Answer:
(a) Profit function P(x) = 0.02x^2+60x-80
(b) Average profit P(x)/x = P/x = 0.02x+60-80/x
Marginal profit dP/dx = 0.04x+60
Step-by-step explanation:
Cost function: C(x) = -0.02x^2+40x+80
Price function: p(x) = 100
(a) The profit function P(x) = x*p(x)-C(x) can be expressed as:
[tex]P=x*p-C\\P=x*100-(-0.02x^{2} +40x+80)\\P=0.02x^{2}+60x-80[/tex]
(b)Average profit function: P(x)/x
[tex]P/x=(0.02x^{2}+60x-80)/x\\P/x = 0.02x+60-80/x[/tex]
Marginal profit function: dP/dx
[tex]P=0.02x^{2}+60x-80\\dP/dx=0.02*2*x+60+0\\dP/dx=0.04x+60[/tex]
Final answer:
The problem involves calculating the profit, average profit per item, and marginal profit for selling x items based on a given cost and price function. By subtracting the cost function from the revenue, we obtain the profit function P(x) = -0.02x² + 60x + 80. The average profit and marginal profit functions further analyze profitability.
Explanation:
To solve the problem given, we need to start by finding the profit function P(x), which is obtained by subtracting the cost function C(x) from the revenue function, where the revenue is the sale price per item times the number of items sold (xp(x)). Given C(x) = -0.02x² + 40x + 80 and p(x) = 100, the profit function can be determined.
Next, the average profit function is found by dividing the profit function by x, and the marginal profit function, dP/dx, is the derivative of the profit function with respect to x, which provides an approximation of the profit gained by selling one more item after x items have been sold.
Profit Function
Substituting p(x) = 100 into P(x) = xp(x) - C(x), we obtain:
P(x) = x(100) - (-0.02x² + 40x + 80)
P(x) = -0.02x² + 60x + 80
Average Profit Function
The average profit per item for x items sold is:
P(x)/x = (-0.02x² + 60x + 80) / x
Given the graph of the function f(x) below what happens to f x when x is a very small negative number?
Answer:
The function tends to negative infinity
Step-by-step explanation:
The Y and X axes in this case are asymptotes, it means that the function will never touch them. When x is negative and is so small, the function tends to negative infinity, because the function try to cut it but it will never happen.
:)
Answer:
F(x) is a negative number with a large absolute value
Step-by-step explanation:
ap ex
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
Determine if each statement is True or False.
1. Suppose an is an arithmetic sequence with d > 0. Then the sum
of the series a1+a2+a3+...a12 must be positive.
2. All infinite arithmetic series diverge.
3. An infinite sum is equal to the limit of the sequence of partial
sums.
4. An infinite geometric series will converge if r < 1.
Answer:
1) False 2) True 3) True 4) True
Step-by-step explanation:
1)FALSE
We can prove this by giving a counterexample,
Take the arithmetic sequence
[tex]\left \{ a_1,a_2,a_3,... \right \}[/tex]
where
[tex]a_n=(n-1)-15[/tex]
in this case d=1
Then
[tex]a_1+a_2+...+a_12=-15-14-...-4<0[/tex]
2)TRUE
Given that for an arithmetic sequence
[tex]a_n=a_1+(n-1)d[/tex]
Where d is a constant other than 0, then
[tex]\lim_{n \to \infty}a_n\neq 0 [/tex]
and so, the series
[tex]\sum_{n=1}^{\infty}a_n[/tex]
diverges.
3)TRUE
This is the definition of infinite sum.
If [tex]S_n=a_1+a_2+...+a_n[/tex]
then [tex]\sum_{n=1}^{\infty}a_n=\lim_{n \to \infty}S_n[/tex]
4)TRUE
If
[tex]\left \{ a_1,a_2,a_3,... \right \}[/tex]
is a geometric sequence, then the n-th partial sum is given by
[tex]S_n=\frac{a_1r^n-a_1}{r-1}[/tex]
Since r<1
[tex]\lim_{n \to\infty}r^n=0[/tex]
and so, the geometric series
[tex]\sum_{n=1}^{\infty}a_n=\lim_{n \to\infty}S_n=\frac{a_1}{1-r}[/tex]
The statements about arithmetic and geometric sequences and series are evaluated for truthfulness, illustrating key concepts of convergence and divergence in series within mathematics.
Explanation:Let's examine each statement regarding sequences and series in mathematics:
Arithmetic sequence with positive common difference resulting in positive sum: True. If an is an arithmetic sequence with a positive common difference d (> 0), then each term is larger than the previous one. Since a1 is the first term, if a1 is non-negative, then all terms in the sequence will be positive, making the sum positive. If a1 is negative, the sum a1+a2+a3+...+a12 could still be negative depending on how negative a1 is in relation to d.All infinite arithmetic series diverge: True. An infinite arithmetic series where the common difference is non-zero will either positively or negatively diverge because the terms do not approach zero.Infinite sum as a limit: True. An infinite sum is indeed equal to the limit of its sequence of partial sums, which is the definition of convergence for an infinite series. If the limit exists and is finite, the series converges; otherwise, it diverges.Convergence of infinite geometric series with r < 1: True. An infinite geometric series will converge if the absolute value of the common ratio r is less than 1, as the successive terms of the series approach zero.A 50-year-old man has a maximum heart rate of 168 bpm (beats per minute) He plans an exercise program so that his heart rate will be between 75 % and 90% of his maximum heart rate for at least 30 minutes a day, 3 days a week Calculate the desired range of heart beats per minute for this exercise program The minimum rate for this exercise program will be (Round to the nearest whole number as needed ) beats per minute The maximum rate will be beats per minute Enter your answer in each of the answer boxes 757 PM 9/2/2019
Multiply his max heart rate by both 75% and 90%.
168 x 0.75 = 126
168 x 0.90 = 151.2, round to 151
The minimum is 126
The maximum is 151
If cos (50 degrees) = a, express tan (130 degrees) in terms of a.
Answer: The tan 130° is expressed as [tex]\dfrac{\sqrt{1-a^2}}{a}[/tex]
Step-by-step explanation:
Since we have given that
[tex]\cos 50^\circ=a[/tex]
As we know that
cos (π - θ ) = -cos θ
so, cos(180-50)=-cos 130° = -a
so, sin 130° would become
[tex]\sqrt{1-(-a)^2}\\\\=\sqrt{1-a^2}[/tex]
So, tan 130° is given by
[tex]\dfrac{\sin 130^\circ}{\cos 130^\circ}\\\\=\dfrac{\sqrt{1-a^2}}{a}[/tex]
Hence, the tan 130° is expressed as [tex]\dfrac{\sqrt{1-a^2}}{a}[/tex]
To express tan(130 degrees) in terms of cos(50 degrees), use the identities tan(180 - x) = -tan(x) and sin²(x) + cos²(x) = 1. tan(130 degrees) = -tan(50 degrees) simplifies to -√(1 - a²) / a. Therefore, tan(130 degrees) in terms of a is -√(1 - a²) / a.
Given that cos(50 degrees) = a, we need to express tan(130 degrees) in terms of a.
First, recall the relationship between the cosine and tangent of angles. We know that:
tan(180 degrees - x) = -tan(x)Therefore, tan(130 degrees) can be written as:
tan(130 degrees) = tan(180 degrees - 50 degrees) = -tan(50 degrees)Next, using the identity for tangent in terms of cosine, we have:
tan(x) = sin(x) / cos(x)Since sin²(x) + cos²(x) = 1, we can express sin(50 degrees) as:
sin(50 degrees) = √(1 - cos²(50 degrees)) = √(1 - a²)Thus,
tan(50 degrees) = sin(50 degrees) / cos(50 degrees) = √(1 - a²) / aFinally, substituting back, we get:
tan(130 degrees) = -√(1 - a²) / aTherefore, tan(130 degrees) in terms of a is -√(1 - a²) / a.
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]
Explain how the symbols for subset and proper subset are related to the symbols < and for ≤ numbers.
provide an example
Answer:
The line under the symbols will have the same effect for the subset and proper subset symbols and the less than and less than or equal symbols.
By concept, a proper subset is the set that will have some but not all of the values of a given set, for example:
Imagine we had the sets:
A={a,e,i,o,u}
B={a,o,u}
C={a,e,i,ou}
we can say that
B⊂A (B is a proper subset of A)
but we cannot say that:
C⊂A
because B has some elements of A, while C has all the elements in A, so C is not a proper subset of A.
Now, a subset can contain some or all of the elements contained in another set.
we can for sure say that:
B⊆A and also that C⊆A
Because C has all the elements of A, so it fits into the subset definition.
Comparing this to the < and ≤ symbols, the < symbol means that a value will be less than another value. This doesn't include the greater value, for
example, we can say that:
2<5
but we cannot say that 5<5 because they are both the same. That statement is false.
On the other hand, the ≤ stands for, less than or equal to. This symbol can be used when a number is less than another one or equal to it, for example, we can say that:
2≤5 and we can also say that 5≤5 because they are the same and the symbol does include the actual value of 5.
So as you may see, the relation is that the line under the symbol includes the values or sets while if the symbols don't have a line under them, this means that the greater value or the original set is not to be included.
The symbols for subset (⊆) and proper subset (⊂) in sets are similar to ≤ and < in numbers, respectively. A subset includes the possibility of equality, while a proper subset does not, analogous to how ≤ and < function with numbers.
Explanation:The symbols for subset (⊆) and proper subset (⊂) in set theory are analogous to the symbols for ≤ (less than or equal to) and < (less than) for numbers. A subset can be thought of as ≤ because it includes the possibility of being equal to the set it is compared with (similar to how 5 ≤ 5 is true). A proper subset is like < because it does not include the set itself; it must be a strict part of the set, excluding equality (as 5 < 6 is true but 5 is not < 5).
Example: Let's consider two sets A = {1, 2, 3} and B = {1, 2, 3, 4}. Here, A is a proper subset of B, which we denote as A ⊂ B, akin to stating A < B if they were numbers since A does not contain all elements of B. However, if A was {1, 2, 3, 4}, then A ⊆ B, similar to A ≤ B in number terms because A contains all elements of B.
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
Estimate the difference. Use benchmarks with decimal parts of 0, 0.25, 0.50, or 0.75.
5.99–2.3
A. 3.25
B. 3.50
C. 3.75
Answer:
C
Step-by-step explanation:
1. 5.99 = 6
2. 2.3 = 2.25
3. 6 - 2.25 = 3.75
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
In Civil War History (June 2009), historian Jane Flaherty researched the condition of the U.S. Treasury on the eve of the Civil War in 1861. Between 1854 and 1857 (under President Franklin Pierce), the annual surplus/deficit was plus18.8, plus6.7, plus5.3, and plus1.3 million dollars, respectively. In contrast, between 1858 and 1861 (under President James Buchanan), the annual surplus/deficit was minus27.3, minus16.2, minus7.2, and minus25.2 million dollars, respectively. Flaherty used these data to aid in portraying the exhausted condition of the U.S. Treasury when Abraham Lincoln took office in 1861. Does this study represent a descriptive or inferential statistical study? Explain.
Statistics are divided into two main branches: the descriptive one that, as the name implies, seeks to describe the data through mathematical processes that allow them to be analyzed, find numerical patterns, representative data, group values and summarize the information; the other branch is the inferential, which seeks to make estimates or predictions, inferences, on a set of data obtained, usually samples, making use of probabilities and distributions.
Answer
The study described is a descriptive study since it performs a series of mathematical processes on a set of data to be able to analyze them, find representative patterns and data, describe their behavior and make a comparison.
Historian Jane Flaherty's research on the U.S. Treasury's state before the Civil War is a descriptive statistical study, summarizing the surplus/deficit data from 1854 to 1861 to depict the financial condition at the time.
Explanation:The study by historian Jane Flaherty that researched the condition of the U.S. Treasury on the eve of the Civil War represents a descriptive statistical study. This type of study involves summarizing and describing aspects of a specific set of information. Flaherty used information about the annual surplus/deficit figures from the periods 1854-1857 under President Franklin Pierce and 1858-1861 under President James Buchanan to illustrate the financial state of the Union prior to Abraham Lincoln's presidency. The data clearly indicate a shift from consistent surpluses during Pierce's term to substantial deficits under Buchanan, which contributed to the exhausted condition of the Treasury as the nation approached the Civil War. Descriptive statistics, like those presented, help to paint a historical picture without making predictive claims or inferences about other data sets or future outcomes.
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%.
Ronald is distributing writing supplies to campers. He must give the same number of supplies to each camper and use all the supplies. He has 14 pencils and 21 note pads. What is the greatest number of campers he can give writing supplies?
Answer:
Robert can give writing supplies to at most 7 campers.
Step-by-step explanation:
The problem states that Robert must give the same number of supplies to each camper and use all the supplies. It means that the greatest number of campers that Robert can give writing supplies is the greatest common divisor(gcd) between the number of pencils and the number of note pads.
The gcd between two integers is the largest positive number that divides each of the integers. We can find this value by prime factorization.
The problem states that he has 14 pencils and 21 note pads. So we have to find gcd(14,21).
21 is not divisible by 2, so we try factoring by 3
14 is not divisible by 3, so we try factoring by 5
None of them are divisible by 5, so we move to 7
Both are divisible by 7, so
14 - 21 | 7
2 - 3
2<7, 3<7, so gcd(14,21) = 7.
Robert can give writing supplies to at most 7 campers.
A survey of 1,168 tourists visiting Orlando was taken. Of those surveyed:
266 tourists had visited LEGOLAND
295 tourists had visited Universal Studios
87 tourists had visited both the Magic Kingdom and LEGOLAND
68 tourists had visited both the Magic Kingdom and Universal Studios
91 tourists had visited both LEGOLAND and Universal Studios
16 tourists had visited all three theme parks
74 tourists did not visit any of these theme parks
How many tourists only visited the Magic Kingdom (of these three)?
Answer:
624 tourists only visited the Magic Kindgom.
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 tourists that visited LEGOLAND
-The set B represents the tourists that visited Universal Studios
-The set C represents the tourists that visited Magic Kingdown.
-The value d is the number of tourists that did not visit any of these parks, so: [tex]d = 74[/tex]
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 tourists that only visited LEGOLAND, [tex]A \cap B[/tex] is the number of tourists that visited both LEGOLAND and Universal Studies, [tex]A \cap C[/tex] is the number of tourists that visited both LEGOLAND and the Magic Kingdom. and [tex]A \cap B \cap C[/tex] is the number of students that visited all these parks.
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 1,168 tourists suveyed. This means that:
[tex]a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1,168[/tex]
We start finding the values from the intersection of three sets.
The problem states that:
16 tourists had visited all three theme parks. So:
[tex]A \cap B \cap C = 16[/tex]
91 tourists had visited both LEGOLAND and Universal Studios. So:
[tex](A \cap B) + (A \cap B \cap C) = 91[/tex]
[tex](A \cap B) = 91-16[/tex]
[tex](A \cap B) = 75[/tex]
68 tourists had visited both the Magic Kingdom and Universal Studios. So
[tex](B \cap C) + (A \cap B \cap C) = 68[/tex]
[tex](B \cap C) = 68-16[/tex]
[tex](B \cap C) = 52[/tex]
87 tourists had visited both the Magic Kingdom and LEGOLAND
[tex](A \cap C) + (A \cap B \cap C) = 87[/tex]
[tex](A \cap C) = 87-16[/tex]
[tex](A \cap C) = 71[/tex]
295 tourists had visited Universal Studios
[tex]B = 295[/tex]
[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]
[tex]295 = b + 52 + 75 + 16[/tex]
[tex]b + 143 = 295[/tex]
[tex]b = 152[/tex]
266 tourists had visited LEGOLAND
[tex]A = 266[/tex]
[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]
[tex]266 = a + 75 + 71 + 16[/tex]
[tex]a + 162 = 266[/tex]
[tex]a = 104[/tex]
How many tourists only visited the Magic Kingdom (of these three)?
This is the value of c, the we can find in the following equation:
[tex]a + b + c + d + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1,168[/tex]
[tex]104 + 152 + c + 74 + 75 + 71 + 52 + 16 = 1,168[/tex]
[tex]c + 544 = 1,168[/tex]
[tex]c = 624[/tex]
624 tourists only visited the Magic Kindgom.
Help! logarithmic differentiation
Recall that we can write
[tex]f(x)^{g(x)}=e^{\ln f(x)^{g(x)}}=e^{g(x)\ln f(x)}[/tex]
so that when we take the derivative, we get by the chain rule
[tex]\left(f(x)^{g(x)}\right)'=(g(x)\ln f(x))' e^{g(x)\ln f(x)}=(g(x)\ln f(x))'f(x)^{g(x)}[/tex]
Here, we have [tex]f(x)=3-\sin2x[/tex] and [tex]g(x)=\sqrt[3]{x}=x^{1/3}[/tex]. By the product rule,
[tex]\left(x^{1/3}\ln(3-\sin2x)\right)'=\left(x^{1/3}\right)\ln(3-\sin2x)+x^{1/3}(\ln(3-\sin2x))'[/tex]
[tex]=\dfrac13x^{-2/3}\ln(3-\sin2x)+x^{1/3}\dfrac{(3-\sin2x)'}{3-\sin2x}[/tex]
[tex]=\dfrac13x^{-2/3}\ln(3-\sin2x)+x^{1/3}\dfrac{2\cos2x}{\sin2x-3}[/tex]
[tex]=\dfrac1{3x^{2/3}}\left(\ln(3-\sin2x)+3x\dfrac{2\cos2x}{\sin2x-3}\right)[/tex]
[tex]=\dfrac1{3\sqrt[3]{x^2}}\left(\ln(3-\sin2x)+\dfrac{6x\cos2x}{\sin2x-3}\right)[/tex]
So we have
[tex]y'=\dfrac{(3-\sin2x)^{\sqrt[3]{x}}}{3\sqrt[3]{x^2}}\left(\ln(3-\sin2x)+\dfrac{6x\cos2x}{\sin2x-3}\right)[/tex]
By means of logarithmic differentiation, the derivative of function [tex]y = (3 - \sin 2x)^{\sqrt[3]{x}}[/tex] is equal to [tex]y' = \left[\frac{\ln (3 - \sin 2x)}{\sqrt[3]{x^2} } + \frac{2\cdot \sqrt[3]{x}\cdot \cos 2x}{3 - \sin 2x} \right]\cdot (3 - \sin 2x)^{\sqrt[3]{x} }[/tex].
How to use logarithmic differentiation in a function containing trascendent functions
In this problem we find the case of a function that cannot be differentiated easily by direct differentiation, due to presence of trascendent function, we shall use an alternative approach known as logarithmic differentiation. First, write the entire expression described in the statement:
[tex]y = (3 - \sin 2x)^{\sqrt[3]{x}}[/tex]
Second, use logarithms and its properties to modify the expression:
[tex]\ln y = \ln (3 - \sin 2x)^{\sqrt[3]{x}}[/tex]
[tex]\ln y = \sqrt[3]{x}\cdot \ln (3 - \sin 2x)[/tex]
Third, use derivative rules and clear variable y':
[tex]\frac{y'}{y} = \frac{\ln (3 - \sin 2x)}{\sqrt[3]{x^2} } + \frac{2 \cdot \sqrt[3]{x} \cdot \cos 2x}{3 - \sin 2x}[/tex]
[tex]y' = \left[\frac{\ln (3 - \sin 2x)}{\sqrt[3]{x^2} } + \frac{2\cdot \sqrt[3]{x}\cdot \cos 2x}{3 - \sin 2x} \right]\cdot y[/tex]
Fourth, substitute on y and write the resulting expression:
[tex]y' = \left[\frac{\ln (3 - \sin 2x)}{\sqrt[3]{x^2} } + \frac{2\cdot \sqrt[3]{x}\cdot \cos 2x}{3 - \sin 2x} \right]\cdot (3 - \sin 2x)^{\sqrt[3]{x} }[/tex]
Thus, the derivative of function [tex]y = (3 - \sin 2x)^{\sqrt[3]{x}}[/tex] is equal to [tex]y' = \left[\frac{\ln (3 - \sin 2x)}{\sqrt[3]{x^2} } + \frac{2\cdot \sqrt[3]{x}\cdot \cos 2x}{3 - \sin 2x} \right]\cdot (3 - \sin 2x)^{\sqrt[3]{x} }[/tex].
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 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
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]
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.
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
help me solve this ok..
3x+3x-5= 1
6x-5=1
Whenever moving a number, the sign always changes.
6x-5+5= 1+5
6x= 1+5
6x= 6
divide both sides by 6
6x/6= 6/6
x= 1
Check solution by using the substitution method
3(1)+3(1)-5=1
3+3-5=1
6-5= 1
1=1
Answer: x=1
Answer is provided in the image attached.
How much interest will you have to pay for a credit card balance of $1,034 that is 1 month overdue, if a 17% annual rate is charged? You will have to pay $1224.43| in interest. (Round to two decimal places.)
Answer:
Ans. Since the annual rate is not compunded (for example, compounded monthly) you will have to pay in interest $27.41, and the total payment is $1,061.41
Step-by-step explanation:
Hi, since the balance is 1 month overdue, it means that you owe 2 months of interest to this obligation, but before we start finding the interest of your credit card, first let´s find the effective monthly equivalent rate for that 17% annual interest rate.
The formula is as follows.
[tex]r(monthly)=(1+r(annual))^{\frac{1}{12} }-1[/tex]
Therefore
[tex]r(monthly)=(1+0.17)^{\frac{1}{12} }-1 =0.01317[/tex]
So your monthly interest rate is 1.317%. Now let´s find the amount of interests that you have to pay for 2 months. This is the formula.
[tex]Interest=Present Value(1+r(monthly))^{n} )-PresentValue[/tex]
Where "n" is the period of time in months that you owe to the financial institution. The result of that is:
[tex]Interest=1,034(1+0.01317)^{2} -1,034=24.41[/tex]
This way, interest are = $27.41 and the total amount that you will have to pay is:
[tex]Payment=Present Value+Interest=1,034+27.41=1,061.41[/tex]
Best of luck.
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.
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
Show your work:
Express 160 pounds (lbs) in kilograms (kg). Round to the nearest hundredths.
Final answer:
To express 160 pounds (lbs) in kilograms (kg), multiply 160 pounds by the conversion factor of 1 kilogram to 2.205 pounds. The result is approximately 72.6 kilograms.
Explanation:
To express 160 pounds (lbs) in kilograms (kg), we can use the conversion factor that 1 kg is equal to 2.205 pounds (lb). We can set up a proportion to solve for the weight in kilograms:
160 lb = x kg
1 kg = 2.205 lb
By multiplying both sides of the equation by 1 kg, we get:
160 lb * 1 kg / 2.205 lb = x kg
Simplifying the expression, we find that x is approximately 72.6 kg when rounded to the nearest hundredths.
The result is approximately 72.56 kg when rounded to the nearest hundredths place. To convert 160 pounds to kilograms, you divide by 2.205.
To convert pounds (lbs) to kilograms (kg), you can use the conversion factor where 1 kg is approximately equal to 2.205 lbs.
Given:
Weight in pounds (lbs): 160 lbsWe use the formula:[tex]\[ \text{Weight (kg)} = \frac{\text{Weight (lbs)}}{2.205} \][/tex]Substituting the given value:
[tex]\[ \text{Weight (kg)} = \frac{160 \text{ lbs}}{2.205} \approx 72.5624 \text{ kg} \][/tex]Rounding to the nearest hundredths place, we get:
Weight in kilograms: 72.56 kgThus, 160 pounds is approximately equal to 72.56 kilograms.
Among a random sample of 500 college students, the mean number of hours worked per week at non-college-related jobs is 14.6. This mean lies 0.4 standard deviations below the mean of the sampling distribution. If a second sample of 500 students is selected, what is the probability that for the second sample, the mean number of hours worked will be less than 14.6?
Answer:
The probability that for the second sample of 500 college students, the mean number of hours worked will be less than 14.6 is 0.6554
Step-by-step explanation:
The sampling distribution of the sample mean is given by a normal distribution with mean [tex]\mu[/tex] and variance [tex]\frac{\sigma^2}{n}[/tex], where [tex]\mu[/tex] is the mean and [tex]\sigma^2[/tex] is the variance of the population that generates the data. In this way the random variable;
[tex]Z=\frac{\bar x - \mu_{\bar x}}{\sigma_{\bar x}}[/tex] is a standard normal variable. As [tex]\bar {x}-\mu_{\bar x} = 0.4\sigma_{\bar x}[/tex], then [tex]Z = 0.4[/tex].
[tex]P (X <14.6) = P (Z <0.4) = 0.6554[/tex]
The probability that the mean number of hours worked per week in the second sample of 500 students will be less than 14.6 hours is approximately [tex]\( {0.345} \).[/tex]
To solve this problem, we need to understand the relationship between the sample mean, the population mean, and the standard deviation of the sampling distribution. Here’s the breakdown of the steps needed:
1. Determine the z-score for the sample mean in the first sample:
The mean number of hours worked per week in the first sample is 14.6, which lies 0.4 standard deviations below the mean of the sampling distribution. This means the z-score is -0.4.
2. Find the corresponding probability:
The z-score tells us how many standard deviations away from the mean our sample mean is. We need to find the probability that a second sample will have a mean number of hours worked that is less than 14.6.
3. Use the standard normal distribution:
The z-score formula for the sampling distribution is given by:
[tex]\[ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \][/tex]
where [tex]\(\bar{x}\)[/tex] is the sample mean, [tex]\(\mu\)[/tex] is the population mean, and [tex]\(\sigma_{\bar{x}}\)[/tex] is the standard error of the mean.
In this case, we know that the mean of the sampling distribution [tex](\(\mu_{\bar{x}}\))[/tex] is such that:
[tex]\[ z = \frac{14.6 - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = -0.4 \][/tex]
4. Determine the probability:
To find the probability that the mean number of hours worked in the second sample is less than 14.6, we look up the z-score of -0.4 in the standard normal distribution table or use a cumulative distribution function (CDF) for the normal distribution.
The z-score of -0.4 corresponds to a cumulative probability (or area to the left of z) of approximately 0.3446.
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.
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 S = sin(x) + sin(x + α) + sin(x + 2α) + ... + sin(x + nα), n ∈N. What is the value of S?
Answer:
[tex]S=\frac{cos(x-\frac{\alpha}{2})-cos(x+n\alpha-\frac{\alpha}{2})}{2sin\frac{\alpha}{2}}[/tex]
Step-by-step explanation:
We are given that [tex]S=sin(x) +sin(x+\alpha)+sin(x+2\alpha)+....+sin(x+n\alpha),n\in N[/tex]
We have to find the value of S
We know that
[tex]\sum_{k=0}^{n-1}sin(x+k.d)=\frac{sinn\times \frac{d}{2}}{sin\frac{d}{2}}\times sin(\frac{2x+(n-1)d}{2})[/tex]
We have d=[tex]\alpha[/tex]
Substitute the values then we get
[tex]\sum_{k=0}^{n-1}sin(x+k.\alpha)=\frac{sin\frac{n\alpha}{2}}{sin\frac{\alpha}{2}}\times sin(\frac{2x+(n-1)\alpha}{2})[/tex]
[tex]\sum_{k=0}^{n-1}sin(x+k.\alpha)=\frac{sin\frac{n\alpha}{2}\cdot sin(\frac{2x+(n-1)\alpha}{2})}{sin\frac{\alpha}{2}}[/tex]
[tex]S=\frac{sin\frac{n\alpha}{2}\cdot sin(\frac{2x+(n-1)\alpha}{2})}{sin\frac{\alpha}{2}}[/tex]
[tex]S=\frac{2sin\frac{n\alpha}{2}\cdot sin(\frac{2x+(n-1)\alpha}{2})}{2sin\frac{\alpha}{2}}[/tex]
[tex]S=\frac{cos(x+\frac{n\alpha}{2}-\frac{\alpha}{2}-\frac{n\alpha}{2})-cos(x+\frac{n\alpha}{2}-\frac{\alpha}{2}+\frac{n\alpha}{2})}{2sin\frac{\alpha}{2}}[/tex]
Because [tex]cos(x-y)-cos(x+y)=2 sinxsiny[/tex]
[tex]S=\frac{cos(x-\frac{\alpha}{2})-cos(x+n\alpha-\frac{\alpha}{2})}{2sin\frac{\alpha}{2}}[/tex]
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].