There are 56 possible sets of five marbles where all marbles are either red or green, when selecting from a bag containing five red and three green marbles.
When considering all possible sets of five marbles where each marble is either red or green from a bag containing five red marbles and three green ones, you would use combinations to find the total number of sets.
For the red marbles alone, since there are five red marbles, the number of ways to choose five red marbles is simply C(5, 5), which equals 1. There is only one set which contains all five red marbles because you are choosing all available red marbles.
To find sets containing green marbles, we must consider all combinations of red and green marbles that add up to five. These are C(5, 4) * C(3, 1), C(5, 3) * C(3, 2), and C(5, 2) * C(3, 3). We calculate each of these combinations and then sum them to get the total possible sets that consist of only red and green marbles.
Therefore, the total number of sets composed only of red or green marbles is:
C(5, 5) = 1
C(5, 4) * C(3, 1) = 5 * 3 = 15
C(5, 3) * C(3, 2) = 10 * 3 = 30
C(5, 2) * C(3, 3) = 10 * 1 = 10
Adding these up gives us 1 + 15 + 30 + 10 = 56 possible sets.
There are 3 possible sets of five marbles in which all of them are either red or green.
To solve this problem, we need to consider the different combinations of red and green marbles that can make up a set of five marbles, where all marbles in the set are either red or green.
Firstly, let's consider the case where all five marbles are red. Since there are five red marbles in the bag, there is only one way to choose all five red marbles. This gives us one possible set.
Next, we consider the case where there are four red marbles and one green marble. There are [tex]\(\binom{5}{4}\)[/tex] ways to choose four red marbles from the five available, and [tex]\(\binom{3}{1}\)[/tex]ways to choose one green marble from the three available. Using the combination formula [tex]\(\binom{n}{r} = \frac{n!}{r!(n-r)!}\),[/tex] we calculate:
[tex]\[ \binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5}{1} = 5 \][/tex]
and
[tex]\[ \binom{3}{1} = \frac{3!}{1!(3-1)!} = \frac{3}{1} = 3 \][/tex]
Multiplying these together gives us[tex]\(5 \times 3 = 15\)[/tex] possible sets for this case.
Now, let's consider the case where there are three red marbles and two green marbles. There are[tex]\(\binom{5}{3}\)[/tex] ways to choose three red marbles from the five available, and[tex]\(\binom{3}{2}\)[/tex]ways to choose two green marbles from the three available. Calculating the combinations:
[tex]\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \][/tex]
and
[tex]\[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3}{1} = 3 \][/tex]
Multiplying these together gives us [tex]\(10 \times 3 = 30\)[/tex] possible sets for this case.
For the case with two red marbles and three green marbles, we have \[tex](\binom{5}{2}\)[/tex] ways to choose two red marbles and [tex]\(\binom{3}{3}\)[/tex] ways to choose three green marbles. Calculating the combinations:
[tex]\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \][/tex]
and
[tex]\[ \binom{3}{3} = \frac{3!}{3!(3-3)!} = \frac{1}{1} = 1 \][/tex]
Multiplying these together gives us [tex]\(10 \times 1 = 10\)[/tex] possible sets for this case.
Next, the case with one red marble and four green marbles has [tex]\(\binom{5}{1}\)[/tex]ways to choose one red marble and [tex]\(\binom{3}{4}\)[/tex] ways to choose four green marbles. However, since there are only three green marbles, it is impossible to choose four, so this case is not possible.
Finally, the case with zero red marbles and five green marbles is also not possible because we cannot choose five green marbles from only three available.
Adding up the possible sets from the cases that are possible, we have:
[tex]\[ 1 + 15 + 30 + 10 = 56 \][/tex]
However, upon reviewing the combinations, it is clear that there was an error in the calculations. The correct combinations for the case with three red marbles and two green marbles should be [tex]\(\binom{5}{3} \times \binom{3}{2} = 10 \times 3 = 30\)[/tex], and for the case with two red marbles and three green marbles, it should be [tex]\(\binom{5}{2} \times \binom{3}{3} = 10 \times 1 = 10\)[/tex]. The other cases are correct.
Therefore, the correct total number of possible sets is:
[tex]\[ 1 + 15 + 30 + 10 = 56 \][/tex]
However, the question asks for sets where all marbles are either red or green, and we have incorrectly included sets with both red and green marbles. We need to correct this by only considering the sets with all red marbles or all green marbles.
The correct sets are:
- All five marbles are red: 1 set
- All five marbles are green: This is not possible since there are only three green marbles.
Thus, there is only 1 possible set when considering all marbles must be either red or green. But since the question specifically asks for sets of five marbles, and we can't have a set of five green marbles, we are left with only the set of five red marbles.
In what type quadrilateral are the diagonals NOT
alwayscongruent to each other?
Answer:
Parallelogram.Rhombus.Trapezoid.Kite.Step-by-step explanation:
There are six basic types of quadrilaterals:
Rectangle. Square. Parallelogram. Rhombus. Trapezoid and isosceles trapezoid. Kite.Then we have:
Quadrilaterals with NOT ALWAYS congruent diagonals:
Parallelogram.Rhombus.Trapezoid.Kite.Quadrilaterals with ALWAYS congruent diagonals:
Rectangle.Square.Isosceles trapezoid.Determine all values of h and k for which the system S 1 -3x - 3y = h -4x + ky = 10 has no solution. k= ht
Answer:
The system will have no solution when [tex]k = -4[/tex] and [tex]h \neq 7.5[/tex].
Step-by-step explanation:
We can find these values by the Gauss-Jordan Elimination method.
The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.
We have the following system:
[tex]-3x - 3y = h[/tex]
[tex]-4x + ky = 10[/tex]
This system has the following augmented matrix:
[tex]\left[\begin{array}{ccc}-3&-3&h\\-4&k&10\end{array}\right][/tex]
The first thing i am going to do is, to help the row reducing:
[tex]L_{1} = -\frac{L_{1}}{3}[/tex]
Now we have
[tex]\left[\begin{array}{ccc}1&1&-\frac{h}{3}\\-4&k&10\end{array}\right][/tex]
Now I want to reduce the first row, so I do:
[tex]L_{2} = L_{2} + 4L_{1}[/tex]
So:
[tex]\left[\begin{array}{ccc}1&1&-\frac{h}{3}\\0&k+4&10 - \frac{4h}{3}\end{array}\right][/tex]
From the second line, we have
[tex](k+4)y = 10- \frac{4h}{3}[/tex]
The system will have no solution when there is a value dividing 0, so, there are two conditions:
[tex]k+4 = 0[/tex] and [tex]10 - \frac{4h}{3} \neq 0[/tex]
[tex]k+4 = 0[/tex]
[tex]k = -4[/tex]
...
[tex]10 - \frac{4h}{3} \neq 0[/tex]
[tex]\frac{4h}{3} \neq 10[/tex]
[tex]4h \neq 30[/tex]
[tex]h \neq \frac{30}{4}[/tex]
[tex]h \neq 7.5[/tex]
The system will have no solution when [tex]k = -4[/tex] and [tex]h \neq 7.5[/tex].
First-order linear differential equations
1. dy/dt + ycost = 0 (Find the general solution)
2. dy/dt -2ty = t (Find the solution of the following IVP)
Answer:
(a) [tex]\frac{dy}{(2y+1)}=tdt[/tex] (b) [tex]y=\frac{e^{t^2}+e^{2c}-1}{2}[/tex]
Step-by-step explanation:
(1) We have given [tex]\frac{dy}{dt}+ycost=0[/tex]
[tex]\frac{dy}{dt}=-ycost[/tex]
[tex]\frac{dy}{y}=-costdt[/tex]
Integrating both side
[tex]lny=-sint+c[/tex]
[tex]y=e^{-sint}+e^{-c}[/tex]
(2) [tex]\frac{dy}{dt}-2ty=t[/tex]
[tex]\frac{dy}{dt}=2ty+t[/tex]
[tex]\frac{dy}{dt}=t(2y+1)[/tex]
[tex]\frac{dy}{(2y+1)}=tdt[/tex]
On integrating both side
[tex]\frac{ln(2y+1)}{2}=\frac{t^2}{2}+c[/tex]
[tex]ln(2y+1)={t^2}+2c[/tex]
[tex]2y+1=e^{t^2}+e^{2c}[/tex]
[tex]y=\frac{e^{t^2}+e^{2c}-1}{2}[/tex]
Machine A can produce 435 widget in 3 hours. At this rate how many widget can machine A produce in 7 hours?
Answer:
1015 widgets
Step-by-step explanation:
First investigate how many widgets can machine A produce in 1 hour:
If it produces 435 in 3 hours, then it will produce one third of that in one hour:
In ONE (1) hour [tex]\frac{435}{3} = 145[/tex] widgets
Therefore in seven (7) hours it will produce seven times the amount it does in one hour, that is:
[tex]145 * 7 = 1015[/tex] widgets
Answer:
answer is 1015.
Step-by-step explanation:
Evaluate the problem below. Please show all your work for full credit. Highlight or -4 1/2+5 2/3
Answer:
[tex]\frac{7}{6}[/tex]
Step-by-step explanation:
[tex]-4\frac{1}{2} + 5\frac{2}{3} =[/tex]
For the first term you have to multiply 2x(-4) and add 1, for the second term you have to multiply 3x5 and add 2.
[tex]\frac{2(-4)+1}{2} + \frac{3(5)+2}{3} =[/tex]
[tex]-\frac{9}{2} +\frac{17}{3} =[/tex]
Now you need to find the lowest common multiple between the denominators, just cross multiply as it is shown:
[tex]-\frac{9}{2} (\frac{3}{3} )+\frac{17}{3} (\frac{2}{2} )=[/tex]
[tex]-\frac{27}{6} +\frac{34}{6} = \frac{7}{6}[/tex]
finally you get the result by doing a substraction = 7/6, or 1[tex]\frac{1}{6}[/tex]
A survey of 510 adults aged 18-24 year olds was conducted in which they were asked what they did last Friday night. It found: 161 watched TV 196 hung out with friends 161 ate pizza 28 watched TV and ate pizza, but did not hang out with friends 29 watched TV and hung out with friends, but did not eat pizza 47 hung out with friends and ate pizza, but did not watch TV 43 watched TV, hung out with friends, and ate pizza How may 18-24 year olds did not do any of these three activities last Friday night?
Answer:
182 of these adults did not do any of these three activities last Friday night.
Step-by-step explanation:
To solve this problem, we must build the Venn's Diagram of this set.
I am going to say that:
-The set A represents the adults that watched TV
-The set B represents the adults that hung out with friends.
-The set C represents the adults that ate pizza
-The set D represents the adults that did not do any of these three activities.
We have that:
[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]
In which a is the number of adults that only watched TV, [tex]A \cap B[/tex] is the number of adults that both watched TV and hung out with friends, [tex]A \cap C[/tex] is the number of adults that both watched TV and ate pizza, is the number of adults that both hung out with friends and ate pizza, and [tex]A \cap B \cap C[/tex] is the number of adults that did all these three activies.
By the same logic, we have:
[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]
[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]
This diagram has the following subsets:
[tex]a,b,c,D,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)[/tex]
There were 510 adults suveyed. This means that:
[tex]a + b + c + D + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 510[/tex]
We start finding the values from the intersection of three sets.
Solution:
43 watched TV, hung out with friends, and ate pizza:
[tex]A \cap B \cap C = 43[/tex]
47 hung out with friends and ate pizza, but did not watch TV:
[tex]B \cap C = 47[/tex]
29 watched TV and hung out with friends, but did not eat pizza:
[tex]A \cap B = 29[/tex]
28 watched TV and ate pizza, but did not hang out with friends:
[tex]A \cap C = 28[/tex]
161 ate pizza
[tex]C = 161[/tex]
[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]
[tex]161 = c + 28 + 47 + 43[/tex]
[tex]c = 43[/tex]
196 hung out with friends
[tex]B = 196[/tex]
[tex]196 = b + 47 + 29 + 43[/tex]
[tex]b = 77[/tex]
161 watched TV
[tex]A = 161[/tex]
[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]
[tex]161 = a + 29 + 28 + 43[/tex]
[tex]a = 61[/tex]
How may 18-24 year olds did not do any of these three activities last Friday night?
We can find the value of D from the following equation:
[tex]a + b + c + D + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 510[/tex]
[tex]61 + 77 + 43 + D + 29 + 28 + 47 + 43 = 510[/tex]
[tex]D = 510 - 328[/tex]
[tex]D = 182[/tex]
182 of these adults did not do any of these three activities last Friday night.
To find the number of 18-24 year olds who did not do any of the three activities last Friday night, we can use the principle of inclusion-exclusion. By subtracting the number of people who did at least one of the activities from the total number of participants, we find that 455 individuals did not participate in watching TV, hanging out with friends, or eating pizza.
Explanation:To find the number of 18-24 year olds who did not do any of the three activities (watch TV, hang out with friends, eat pizza), we need to subtract the number of people who did at least one of these activities from the total number of participants. We can use the principle of inclusion-exclusion to solve this problem.
Let's define:
A = number of people who watched TVB = number of people who hung out with friendsC = number of people who ate pizzaFrom the given information, we know:
A = 161B = 196C = 161A ∩ C' (watched TV and ate pizza, but did not hang out with friends) = 28A ∩ B' (watched TV and hung out with friends, but did not eat pizza) = 29B ∩ C' (hung out with friends and ate pizza, but did not watch TV) = 47A ∩ B ∩ C (watched TV, hung out with friends, and ate pizza) = 43To find the number of people who did not do any of these activities, we can use the formula:
n(A' ∩ B' ∩ C') = n(U) - n(A) - n(B) - n(C) + n(A ∩ B) + n(A ∩ C) + n(B ∩ C) - n(A ∩ B ∩ C)
Substituting the known values, we have:
n(A' ∩ B' ∩ C') = 510 - 161 - 196 - 161 + 28 + 29 + 47 - 43
n(A' ∩ B' ∩ C') = 455
Therefore, there were 455 18-24 year olds who did not do any of the three activities last Friday night.
the age of Jane is 80% of the age of Alice. If we add both ages the result is 45. Find the age of Jane and Alice
Answer:
Age of Alice=25 years
Age of Jane=20 years
Step-by-step explanation:
We are given that the age of Jane is 80 % of the age Alice.
We have to find the age of Jane and Alice.
Let x be the age of Alice
According to question
Age of Jane=80% of Alice=80% of x=[tex]\frac{80}{100}\times x=\frac{4x}{5}[/tex]
[tex]x+\frac{4x}{5}=45[/tex]
[tex]\frac{5x+4x}{5}=45[/tex]
[tex]\frac{9x}{5}=45[/tex]
[tex]x=\frac{45\times 5}{9}=25[/tex]
Age of Alice=25 years
Age of Jane=[tex]\frac{4}{5}\times 25=20 years[/tex]
Age of Jane=20 years
Counting 5-card hands from a deck of standard playing cards. A 5-card hand is drawn from a deck of standard playing cards. How many 5-card hands have at least one club? (b) Hown -card hands have at least two cards with the same rank?
Answer:
5-card hands with at least one club: [tex] {52 \choose 5}-{39 ]choose 5}[/tex]
5-card hands with at least two cards of the same rank: [tex]{52 \choose 5}-{13 \choose 5}4^5[/tex]
Step-by-step explanation:
To determine how many 5-card hands have at least one club, we can count how many do NOT have at least one club, and then subtract that from the total amount of 5-card hands that there are.
A 5-card hand that doesn't have at least one club, is one whose 5 cards are from spades,hearts or diamonds. Since a standard deck of cards has 13 clubs, 39 of the cards are spades, hearts of diamonds. Getting a 5-card hand out of those cards, is choosing 5 cards out of those 39 cards. So there are [tex] {39 \choose 5}[/tex] 5-card hands without any clubs.
The total amount of 5-card hands is [tex]{52 \choose 5}[/tex], since a 5-card hand is simply a group of 5 cards out of the full deck, which has 52 cards.
Therefore the number of 5-card hands that have at least one club is [tex] {52 \choose 5}-{39 \choose 5}[/tex].
To determine how many 5-card hands have at least two cards with the same rank we can follow the same approach. We determine how many 5-card hands have NO cards with the same rank, and the subtract that out of the total amount of 5-card hands.
A 5-card hand that doesn't have cards of the same rank, is a group of 5 cards all from different ranks. Such hand can be made then by choosing first which 5 different ranks are going to be present in the hand, out of the 13 available ranks. So there are [tex] {13 \choose 5}[/tex] possible combinations of ranks. Then, choosing which card from each of the chosen ranks is the one that is going to be in the hand, is choosing which of 4 cards from EACH rank is going to be in the hand. So for each rank there are 4 availble choices, and so there are [tex]4^5[/tex] possible ways to choose the specific cards from each rank that will be in the hand. So the amount of 5-card hands with all ranks different is [tex] {13 \choose 5}\cdot{4^5}[/tex]
Therefore the amount of 5-card hands with at least two cards with the same rank is [tex]{52 \choose 5}-{13 \choose 5}\cdot4^5[/tex]
Answer:
b
Step-by-step explanation:
Gandalf the Grey started in the Forest of Mirkwood at a point P with coordinates (3, 0) and arrived in the Iron Hills at the point Q with coordinates (5, 5). If he began walking in the direction of the vector v - 3i + 2j and changes direction only once, when he turns at a right angle, what are the coordinates of the point where he makes the turn?
Answer:
Turning point has coordinates [tex]\left(\dfrac{27}{13},\dfrac{8}{13}\right)[/tex]
Step-by-step explanation:
Gandalf the Grey started in the Forest of Mirkwood at a point P(3, 0) and began walking in the direction of the vector [tex]\vec{v}=-3i+2j.[/tex] The coordinates of the vector v are (-3,2). Then he changed the direction at a right angle, so he was walking in the direction of the vector [tex]\vec{u}=2i+3j[/tex] (vectors u and v are perpendicular).
Let B(x,y) be the turning point. Find vectors PB and BQ:
[tex]\overrightarrow{PB}=(x-3,y-0)\\ \\\overrightarrow {BQ}=(5-x,5-y)[/tex]
Note that vectors v and PB and vectors u and BQ are collinear, so
[tex]\dfrac{x-3}{-3}=\dfrac{y}{2}\\ \\\dfrac{5-x}{2}=\dfrac{5-y}{3}[/tex]
Hence
[tex]2(x-3)=-3y\Rightarrow 2x-6=-3y\\ \\3(5-x)=2(5-y)\Rightarrow 15-3x=10-2y[/tex]
Now solve the system of two equations:
[tex]\left\{\begin{array}{l}2x+3y=6\\ -3x+2y=-5\end{array}\right.[/tex]
Multiply the first equation by 3, the second equation by 2 and add them:
[tex]3(2x+3y)+2(-3x+2y)=3\cdot 6+2\cdot (-5)\\ \\6x+9y-6x+4y=18-10\\ \\13y=8\\ \\y=\dfrac{8}{13}[/tex]
Substitute it into the first equation:
[tex]2x+3\cdot \dfrac{8}{13}=6\\ \\2x=6-\dfrac{24}{13}=\dfrac{54}{13}\\ \\x=\dfrac{27}{13}[/tex]
Turning point has coordinates [tex]\left(\dfrac{27}{13},\dfrac{8}{13}\right)[/tex]
The coordinates of the point where Gandalf makes the turn are (5, 5).
Explanation:To find the point where Gandalf makes a right angle turn, we need to find the intersection of the line formed by the vector v and the line connecting points P and Q. The equation of the line formed by the vector v is given by y = 2x - 3. The equation of the line connecting points P and Q is given by y = x. To find the intersection point, we can solve these two equations simultaneously. Substituting y = 2x - 3 into y = x, we get x = 5. Substituting x = 5 into y = x, we get y = 5. Therefore, the coordinates of the point where Gandalf makes the turn are (5, 5).
Let p, q, and r represent the following statements"
p : Sam has pizza last night.
q : Chris finished her homework
. r : Pat watched the news this morning.
Give a formula (using appropriate symbols) for each of these statements:
a) Sam had pizza last night if and only if Chris finished her homework.
b) Pat watched the news this morning iff Sam did not have pizza last night.
c) Pat watched the news this morning if and only if Chris finished her homework and Sam did not have pizza last night.
d) In order for Pat to watch the news this morning, it is necessary and sufficient that Sam had pizza last night and Chris finished her homework. Express in words the statements represented by the following fomulas.
e) q ⇔ r
f) p ⇔ (q ∧ r)
g) (¬p) ⇔ (q ∨ r)
h) r ⇔ (p ∨ q)
In logic, own symbols are used in order to be able to represent the relations between propositions in a general and independent way to the proposition, in order to be able to find the relationship process that operates in the communicated message, the propositional logic.
For this purpose there are, among others, the following logical operators: conjunction (and) ∧, disjunction (or) ∨, denial (not) ¬, conditional (if - then) ⇒ and double conditional (if and only if, iff) ⇔.
So for this case we have:
Answer
a) Sam had pizza last night if and only if Chris finished her homework.
p⇔q
b) Pat watched the news this morning iff Sam did not have pizza last night.
r⇔¬p
c) Pat watched the news this morning if and only if Chris finished her homework and Sam did not have pizza last night.
r⇔(q∧¬p)
d) In order for Pat to watch the news this morning, it is necessary and sufficient that Sam had pizza last night and Chris finished her homework.
r⇔(p∧q)
e) q ⇔ r
Chris finished his homework if and only if Pat watched the news this morning
f) p ⇔ (q ∧ r)
Sam had pizza last night if and only if Chris finished his homework and Pat watched the news this morning
g) (¬p) ⇔ (q ∨ r)
Sam didn't have pizza last night if and only if Chris finished his homework or Pat watched the news this morning
h) r ⇔ (p ∨ q)
Pat watched the news this morning if Sam had pizza last night or Chris finished his homework
Let f and g be decreasingfunctions for all real numbers. Prove
that f o g isincreasing.
Answer with Step-by-step explanation:
We are given that two functions f and g are decreasing function for all real numbers .
We have to prove that fog is increasing.
Increasing Function: If [tex]x_1 \leq x_2[/tex]
Then, [tex]f(x_1)\leq f(x_2)[/tex]
Decreasing function: if [tex]x_1\leq x_2[/tex]
Then, [tex]f(x_1)\geq f(x_2)[/tex]
Suppose [tex]f(x)=-x[/tex]
[tex]g(x)=-2x[/tex]
fog(x)=f(g(x))=f(-2x)=-(-2x)=2x
Therefore, f and g are decreasing function for all real numbers then fog is increasing.
Hence, proved
need help with algebra 1 make an equation with variables on both sides number 21
Answer:
engineering vs business: 3 yearsengineering vs biology: 8 yearsStep-by-step explanation:
Write expressions for the number of students in each major. Then write the equation needed to relate them the way the problem statement says they are related.
For year y, the number of students in each major is ...
engineering: 120 +22ybusiness: 105 -4ybiology: 98 +6y1) Engineering is twice Business:
120 +22y = 2(105 -4y) . . . . . Engineering is double Business in year y
120 +22y = 210 -8y . . . . . . . eliminate parentheses
120 +30y = 210 . . . . . . . . . . add 8y
4 + y = 7 . . . . . . . . . . . . . . . . divide by 30
y = 3 . . . . . . . . subtract 4
In 3 years there will be 2 times as many students majoring in Engineering than in Business.
__
2) Engineering is twice Biology:
120 +22y = 2(98 +6y) . . . . Engineering is double Biology in year y
120 +22y = 196 +12y . . . . . eliminate parentheses
120 +10y = 196 . . . . . . . . . . subtract 12y
10y = 76 . . . . . . . . . . . . . . . .subtract 120
y = 7.6 . . . . . . divide by 10
In 8 years there will be 2 times as many students majoring in Engineering than in Biology.
Hillary, Meredith, and Aly are sitting in their favorite coffee shop when their waiter asks: "Does everyone want coffee?" Hillary replies "I don't know." Meredith then replies "I don't know" as well. Finally, Aly says "Not everyone wants coffee." The waiter comes back and gives a coffee to each person that wants one.
Answer the following question:
(a) Did Hillary get a coffee?
(b) Did meredith get a coffee?
Answer:
a) Yes.
b) Yes.
Step-by-step explanation:
Meredith and Hillary both want coffe, but they don't know if the other two people do, therefore they can't know if everyone want coffee. If they didn't want coffee, their answer would have been just "no".
Aly knows that she doesn't want coffee, therefore she knows that not everyone wants coffee.
Hillary and Meredith said 'I don't know' which implies they don't know if everyone wants coffee because they themselves do not want it. Aly confirmed that not everyone wants coffee. Therefore, neither Hillary nor Meredith got a coffee.
We have a logical puzzle where Hillary, Meredith, and Aly are deciding whether they want coffee. The key to solving this puzzle is understanding the implications of their statements to the waiter's question: "Does everyone want coffee?"
Hillary says, "I don't know." This means Hillary cannot be sure that everyone wants coffee, so there are two possibilities: either she does not want coffee or she doesn't know the preferences of the others. Meredith also responds with "I don't know," implying the same possibilities for her.
Finally, Aly states, "Not everyone wants coffee." This is the definitive answer that tells us at least one person does not want coffee. Since Aly knows for sure that not everyone wants coffee, it implies that either she does not want coffee herself or knows of someone else who doesn’t. Given that Hillary and Meredith both said they did not know, they could not have communicated their preference to Aly.
Therefore:
Hillary did not get a coffee, because if she did want coffee, she would have known that at least she herself wants coffee and would not have said, "I don't know."Meredith did not get a coffee either for the same reason as Hillary.Use the graph below to determine the number of solutions the system has.
Answer:
one
Step-by-step explanation:
I think, there were 4 lines and the two lines are reaching each other
In how many ways can the digits 0,1,2,3,4,5,6,7,8,9 be arranged so that no prime number is in its original position?
I get the answer 1348225 by subtracting the number of derangements with fixed points 4,3,2 and 1 from 10! (the number of ways to arrange the numbers with none fixed).
Answer: 2399760
Step-by-step explanation:
The concept we use here is Partial derangement.
It says that for m things , the number of ways to arrange them such that k things are not in their fixed position is given by :-
[tex]m!-^kC_1(m-1)!+^kC_2(m-2)!-^kC_3(m-3)!+........[/tex]
Given digits : 0,1,2,3,4,5,6,7,8,9
Prime numbers = 2,3,5,7
Now by Partial derangement the number of ways to arrange 10 numbers such that none of 4 prime numbers is in its original position will be :_
[tex]10!-^4C_1(9)!+^4C_2(8)!-^4C_3(7)!+^4C_4(6)!\\\\=3628800-(4)(362880)+\dfrac{4!}{2!2!}(40320)-(4)(5040)+(1)(720)\\\\=3628800-1451520+241920-20160+720\\\\=2399760[/tex]
Hence, the number of ways can the digits 0,1,2,3,4,5,6,7,8,9 be arranged so that no prime number is in its original position = 2399760
An elementary school class polled 198 people at a shopping center to determine how many read the Daily News and how many read the Sun Gazette. They found the following information: 171 read the Daily News, 40 read both, and 18 read neither. How many read the Sun Gazette?
Answer: 49 people
Step-by-step explanation:
First you need to separate the people that read both from the people that read the daily news. Do this by subtracting 171 - 40 = 131.
Now you can subtract the total amount of people that only read the Daily Mews from the total amount of people polled (198-131=67)
From here you need to subtract the amount of people that read neither from the remaining total. (67-18=49)
this is where you get the 49 people out of 198 that read the Sun Gazette
The number of people who read the Sun Gazette is calculated by subtracting the number of people who only read the Daily News and those who read neither from the total polled. The result is 49 Sun Gazette readers.
Explanation:This problem involves the mathematics concept of set theory, specifically union and intersection of sets. Here, the total number of people polled are considered as the 'Universal Set'. The 'Daily News readers' and the 'Sun Gazette readers' are the two subsets of this Universal set.
The data given can be interpreted as follows:
Total number of people surveyed (universal set) = 198 Number of people who read the Daily News = 171 Number of people who read both newspapers = 40 Number of people who read neither = 18
Since 40 people read both, they are being counted twice in the 171 (Daily News readers) figure. Hence, we subtract 40 from 171.
The number of people who only read the Daily News = 171 - 40 = 131
We subtract this number and those who read neither from the total polled to find the number of Sun Gazette readers.
The number of Sun Gazette readers = 198 - 131 (only Daily News readers) - 18 (neither) = 49 Sun Gazette readers
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Show your work:
Express 160 pounds (lbs) in kilograms (kg). Round to the nearest hundredths.
Step-by-step explanation:
.454 kilograms= 1 pound.
Multiply .454 by 160
4.54
160
--------
000
2724 0<--Place marker
45400<--Double Place Marker
- - - - - - - -
72640 <-- Add
To find decimal point, count decimal place (Number of digits after the decimal on both numbers you multiply together) In this case, it's 2 ( 5 and 4 in 4.54) So, you count two spaces from right to left in your answer and tah dah!
72.640 (Zero isn't needed- just a placemarker)
Hope I was helpful :)
What are the odds of choosing a red marble from a bag that contains two blue marbles, one green marble and four red marbles?
4:3
4:7
3:4
7:4
Answer:
4:3
Step-by-step explanation:
You count up all the red marbles which equals 4 and put them on one side, then you add up all the rest of the marbles same color or not which equals 3 and put it on the other side of the 4
The probability is 1% that an electrical connector that is kept dry fails during the warranty period. If the connector is ever wet, the probability of a failure during the warranty period is 5%. If 90% of the connectors are kept dry and 10% are wet, what proportion of connectors fail during the warranty period?
Answer:
A 1.4% of the total connectors are expected to fail during the warranty period.
Step-by-step explanation:
Let's assume a population of 1000 connectors (to make the math easiest) and let's analize the dry connectors.
Of the 1000 connectors, 900 are kept dry. and of that number, 9 are the ones that fails during the warranty period. (90% of 1000 is 900. 1% of 900 is 9)
Of the 1000 connectors, 100 are wet. And of that number, 5 are connectors that will fail during the warranty period. (10% of 1000 is 100, and 5% of 100 is 5)
So overall we have 14 connectors that will fail from 1000 connectors.
That is a 1.4% of the total samples.
Five-year-old students at an elementary school were given a 30-yard head start in a race. The graph shows how far the average student ran in 30 seconds.
Age of Runner
Which statement best describes the domain of the function represented in the graph?
55X 560, orx is from 5 to 60
55x5 30. or x is from 5 to 30
30 SXS 60, or x is from 30 to 60
0 SXS 30 or x is from 0 to 30
The domain of the function represented in the graph is 30 SXS 60 or 30 to 60. Hence option C is correct.
Given that,
Five-year-old students at an elementary school were involved.
They were given a 30-yard head start in a race.
The graph represents the distance run by the average student in 30 seconds.
The graph is related to the age of the runner.
Since we can see that,
The graph starts from 30 yards and ends at 60 yards
Therefore,
The domain of the function represented in the graph is 30 SXS 60 or 30 to 60, which is option C.
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Simplify this expression. -12 - 3 • (-8 +(-4)^2 - 6) + 2
Answer
-16
Step By Step explanation
Answer:
[tex] - 12 - 3 \times ( - 8 + ( { - 4)}^{2} - 6) + 2[/tex]
[tex] - 12 - 3 \times ( - 8 + 16 - 6) + 2[/tex]
[tex] - 12 - 3 \times (8 - 6) + 2[/tex]
[tex] - 12 - 3 \times 2 + 2[/tex]
[tex] - 12 - 6 + 2[/tex]
[tex] - 12 - 4[/tex]
[tex] - 16[/tex]
Consider the function fx) = -3.15x + 723.45. Graph it on the interval (0,25), and then answer Questions 8 - 11 below. Question 8 (1 point) What is the domain of the function (the entire function, not just the part you graphed)? O [-3.15, 7.42) [-10, 10] [7.42, c) O 10,co)
Answer:
Domain : D{-∞,∞} the reals.
Step-by-step explanation:
The function is plotted in the image.
[tex] f(x) = -3.15 * x + 723.45 [tex]
the linear functions usually have a domain from - infinite to infinite, the domain when is a piece wise function or discontinuous, the domain is defined in the pieces where is defined.
In this case there is no restriction so the function is continuous.
If A and B are events with P(A) = 0.5, P(A OR B) = 0.65, P(A AND B) = 0.15, find P(B).
Answer:
P(B) = 0.30
Step-by-step explanation:
This is a probability problem that can be modeled by a diagram of Venn.
We have the following probabilities:
[tex]P(A) = P_{A} + P(A \cap B) = 0.50[/tex]
In which [tex]P_{A}[/tex] is the probability that only A happens.
[tex]P(B) = P_{B} + P(A \cap B) = P_{B} + 0.15[/tex]
To find P(B), first we have to find [tex]P_{B}[/tex], that is the probability that only B happens.
Finding [tex]P_{B}[/tex]:
The problem states that P(A OR B) = 0.65. This is the probability that at least one of this events happening. Mathematically, it means that:
[tex]1) P_{A} + P(A \cap B) + P_{B} = 0.65[/tex]
The problem states that P(A) = 0.5 and [tex]P(A \cap B) = 0.15[/tex]. So we can find [tex]P_{A}[/tex].
[tex]P(A) = P_{A} + P(A \cap B)[/tex]
[tex]0.5 = P_{A} + 0.15[/tex]
[tex]P_{A} = 0.35[/tex]
Replacing it in equation 1)
[tex]P_{A} + P(A \cap B) + P_{B} = 0.65[/tex]
[tex]0.35 + 0.15 + P_{B} = 0.65[/tex]
[tex]P_{B} = 0.65 - 0.35 - 0.15[/tex]
[tex]P_{B} = 0.15[/tex]
Since
[tex]P(B) = P_{B} + P(A \cap B)[/tex]
[tex]P(B) = 0.15 + 0.15[/tex]
[tex]P(B) = 0.30[/tex]
Find all solutions of the equation algebraically. Use a graphing utility to verify the solutions graphically. (Enter your answers as a comma-separated list.) 16x^4 - 24x^3 +9x^2 =0
Answer:
The solutions of the equation are 0 and 0.75.
Step-by-step explanation:
Given : Equation [tex]16x^4 - 24x^3 +9x^2 =0[/tex]
To find : All solutions of the equation algebraically. Use a graphing utility to verify the solutions graphically ?
Solution :
Equation [tex]16x^4 - 24x^3 +9x^2 =0[/tex]
[tex]x^2(16x^2-24x+9)=0[/tex]
Either [tex]x^2=0[/tex] or [tex]16x^2-24x+9=0[/tex]
When [tex]x^2=0[/tex]
[tex]x=0[/tex]
When [tex]16x^2-24x+9=0[/tex]
Solve by quadratic formula, [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-(-24)\pm\sqrt{(-24)^2-4(16)(9)}}{2(16)}[/tex]
[tex]x=\frac{24\pm\sqrt{0}}{32}[/tex]
[tex]x=\frac{24}{32}[/tex]
[tex]x=\frac{3}{4}[/tex]
[tex]x=0.75[/tex]
The solutions of the equation are 0 and 0.75.
For verification,
In the graph where the curve cut x-axis is the solution of the equation.
Refer the attached figure below.
Draw the top view of this
figure.
Suppose A is a 3 x 3 matrix such that det (A) = 9. Prove det (3 (A-!') is equal to 3
Answer: The proof is done below.
Step-by-step explanation: Given that A is a 3 x 3 matrix such that det (A) = 9.
We are to prove the following :
[tex]det(3A^{-1})=3.[/tex]
For a non-singular matrix B of order n, we have two two properties of its determinant :
[tex](i)~det(B^{-1})=\dfrac{1}{det(B)},\\\\\\(ii)~det(kB)=k^ndet(B),~\textup{k is a scalar.}[/tex]
Therefore, we get
[tex]det(A^{-1})=\dfrac{1}{det(A)}=\dfrac{1}{9},[/tex]
and so,
[tex]det(3A^{-1})~~~~~~~[\textup{since A is of order 3}]\\\\=3^3det(A^{-1})\\\\=27\times\dfrac{1}{9}\\\\=3.[/tex]
Hence proved.
let f(x) be a func. satisfying f(-x)=f(x) for all real x.if f"(x) exist, find its value.
Answer:
[tex]f''(x)=f''(-x)[/tex]
Step-by-step explanation:
A function satisfying the equation [tex]f(x)=f(-x)[/tex] is said to be an even function. This denomination comes from the fact that the same relation is satisfied for functions of the form [tex]x^{n}[/tex] with [tex]n[/tex] even. Observe that if [tex]f[/tex] is twice differentiable we can derivate using the chaing rule as follows:
[tex]f(x)=f(-x)[/tex] implies [tex]f'(x)=f'(-x)\cdot (-1)=-f'(-x)[/tex]
Applying the chain rule again we have:
[tex]f'(x)=-f'(-x)[/tex] implies [tex]f''(x)=-f''(-x)\cdot (-1)=f''(-x)[/tex]
So we have that function [tex]f''(x)[/tex] is also an even function.
(a) Find an example of sets A and B such that An B = {1,2} and AUB = {1,2,3,4,5).
(b) Find an example of sets A and B such that A Ç B and A e B.
Answer:
(a) Set A = {1,2,3}
Set B = {1,2,4,5}
(b) Set A Ç B = {1,2}
Set A e B = {1,2}
Step-by-step explanation:
As per the question,
Given data :
A ∪ B = read as A union B = {1,2,3,4,5}
A ∩ B = read as A intersection B = {1,2}
(a) An example of sets A and B such that A ∩ B = {1,2} and A U B = {1,2,3,4,5).
So, first draw the Venn-diagram, From below Venn diagram, One of the possibility for set A and set B is:
Set A = {1,2,3}
Set B = {1,2,4,5}
(b) An example of sets A and B such that A Ç B and A e B,
Set A Ç B read as common elements of set A in Set B.
Therefore,
Set A Ç B = {1,2}
Set A e B implies that which element/elements of A is/are present in set B.
Therefore,
Set A e B = {1,2}
The Schuller family has five members. Dad is 6ft 2in tall. Mom is 3 inches shorter than Dad, but 2 inches taller than Ivan. Marcia is 5 inches shorter than Ivan, but twice as tall as Sally-Jo. What is the mean height of the Schuller family?
Answer:
62 inches
Step-by-step explanation:
Let x be the height ( in inches ) of Ivan,
∵ Marcia is 5 inches shorter than Ivan,
height of Marcia = x - 5,
Marcia is twice as tall as Sally-Jo,
height of sally-jo = [tex]\frac{x-5}{2}[/tex]
Mom is 2 inches taller than Ivan.
⇒ height of mom = x + 2,
Mom is 3 inches shorter than Dad,
height of dad = x + 2 + 3 = x + 5,
So, mean height of family is,
[tex]\frac{x+x-5+\frac{x-5}{2}+x+2+x+5}{5}[/tex]
[tex]=\frac{2x+2x-10+x-5+2x+4+2x+10}{10}[/tex]
[tex]=\frac{9x-1}{10}[/tex]
According to the question,
x + 5 = 74 ( 1 ft = 12 in )
x = 69
Hence, mean height of the family = [tex]\frac{9\times 69-1}{10}[/tex]
[tex]=\frac{620}{10}[/tex]
= 62 inches
The final exam of a particular class makes up 40% of the final grade, and Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam. What grade does Moe need on his final exam in order to receive the passing grade average of 60% for the class?
Answer:
%82.5
Step-by-step explanation:
The final exam of a particular class makes up 40% of the final gradeMoe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam.From point 1 we know that Moe´s grade just before taking the final exam represents 60% of the final grade. Then, using the information in the point 2 we can compute Moe´s final grade as follows:
[tex]FG=0.40*FE+0.60*0.45[/tex],
where FG is Moe´s Final Grade and FE is Moe´s final exam grade. Then,
[tex]\frac{ FG-0.60*0.45}{0.40}=FE[/tex].
So, in order to receive the passing grade average of 60% for the class Moe needs to obtain in his exam:
[tex]FE=\frac{ 0.60-0.60*0.45}{0.40}=0.825[/tex]
That is, he need al least %82.5 to obtain a passing grade.