Answer:
Cost function C(x) == FC + VC*Q
Revenue function R(x) = Px * Q
Profit function P(x) =(Px * Q)-(FC + VC*Q)
P(12000) = -38000 Loss
P(23000) = 28000 profit
Step-by-step explanation:
Total Cost is Fixed cost plus Variable cost multiplied by the produce quantity.
(a)Cost function
C(x) = FC + vc*Q
Where
FC=Fixed cost
VC=Variable cost
Q=produce quantity
(b)
Revenue function
R(x) = Px * Q
Where
Px= Sales Price
Q=produce quantity
(c) Profit function
Profit = Revenue- Total cost
P(x) =(Px * Q)-(FC + vc*Q)
(d) We have to replace in the profit function
at 12,000 units
P(12000) =($20 * 12,000)-($110,000 + $14*12,000)
P(12000) = -38000
at 23,000 units
P(x) =($20 * 23,000)-($110,000 + $14*23,000)
P(23000) = 28000
Use properties of limits and algebraic methods to find the limit, if it exists. (If the limit is infinite, enter '[infinity]' or '-[infinity]', as appropriate. If the limit does not otherwise exist, enter DNE.) lim x→3 f(x),
where f(x) = 9 − 3x if x < 3 ;
and x^2 − x if x ≥ 3
Answer:
The limit of this function does not exist.
Step-by-step explanation:
[tex]\lim_{x \to 3} f(x)[/tex]
[tex]f(x)=\left \{ {{9-3x} \quad if \>{x \>< \>3} \atop {x^{2}-x }\quad if \>{x\ \geq \>3 }} \right.[/tex]
To find the limit of this function you always need to evaluate the one-sided limits. In mathematical language the limit exists if
[tex]\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) =L[/tex]
and the limit does not exist if
[tex]\lim_{x \to a^{-}} f(x) \neq \lim_{x \to a^{+}} f(x)[/tex]
Evaluate the one-sided limits.
The left-hand limit
[tex]\lim_{x \to 3^{-} } 9-3x= \lim_{x \to 3^{-} } 9-3*3=0[/tex]
The right-hand limit
[tex]\lim_{x \to 3^{+} } x^{2} -x= \lim_{x \to 3^{+} } 3^{2}-3 =6[/tex]
Because the limits are not the same the limit does not exist.
How many subsets does the set D={c,a,t}D={c,a,t} have?
Answer:
The number of all possible subsets of D is 8.
Step-by-step explanation:
Consider the provided set D={c,a,t}
The subset of D contains no elements: { }
The subset of D contains one element each: {c} {a} {t}
The subset of D contains two elements each: {c, a} {a, t} {c, t}
The subset of D contains three elements: {c, a, t)
Hence, all possible subsets of D are { }, {c}, {a}, {t}, {c, a}, {a, t}, {c, t}, {c, a, t}
Therefore, number of all possible subsets of D is 8.
Or we can use the formula:
The number of subsets of the set is [tex]2^n[/tex] If the set contains ‘n’ elements.
There are 3 elements in the provided set, thus use the above formula as shown:
2³=8
Hence, the number of all possible subsets of D is 8.
Two players A and B play a marble game. Each player has both a red and blue marble. They present one marble to each other. If both present red, A wins $3. If both present blue, A wins $1. If the colors of the two marbles do not match, B wins $2. Is it better to be A, or B, or does it matter?
Answer:
Step-by-step explanation:
A has one red and blue marble and B has one red and blue marble.
Hence selecting one marble is equally likely with prob = 0.5
Since A and B are independent the joint event would be product of probabilities.
Let A be the amount A wins.
If each selects one, the sample space would be
(R,R) (R,B) (B,R) (B,B)
Prob 0.25 0.25 0.25 0.25
A 3 -2 -2 1
E(A) 0.75 -0.5 -0.5 0.25 = 0
The game is a fair game with equal expected values for A and B.
It does not matter whether to be A or B
The game described is fair, as both A and B have an expected value of $1. This is calculated using the probability of each outcome times its respective payoff. There is no advantage in being either A or B.
Explanation:The marble game presented in the question involves probability and expectation of winnings. First, we must figure out the possible outcomes. There are four possibilities: (1) both A and B present red marbles, (2) both present blue marbles, (3) A presents red and B presents blue, and (4) A presents blue and B presents red.
When calculating the expected payoff for each player, we assume that each possibility carries an equal weight, as the question doesn't offer any bias towards a certain color. Therefore, the expected payoff (E) for each player is calculated by adding the product of each possible payout and its probability. For A, [tex]E(A) = (1/4)*$3 + (1/4)*$1 + (1/4)*$0 + (1/4)*$0 = $1.\\ For B, E(B) = (1/4)*$0 + (1/4)*$0 + (1/4)*$2 + (1/4)*$2 = $1.[/tex]
Given these calculations, we can see that the expected value of the game is equal for A and B, hence it doesn't matter who is A and who is B.
Learn more about Probability and Expectation here:https://brainly.com/question/23429157
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Suppose an experiment has 3 stages: A, B, and C. If stage A has 6 outcomes, stage B has 4 outcomes, and stage C has 3 outcomes. how many outcomes does the entire experiment have?
Answer: 72
Step-by-step explanation:
Given : An experiment has 3 stages: A, B, and C.
If stage A has 6 outcomes, stage B has 4 outcomes, and stage C has 3 outcomes.
Then, by using the fundamental principle of counting (Total outcomes= product of all outcomes of each stage ) , we have
The number of outcomes the entire experiment have:-
[tex]6\times4\times3=72[/tex]
Hence, the number of outcomes the entire experiment =72
Final answer:
To find the total number of outcomes for an experiment with 3 stages having 6, 4, and 3 outcomes respectively, you multiply the outcomes of each stage together, resulting in 6 × 4 × 3 = 72 possible outcomes.
Explanation:
To determine how many outcomes the entire experiment has when an experiment has stages A, B, and C with different numbers of outcomes, you multiply the number of outcomes for each stage. Stage A has 6 outcomes, stage B has 4 outcomes, and stage C has 3 outcomes. Therefore, the total number of outcomes for the entire experiment is calculated as follows:
Stage A outcomes: 6
Stage B outcomes: 4
Stage C outcomes: 3
Multiply the outcomes of each stage:
Total outcomes = 6 (Stage A) × 4 (Stage B) × 3 (Stage C) = 72 possible outcomes.
This is similar to how the sample space is determined in other contexts, such as flipping a coin and rolling a die, where you would also multiply the number of outcomes to get the size of the sample space.
A company has determined that it must increase production of a certain line of goods by 112 times last years production. The new output will be 2885 items. What was last year's output?
Answer: 26 items
Step-by-step explanation:
Let x denotes the last year's output.
Given : A company has determined that it must increase production of a certain line of goods by 112 times last years production.
The new output will be 2885 items.
According to the above statement we have the following equation :-
[tex]112x=2885[/tex]
Divide both sides by 112, we get
[tex]x=25.7589285714\approx26[/tex]
Hence, the last year's output =26 items
The "absorption law" (theorem 2.1.1 in our book) states that p V (p Aq) is logically equivalent to p. Construct a truth table to show these statements are equivalent.
Answer:
According to the Law of Absorption, these 2 expressions are equivalent:
p ∨ (p ∧ q) = p
Truth Table:
(see the image attached)
Step-by-step explanation:
To construct the Truth Table you can consider the 4 possible combinations of states that p and q could have, that is
1. p=T, q=T
2. p=T, q=F
3. p=F, q=T
4. p=F, q=F
Then you can calculate p ∨ (p ∧ q) = p for each combination
1. T ∨ (T ∧ T) = T
2. T ∨ (T ∧ F) = T
3. F ∨ (F ∧ T) = F
4. F ∨ (F ∧ F) = F
You can see that the previous values are the same states that p has, you can also see it in the table attached.
How many grams of glucose would you use to make 200ml of a 1m sugar solution
Answer:
Mass of glucose will be 36 gram
Step-by-step explanation:
We have given volume of solution V = 200 ml =0.2 L
Morality = 1 M
Molar mass of glucose = 180 g/mol
We know that morality is given by
[tex]molarity=\frac{number\ of\ moles}{volume\ of\ solution\ in\ liters}[/tex]
[tex]1=\frac{number\ of\ moles}{0.2}[/tex]
[tex]number\ of\ moles\ n =0.2[/tex]
We know that number of moles is given by
[tex]n=\frac{mass\ in\ gram}{molar\ mass}[/tex]
[tex]0.2=\frac{mass\ in\ gram}{180}[/tex]
Mass = 36 gram
Find the average cost per item when the required number of items are produced. C(x) = 19x + 1900, 1000 items What is the average cost per item?
Answer:
$20.9
Step-by-step explanation:
We have been given a formula [tex]C(x)=19x+1900[/tex], which represents the cost of x items.
First of all, we will find cost of 1000 items by substituting [tex]x=1000[/tex] in our given formula as:
[tex]C(1000)=19(1000)+1900[/tex]
[tex]C(1000)=19,000+1900[/tex]
[tex]C(1000)=20,900[/tex]
To find average cost per item, we will divide total cost by number of items as:
[tex]\text{Average cost per item}=\frac{\$20,900}{1000}[/tex]
[tex]\text{Average cost per item}=\$20.9[/tex]
Therefore, the average cost per item would be $20.9.
Calculate:
(Round two decimal places for final answer)
3gallons (gal) =_____liters (L)
Answer:
3gallons (gal) = 11.35 liters
Step-by-step explanation:
This can be solved as a rule of three problem.
In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.
When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.
When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.
Unit conversion problems, like this one, is an example of a direct relationship between measures.
Each gallon has 3.78L. So
1 gallon - 3.78L
3 gallons - xL
[tex]x = 3*3.78[/tex]
[tex]x = 11.35[/tex] liters
3gallons (gal) = 11.35 liters
Let A be as above, consider Ax = b where b = (31, 2, 21, 11). Find x1 using Cramer’s rule. (You may use MATLAB/Octave to compute the determinants, but write out what you are computing.).
Matrix a= 8 6 -3 20
4 2 -5 -7
8 2 7 20
4 2 -11 -4
Answer:
x1= 1
Step-by-step explanation:
The Cramer's rule say that x1=[tex]\frac{det(A1)}{det(A)}[/tex] where A1 is the matrix A change the column 1 by the vector b.
Then A1= [tex]\left[\begin{array}{cccc}31&6&-3&20\\2&2&-5&-7\\21&2&7&20\\11& 2&-11&-4\end{array}\right][/tex].
Using Octave we have that det(A1)=-3840 and det(A)=-3840.
Then x1=[tex]\frac{-3840}{-3840}=1[/tex].
A and B are bounded non-empty subsets of R. For inf(A) to be less than or equal to inf(B), which of the following conditions must be met?
a) For every b in B and epsilon > 0, there exists a in A, such that a < b + epsilon.
b) There exists a in A, and b in B such that a < b.
If neither of these conditions are appropriate, what would be appropriate conditions for inf(A) to be less than or equal to inf(B)?
Answer:
a) must be met
Step-by-step explanation:
We have two conditions:
a) For every [tex]b\in B[/tex] and [tex]\epsilon>0[/tex], there exists [tex]a\in A[/tex], such that [tex]a<b+\epsilon[/tex].
b) There exists [tex]a\in A[/tex] and [tex]b\in B[/tex] such that [tex]a<b[/tex].
We will prove that conditon a) is equivalent to [tex]inf(A)\leq inf(B)[/tex]
If a) is not satisfied, then it would exist [tex]b\in B[/tex] and [tex]\epsilon >0[/tex] such that, for every [tex]a\in A[/tex], [tex]a\geq b+\epsilon[/tex]. This implies that [tex]b+\epsilon[/tex] is a lower bound for A and in consequence
[tex]inf(A)\geq b+\epsilon > b\geq inf(B)[/tex]
Then, [tex]inf(A) \leq inf(B)[/tex] implies a).
If [tex]inf(A) \leq inf(B)[/tex] is not satisfied then, [tex]inf(A) > inf(B)[/tex] and in consequence exists [tex]b\inB[/tex] such that [tex]b-inf(A)=\epsilon >0[/tex]. Then [tex]b-\epsilon=inf(A)[/tex] and, for every [tex]a\in A[/tex],
[tex]b-\epsilon =inf(A)\leq a[/tex].
So, a) is not satisfied.
In conclusion, a) is equivalent to [tex]inf(A)\leq inf(B)[/tex]
Finally, observe that condition b) is not an appropiate condition to determine if [tex]inf(A)\leq inf(B)[/tex] or not. For example:
A={0}, B={1}. b) is satisfied and [tex]inf(A)=0<1=inf(B)[/tex]A={0}. B={-1,1}. b) is satisfied and [tex]inf(A)=0>-1=inf(B)[/tex]A company’s total revenue from manufacturing and selling x units of their product is given by: y = –3x2 + 900x – 5,000. How many units should be sold in order to maximize revenue, and what is the maximum revenue
Answer:
150 units;
Maximum revenue: $62,500.
Step-by-step explanation:
We have been given that a company’s total revenue from manufacturing and selling x units of their product is given by [tex]y=-3x^2+900x-5,000[/tex]. We are asked to find the number of units sold that will maximize the revenue.
We can see that our given equation in a downward opening parabola as leading coefficient is negative.
We also know that maximum point of a downward opening parabola is ts vertex.
To find the number of units sold to maximize the revenue, we need to figure our x-coordinate of vertex.
We will use formula [tex]\frac{-b}{2a}[/tex] to find x-coordinate of vertex.
[tex]\frac{-900}{2(-3)}[/tex]
[tex]\frac{-900}{-6}[/tex]
[tex]150[/tex]
Therefore, 150 units should be sold in order to maximize revenue.
To find the maximum revenue, we will substitute [tex]x=150[/tex] in our given formula.
[tex]y=-3(150)^2+900(150)-5,000[/tex]
[tex]y=-3*22,500+135,000-5,000[/tex]
[tex]y=-67,500+135,000-5,000[/tex]
[tex]y=62,500[/tex]
Therefore, the maximum revenue would be $62,500.
If angle 1 has a measure of 56° and angle 2 has a measure of 124°, the two angles are complementary.
Question 1 options:
True
False
Answer:
False.
Step-by-step explanation:
Two angles are complementary when added up, they give a result of 90°.
So, to this question to be true we have to do:
Angle 1 + Angle 2 = 90
But if we resolve 56° + 124° = 180, so this means that this question is false, as the addition of both angles doesn't have a result of 90°.
A basic cellular phone plan costs $30 per month for 50 calling minutes. Additional time costs $0.50 per minute. The formula C 30 +0.50(x - 50) gives the monthly cost for this plan, C, for x calling minutes, where x>50. How many calling minutes are possible for a monthly cost of at least $35 and at most $40? For a monthly cost of at least $35 and at most $40 sxs calling minutes are possible S
Answer:
Step-by-step explanation:
Given that a basic cellular phone plan costs $30 per month for 50 calling minutes
i.e. C(x) = [tex]30+0.50(x-50)[/tex], where x = calling minutes
Here 30 is fixed upto 50 calls after that cost increases at 0.50 per minute talk time.
[tex]C(x) = 30+0.5x-25 = 0.5x+5\\[/tex]
When monthly cost is atleast 35 and atmost 40 we have
[tex]35\leq C(x)\leq 40\\35\leq 0.5x+5\leq 40\\30\leq 0.5x\leq 35\\60\leq x\leq 70[/tex]
i.e. talking time must be atleast 60 minutes and atmost 70 minutes
develop an explicit formula in terms of n for the nth term of the following sequence:
0,4,18,48,100,180,......
show that it "works" for the sixth term and use it to find the seventh term.
The given sequence : 0,4,18,48,100,180,......
We can write the terms of the sequence as :
[tex]\text{Ist term }:a_1=1(1^2-1)=0\\\\\text{IInd term }: a_2=2^2(2-1)=4(2-1)=4\\\\\text{IIIrd term }:a_3=3^2(3-1)=9(2)=18\\\\\text{IVth term }:a_4=4^2(4-1)=(16)(3)=48\\\\\text{Vth term }:a_5=5^2(5-1)=25(4)=100\\\\\text{VIth term }:a_6=6^2(6-1)=36(5)=180[/tex]
From the above presentation of the terms, the explicit formula in terms of n for the nth term will be :-
[tex]a_n=n^2(n-1)[/tex]
Put n= 7 , we get
[tex]a_7=7^2(7-1)=49(6)=294[/tex]
Therefore, the seventh term of the given sequence = 294
The function s(t) represents the position of an object at time t moving along a line. Suppose s(2)=150 and s(5)=237. Find the average velocity of the object over the interval of time [1, 3].
Answer:
29 is answer.
Step-by-step explanation:
Given that the function s(t) represents the position of an object at time t moving along a line. Suppose s(2)=150 and s(5)=237.
To find average velocity of the object over the interval of time [1,3]
We know that derivative of s is velocity and antiderivative of velocity is position vector .
Since moving along a line equation of s is
use two point formula
[tex]\frac{s-150}{237-150} =\frac{t-2}{5-2} \\s=29t-58+150\\s=29t+92[/tex] gives the position at time t.
Average velocity in interval (1,3)
=[tex]\frac{1}{3-1} (s(3)-s(1))\\=\frac{1}{2} [87+58-29-58]\\=29[/tex]
The average velocity of the object over the interval [1, 3] is 43.5 units of distance per unit of time.
Explanation:The average velocity of an object over an interval of time is found by dividing the change in position by the change in time. In this case, we want to find the average velocity of the object over the interval [1, 3].
To do this, we first need to find the change in position during this interval. Given that s(2) = 150 and s(5) = 237, we can calculate the change in position as follows:
s(3) - s(1) = (s(5) - s(3)) + (s(3) - s(1))
Therefore, the average velocity is (s(3) - s(1)) / (3 - 1). By substituting the given values, we find that the average velocity is (237 - 150) / (3 - 1) = 87 / 2 = 43.5 units of distance per unit of time.
Steve reads 80 pages in 2 hours and 40 minutes. If Cassandra reads twice as fast as Steve, how long will it take her to read a 300 page book? O A 4 hours 40 minutes O B. 5 hours C.5 hours 20 minutes O D. 5 hours 40 minutes
Answer:
Option B. 5 hours.
Step-by-step explanation:
Steve reads 80 pages in 2 hours and 40 minutes.
1 hour = 60 minutes
2 hour 40 minutes = (2 × 60) + 40 = 160 minutes
Speed of reading of Steve = [tex]\frac{80}{160}[/tex]
= 0.5 page per minute
Cassandra reads twice as fast as Steve.
Therefore, speed of reading of Cassandra = 0.5 × 2
= 1 page per minute
Therefore, time to read 300 pages by Cassandra = 300 × 1
= 300 minutes
Now convert minutes to hours
[tex]\frac{300}{60}[/tex]
= 5 hours
Cassandra will take 5 hours to read 300 page book.
5. Determine whether each of the following statements is true or false a. 1 + 1 = 3 if and only if monkeys can fly. b. If birds can fly, then 1 + 1 = 3. c. If 1 1- 3, then pigs can fly. 目view as Text A1 2
Answer:
Part (A) True
Part (B) False
Part (C) True
Step-by-step explanation:
Consider the provided information.
If both the statements are either true or false then the biconditionals are true. Otherwise biconditionals are false.
Part (A) 1 + 1 = 3 if and only if monkeys can fly.
Consider the first statement: 1+1=3 (This is a False statement)
Consider the second statement: "monkeys can fly" (This is also a False statement)
True: First statement is false and the second statement is also false, Thus, making the biconditional true.
Part (B) If birds can fly, then 1 + 1 = 3.
Consider the first statement: "birds can fly" (This is true statement)
Consider the second statement: 1 + 1 = 3 (This is a False statement)
False: First statement is true, but second statement is false, making everything false.
Part (C) If 1 + 1= 3, then pigs can fly.
Consider the first statement: 1+1=3 (This is a False statement)
Consider the second statement: "pigs can fly" (This is also a False statement)
True: First statement is false and the second statement is also false, Thus, making the biconditional true.
There are 143076 books in the library. The 3rd grade class walked over and each kid checked out 2 books. There are 21 students in the class. How many books are in left in the library after the students return to their classroom?
Answer:
The answer is: There are only 143034 books left after the students return to their classroom
Step-by-step explanation:
Lets call X= Number of books left in the library after the students return to their classroom.
B=Total of Books in the library
S=Total of books taken by the 3rd grade students
If each student of the 3rd grage class take 2 books, the total of book taken by the 3rd grade students would be:
S=2 books per student * Total of students of the 3rdgrade
S=2*21=42
Then we must substract Total of Books in the library minus Total of books taken by the 3rd grade students. That would be:
X=B-S
X=143076-42
X=143034
I hope that this answer will help you
There are four candidates (A. B, C, and D) and 115 voters. When the points were tallied (using 4 points for frst, 3 points for second, 2 points for third, and 1 point for fourth) candidates. (Hint: Figure out how many points are packed in each ballot) ). A had 320 points, B had 330 points, and C had 190 points. Find how many points D had and give ranking of the Find the complete ranking of the candidates Choose the correct answer below
Answer:
D had 310 points.
The final ranking is: B,A,D,C
Step-by-step explanation:
The problems states that:
There are 115 votes.
Each ballot has a vote for first place, that counts 4 points, a vote for second place, that counts 3 points, a vote for third place, that counts 2 points and a vote for fourth place that counts 1 point. This means that each ballot has 4+3+2+1 = 10 points.
Since there are 115 voters, there are 115 ballots. This means that the total points of A,B,C and D combined must be equal to 115*10 = 1150. So:
[tex]A + B + C + D = 1150[/tex]
We already know that:
A had 320 points, B had 330 points, and C had 190 points.
So:
[tex]A + B + C + D = 1150[/tex]
[tex]320 + 330 + 190 + D = 1150[/tex]
[tex]D = 310[/tex]
B had the most points, followed by A, D and C. So the final ranking is: B,A,D,C
Cindy worked for 15 consecutive days, earning an average wage of $91 per day. During the first 7 days her average was $87/day, and her average during the last 7 days was $93/day. What was her wage on the 8th day?
A. $83 B. $92 C. $97 D. $105
(I believe it's D. But feel free to correct me if i'm wrong)
Answer:
D. $105
Step-by-step explanation:
Use definition:
[tex]\text{Average of }n\text{ numbers}=\dfrac{\text{The sum of }n\text{ numbers}}{n}[/tex]
The average wage for 15 consecutive days is $91, then by definition
[tex]\$91=\dfrac{\text{The sum of wages for 15 days}}{15}\Rightarrow \\ \\\text{The sum of wages for 15 days}=\$91\cdot 15=\$1,365[/tex]
The average wage for first 7 consecutive days is $87, then by definition
[tex]\$87=\dfrac{\text{The sum of wages for first 7 days}}{7}\Rightarrow \\ \\\text{The sum of wages for first 7 days}=\$87\cdot 7=\$609[/tex]
The average wage for last 7 consecutive days is $93, then by definition
[tex]\$93=\dfrac{\text{The sum of wages for last 7 days}}{7}\Rightarrow \\ \\\text{The sum of wages for last 7 days}=\$93\cdot 7=\$651[/tex]
Now,
[tex]\text{The sum of wages for 15 days}=\text{The sum of wages for first 7 days }+\\ \\+\text{ The wage for 8th day}+\text{The sum of wages for last 7 days}\\ \\\$1,365=\$609+\text{ The wage for 8th day}+\$651\\ \\\text{ The wage for 8th day}=\$1,365-\$609-\$651=\$105[/tex]
The marketing department at Cable TV (CTV) wants to know how promotional advertising affects the number of viewers for the Saturday Night Movie. Research shows that 10 million viewers watched the movie when CTV ran 15 one-minute ads on Friday. When they ran 25 one-minute ads on Friday, the movie had 18 million viewers. Use linear interpolation to estimate the number of viewers if CTV runs 23 one-minute ads on Friday.
Answer:
16.4 million viewers
Step-by-step explanation:
The number of viewers increased by 8 million from 10 to 18 million when the number of ads increased by 10 ads from 15 to 25. If 23 ads are run, that represents an increase of 8 ads from 15, so we expect 8/10 of the increase in viewers.
8/10 × 8 million = 6.4 million
The number we expect with 23 ads is 6.4 million more viewers than 10 million viewers, so is 16.4 million.
_____
Alternate solution
We can write a linear equation in 2-point form for the number of viewers expected for a given number of ads:
y = (18 -10)/(25 -15)(x -15) +10
y = (8/10)(x -15) +10
y = 0.8x -2 . . . . . million viewers for x ads
For 23 ads, this gives ...
y = 0.8×23 -2 = 18.4 -2 = 16.4 . . . . million viewers, as above
_____
Comment on 8/10
I consider it coincidence that the number 23 is 8/10 of the difference between 25 and 15, and the slope of the line is 8/10. The point we're trying to interpolate has no relationship to the slope of the line, and vice versa.
Linear interpolation illustrates the use of linear equation of several points
The number of viewers is 16.4 million, if a 23 one-minute ads runs on Friday.
Linear interpolation is represented as:
[tex]\frac{y_2 - y_1}{x_2 - x_1} = \frac{y - y_1}{x - x_1}[/tex]
Let:
[tex]x \to[/tex] Time
[tex]y \to[/tex] Viewers
So, we have:
[tex](x_1,y_1) = (15,10m)[/tex]
[tex](x_2,y_2) = (25,18m)[/tex]
[tex](x,y) = (23,y)[/tex]
Substitute the above points in:
[tex]\frac{y_2 - y_1}{x_2 - x_1} = \frac{y - y_1}{x - x_1}[/tex]
So, we have:
[tex]\frac{y_2 - y_1}{x_2 - x_1} = \frac{y - y_1}{x - x_1}[/tex]
[tex]\frac{18m - 10m}{25 -15} = \frac{y - 10m}{23 -15}[/tex]
[tex]\frac{8m}{10} = \frac{y - 10m}{8}[/tex]
Multiply both sides by 8
[tex]\frac{64m}{10} = y - 10m[/tex]
[tex]6.4m = y - 10m[/tex]
Collect like terms
[tex]y =10m + 6.4m[/tex]
[tex]y =16.4m[/tex]
Hence, the number of viewers is 16.4 million, if a 23 one-minute ads runs on Friday.
Read more about linear interpolation at:
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Problem Solving REAL WORLD 14. 13. Jerome is making prizes for a game at the school fair. He has two bags of different candies, one with 15 pieces of candy and one with 20 pieces. Every prize will have one kind of candy, the same number of pieces, and the greatest number of pieces possible. How many candies should be in each prize? 2 candies
Answer: 5
Step-by-step explanation:
Given : Jerome has two bags of different candies, one with 15 pieces of candy and one with 20 pieces.
Every prize will have one kind of candy, the same number of pieces.
Then to find the greatest number of pieces of candies possible in each price , we need to find the greatest common factor of 15 and 20.
Prime factorization of 15 and 20 :-
[tex]15=3\times5\\\\20=2\times2\times5[/tex]
∴ Greatest common factor of 15 and 20 = 5
Hence , the greatest number of pieces of candies possible in each prize = 5
On three examinations, you have grades of 85, 78, and 84. There is still a final examination, which counts as one grade In order to get an A your average must be at least 90. If you get 100 on the final, what is your numerical average? 86.75 (Type an integer or a decimal.) Is an A in the course possible? Yes No To eam a B in the course, you must have a final average of at least 80 What grade must you got on the final to earn a B in the course? Type an integer or a decimal) Enter your answer in the answer box and then click Check Answer All parts showing Clear All Pearson Copyright © 2019 Pearson Education Inc. All rights reserved. Terms of Use Privacy
Answer:
It's not possible to earn an A in the courseI must have a 73 or MORE to earn a B in the courseStep-by-step explanation:
The average obtained at the end of the course will be:
[tex]\frac{85+78+84+x}{4} = av[/tex]
Where x is the grade obtained in the final examination and av is the final average. To obtain an A, av has to be at least 90, av≥90, and to obtain an B, av has to be at least 80, av≥80
Is an A in the course possible?So, if we get 100 on the final average:
x = 100,
av = (85+78+84+100)/4 = 86,75 and 86,75∠90.
Answer: No, the higher grade obtained would be 86,75.
What grade you must have in the final to earn a B in the courseTo earn a B, av≥80:
[tex]av= \frac{85+78+84+x}{4 \ }\geq 80\\ \frac{247+x}{4}\geq80 \\247+x\geq 80*4\\ x\geq 320-247\\ x\geq 73[/tex]
Answer: I must have a 73 or MORE to earn a B in the course
If the area of polygon P is 72 square units, what is the scaled factor did deigo use to go from P to Q?
Answer:
The scale factor used to go from P to Q is 1/4
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
If two figures are similar, then the ratio of its areas i equal to the scale factor squared
Let
z ----> the scale factor
x -----> area of polygon Q
y -----> area of polygon P
[tex]z^{2}=\frac{x}{y}[/tex]
we have
[tex]y=72\ units^2[/tex]
Find the area of polygon Q
Divide the the area of polygon Q in two triangles and three squares
The area of the polygon Q is equal to the area of two triangles plus the area of three squares
see the attached figure N 2
Find the area of triangle 1
[tex]A=(1/2)(1)(2)=1\ units^2[/tex]
Find the area of three squares (A2,A3 and A4)
[tex]A=3(1)^2=3\ units^2[/tex]
Find the area of triangle 5
[tex]A=(1/2)(1)(1)=0.5\ units^2[/tex]
The area of polygon Q is
[tex]x=1+3+0.5=4.5\ units^2[/tex]
Find the scale factor
[tex]z^{2}=\frac{x}{y}[/tex]
we have
[tex]y=72\ units^2[/tex]
[tex]x=4.5\ units^2[/tex]
substitute and solve for z
[tex]z^{2}=\frac{4.5}{72}[/tex]
[tex]z^{2}=\frac{1}{16}[/tex]
square root both sides
[tex]z=\frac{1}{4}[/tex]
therefore
The scale factor used to go from P to Q is 1/4
The scaled factor did Diego used to go from P to Q is 1/4.
Given
The area of polygon P is 72 square units.
What is the scale factor?If two figures are similar, then the ratio of its areas is equal to the scale factor squared.
[tex]\rm z^2=\dfrac{x}{y}[/tex]
Where; z is the scale factor, x = area of polygon Q, y = area of polygon P.
Therefore,
The area of the first triangle is;
[tex]\rm Area = \dfrac{1}{2} \times base \times height\\\\Area = \dfrac{1}{2} \times 1 \times 2\\\\Area = 1 \ units[/tex]
The area of three squares (A2, A3, and A4)
[tex]\rm Area = 3(1)^2\\\\Area = 3\ square \ units[/tex]
The area of the 5th triangle is;
[tex]\rm Area = \dfrac{1}{2} \times base \times height\\\\Area = \dfrac{1}{2} \times 1 \times 1\\\\Area = 0.5 \ units[/tex]
Then,
The area of the polygon is;
x = 1 + 3 + 0.5 = 4.5 units
Therefore,
The scaled factor did Diego used to go from P to Q is;
[tex]\rm z^2=\dfrac{4.5}{72}\\\\z^2=\dfrac{1}{16}\\\\z=\dfrac{1}{4}\\\\[/tex]
Hence, the scaled factor did Diego used to go from P to Q is 1/4.
To know more about the Scale factor click the link given below.
https://brainly.com/question/22644306
Solve each formula for the given variable. State the restrictions, if any, for the formula
obtained to be meaningful.
F = ma, for a
d = [tex]\frac{m}{v}[/tex], for v
A = P + Prt, for t
A = [tex]\frac{1}{2}[/tex]h a + b, for h
P = 2(L +W ), for W
m = [tex]\frac{x+y}{2}[/tex], for y
3x + 2y = 8,for y
a = [tex]\frac{v- u}{t}[/tex], for t
It's the even numbers on this pdf:
Answer:[tex]a = \frac{F}{m}[/tex]
[tex]v = \frac{m}{d}[/tex]
[tex]t = \frac{A - P}{Pr}[/tex]
[tex]h = \frac{2A}{a + b}[/tex]
[tex]W = \frac{P - 2L}{2}[/tex]
[tex]y = 2m - x[/tex]
[tex]y = -1.5x + 4[/tex]
[tex]t = \frac{v - u}{a}[/tex]
Step-by-step explanation:
[tex]F = ma[/tex]
[tex]\frac{F}{m} = \frac{ma}{m}[/tex]
[tex]\frac{F}{m} = a[/tex]
[tex]d = \frac{m}{v}[/tex]
[tex]vd = m[/tex]
[tex]\frac{vd}{d} = \frac{m}{d}[/tex]
[tex]v = \frac{m}{d}[/tex]
[tex]A = P + Prt[/tex]
[tex]A - P = Prt[/tex]
[tex]\frac{A - P}{Pr} = \frac{Prt}{Pr}[/tex]
[tex]\frac{A - P}{Pr} = t[/tex]
[tex]A = \frac{1}{2}h(a + b)[/tex]
[tex]2A = h(a + b)[/tex]
[tex]\frac{2A}{a + b} = \frac{h(a + b)}{a + b}[/tex]
[tex]\frac{2A}{a + b} = h[/tex]
[tex]P = 2(L + W)[/tex]
[tex]P = 2(L) + 2(W)[/tex]
[tex]P = 2L + 2W[/tex]
[tex]P - 2L = 2W[/tex]
[tex]\frac{P - 2L}{2} = \frac{2W}{2}[/tex]
[tex]\frac{P - 2L}{2} = W[/tex]
[tex]m = \frac{x + y}{2}\\2m = x + y\\2m - x = y[/tex]
[tex]3x + 2y = 8\\2y = -3x + 8\\\frac{2y}{2} = \frac{-3x + 8}{2}\\y = -1.5x + 4[/tex]
[tex]a = \frac{v - u}{t}\\at = v - u\\\frac{at}{a} = \frac{v - u}{a}\\t = \frac{v - u}{a}[/tex]
Suppose that coin 1 hasprobability 0.7 of coming up
heads, and coin 2 has probability 0.6of coming up heads. If the
coin flipped today comes up heads, thenwe select coin 1 to flip
tomorrow, and if it comes up tails, thenwe select coin 2 to flip
tomorrow. If the coin initially flipped isequally likely to be coin
1 or coin 2, then what is the probabilitythat the coin flipped on
the 3rd day after the initial flip is coin1?
Answer:
[tex]\frac{1333}{2000}[/tex]
Step-by-step explanation:
We want to compute the probabily of being flipping coin1 in the third day. Observe that in day zero (the day were the coin to be flipped is chosen randomly with equal probability to be coin1 or coin2), day 1 and day 2 we will flip coin1 or coin2. So, there are 8 possible scenarios to consider:
[tex]S1=(1,1,1,1)\\S2=(1,1,2,1)\\S3=(1,2,1,1)\\S4=(1,2,2,1)\\S5=(2,1,1,1)\\S6=(2,1,2,1)\\S7=(2,2,1,1)\\S8=(2,2,2,1)[/tex]
Where S1 is the scenario where we flip coin1 everyday. S2 is the scenario where we flip coin1 the day zero and first day, coin2 the second day, and again coin1 the third day. S3,...,S8 are defined the same.
Observe that the probability to flip coin1 the third day is equal to the sum of [tex]P(S1)+P(S2)+...+P(S8).[/tex] To compute this probabilities we will define:
[tex]P(1,1)=[/tex]Probability to flip coin1 one day given that coin1 was flipped the day before.
[tex]P(1,2)=[/tex]Probability to flip coin2 one day given that coin1 was flipped the day before.
[tex]P(2,1)=[/tex]Probability to flip coin1 one day given that coin2 was flipped the day before.
[tex]P(2,2)=[/tex]Probability to flip coin2 one day given that coin2 was flipped the day before.
Then, using the question information, we can conclude that
[tex]P(1,1)=0.7, P(1,2)=0.3, P(2,1)=0.6, P(2,2)=0.4[/tex]
With this we can compute P(S1),...,P(S8) as follows:
[tex]P(S1)=\frac{1}{2}P(1,1)*P(1,1)*P(1,1)=\frac{1}{2}*(\frac{7}{10})^3=\frac{343}{2000}\\\\P(S2)=\frac{1}{2}P(1,1)*P(1,2)*P(2,1)=\frac{1}{2}*\frac{7}{10}*\frac{3}{10}*\frac{6}{10}=\frac{126}{2000}\\\\P(S3)=\frac{1}{2}P(1,2)*P(2,1)*P(1,1)=\frac{1}{2}*\frac{3}{10}*\frac{6}{10}*\frac{7}{10}=\frac{126}{2000}\\\\P(S4)=\frac{1}{2}P(1,2)*P(2,2)*P(2,1)=\frac{1}{2}*\frac{3}{10}*\frac{4}{10}*\frac{6}{10}=\frac{72}{2000}\\\\[/tex]
[tex]P(S5)=\frac{1}{2}P(2,1)*P(1,1)*P(1,1)=\frac{1}{2}*\frac{6}{10}*\frac{7}{10}*\frac{7}{10}=\frac{294}{2000}\\\\P(S6)=\frac{1}{2}P(2,1)*P(1,2)*P(2,1)=\frac{1}{2}*\frac{6}{10}*\frac{3}{10}*\frac{6}{10}=\frac{108}{2000}\\\\P(S7)=\frac{1}{2}P(2,2)*P(2,1)*P(1,1)=\frac{1}{2}*\frac{4}{10}*\frac{6}{10}*\frac{7}{10}=\frac{168}{2000}\\\\P(S8)=\frac{1}{2}P(2,2)*P(2,2)*P(2,1)=\frac{1}{2}*\frac{4}{10}*\frac{4}{10}*\frac{6}{10}=\frac{96}{2000}\\\\[/tex]
Finally, the probability to flip coin1 the third day is
[tex]P(S1)+...+P(S8)=\frac{1333}{2000}[/tex]
The probability that the coin flipped on the 3rd day after the initial flip is coin 1 is 65%.
To determine the probability that the coin flipped on the 3rd day after the initial flip is coin 1, let's analyze the scenario:
There is a 0.5 chance that we start with either coin 1 or coin 2 on the first day.
If we start with coin 1 and flip heads (with probability 0.7), we will flip coin 1 again on the second day.
If we start with coin 1 and flip tails (with probability 0.3), we will flip coin 2 on the second day.
If we start with coin 2 and flip heads (with probability 0.6), we will flip coin 1 on the second day.
If we start with coin 2 and flip tails (with probability 0.4), we will flip coin 2 on the second day.
To get to coin 1 on the third day, we have the following cases:
Start with coin 1, flip heads (0.5*0.7).
Start with coin 2, flip heads (0.5*0.6).
The total probability of flipping coin 1 on the third day is the sum of the probabilities of these two independent events: (0.5*0.7) + (0.5*0.6) = 0.35 + 0.3 = 0.65 or 65%.
Determine the payment to amortize the debt. (Round your answer to the nearest cent.) Quarterly payments on $19,500 at 3.9% for 6 years.
Answer:
$925.20
Step-by-step explanation:
Loan Amount, P = $19,500
Rate of interest, r = 3.9%
Time, t = 6 years
Payment mode, n = Quarterly (4)
payment to amortize, EMI = ?
Formula: [tex]EMI=\dfrac{P\cdot \frac{r}{n}}{1-(1+\frac{r}{n})^{-n\cdot t}}[/tex]
where,
n = 4 , Rate of interest , r = 0.039
Put the values into formula
[tex]EMI=\dfrac{19500\cdot \frac{0.039}{4}}{1-(1+\frac{0.039}{4})^{-4\cdot 6}}[/tex]
[tex]EMI=915.20[/tex]
Hence, The payment to amortize the debt is $915.20
Find all the square roots of x^2 = 53 (mod 77) by hand. 2 marks
Answer:
[tex]x=\pm\sqrt{77n+53}[/tex]
Step-by-step explanation:
Given : [tex]x^2\equiv 53\mod 77[/tex]
To find : All the square roots ?
Solution :
The primitive roots modulo is defined as
[tex]a\equiv b\mod c[/tex]
Where, a is reminder
b is dividend
c is divisor
Converting equivalent into equal,
[tex]a-b=nc[/tex]
Applying in [tex]x^2\equiv 53\mod 77[/tex],
[tex]x^2\equiv 53\mod 77[/tex]
[tex]x^2-53=77n[/tex]
[tex]x^2=77n+53[/tex]
[tex]x=\pm\sqrt{77n+53}[/tex]
We have to find the possible value in which the x appear to be integer.
The possible value of n is 4.
As [tex]x=\pm\sqrt{77(4)+53}[/tex]
[tex]x=\pm\sqrt{308+53}[/tex]
[tex]x=\pm\sqrt{361}[/tex]
[tex]x=\pm 9[/tex]
Write an equation of an hyperbola whose vertices are
(0,0)and(16,0), and whose foci are (18,0) and (-2,0).
Answer:
[tex]\frac{(x-8)^2}{8^2}-\frac{(y-0)^2}{6^2}=1[/tex]
Step-by-step explanation:
∵ The equation of a hyperbola along x-axis is,
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]
Where,
(h, k) is the center,
a = distance of vertex from the center,
b² = c² - a² ( c = distance of focus from the center ),
Here,
vertices are (0,0) and (16,0), ( i.e. hyperbola is along the x-axis )
So, the center of the hyperbola = midpoint of the vertices (0,0) and (16,0)
[tex]=(\frac{0+16}{2}, \frac{0+0}{2})[/tex]
= (8,0)
Thus, the distance of the vertex from the center, a = 8 unit
Now, foci are (18,0) and (-2,0).
Also, the distance of the focus from the center, c = 18 - 8 = 10 units,
[tex]\implies b^2=10^2-8^2=100-64=36\implies b = 6[/tex]
( Note : b ≠ -6 because distance can not be negative )
Hence, the equation of the required hyperbola would be,
[tex]\frac{(x-8)^2}{8^2}-\frac{(y-0)^2}{6^2}=1[/tex]