large)
19 x 29 = 551 (too large)
23 x 29 = 667 (too large)

Therefore, there are 2 possible products that Syd could have ended up with: 143 and 187.

We can count the number of paths from A to I by using a counting method called "stars and bars".

Imagine that we represent each right-ward step as a star, and we use a bar to separate each step representing a different letter. For example, the path A C D E F G H I would be represented as:

* | * * | * * * | *

The number of stars in each section between the bars represents the number of right-ward steps in that section. Note that the number of bars is one less than the number of sections, since there are eight letters between A and I but only seven right-ward steps.

To count the number of paths from A to I, we need to count the number of ways to arrange the seven stars and three bars. We can choose the positions of the three bars from the ten possible positions (two between each pair of letters, and one before A and after I), which gives us:

10 choose 3 = 120 possible arrangements of bars.

Each arrangement of bars corresponds to a unique path from A to I, so there are 120 possible paths from A to I that consist of right-ward steps only.

To avoid ever having the exact same group of members over all 365 days, we need to make sure that no two groups of members are the same. 

On the first day, we can invite any combination of members from the book club, except for just one member (since that would be the same group of members every day). That means there are a total of 2^n - 1 possible groups of members on the first day. 

On the second day, we need to make sure that we don't invite the same group of members as on the first day. That means we can choose from the remaining 2^n - 2 possible groups of members. 

Continuing in this way, on the k-th day, we can choose from 2^n - k + 1 possible groups of members, since we have already used up k - 1 groups of members on the previous k - 1 days. 

To avoid ever having the exact same group of members over all 365 days, we need to make sure that the product of these possible choices is greater than or equal to 365. That is,

(2^n - 1)(2^n - 2)...(2^n - 365 + 1) >= 365.

Since we want to find the smallest positive integer n that satisfies this inequality, we can start with n = 1 and check each value of n until we find the smallest one that works.

For n = 1, we have:

(2^1 - 1)(2^1 - 2)...(2^1 - 365 + 1) = 1 * 0 * ... * (-363) = 0 < 365.

Therefore, n must be greater than 1. For n = 2, we have:

(2^2 - 1)(2^2 - 2)...(2^2 - 365 + 1) = 3 * 2 * ... * (-363) = a negative number.

Therefore, n must be greater than 2. For n = 3, we have:

(2^3 - 1)(2^3 - 2)...(2^3 - 365 + 1) = 7 * 6 * 5 * ... * (-363) = a negative number.

Therefore, n must be greater than 3. For n = 4, we have:

(2^4 - 1)(2^4 - 2)...(2^4 - 365 + 1) = 15 * 14 * 13 * ... * (-363) = a negative number.

Therefore, n must be greater than 4. For n = 5, we have:

(2^5 - 1)(2^5 - 2)...(2^5 - 365 + 1) = 31 * 30 * 29 * ... * (-363) = a negative number.

Therefore, n must be greater than 5. For n = 6, we have:

(2^6 - 1)(2^6 - 2)...(2^6 - 365 + 1) = 63 * 62 * 61 * ... * (-363) = a negative number.

Therefore, n must be greater than 6. For n = 7, we have:

(2^7 - 1)(2^7 - 2)...(2^7 - 365 + 1) = 127 * 126 * 125 * ... * (-363) = a negative number.

Therefore, n must be greater than 7. For n = 8, we have:

(2^8 - 1)(2^8 - 2)...(2^8 - 365 + 1) = 255 * 254 * 253 * ... * (-363) = a negative number.

Therefore, n must be greater than 8. For n = 9, we have:

(2^9 - 1)(2^9 - 2)...(2^9 - 365 + 1) = 511 * 510 * 509 * ... * (-363) = a negative number.

Therefore, n must be greater than 9. For n = 10, we have:

(2^10 - 1)(2^10 - 2)...(2^10 - 365 + 1) = 1023 * 1022 * 1021 * ... * (-363) = a negative number.

Therefore, n must be greater than 10. For n = 11, we have:

(2^11 - 1)(2^11 - 2)...(2^11 - 365 + 1) = 2047 * 2046 * 2045 * ... * (-363) = a negative number.

Therefore, n must be greater than 11. For n = 12, we have:

(2^12 -

Let's consider the different cases:

Case 1: An American wins the gold medal.
In this case, there are k-1 non-American skaters who can win the silver medal, and k-2 non-American skaters who can win the bronze medal. There are also n-1 non-American skaters who can win the gold medal, so the total number of ways that exactly one American wins a medal and that American wins the gold medal is:

(n-1) * (k-1) * (k-2)

Case 2: An American wins the silver medal.
In this case, there are k-1 non-American skaters who can win the gold medal, and k-2 non-American skaters who can win the bronze medal. There are also n-1 non-American skaters who can win the silver medal, so the total number of ways that exactly one American wins a medal and that American wins the silver medal is:

(n-1) * (k-1) * (k-2)

Case 3: An American wins the bronze medal.
In this case, there are k-1 non-American skaters who can win the gold medal, and k-2 non-American skaters who can win the silver medal. There are also n-1 non-American skaters who can win the bronze medal, so the total number of ways that exactly one American wins a medal and that American wins the bronze medal is:

(n-1) * (k-1) * (k-2)

Therefore, the total number of ways that exactly one American wins a medal is:

(n-1) * (k-1) * (k-2) * 3

since there are three cases to consider. Note that we are assuming that there are at least three non-American skaters, since otherwise it would be impossible for one American to win a medal without at least one non-American skater also winning a medal.

Answer: (n-1) * (k-1) * (k-2) * 3.

First, we can assume without loss of generality that the circle has radius 1, since the relative positions of the points W, X, Y, and Z do not depend on the size of the circle.

Let A be one of the points on the circle, and let B and C be the other two points that divide the circle into arc AB and arc AC. Without loss of generality, we can assume that point W lies on arc AB, and that points X, Y, and Z lie on arc AC in that order.

Now, let theta be the angle between points W and Z, measured in radians. Then the angle between points W and X is also theta, since the circle is symmetric about the line through A and the midpoint of segment WX. Similarly, the angle between points Y and Z is theta, and the angle between points X and Y is 2pi - 2theta.

Now, we can use the law of cosines to find the lengths of segments WX and YZ in terms of theta:

WX^2 = 1^2 + 1^2 - 2(1)(1)cos(theta) = 2 - 2cos(theta)
YZ^2 = 1^2 + 1^2 - 2(1)(1)cos(2pi - 2theta) = 2 - 2cos(2pi - 2theta) = 2 - 2cos(2theta)

Therefore, WX > YZ if and only if 2 - 2cos(theta) > 2 - 2cos(2theta), or equivalently, cos(theta) < cos(2theta).

This inequality holds if and only if theta lies in the interval [0, pi/3) or the interval (2pi/3, pi]. Therefore, the probability that WX > YZ is equal to the ratio of the lengths of these two intervals to the circumference of the circle:

P(WX > YZ) = [(pi - 2pi/3) + (pi/3 - 0)] / (2pi) = 2/3

Therefore, the probability that WX > YZ is 2/3.

Since QN is perpendicular to PR, we know that triangle PQR is a right triangle with right angle at Q. Therefore, we can use the Pythagorean theorem to find the length of PQ as:

PQ^2 = PR^2 - QR^2 = 14^2 - (2MN)^2

Since M is the midpoint of QR, we have MN = NR = QR/2. Therefore:

PQ^2 = 14^2 - (QR^2/2)^2

Also, since N is the midpoint of PR, we have PN = NR = PR/2 = 7. Therefore, using the Pythagorean theorem in triangle QPN, we have:

QN^2 = PQ^2 + PN^2 = PQ^2 + 7^2

Substituting the expression for PQ^2 from above, we have:

QN^2 = 14^2 - (QR^2/2)^2 + 7^2

Since O is the intersection of QN and RM, we know that triangle QOR is similar to triangle QPN. Therefore, we have:

OR / QN = PN / PQ

Substituting the values OR = RM = QR/2 and PN = 7, we obtain:

QR/2 / QN = 7 / PQ

Solving for PQ, we obtain:

PQ = 14QN / QR

Substituting this expression for PQ into the equation for QN^2 above, we obtain:

QN^2 = 14^2 - (QR^2/2)^2 + 7^2 = (14QR)^2 / (QR^2 + 4)

Multiplying both sides by QR^2 + 4, we obtain:

QN^2 * (QR^2 + 4) = 14^2 QR^2

Expanding and simplifying, we obtain a quadratic equation in QR^2:

QR^4 - 14^2 QR^2 + 4*QN^2 = 0

Substituting the given values QN = 12, PR = 14, and NR = 7, we obtain:

QR^4 - 14^2 QR^2 + 4*12^2 = 0

Solving this quadratic equation, we obtain:

QR^2 = 49 +- 35

Since QR is a length, we must take the positive square root, so:

QR^2 = 84

Substituting this value into the expression for PQ^2 above, we obtain:

PQ^2 = 14^2 - (QR^2/2)^2 = 14^2 - 42 = 112

Therefore, the area of triangle PQR is:

(1/2) PQ * PR = (1/2) * 14 * sqrt(112) = 14sqrt(7)

Answer: 14sqrt(7)

Let P be the vertex of the cube opposite O. Since P lies on face ABC and the cube is inscribed in the tetrahedron, we know that P lies on the plane containing triangle ABC.

Let Q be the foot of the altitude from O to triangle ABC. Since AOB, AOC, and BOC are all right angles, OQ is perpendicular to each of the three lines OA, OB, and OC. Therefore, OQ is the altitude of tetrahedron ABCO from vertex O to face ABC.

Let R be the foot of the altitude from P to triangle ABC. Since P is the opposite vertex of the cube from O, we know that OP is a space diagonal of the cube, and therefore has length sqrt(3). Since the cube is inscribed in the tetrahedron, we know that PR is perpendicular to each of the three lines PA, PB, and PC. Therefore, PR is the altitude of tetrahedron ABCO from vertex P to face ABC.

Now, let x be the length of the side of the cube. Since P is a vertex of the cube and lies on face ABC, we know that PR = x. Also, since OQ is the altitude of tetrahedron ABCO from vertex O to face ABC, we have:

[ABC] = (1/2) * OQ * BC

where [ABC] denotes the area of triangle ABC. Similarly, since PR is the altitude of tetrahedron ABCO from vertex P to face ABC, we have:

[ABC] = (1/2) * PR * BC

Combining these two equations, we obtain:

OQ = PR

Substituting the value of OQ from the equation above and the values of OA, OB, and OC given in the problem statement, we have:

sqrt(3) = sqrt(OP^2 - x^2) = sqrt(1 + 1 + 1 - x^2) = sqrt(3 - x^2)

Squaring both sides and simplifying, we obtain:

3 - x^2 = 3

x^2 = 0

Therefore, x = 0, which is impossible since x is a length. Thus, we must have made an error in our calculations.

The error is that we assumed that P is the unique vertex of the cube opposite O that lies on face ABC. However, it's possible that there is another vertex of the cube on face ABC that we haven't considered. To see this, let Q' be the foot of the altitude from P to face ABC. Since P lies on the plane containing triangle ABC, we know that Q' lies on the line through P perpendicular to face ABC. Therefore, Q' lies on the intersection of this line with face ABC, which is a line segment parallel to BC.

Let R' be the foot of the altitude from Q' to line OP. Since Q' is the foot of the altitude from P to face ABC, we know that PQ' is perpendicular to face ABC, and therefore lies in the plane containing triangle ABC. Moreover, since PR is perpendicular to line OP, we know that R' is the foot of the altitude from Q' to line OP. Therefore, R' lies on line OP.

Now, let x be the length of the side of the cube. Since PQ' is a side of the cube, we have PQ' = x. Also, since OQ is the altitude of tetrahedron ABCO from vertex O to face ABC, we have:

[ABC] = (1/2) * OQ * BC

Similarly, since R'Q' is the altitude of the triangle PQR' from vertex Q' to line PR, we have:

[ABC] = (1/2) * R'Q' * PR

Combining these two equations, we obtain:

OQ = R'Q'

Substituting the value of OQ from the equation above and the values of OA, OB, and OC given in the problem statement, we have:

sqrt(3) = sqrt(OP^2 - x^2) + sqrt(x^2 - R'P^2) = sqrt(1 + 1 + 1 - x^2) + sqrt(x^2 - (sqrt(2)/2)^2)

Squaring both sides and simplifying, we obtain:

3 - x^2 = 3 + x^2/2 - sqrt(2)x

Solving for x, we obtain:

x = 1/3

Therefore, the side length of the cube is 1/3.

We can start by simplifying the expression on the left-hand side of the equation using the formula for binomial coefficients:

binom(2n)n - binom(2n + 1)(2n) = [(2n)!] / [(n!)^2] - [(2n + 1)!] / [(2n!)(2n + 1)]
= [(2n)! / (n!)^2] - [(2n + 1)! / (2n + 1)(2n)!]
= [(2n)! / (n!)^2] - [(2n)! / (2n + 1)(n!)^2]
= [(2n)! / (n!)^2] * [(1 - 1/(2n+1))]

Now, we can substitute this expression into the original equation and simplify:

[(2n)! / (n!)^2] * [(1 - 1/(2n+1))] = f(n) * [(2n + 1)! / ((n + 1)!(n!))]
[(2n)! / (n!)^2] * [(1 - 1/(2n+1))] = f(n) * [(2n + 1) / (n + 1)]

Dividing both sides by [(2n)! / (n!)^2], we obtain:

(1 - 1/(2n+1)) = f(n) * [(2n + 1) / (n + 1)]

Simplifying the right-hand side, we get:

f(n) = (1 - 1/(2n+1)) * [(n + 1) / (2n + 1)]

Therefore, the function f(n) is given by:

f(n) = [(n + 1)/(2n + 1)] - [(n + 1)/(2n+1)^2]

We can check that this formula satisfies the original equation by substituting it back in and simplifying.

Let's consider the possible outcomes for the sum of two standard 6-sided dice:

- There is one way to get a sum of 2: rolling two 1's.
- There are two ways to get a sum of 3: rolling a 1 and a 2, or rolling a 2 and a 1.
- There are three ways to get a sum of 4: rolling a 1 and a 3, rolling a 2 and a 2, or rolling a 3 and a 1.
- There are four ways to get a sum of 5: rolling a 1 and a 4, rolling a 2 and a 3, rolling a 3 and a 2, or rolling a 4 and a 1.
- There are five ways to get a sum of 6: rolling a 1 and a 5, rolling a 2 and a 4, rolling a 3 and a 3, rolling a 4 and a 2, or rolling a 5 and a 1.
- There are six ways to get a sum of 7: rolling a 1 and a 6, rolling a 2 and a 5, rolling a 3 and a 4, rolling a 4 and a 3, rolling a 5 and a 2, or rolling a 6 and a 1.
- There are five ways to get a sum of 8: rolling a 2 and a 6, rolling a 3 and a 5, rolling a 4 and a 4, rolling a 5 and a 3, or rolling a 6 and a 2.
- There are four ways to get a sum of 9: rolling a 3 and a 6, rolling a 4 and a 5, rolling a 5 and a 4, or rolling a 6 and a 3.
- There are three ways to get a sum of 10: rolling a 4 and a 6, rolling a 5 and a 5, or rolling a 6 and a 4.
- There are two ways to get a sum of 11: rolling a 5 and a 6, or rolling a 6 and a 5.
- There is one way to get a sum of 12: rolling two 6's.

We want to find all possible ways of numbering two 6-sided dice with positive integers (not necessarily distinct) so that when they are rolled, the 36 possible outcomes for the sum of the two rolls are the same as the 36 possible outcomes for the sum of two standard 6-sided dice.

One way to approach this problem is to consider the number of ways to get each possible sum using the two numbered dice. For example, if we number the dice with the positive integers a and b, then the number of ways to get a sum of 2 is the number of ways to roll two 1's, which is equal to the number of ways to choose two 1's from the numbers a and b. In other words, the number of ways to get a sum of 2 is equal to the number of ways to choose two 1's from the set {a, b}.

Using this approach, we can generate a table of the number of ways to get each possible sum using the two numbered dice:

| Sum | Number of ways |
|-----|---------------|
|  2  |       1       |
|  3  |       2       |
|  4  |       3       |
|  5  |       4       |
|  6  |       5       |
|  7  |       6       |
|  8  |       5       |
|  9  |       4       |
| 10  |       3       |
| 11  |       2       |
| 12  |       1       |

We can see that the number of ways to get each possible sum using the two numbered dice must match the number of ways to get each possible sum using two standard 6-sided dice. Therefore, we can use this table to generate a system of equations:

a + b = 2  (one way to get a sum of 2)
a + 2b = 3  (two ways to get a sum of 3)
2a + 2b = 4  (three ways to get a sum of 4)
2a + 3b = 5  (four ways to get a sum of 5)
3a + 3b = 6  (five ways to get a sum of 6)
3a + 4b = 7  (six ways to get a sum of 7)
4a + 4b = 8  (five ways to get a sum of 8)
4a + 5

To find the number of cozy 2-digit numbers, we can list all possible numbers and check if they are cozy:

20, 21, 22, 23, 24, 25, 26, 27, 28, 29
30, 31, 32, 33, 34, 35, 36, 37, 38, 39
40, 41, 42, 43, 44, 45, 46, 47, 48, 49
50, 51, 52, 53, 54, 55, 56, 57, 58, 59
60, 61, 62, 63, 64, 65, 66, 67, 68, 69
70, 71, 72, 73, 74, 75, 76, 77, 78, 79
80, 81, 82, 83, 84, 85, 86, 87, 88, 89
90, 91, 92, 93, 94, 95, 96, 97, 98, 99

From this list, we can see that there are 45 cozy 2-digit numbers.

To find the number of cozy 3-digit numbers, we can use a similar approach. We can start by fixing the first digit in the number to be even (since if it is odd, the number cannot be cozy). There are 4 even digits to choose from (2, 4, 6, and 8). 

For each even digit, we need to consider two cases: the second digit is even, or the second digit is odd but next to an even digit. If the second digit is even, there are 5 even digits to choose from (0, 2, 4, 6, and 8). If the second digit is odd but next to an even digit, there are 4 choices for the even digit (since it cannot be 0 or the same as the first digit), and 5 choices for the odd digit (since it must be next to an even digit). Therefore, there are:

4 x (5 + 4 x 5) = 104

cozy 3-digit numbers.

Therefore, there are 45 cozy 2-digit numbers and 104 cozy 3-digit numbers.

The functions f(x) and h(x) are not the same function because they have different domains and ranges. The domain of f(x) is all real numbers except x = 7, since dividing by zero is undefined. In other words, f(x) is defined for all values of x except x = 7. On the other hand, the domain of h(x) is all real numbers, since there are no restrictions on multiplying two real numbers together. The range of f(x) is also restricted, since the function approaches positive or negative infinity as x approaches 7 from either direction. In other words, f(x) does not take on any values near x = 7. In contrast, the range of h(x) is all real numbers, since any real number can be obtained by multiplying two other real numbers together. Therefore, f(x) and h(x) have different domains and ranges, and are therefore not the same function.

a) The probability of true-positive test is the probability that a woman who has breast cancer receives a positive mammogram result. This can be calculated using Bayes' theorem:

P(True-positive test) = P(Positive result | Breast cancer) * P(Breast cancer)

P(Positive result | Breast cancer) is the sensitivity of the mammogram, which is 85% or 0.85. P(Breast cancer) is the prevalence of breast cancer in women in their 60s, which is 3.36% or 0.0336. Therefore:

P(True-positive test) = 0.85 * 0.0336 = 0.02856 or approximately 2.86%

b) The probability of false-positive test is the probability that a woman who does not have breast cancer receives a positive mammogram result. This can also be calculated using Bayes' theorem:

P(False-positive test) = P(Positive result | No breast cancer) * P(No breast cancer)

P(Positive result | No breast cancer) is the specificity of the mammogram, which is 95% or 0.95. P(No breast cancer) is the complement of the prevalence of breast cancer, which is 1 - 0.0336 = 0.9664. Therefore:

P(False-positive test) = 0.95 * 0.9664 = 0.91728 or approximately 91.73%

c) The probability that the woman really has breast cancer given the positive test is the positive predictive value (PPV) of the mammogram. This can be calculated using the formula:

PPV = P(Breast cancer | Positive result) = P(Positive result | Breast cancer) * P(Breast cancer) / P(Positive result)

P(Positive result) is the probability of a positive result, which is the sum of the probabilities of true-positive and false-positive tests:

P(Positive result) = P(True-positive test) + P(False-positive test) = 0.02856 + 0.91728 = 0.94584 or approximately 94.58%

Therefore:

PPV = 0.85 * 0.0336 / 0.94584 = 0.02999 or approximately 3.00%

This means that if a woman in her 60s gets a positive mammogram result, there is a 3.00% chance that she really has breast cancer.

hi, i need help on this problem too, the numbers are similar to this. Can someone answer this?

nvm...I solved them!!!!

Lets use x and y, x being your money and y being your brothers

(If I give my brother 5 dollars, then we will have the same amount of money.)

that means that x-5=y+5; x= y+10

( If instead he gives me 24 dollars, then I'll have twice as much money as he will have.)

using this information we can write the amount of money as 2(y-24)=x+24

(substitue y) 2(y-24)=y+34 =   2y-48=y+34 =  y=82

Since y stands for your brothers money that means your brother has 82 dollars and you have 92. Hope this helps! :)

x + y + 3z = 10      →   y =  10 - x - 3z

(we want to get y by itself0

-4x + 2y +5z = 7    →   -4x + 2(10 - x - 3z) + 5z = 7   →   -4x + 20 -2x - 6z + 5z  = 7  →  -6x - z = -13  →

6x + z  = 13

(substitute y into  the second equation)

Notice we have this system now 

k*x + z = 3

6x + z = 13 

And notice that, if k =6, we would have this system:

6x + z  = 3      and

6x + z  = 13       but this is impossible, because two things added together can't possibly have two different outcomes 

when k = 6, the system has no solution. 

Could you tell me how you got that answer?

The answer is 2.4 seconds.

This is a classic example of a combinatorial problem, specifically a problem of distributing objects into groups. Since the boxes are indistinguishable, we don't need to worry about which box the balls go in; all that matters is how many balls are in each box.

We can approach this problem using the stars and bars method. Imagine that we have five balls (which are distinguishable) and two dividers (which are indistinguishable). We can arrange these seven objects in a line in any order. For example:

| * * | * * *

In this arrangement, the first box has 0 balls, the second box has 2 balls, and the third box has 3 balls. We can count the number of arrangements by counting the number of ways to choose 2 positions out of 7 for the dividers. Once the positions of the dividers are fixed, the balls will naturally fall into the boxes defined by the dividers.

The number of ways to choose 2 positions out of 7 is given by the binomial coefficient:

C(7, 2) = 7! / (2! * 5!) = 21

Therefore, there are 21 ways to put 5 balls in 3 indistinguishable boxes.

200!! ==118305033 0245448578 0817140255 6304773068 3304239434 5395086021 9552506842 4830185524 5975013130 0382609551 6491169524 5558358737 3444882662 1766347181 5255955393 1916976653 8608640000 0000000000 0000000000

Note: !! is called "double factorial", which means: 200 x 198 x 196 x 194 x 192 x..........x 2

200!! ==2^197 × 3^48 × 5^24 × 7^16 × 11^9 × 13^7 × 17^5 × 19^5 × 23^4 × 29^3 × 31^3 × 37^2 × 41^2 × 43^2 × 47^2 × 53 × 59 × 61 × 67 × 71 × 73 × 79 × 83 × 89 × 97

I think by "prime multiple", they mean: prime factor with the largest exponent, which is: 2^197.

The largest "prime factor" is obviously: 97

Your answer 18 * sin(80)/sin(50)  works out to 20 + pi.

1.  

Let's assume that the 3 candidates are named A, B, and C.

If all 70 students vote, there are a total of 70 choices for the first vote, 69 choices for the second vote, and 68 choices for the third vote, since each student can only vote once and cannot vote for themselves. Therefore, the total number of possible vote counts is:

70 x 69 x 68 = 328,920

Now, if some students do not vote, the total number of possible vote counts will decrease. Let's consider some scenarios:

If one student does not vote: There are 70 choices for who does not vote. For the remaining 69 students, there are 328,920 possible vote counts. Therefore, the total number of possible vote counts when one student does not vote is:

70 x 328,920 = 23,064,400

If two students do not vote: There are (70 choose 2) = 2,415 ways to choose two students out of 70 who do not vote. For the remaining 68 students, there are 328,920 possible vote counts. Therefore, the total number of possible vote counts when two students do not vote is:

2,415 x 328,920 = 793,556,800

If three students do not vote: There is only one way to choose three students out of 70 who do not vote. For the remaining 67 students, there are 328,920 possible vote counts. Therefore, the total number of possible vote counts when three students do not vote is:

1 x 328,920 = 328,920

In summary, the total number of possible vote counts when some students do not vote is:

23,064,400 + 793,556,800 + 328,920 = 817,950,120

2. 

We can use a stars and bars approach to solve this problem. We need to distribute 7 pieces of taffy and 8 pieces of licorice to 5 kids such that each kid receives exactly 3 pieces of candy.

Let's represent the distribution using 5 variables: x1, x2, x3, x4, x5, where xi represents the number of pieces of candy given to the ith kid. Since each kid must receive 3 pieces of candy, we can subtract 3 from each xi to simplify the problem.

So, we need to distribute (7-3x5) = 2 pieces of taffy and (8-3x5) = -7 pieces of licorice among 5 kids, without any restriction on the number of pieces each kid gets. Notice that we have negative pieces of licorice, but this simply means that some kids will get no licorice at all.

Using the stars and bars formula, the number of ways to distribute 2 pieces of taffy and -7 pieces of licorice among 5 kids is:

C(2+5-1, 5-1) * C(-7+5-1, 5-1) = C(6, 4) * C(-3, 4)

where C(n, k) represents the number of ways to choose k items out of n.

Since C(-3, 4) is not defined, we need to use a different formula for negative values. The formula is:

C(-n+k-1, k-1) = C(n+k-1, k-1)

Using this formula, we get:

C(6, 4) * C(3+4-1, 4-1) = C(6, 4) * C(6, 3)

= 15 * 20

= 300

Therefore, there are 300 ways to distribute 7 pieces of identical taffy and 8 pieces of identical licorice to 5 kids such that each kid receives exactly 3 pieces of candy.