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Out of 200 students, 75% study math, which is 0.75200=<<75.01200=150>>150 students. 15% study physics, which is 0.15200=<<15*.01*200=30>>30 students. Thus, the rest study chemistry, which is 200-150-30=<<200-150-30=20>>20 students.

On Monday, 20% of physics students leave, which means 0.230=<<0.230=6>>6 students leave. Thus, the number of physics students remaining after Monday is 30-6=<<30-6=24>>24 students. Additionally, 6 new chemistry students join, increasing the number of chemistry students to 20+6=<<20+6=26>>26 students.

On Tuesday, 4 physics students decide to study math instead, which means the number of physics students decreases to 24-4=<<24-4=20>>20 students. Furthermore, 4 math students join, increasing the number of math students to 150+4=<<150+4=154>>154 students.

After Tuesday, the number of students who do not study physics is 200-154=<<200-154=46 students. Therefore, the percentage of students who do not study physics is (46/200)100%=23%.

The answer is 23%.

 " if the number of heads was at least........... ?".  What number was it?

I think your question maybe stated incorrectly! You say " A 100 ounce jug of very sour lemonade initially consists of 60% water and 40% lemon juice." Very sour lemonade implies that of the 100 ounces, 60% is lemon juice and 40% water! If that is the case, then the solution should be:

At the beginning, the jug contains 60% of lemon juice, which means there are 60 ounces of lemon juice in the jug, and 40% of water, which means there are 40 ounces of water in the jug.

When 60% of the lemonade is poured out, the remaining lemonade in the jug weighs 40 ounces. This means that 60 ounces x (40/100) = 24 ounces of lemon juice are left in the jug, and 16 ounces of water are left in the jug.

When 8 ounces of lemon juice is added to the jug, the total amount of lemon juice in the jug becomes 24 + 8 = 32 ounces, and the total amount of liquid in the jug becomes 32 + 16 = 48 ounces.

To find the percentage of the mixture that is now lemon juice, we can divide the amount of lemon juice in the jug by the total amount of liquid in the jug and multiply by 100:

Percentage of lemon juice = (32/48) x 100% = 66.67%

Therefore, the percentage of the mixture that is now lemon juice is 66.67

\(e^{i \theta} = \frac{5}{12} + \frac{13}{12} i\)

I get 13 such numbers:

I think you got carried away with excitement Justingavriel1233!!

"Therefore, the polynomial f(x) = (1/6) x^3 satisfies both properties."  Note that (1/6)x^3 is not a degree 5 polynomial.

Looks good to me Stereo.  

Give yourself a thumbs up :)

Since f(x) is divisible by x^3, we can write:

f(x) = x^3 g(x)

where g(x) is some polynomial of degree 2. We want f(x) + 1 to be divisible by (x + 1)(x + 2)(x + 3), so we can write:

f(x) + 1 = x^3 g(x) + 1 = (x + 1)(x + 2)(x + 3) h(x)

where h(x) is some polynomial of degree 2.

Expanding both sides of the equation, we get:

x^3 g(x) + 1 = (x + 1)(x + 2)(x + 3) h(x)

x^3 g(x) + 1 = (x^2 + 3x + 2)(x + 3) h(x)

x^3 g(x) + 1 = (x^3 + 3x^2 + 2x + 3x^2 + 9x + 6) h(x)

x^3 g(x) + 1 = x^3 h(x) + 6x^2 h(x) + 11x h(x) + 6h(x)

Comparing coefficients of the same powers of x on both sides of the equation, we get:

g(x) = h(x)

6g(x) = 11h(x)

6h(x) = 1

Solving for h(x), we get:

h(x) = 1/6

Therefore, g(x) = h(x) = 1/6, and we have:

f(x) = x^3 g(x) = x^3/6

Expanding this polynomial, we get:

f(x) = (1/6) x^3

Therefore, the polynomial f(x) = (1/6) x^3 satisfies both properties.

First, we will prove the left-hand inequality:

(x + abc)^3 >= abc(x + a)(x + b)(x + c)

Expanding both sides of the inequality, we get:

x^3 + 3x^2(abc) + 3x(abc)^2 + (abc)^3 >= abc(x^3 + (a+b+c)x^2 + (ab+bc+ac)x + abc)

Simplifying, we get:

x^3(1-abc) + x^2(3abc + a^2bc + ab^2c + abc^2) + x(3a^2b^2c + 3ab^2c^2 + 3a^2bc^2) + abc^3 >= 0

Since a, b, and c are nonnegative, all the coefficients in this polynomial are nonnegative. Therefore, the polynomial is nonnegative for all x >= 0, which proves the left-hand inequality.

Next, we will prove the right-hand inequality:

abc(x + a)(x + b)(x + c) >= (x + a + b + c)^3

Expanding both sides of the inequality, we get:

abc(x^3 + (a+b+c)x^2 + (ab+bc+ac)x + abc) >= x^3 + 3x^2(a+b+c) + 3x(ab+bc+ac) + (a+b+c)^3

Simplifying, we get:

x^3(1-abc) + x^2(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2-3abc) + x(abc(a+b+c)-(a+b+c)^2) + (a+b+c)^3 >= 0

Since a, b, and c are nonnegative, we have:

ab^2+ac^2+ba^2+bc^2+ca^2+cb^2 >= 6abc

and

abc(a+b+c) >= 3abc(x+y+z)

where x = a+b, y = b+c, and z = c+a. Therefore, we can rewrite the inequality as:

x^3(1-abc) + x^2(6abc-3abc) + x(3abc-3abc) + (a+b+c)^3 >= 0

Simplifying, we get:

x^3(1-abc) + 3(a+b+c)^3 >= 0

Since x >= 0 and a, b, and c are nonnegative, both terms in this expression are nonnegative. Therefore, the inequality is true, which proves the right-hand inequality.

Therefore, we have proven both inequalities:

(x + abc)^3 >= abc(x + a)(x + b)(x + c) >= (x + a + b + c)^3

for all nonnegative real numbers a, b, and c and all x >= 0. 

We can start by using the Pythagorean theorem to find the length of the hypotenuse XZ of triangle XYZ:

XZ^2 = XY^2 + YZ^2 = 12^2 + 6^2 = 144 + 36 = 180

Taking the square root of both sides, we get XZ = 6√5.

The area of triangle XYZ is (1/2) * XY * YZ = 36, and the area of rectangle XYX'Z, where X' is the midpoint of XZ, is XY * X'Z = 36√5.

Since triangle XYD is a right triangle with legs of length 6 and a and hypotenuse XY, its area is (1/2) * 6 * a = 3a, where a is the length of the altitude from D to XY.

If we choose D randomly within triangle XYZ, the probability that it lies within a certain region of the triangle is proportional to the area of that region. Therefore, we can find the probability that the area of triangle XYD is at most 20 by finding the ratio of the area of the region where this is true to the area of triangle XYZ.

Let H be the foot of the altitude from X to YZ, and let E be the point where the line parallel to YZ through D intersects XY. Then, triangle XYD has area at most 20 if and only if DE is at most 10.

The area of region XYZE is (1/2) * YZ * XE = 3XE, and the area of region XHY is (1/2) * XY * YH = 36/5. Therefore, the area of the region where the area of triangle XYD is at most 20 is:

36 - 3XE - (36/5) = 144/5 - 3XE

We want this area to be proportional to the area of triangle XYZ, which is 36, so we want:

(144/5 - 3XE)/36 = P

where P is the probability we are looking for.

Solving for XE, we get:

XE = (144/5 - 36P)/3

We need XE to be at most 6 in order for DE to be at most 10. Therefore, we have:

(144/5 - 36P)/3 ≤ 6

Multiplying both sides by 3, we get:

144/5 - 36P ≤ 18

Subtracting 144/5 from both sides, we get:

-36P ≤ -126/5

Dividing both sides by -36 and reversing the inequality, we get:

P ≥ 7/60

Therefore, the probability that the area of triangle XYD is at most 20 is at least 7/60. 

(a) The expected value of the number shown when we draw a single slip of paper is the weighted average of the possible outcomes, where the weights are the probabilities of each outcome. There are 8 slips with a 3 on them and 2 slips with a 5 on them, so the probability of drawing a 3 is 8/10 = 4/5 and the probability of drawing a 5 is 2/10 = 1/5. Therefore, the expected value is:

E(X) = (4/5) * 3 + (1/5) * 5 = 12/5 = 2.4

(b) If we add one additional 3 to the bag, there are now 9 slips with a 3 on them and 1 slip with a 5 on it. The probability of drawing a 3 is now 9/10 and the probability of drawing a 5 is 1/10. Therefore, the expected value is:

E(X) = (9/10) * 3 + (1/10) * 5 = 27/10 = 2.7

(c) If we add two additional 3's to the bag, there are now 10 slips with a 3 on them and 0 slips with a 5 on them. The probability of drawing a 3 is now 10/10 = 1. Therefore, the expected value is:

E(X) = 1 * 3 = 3

Therefore, the expected value of the number shown when we draw a single slip of paper is 2.4, the expected value if we add one additional 3 to the bag is 2.7, and the expected value if we add two additional 3's to the bag is 3. 

(a) This is a classic stars and bars problem, where we have 20 identical objects (stickers) that we want to distribute among 5 people. We can model this by placing 4 dividers (bars) among the stickers to separate the stickers into 5 groups. For example, if we have 20 stickers and we place the dividers like this:

****|***|***|****|***

then the first friend gets 4 stickers, the second friend gets 3 stickers, the third friend gets 3 stickers, the fourth friend gets 4 stickers, and the fifth friend gets 6 stickers.

Thus, we need to count the number of ways to place 4 dividers among 20 stickers, which can be done in (20 + 4) choose 4 = 24,024 ways.

(b) This is similar to part (a), but now we need to ensure that every friend gets at least one sticker. We can do this by giving each friend one sticker to start with, and then distributing the remaining 15 stickers using stars and bars. Specifically, we can place 4 dividers among the 15 stickers to separate them into 5 groups, where each group corresponds to the additional stickers given to each friend. For example, if we have 20 stickers and we give each friend one sticker, we are left with 15 stickers, which we can distribute as follows:

*|**|***|****|*****

This corresponds to the first friend getting 2 stickers, the second friend getting 3 stickers, the third friend getting 4 stickers, the fourth friend getting 5 stickers, and the fifth friend getting 6 stickers.

Thus, we need to count the number of ways to place 4 dividers among 15 stickers, which can be done in (15 + 4) choose 4 = 3,432 ways.

Therefore, the number of ways to distribute 20 identical stickers to 5 people, where not everyone has to get a sticker, is 24,024, while the number of ways to distribute 20 identical stickers to 5 people, where every person gets at least one sticker, is 3,432. 

First, we can simplify the equation x^3 - 2x + 5 = x^3 - x^2 + 9 by subtracting x^3 from both sides:

-2x + 5 = -x^2 + 9

Rearranging, we get:

x^2 - 2x + 4 = 0

Solving for x using the quadratic formula, we get:

x = 1 ± i√3

So, the roots of the equation x^3 - x^2 + 9 are 1 + i√3, 1 - i√3, and an unknown third root r.

We can now show that r, r^2, and r^3 are not irrational. Since the coefficients of the polynomial x^3 - x^2 + 9 are all integers, any rational root of the equation must be an integer. (This follows from the rational root theorem.) However, 1 ± i√3 are not integers, so r cannot be rational.

Next, we can show that r^2 and r^3 are also not irrational. Since r is a root of x^3 - x^2 + 9, we have:

r^3 - r^2 + 9 = 0

Multiplying both sides by r, we get:

r^4 - r^3 + 9r = 0

Substituting r^3 - 9 for -r^2 in the equation x^3 - 2x + 5 = x^3 - x^2 + 9, we get:

x^3 - 2x + 5 = (x^3 - 9) + 8

Rearranging, we get:

x^3 - 9 = -x^3 + 2x - 3

Substituting r^3 for x, we get:

r^3 - 9 = -r^3 + 2r - 3

Solving for r^3, we get:

r^3 = (r^2 - 3)/2

Substituting this into the equation for r^4 - r^3 + 9r = 0, we get:

r^4 - (r^2 - 3)/2 + 9r = 0

Multiplying both sides by 2, we get:

2r^4 - r^2 + 18r = 0

Substituting r^2 - 2r + 4 for 0, we get:

2(r^2 - 2r + 4)^2 - (r^2 - 2r + 4) + 18(r^2 - 2r + 4) = 0

Expanding, we get:

2r^4 - 5r^3 + 10r^2 - 17r + 40 = 0

Substituting r^3 - 9 for -r^2 + 1 in the equation x^3 - x^2 + 9, we get:

x^3 - x^2 + 9 = (x^3 - 9) + 8

Rearranging, we get:

x^3 - 9 = x^2 - 8

Substituting r^2 for x, we get:

r^2 - 8 = r^3 - 9

Solving for r^3, we get:

r^3 = r^2 - r + 1

Substituting this into the equation for 2r^4 - 5r^3 + 10r^2 - 17r + 40 = 0, we get:

2r^4 - 5(r^2 - r + 1) + 10r^2 - 17r + 40 = 0

Simplifying, we get:

2r^4 + 5r^2 - 22r + 35 = 0

This equation factors as:

(r - 1)(2r^3 + 7r^2 - 15r - 35) = 0

Since r is a root of x^3 - x^2 + 9, it cannot be 1. Therefore, the other factor, 2r^3 + 7r^2 - 15r - 35, must be 0. This equation factors as:

(r - 1)(2r^2 + 9r + 35) = 0

Since r cannot be 1, we must have:

2r^2 + 9r + 35 = 0

This quadratic equation has discriminant:

9^2 - 4(2)(35) = -131 < 0

Therefore, the quadratic equation has no real roots, which means that r^2 cannot be irrational.Finally, we can conclude that none of r, r^2, or r^3 is irrational. We showed that if r is a root of x^3 - 2x + 5 = x^3 - x^2 + 9, then r, r^2 and r^3 are all algebraic integers, and therefore none of them can be irrational.

Because the balls and the boxes are both indistinguishable it is way simpler than it looks and the only way to this question is by listing all the possible cases.

Case 1: 4 balls in one box

Case 2: 3 balls in one box and 1 ball in a other box

Case 3: 2 balls in one box and 2 balls in a other box

Case 4: 2 balls in one box and 1 ball in a other box and 1 ball in a other box

Case 5: 1 ball in one box and 1 ball in a other box and 1 ball in a other box and 1 ball in a other box

Which shows that there are only 5 ways to distribute.

To solve this problem, we can use the angle bisector theorem and the properties of cyclic quadrilaterals.

First, let's draw a diagram:B / \ / \ / \ / \ / \ / \ A-------------C

Since AB is a diameter of the circle, angle ACB is a right angle (90 degrees).

Let AM be the angle bisector of angle ACB. Then, by the angle bisector theorem, we have:

AC AM -- = ----- BC BM

Substituting the given values, we get:

8 AM -- = --- 4 BM

Simplifying, we get:

AM = 2BM

Now, let's draw the line CM and label the point where it intersects the circle as D:B / \ / \ / D \ / \ / \ / \ A-------------C | | M

Since AM is twice BM, we can let BM = x and AM = 2x. Then, using the Pythagorean theorem in triangle ABC, we have:

AC^2 + BC^2 = AB^2 8^2 + 4^2 = AB^2 AB = 2sqrt(5)

Since AD and BD are radii of the circle, we have AD = BD = AB/2 = sqrt(5).

Now, let's use the properties of cyclic quadrilaterals to find the length of CD. Since angle CMB is an inscribed angle that intercepts arc CD, we have:

angle CMB = 1/2 * arc CD

Also, angle CAM and CBM are equal because they are opposite angles of an isosceles triangle. Therefore, angle AMC = 180 - 2CAM = 180 - 2CBM. Since angle AMC is an inscribed angle that intercepts arc CD, we have:

angle AMC = 1/2 * arc CD

Adding these two equations, we get:angle CMB + angle AMC = arc CD 1/2 * arc CD + 1/2 * arc CD = arc CD arc CD = angle ACB = 90 degrees

Therefore, CD is a diameter of the circle, and CM is the radius. Since CM = CD/2, we have:CM = AB/2 - AC/2 = (2sqrt(5))/2 - 8/2 = sqrt(5) - 4

I've answered on th original.

We can use casework to solve this problem. We need to find all the ways to arrange the digits 1, 2, and 3 to sum to 5.

Case 1: One digit is 5
There is only one way to arrange the digits in this case: 5. This is not a valid solution since the digits must be 1, 2, or 3.

Case 2: Two digits sum to 5
The two digits that sum to 5 can be either 2 and 3 or 3 and 2. There are two ways to arrange each pair of digits: 23 or 32. We need to determine how many times each pair appears.

For 23, we can choose any two of the five digits to be 2, and the remaining digit must be 1. There are 5 choose 2 = 10 ways to choose the positions of the 2's, and the remaining digit must be 1. Therefore, there are 10 numbers that have two digits that sum to 5 and contain the digits 1, 2, and 3: 221, 212, 122, 233, 323, 332, 131, 113, 311, and 212.

For 32, we can choose any two of the five digits to be 3, and the remaining digit must be 2. There are 5 choose 2 = 10 ways to choose the positions of the 3's, and the remaining digit must be 2. Therefore, there are 10 numbers that have two digits that sum to 5 and contain the digits 1, 2, and 3: 223, 232, 322, 233, 323, 332, 221, 212, 122, and 131.

Case 3: Three digits sum to 5
The three digits that sum to 5 can only be 1, 2, and 2. There is only one way to arrange these digits: 122. We need to determine how many times this arrangement appears.

We can choose any three of the five digits to be 2, and the remaining digits must be 1. There are 5 choose 3 = 10 ways to choose the positions of the 2's, and the remaining digits must be 1. Therefore, there are 10 numbers that have three digits that sum to 5 and contain the digits 1, 2, and 3: 122, 212, 221, 112, 121, 211, 131, 113, 311, and 313.

Therefore, there are a total of 10 + 10 + 1 = 21 numbers that satisfy the conditions.

To avoid having the exact same group of members over all 365 days, we need to ensure that the combination of members invited on any two days is different. 

We can use the formula for combinations to calculate the total number of unique combinations of n people that can be invited on any given day. The formula for combinations is:

nCk = n! / (k! * (n-k)!)

where n is the number of people in the book club, k is the number of people invited on any given day, and ! denotes the factorial function.

We want to ensure that the combination of members invited on any two days is different, so we need to find the smallest n such that:

nCk1 ≠ nCk2 for all k1 and k2 where 1 ≤ k1 < k2 ≤ n

In other words, we need to find the smallest n such that the number of unique combinations of k people that can be invited on any given day is greater than or equal to 100 (the number of days).

We can start by trying different values of n and calculating the number of unique combinations of k people that can be invited on any given day for each value of k from 1 to n. Once we find an n that satisfies the condition above, we know that it is the smallest n that will work. 

For example, if we try n = 5, we can calculate the number of unique combinations of k people that can be invited on any given day for k = 1, 2, 3, 4, and 5:

nC1 = 5C1 = 5
nC2 = 5C2 = 10
nC3 = 5C3 = 10
nC4 = 5C4 = 5
nC5 = 5C5 = 1

The total number of unique combinations is 31, which is greater than 100. Therefore, n = 5 is a valid solution.

We could continue trying different values of n until we find the smallest one that works, but a more efficient approach is to use the principle of inclusion-exclusion. This principle states that the number of combinations that satisfy at least one of several conditions is equal to the sum of the number of combinations that satisfy each individual condition, minus the sum of the number of combinations that satisfy each pair of conditions, plus the sum of the number of combinations that satisfy each triple of conditions, and so on.

In this case, we can apply the principle of inclusion-exclusion to count the number of combinations that are repeated over the 100 days. Let N be the total number of combinations of n members that can be invited on any given day. Then, the number of combinations that are repeated at least once over the 100 days is:

N - (N choose 2) + (N choose 3) - ... + (-1)^99 (N choose 100)

We want to find the smallest n such that this number is less than or equal to 0 (i.e., all 100 combinations are unique). We can start with n = 1 and calculate this expression for each value of n until we find the smallest n that works. This approach is more efficient than trying all possible values of n.

I get 50% also and I think guest's answer was quite well explained.

You sort of did, you voted it up when you first saw it and then voted it down when you decided it was wrong.

Please do not to this, if a person makes a serious attempt to help you you should always thank them with an upvote, as well as a comment.

You are upvoting them for their time, interest and effort. Not necessarily for their correct answer.

Your comments are quite nice although it is more tactful to say "I do not think it is right because ...(give reason).. " Rather than just  "Sorry, it's incorrect"

2. In how many ways can we distribute 7 pieces of identical taffy and 8 pieces of identical licorice to 5 kids such that each kid recieves exactly 3 pieces of candy.

There are 15 peices of candy of 2 types 

I am only going to concern myself with the 7 pieces of Taffy because for each taffy allocation the licorice allocation is automatically set.

Please just post one question at a time. 

1. 3 students are running for a class president in a class of 70 students. How many different vote counts are possible if some student(s) do not vote.

I still do not understand this question.  

You explanation does not make sense to me. Just for starters the question clearly states that 1 or more students DO NOT vote. So it is impossible for 70 students to vote.

If there are 70 students and some do not vote then there can be a maximum of 69 possible votes to count.

Each of the three students can get between 0 and 69 votes, the sum of the 3 votes is obviously 69 or less.

I do not suppose that this is what the question means but it is impossible to know what the intended meaning is.

How do you know? 

You should always give a reason. even if it is just "I put it in the answer box and was told it was wrong"

well for starters function f is (x+5)/(x-7) which is division and function h is (x+5)(x-7) which is multiplication. 

You can also prove it by inputing numbers. For example, if I input 10 as x function f will become 5 and function h will become 45 which shows that the two functions are not the same.