The way to transform the function such that the bond goes from quarterly compounding interest to annually compounding interest is = 1 + ( 1 + rate / 4 )ⁿ ˣ ⁴.
How to find the value ?The bond is said to be compounded quarterly. There would therefore be four compounding periods in a single year because there are four quarters in a year.
The formula would have been:
= Initial investment + ( 1 + rate ) ^ number of years
The adjusted formula to account for the quarters becomes :
= 1 + ( 1 + rate / 4 )ⁿ ˣ ⁴
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Describe a compound event that has a probability between 50% and 80%.
A compound event can be described for flipping a coin two times and getting at least 1 head.
What is Compound Event in Probability?Compound event is defined as the events in which there are more than one event happens together.
All the outcomes of the events sum up the probability to 1.
We have to describe a compound event where the probability is in between 50% and 80%.
That is, probability is in between 0.5 and 0.8.
That is, there is more than half of the chance to occur the event.
Suppose that we flip a coin two times.
Sample space = {HH, HT, TH, TT}
Find the probability of getting at least 1 head.
There are 3 outcomes out of 4 of getting at least 1 head.
Probability = 3/4 = 75%
Hence there is a 75% probability for getting at least 1 head when flipped a coin two times.
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Describe a situation that can be represented by using this system of equations*. Determine and interpret the system within the context of your situation. *y= 10x+20 and y=15x+5
hurryyyy
2.5 hours spent in each car wash would cost $40
How to describe a situationFrom the question, we have the following parameters that can be used in our computation:
y = 10x + 20
y = 15x + 5
A situation for the system of equations is the total charges of different car wash per hour
So, the solution would be
The number of hours spent in each car wash for the total amount to be the same
From the graph, the solution is (2.5, 40)
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2. this data came from a pre-employment drug screening. positive test result negative test result (drug use is indicated) (drug use is not indicated) subject uses drugs 44 6 (true positive) (false negative) subject is not a drug user 90 860 (false positive) (true negative) (a) false positive find the probability of selecting a subject with a positive test result given that the subject does not use drugs. (b) false negative find the probability of selecting a subject with a negative test result given that the subject uses drugs.
a) The probability of selecting a subject with a positive test result given that the subject does not use drugs is equals to the 0.09473.
b) The probability of selecting a subject with a negative test result given that the subject uses drugs is equals to the 0.12.
The confusion matrix of the pre employment drug screening data is present above. It contains positive test result, Negative test result and Total.
Subject uses drugs 44 6 = 50
Subject does not use drugs 90 860 = 950
Similarly , 90 + 44 = 134 , 860 + 6 = 886 and 950 + 50 = 1000.
a) We have to determine the probability of selecting a subject with a positive test result that the subject does not use drug. Let A be the event that the subject with a positive test result and B be the event that subject does not use drug. We have to determine P(A/B). From conditional probability, P(A/B) = P(A and B)/P(B)
= 90/950 = 0.09473
Hence, the probability of selecting a subject with a positive test result given that the subject does not use drug is 0.09473.
b) We have to calculate the probability of selecting a subject with a negative test result that the subject use drugs. Let R be the event that the subject with a negative test result and S be the event that subject uses drug. We have to calculate the value of P(R/S). By using Bayes theorem, P(R/S) = P(R and S) /P(S)
= 6/50 = 0.12
Hence, the probability of selecting a subject with a negative test result given that the subject uses drug is 0.12.
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Complete question:
The above table complete the question.
2. this data came from a pre-employment drug screening. positive test result negative test result (drug use is indicated) (drug use is not indicated) subject uses drugs 44 6 (true positive) (false negative) subject is not a drug user 90 860 (false positive) (true negative) (a) false positive find the probability of selecting a subject with a positive test result given that the subject does not use drugs. (b) false negative find the probability of selecting a subject with a negative test result given that the subject uses drugs.
I'm not dead lol
I'm still here
Answer:
29
Step-by-step explanation:
trust
Answer: okay good!!!!!!
Using relational algebraa. list the flights that cost more than 800 - report their ID, airport code, destination code, and fareb. Report the airports city, code, and departure time for thr airports that have departing flights in the morningc. List the names and hometown of the passengers that do not have an airport in their hometownd. What airlines fly from Toronto, report the airline namee. what aurlines do not fly from toronto, report the airline namef. what airlines fly from toronto to vancouver? report the airline nameg. list the passangers flying to vancouver, report their name, origin, and destination airport codes, and arrival time
Using relational algebra:
a. flight ⋈ airport ⋈ ρ destination airport
b. flight ⋈ airport
c. passenger ⋈ airport
d. airline ⋈ airport ⋈ flight
e. airline ⋈ airport ⋈ flight
f. airline ⋈ airport ⋈ flight ⋈ ρ destination airport
g. flight ⋈ passenger ⋈ booking
QUESTION (a):
- using EQUI join (⋈) join relations "Flight, Airport and Airport as destination"
- and using select Operation (σ), relational operators (=, >) and connector (and) select tuples from the joined relations
- Then using Project Operation (∏) projects columns "Flight.fID, Airport.code, destination.code, Flight.fare" from the relation
- Below shows the SQL query, relation algebra query and relation algebra tree
RELATIONAL ALGEBRAπ flight . fid, airport . code, destination . code , flight.fare σ flight . airport = airport . code
AND flight . destination = destination . code (flight ⋈ airport ⋈ ρ destination airport)
QUESTION (b):
- using EQUI join (⋈) join relations "Flight, Airport"
- and using select Operation (σ), relational operators (=, >=, <=) and connector (and) select tuples from the joined relations
- Then using Project Operation (∏) projects columns "airport . city, airport . code, flight . departure" from the relation
- Below shows the SQL query, relation algebra query and relation algebra tree
RELATIONAL ALGEBRAπ airport . city, airport . code, flight . departure σ flight . airport = airport . code
AND (4 <= flight . departure AND flight . departure <= 10) (flight ⋈ airport)
QUESTION (c):
- using EQUI join (⋈) join relations "passenger, Airport"
- and using select Operation (σ), relational operators (<>) and connector (and) select tuples from the joined relations
- Then using Project Operation (∏) projects columns "passenger . name, passenger . hometown" from the relation
- Below shows the SQL query, relation algebra query and relation algebra tree
RELATIONAL ALGEBRAπ passenger . name, passenger . hometown σ passenger . hometown <> airport . city (passenger ⋈ airport)
QUESTION (d):
- using EQUI join (⋈) join relations "airline, airport, flight"
- and using select Operation (σ), relational operators (=) and connector (and) select tuples from the joined relations
- Then using Project Operation (∏) projects columns "airline . name" from the relation
- Below shows the SQL query, relation algebra query and relation algebra tree
RELATIONAL ALGEBRAπ airline . name σ airline . aid = flight . airline
AND flight . airport = airport . code
AND airport . city = "Toronto" (airline ⋈ airport ⋈ flight)
QUESTION (e):
- using EQUI join (⋈) join relations "airline, airport, flight"
- and using select Operation (σ), relational operators (<>, =) and connector (and) select tuples from the joined relations
- Then using Project Operation (∏) projects columns "airline . name" from the relation
- Below shows the SQL query, relation algebra query and relation algebra tree
RELATIONAL ALGEBRAπ airline . name σ airline . aid = flight . airline
AND flight . airport = airport . code
AND airport . city <> "Toronto" (airline ⋈ airport ⋈ flight)
QUESTION (f):
- using EQUI join (⋈) join relations "airline, airport, flight and airport as destination"
- and using select Operation (σ), relational operators (<>, =) and connector (and) select tuples from the joined relations
- Then using Project Operation (∏) projects columns "airline . name" from the relation
- Below shows the SQL query, relation algebra query and relation algebra tree
RELATIONAL ALGEBRAπ airline . name σ airline . aid = flight . airline
AND flight . airport = airport . code
AND flight . destination = destination . code
AND airport . city = "Toronto"
AND destination . city = "vancouver" (airline ⋈ airport ⋈ flight ⋈ ρ destination airport)
QUESTION (g):
- using EQUI join (⋈) join relations "flight, passenger, Booking"
- and using select Operation (σ), relational operators (=) and connector (and) select tuples from the joined relations
- Then using Project Operation (∏) projects columns "passenger . name, airport, destination" from the relation
- Below shows the SQL query, relation algebra query and relation algebra tree
RELATIONAL ALGEBRAπ passenger . name, airport, destination σ passenger . pid = booking . pid
AND booking . fid = flight . fid
AND destination . city = "vancouver" (flight ⋈ passenger ⋈ booking)
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Question correction:
See on the attached image.
a spam filer is designed by screening commonly occuring phrases in spam suppose that 60 of email is spam in 20 o the spam emails the phprase fee gift card is used
The probability that a newly arrived email which does mention "free gift card", is not spam is 0.0165.
Let S be the event that an email is spam, and let F be the event that an email mentions "free gift card". We want to find P(S' | F), the probability that the email is not spam given that it mentions "free gift card".
We can use Bayes' theorem to find P(S' | F):
P(S' | F) = P(F | S') P(S') / P(F)
We can calculate each of these probabilities as follows:
P(F | S') = the probability that a non-spam email mentions "free gift card". From the problem, we know that this probability is 0.005 (i.e., 99.5% of non-spam emails do not mention "free gift card").
P(S') = the probability that an email is not spam. From the problem, we know that this probability is 0.4 (i.e., 60% of emails are spam, so 40% are not spam).
P(F) = the probability that an email mentions "free gift card". This can be calculated using the law of total probability:
P(F) = P(F | S) P(S) + P(F | S') P(S')
= 0.20 × 0.60 + 0.005 × 0.40
= 0.121
In the first term, we use the given probability that 20% of spam emails mention "free gift card". In the second term, we use the probability that 0.5% of non-spam emails mention "free gift card".
Plugging these values into Bayes' theorem, we get:
P(S' | F) = 0.005 × 0.4 / 0.121 ≈ 0.0165
Therefore, the probability that a newly arrived email which does mention "free gift card" is not spam is approximately 0.0165 or 1.65%.
Correct Question :
A spam filer is designed by screening commonly occurring phrases in spam. Suppose that 60% of email is spam. In 20% of the spam emails, the phrase “free gift card” is used. In non-spam emails, 99.5% of them do not mention “free gift card”. What is the probability that a newly arrived email which does mention “free gift card”, is not spam?
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In a certain city, E Street, W Street, C Street, and D
Street are parallel streets that intersect K Street and
M Street. How long is K Street between C Street
and D Street?
600 ft
800 ft
540 ft
E St
W St
M St
C St
D St
1400 ft
K St
The length which lies from K Street between C Street and D Street is 337.5 ft.
Given that,
E Street, W Street, C Street, and D Street are parallel streets that cross K Street and M Street in a particular city. It is unknown how far K Street is from C Street and D Street.
Let the distance of street K between C and D be x,
Now,
Taking the equality of the proportionality expression of triangles,
600 / 400 = 600 + x / 400 + 250
6 / 4 = 600 + x / 625
3750/4 = 600 + x
937.5 = 600 + x
x = 337.5 ft
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(1 point) the temperature, h, in degrees celsius, of a cup of coffee placed on the kitchen counter is given by h
For the function for the temperature H = f(t) , for cup of coffee placed on the Kitchen counter , then the derivative f'(t) will be Negative .
The Temperature Function in degree Celsius for a cup of coffee placed on the kitchen counter will be negative because if cup of coffee is placed on the counter there is a loss of heat from cup of coffee to surroundings as temperature increases .
If the cup of coffee is losing heat to the surroundings, the temperature of the coffee will decrease over time.
The derivative of the temperature function, f'(t), will be negative. it means that as time increases, the rate of change of the temperature will be negative, indicating a decrease in temperature.
Therefore , the derivative function f'(t) is negative .
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The given question is incomplete , the complete question is
The temperature, H , in degrees Celsius, of a cup of coffee placed on the kitchen counter is given by H = f(t) , where t is in minutes since the coffee was put on the counter .
Is f'(t) positive or negative ?
let be the solution to satisfying . (a) use euler's method with time step to approximate . -3 5.03421 5.03942 5.04269 5.04269 0.2(8e^(-5.04269)) (b) use separation of variables to find exactly.
Answer:
Step-by-step explanation:
c
Michael has $16 and wants to buy a mixture of cupcakes and fudge to feed at least 4 siblings. Each cupcake costs $4, and each piece of fudge costs $2. This system of equations models the scenario: 4x + 2y ≤ 16 x + y ≥ 4 Part A: Describe the graph of the system, including shading and the types of lines graphed. Provide a description of the solution set. (4 points) Part B: Is the point (2, 3) included in the solution area for the system? Justify your answer mathematically. (3 points) Part C: Choose a different point in the solution set and interpret what it means in terms of the real-world context. (3 points)
Part A: the system of inequalities has a definite solution to it.
Part B: Yes the point is a part of Solution Area
What is Linear Inequality?
In mathematics a linear inequality is an inequality which involves a linear function. A linear inequality contains one of the symbols of inequality. It shows the data which is not equal in graph form.
Solution:
The inequalities 4x + 2y ≤ 16
And x + y ≥ 4 is attached below please refer to it.
Part A: The graph of both inequalities intersects each other therefore the system of inequalities has a definite solution to it.
Part B: Yes point (2, 3) is included in the solution area, the following is the mathematical proof.
Using substitution 4*2 + 2*3 ≤ 16
8+6 ≤ 16
14 ≤ 16
And 2+3 ≥ 4
5 ≥ 4
Hence, the point is included in the solution area.
Part C:
Let’s assume a point 1,1 in the solution set so it implies that only the inequality 4x + 2y ≤ 16 is satisfied. Thus, not giving a feasible solution.
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7x - 31 + 4y + 27 + 5x - 8 + 63
The solution is, 7x - 31 + 4y + 27 + 5x - 8 + 63 = 12x+4y+51.
What is simplification?Simplify simply means to make it simple. In mathematics, simply or simplification is reducing the expression/fraction/problem in a simpler form. It makes the problem easy with calculations and solving.
here, we have,
now, we have to solve the given expression, using addition,
given that,
7x - 31 + 4y + 27 + 5x - 8 + 63
=12x+4y+51
here, we add the terms with x with each other, then add the terms with y, and, add the numbers & subtract them from given terms with minus sign.
Hence, The solution is, 7x - 31 + 4y + 27 + 5x - 8 + 63 = 12x+4y+51.
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derive the moment generating function of a random variable following a negative binomial distribution. recall that a negative binomial random variable is a sum of independent geometric random variables
The moment generating function of a negative binomial distribution with parameters r and p is[tex]M_X(t) = [(1-p)/(1-pe^t)]^r.[/tex]
In probability theory, the negative binomial distribution is a discrete probability distribution that describes the number of successes in a sequence of independent and identically distributed Bernoulli trials before a specified (non-random) number of failures (denoted r) occurs.
The moment generating function (MGF) of a random variable X is defined as
[tex]M_X(t) = E(e^{tX})[/tex]
To derive the MGF of a negative binomial distribution, we can use the fact that a negative binomial random variable can be expressed as a sum of independent geometric random variables.
Let X ~ NB(r, p) be a negative binomial random variable, where r is the number of failures and p is the probability of success in each trial. Let Y ~ Geo(p) be a geometric random variable representing the number of failures before the first success.
We know that the MGF of a geometric distribution is
[tex]M_Y(t) = E(e^{tY}) = (1-p)/(1-pe^t).[/tex]
Using the fact that a negative binomial random variable is a sum of r independent geometric random variables, we can derive the MGF of X as follows:
[tex]M_X(t) = E(e^{tX}) = E(e^{tY_1 + tY_2 + ... + tY_r})\\\\= E(e^{tY_1} * e^{tY_2} * ... * e^{tY_r})[/tex]
(since the Y's are independent)
[tex]= E(e^{tY_1} * e^{tY_2} * ... * e^{tY_r})[/tex]
(since the Y's have the same distribution)
[tex]= [M_Y(t)]^r\\\\= [(1-p)/(1-pe^t)]^r[/tex]
Therefore, the moment generating function of a negative binomial distribution with parameters r and p is[tex]M_X(t) = [(1-p)/(1-pe^t)]^r.[/tex] This function can be used to derive moments and other statistical properties of the negative binomial distribution.
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7. suppose a student carrying a flu virus returns to an isolated college campus of 1000 students. determine a differential equation for the number of people x(t) who have contracted the flu if the rate at which the disease spreads is proportional to the number of interactions between the number of students who have the flu and the number of students who have not yet been exposed to it. (Use k > 0 for the constant of proportionality and x for x(t).)
Answer: If the rate at which the flu spreads is proportional to the number of interactions between students who have the flu and students who have not been exposed to it, we can write:
dx/dt = kx(1000 - x)
where x is the number of students who have contracted the flu, k is the constant of proportionality, and t is time. The term x(1000 - x) represents the number of interactions between the students who have the flu and the students who have not been exposed to it. The factor of 1000 - x represents the number of students who have not yet been exposed to the flu.
Thus, the differential equation models the spread of the flu on the isolated college campus as the number of people who have contracted the flu changes over time due to the number of interactions between the students who have the flu and the students who have not been exposed to it.
Step-by-step explanation:
17 Answer:
Decide how many solutions this equation has:
x2 - 2x + 1 = 0
18 Answer:
Decide how many solutions this equation has:
x2 + 3 = 0
19 Answer:
The revenue from selling x units of a product is given by
y = -0.0002x2 + 20x. How many units must be sold in
order to have the greatest revenue? (Find the x-coordinate
of the vertex of the parabola.)
If
f(x) = x + 2, what is ƒ(4)?
Answer:6
Step-by-step explanation:
ƒ(4)=4+2=6
Which of the following is NOT a requirement of the Permutations Rule, nPr=n!/(n-r)!, for items that are all different?
a) Order is taken into account (rearrangements of the same items are considered to be different).
b) Exactly r of the n items are selected (without replacement).
c) There are n different items available.
d) Order is not taken into account (rearrangements of the same items are considered to be the same).
The option which is not a requirement for the permutation rule is d) Order is not taken into account (rearrangements of the same items are considered to be the same).
What is permutation?Permutations is the number of arrangements of r objects taken out of n objects.
What are the requirements for permutations?The requirements for permutation include the folowing
a) Order is taken into account (rearrangements of the same items are considered to be different).b) Exactly r of the n items are selected (without replacement).c) There are n different items available.Since these are the requirements for a permutation, the options a, b and c are requirements for a permutation while option d is not.
So, the option which is not a requirement is d) Order is not taken into account (rearrangements of the same items are considered to be the same).
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f the mean is greater than the median in a distribution which of following statements are true? (choose one or more) group of answer choices a. the distribution is positively skewed b. the distribution is negatively skewed c. the right tail is longer than the left d. the left tail is longer than the right
When the mean is higher than the median, the distribution has a right-skewed pattern. Thus the statement 'a. the distribution is positively skewed' is the accurate one.
This indicates that the majority of the data are centered on the left and that the right is covered by a lengthy tail. The mean exceeds the median. In a positively skewed distribution, the mean overestimates the values that are most prevalent. The median is larger than the mean. In a distribution that is positively skewed, the mean—the average of all the values—is higher than the median since the data tends to be lower. The center value of the data is called the median, in comparison. As a result, if the data is more skewed towards the negative, the average will be higher than the middle number.
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If y varies directly as x, and y=4 find y when x=6 find y when x=5
Hence, The value of y = 12 when x = -6.
Given that , Y varies directly with x .
If y=−4 when x=2,
Find y when x=−6.
According to the question,
This is a direct variation , y = kx
If,
y = −4 when x=2,
y = kx
-4 = 2k
k = -2
To find ;
y when x = −6
y = kx
y = (-2) (-6)
y = 12
Which of the following is true of the location of the terminal side of an angle 0 who's sine value is 1/2?- has a reference angle of 30° and is in quadrant one or two- as a reference angle 30° and is in quadrant one or four- has a reference angle of 60° and is in quadrant one or two- as a reference angle of 60° in is in quadrant one or four
Out of the following the statement has a reference angle of 30° and is in quadrant one or two- is true of the location of the terminal side of an angle 0 whose sine value is 1/2.
Since the reason behind this is that the sine value of 1/2 can be written as sin(30°), this means that the terminal side of the angle is located at 30°, which is either quadrant one or two. Quadrant one is the upper right quadrant of the coordinate plane, and quadrant two is the upper left quadrant.
The reference angle of an angle is the angle between the terminal side of the angle and the positive x-axis. In this case, the reference angle would be 30°. The reference angle of an angle is important because it is used to determine the quadrant in which the terminal side of the angle is used to determine the quadrant in which the terminal side of the angle is located. The reference angle must be less than 90° in order for the terminal side to be located in the first and second quadrants.
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James baked some cupcakes. If he gave 3 cupcakes to each cousin, he would have 25 cupcakes left. If he gave 15 cupcakes to each cousin, he would need to bake 71 more cupcakes. (a) How many cousins did he have? (b) How many cupcakes did he bake?
Answer:
8 cousins49 cupcakesStep-by-step explanation:
If James gave each cousin 3 cupcakes, there would be 25 left. James is 71 cupcakes short of being able to give each cousin 15 cupcakes. You want to know the number of cousins and the number of cupcakes.
SetupThe given relations between the number of cupcakes (n) and the number of cousins (c) are ...
n -3c = 25 . . . . . 3 each ⇒ 25 left
n -15c = -71 . . . . . 15 each ⇒ 71 short
SolutionSubtracting the second equation from the first, we have ...
(n -3c) -(n -15c) = (25) -(-71)
12c = 96
c = 8 . . . . . . divide by 12
n = 3c +25 = 24 +25 = 49
(a) James has 8 cousins.
(b) James baked 49 cupcakes.
For each of the following, determine the constant c so that f(x) satisfies the conditions of being a pmf for a random variable X, and then depict each pmf as a line graph: (a) f(x) =x/c, x=1.2.3.4. (b) f(x)-cx, x = 1, 2, 3, . . . , 10. (d) f(x) = c(x + 1)2, x=0.1,2.3. (e) f(x) = x/c, x = 1,2,3, . . . ,n. (f) f(x) = x=0.1.2.3, . . . . (x+1)(x+2), HINT: In part (f), write f(x) 1/(x +1)-1/(x+2).
The problem involves finding a constant c that makes each given function a valid probability mass function given are f(x) = x/c, x=1,2,3,4; f(x) = cx, x=1,2,3,...,10; f(x) = c(x+1)², x=0,1,2,3; f(x) = x/c, x=1,2,3,...,n; f(x) = (x+1)(x+2)/[x(x+3)], x=0,1,2,3,...
(a) We know that the sum of all the probabilities in a pmf must be equal to 1.
Therefore, we have:
1/c + 2/c + 3/c + 4/c = 1
Solving for c, we get:
c = 10
The pmf can be depicted as:
x f(x)
1 0.1
2 0.2
3 0.3
4 0.4
(b) Again, we know that the sum of all the probabilities in a pmf must be equal to 1. Therefore, we have:
c(1 + 2 + 3 + ... + 10) = 1
Solving for c, we get:
c = 1/55
The pmf can be depicted as:
x f(x)
1 1/55
2 2/55
3 3/55
... ...
10 10/55
(d) We can use the same approach as in part (a) to find c:
c(1 + 4 + 9) = 1
Solving for c, we get:
c = 1/14
The pmf can be depicted as:
x f(x)
0 1/14
1 1/6
2 1/3
3 4/14
(e) Similar to part (a), we have:
1/c + 2/c + 3/c + ... + n/c = 1
Solving for c, we get:
c = (n+1)n/2
The pmf can be depicted as:
x f(x)
1 1/(n+1)
2 2/(n+1)
3 3/(n+1)
n n/(n+1)
(f) We can use the hint given in the problem:
f(x) = 1/(x + 1) - 1/(x + 2)
We can see that this is a telescoping series, where all terms cancel out except for the first and the last term. Therefore, we have:
f(0) = 1/1 - 1/2 = 1/2
f(1) = 1/2 - 1/3 = 1/6
f(2) = 1/3 - 1/4 = 1/12
f(3) = 1/4 - 1/5 = 1/20
The pmf can be depicted as:
x f(x)
0 1/2
1 1/6
2 1/12
3 1/20
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A company must decide which of two delivery services they will contract with. During a
recent trial period they shipped numerous packages with each service, keeping track of how
often the deliveries did not arrive on time. Here are the data:
Delivery Service Type of Service Number of Deliveries Number of Late Packages
Pack Rats Regular 400 12
overnight 100 16
Boxes R Us Regular 100 2
Overnight 400 28
a) Compare the two service's overall (total) percentage of late deliveries. [5.6% for Pack
Rats; 6% for Boxes R Us]
b) Based on the results in part (a), the company has decided to hire Pack Rats. Do you
agree that they deliver on time more often? Why or why not? Be specific.
a) To compare the overall percentage of late deliveries for each delivery service, we need to calculate the total number of late packages and total number of deliveries for each service, and then divide the number of late packages by the total number of deliveries, and multiply by 100 to get the percentage.
For Pack Rats, the total number of deliveries is 400 + 100 = 500, and the total number of late packages is 12 + 16 = 28. Therefore, the percentage of late deliveries for Pack Rats is:
percentage of late deliveries for Pack Rats = (28/500) x 100 = 5.6%
percentage of late deliveries for Boxes R Us = (30/500) x 100 = 6%
Therefore, based on these calculations, Pack Rats has a lower overall percentage of late deliveries compared to Boxes R Us.
b) While Pack Rats has a lower overall percentage of late deliveries compared to Boxes R Us, it's important to note that this decision should not be based solely on this one comparison. It's possible that there are other factors that the company needs to consider, such as the cost of each delivery service or the quality of customer service provided by each company.
Furthermore, the sample size in this trial period may not be large enough to draw a definitive conclusion about the reliability of each service. Therefore, while Pack Rats may be a good choice based on the available data, the company should consider other factors and conduct further research before making a final decision.
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set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. y = x , y = 0, x = 4; about x = 8
The integral for the volume of the solid is obtained by rotating the region bounded by the given curves about the specified axis is V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex]
When we rotate a thin vertical strip, about the y-axis.
We get a cylindrical shell with an inner of radius an x and an outer of radius x + dx
The height of the cylinder shell is tan (x) - x
The volume of the cylindrical shell is
dV = π [tex](Outer Radius)^{2} (Height)[/tex] - π[tex](Inner Radius)^{2} (Height)[/tex]
dV = π [tex](x + dx )^{2} (tan(x) - x)[/tex] - π [tex](x )^{2} (tan(x) - x)[/tex]
dV = π[tex](x^{2} + 2 xdx + (dx)^{2} ) (tan(x) - x)[/tex] - π [tex](x)^{2} (tan(x) - x)[/tex]
assume [tex]dx^{2}[/tex]≈ 0
dV = π[tex](x^{2} + 2xdx + 0 - x^{2} ) (tan(x) - x)\\[/tex]
dV = 2πx (tan(x) - x) dx
V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex]
Therefore, the integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis is
V = 2[tex]\int\limits^ π/3_[/tex][tex]_{0}[/tex] [tex]{x} tan(x) - x^{2} dx[/tex] .
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The correct question is:
Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=tanx,y=x,x=π/3; about the y-axis
y-intercept of the line that represents each linear
relationship.
4. The equation K = °C + 273.15 can be used to
convert Celsius to Kelvin.
By rewriting the equation in slope-intercept form and setting Celsius to 0, we can find the y-intercept of the corresponding line, which represents the temperature in Kelvin when Celsius is 0°C.
When studying linear relationships between two variables, it is often useful to identify the y-intercept of the line that represents such relationship.
The equation K = °C + 273.15 expresses the relationship between the temperature in Celsius (°C) and the temperature in Kelvin (K). It tells us that to convert a temperature from Celsius to Kelvin, we need to add 273.15 to the original value.
We can plot the Celsius and Kelvin values as two variables on a graph, with Celsius on the x-axis and Kelvin on the y-axis. Since the equation K = °C + 273.15 is a linear equation (meaning that the graph is a straight line), we can use the slope-intercept form of the equation to find the y-intercept, which is the point where the line intersects the y-axis.
The slope-intercept form of a linear equation is y = mx + b, where m is the slope of the line and b is the y-intercept. In the case of K = °C + 273.15, we can rewrite it as K = 1°C + 273.15, which tells us that the slope of the line is 1. This means that for every 1°C increase in temperature, there is a corresponding 1 K increase in temperature.
To find the y-intercept, we can simply set the value of Celsius to 0, since this is the point where the line intersects the y-axis. Plugging in 0 for °C in the equation K = °C + 273.15, we get K = 273.15. This means that the y-intercept of the line is (0, 273.15), which corresponds to the temperature in Kelvin when the temperature in Celsius is 0°C.
In summary, the equation K = °C + 273.15 provides a linear relationship between the temperature in Celsius and the temperature in Kelvin.
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let u and v be distinct vectors of a vector space v. show that if { u, u} is a basis for v and a and b are nonzero scalars, then both { u v, au} and {au, bu} are also bases for v.
A + B =0 => {u + v, au} is linearly independent set of two vectors
=> {u + v, au}is a basis for V.
Let u and v be distinct vectors of a vector space V and {u, v} is a basis for V and a and b are nonzero scalars.
We have to show: both {u + v, au} and {au, bv} are also bases for V.
[tex]u\neq v,a,b\neq 0[/tex]
{u, v} is a basis for V => {u + v , au} and {au, bu} are bases for V
From basis {u , v} we get dimension of V
{u , v} is a basis for V
=> V is a 2-dimensional vector space and (.) [tex]\alpha u+\beta v=0= > \alpha =\beta =0[/tex]
Show {u + v, au} is a basis for V:
{u + v, au} :
A(u + v) + B(au) = 0 <=> Au + Av + Bau = 0
<=> (A + Ba)u + (B)v = 0
By the step (in second step)
A + Ba = 0 , [tex]a\neq 0[/tex]
B = 0
=> A + B =0 => {u + v, au} is linearly independent set of two vectors
=> {u + v, au}is a basis for V.
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a) Work out the prime factor decomposition of 396. Give your
answer in index form.
b) The prime factor decompositions of five numbers are shown
below. Use your answer to part a) to work out which two of
these numbers are factors of 396.
88=2³×11
121=11²
9=3²
14=2×7
22=2×11
Answer:
a) Prime factorization of 396: 2² × 3¹ × 11¹
b) Factors of 396: 88 and 9.
Step-by-step explanation:
a) To find the prime factorization of 396, we can start by dividing it by the smallest prime number, which is 2:
396 ÷ 2 = 198
Now we can write:
396 = 2 × 198
Next, we can divide 198 by 2 again:
198 ÷ 2 = 99
So we have:
396 = 2 × 2 × 99
Now we need to find the prime factorization of 99. We can start by dividing it by 3:
99 ÷ 3 = 33
So we can write:
396 = 2 × 2 × 3 × 33
Now we need to find the prime factorization of 33. We can start by dividing it by 3 again:
33 ÷ 3 = 11
So we have:
396 = 2 × 2 × 3 × 11 × 1
We don't need to include the factor of 1, so we can write the prime factorization of 396 in index form as:
396 = 2² × 3¹ × 11¹
b) To find which two of the given numbers are factors of 396, we need to check if all the prime factors of each number are also prime factors of 396.
The prime factorization of 88 is 2³ × 11. Both 2 and 11 are factors of 396, so 88 is a factor of 396.
The prime factorization of 121 is 11². 11 is a factor of 396, but there is no factor of 121 that matches the remaining prime factors of 396 (2² × 3¹), so 121 is not a factor of 396.
The prime factorization of 9 is 3². Both 3 and 2 are factors of 396, so 9 is a factor of 396.
The prime factorization of 14 is 2 × 7. 7 is not a factor of 396, so 14 is not a factor of 396.
The prime factorization of 22 is 2 × 11. Both 2 and 11 are factors of 396, so 22 is a factor of 396.
Therefore, the two numbers that are factors of 396 are 88 and 9.
three different methods for assembling a product were proposed by an industrial engineer. to investigate the number of units assembled correctly with each method, employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by workers. the number of units assembled correctly was recorded, and the analysis of variance p
Since the p-value is less than .05, we reject the null hypothesis that the means of the three assembly methods are equal.
Therefore, we can conclude that there is a significant difference in the means of the three assembly methods.
Source Variation | Squares' Sum | Degrees of Freedom | Mean Square | F
Treatments 4560 2 2280 9.87
Error 6240 27 231.11
Total 10800 29
Using Alpha = .05 to test for any significant difference in the means for the three assembly methods.
The value of the test statistic is 9.87
The p-value is: less than .01
Conclusion not all means of the three assembly methods are equal.
Complete Question:
Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST = 10,800; SSTR = 4560.
Set up the ANOVA table for this problem (to 2 decimals, if necessary).
Shelly spent 45 minutes jogging and 30 minutes cycling and burned 1350 calories. The next day, Shelly swapped times, doing 30 minutes of jogging and 45 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?
Number of calories burned per minute =
The number of calories burned per minute be 23.33333.
What is meant by system of equations?Equations simultaneously, or a system of equations Several equations in algebra must be solved concurrently (i.e., the solution must satisfy all the equations in the system). There must be an equal number of equations and unknowns for a system to have a singular solution.
In order to locate the point where the lines intersect when the equations are graphed, systems of equations must be solved. The (x,y) ordered pair of this intersection point is regarded as the system's solution.
Let j be the number of calories burned by Shelly while running and c be the number of calories she burns while cycling.
The system of equations be
45j + 30c = 1350
30j + 45c = 1350
45j + 30(10)=1350
j = 23.33333
Therefore, the value j be 23.33333.
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Given: Parallelogram LMNO; MO ⊥ LN
Prove: LMNO is a rhombus.
Parallelogram L M N O is shown. Diagonals are drawn from point L to point N and from point M to point O and intersect at point P. A square is drawn around point P. Sides L M and O N are parallel and sides L O and M N are parallel.
The proof of LMNO is a rhombus is shown below.
What is Rhombus?A parallelogram is a particular instance of a rhombus. The opposing sides and angles in a rhombus are parallel and equal. A rhombus also has equal-length sides on each side, and its diagonals meet at right angles to form its shape. The rhombus is also referred to as a diamond or rhombus.
Given:
|LO|=|MN| and |LM|=|ON|
Since Opposite sides of a parallelogram are equal.
Now, LN⊥OM
So, ∠LPO = ∠NPO = 90° ( by definition of perpendicular lines)
LPO ≅ ∠NPO (by definition of congruent angles)
|LP|=|PN| (diagonals of a parallelogram bisect each other)
Thus, LMNO is a rhombus
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Answer:
♣: ✔ All right angles are congruent.
♦: ✔ reflexive property
♠: ✔ Opposite sides of a parallelogram are congruent.
You find a great deal on a computer that has been marked down by 45% to be $715. What was the original
price of the computer?
Answer:
$1,300
Step-by-step explanation:
Let original price be $x
Computer has been marked down by 45%
45% = 45/100 = 0.45
So discount = 45% of x = 0.45 x
Original price - Discount = Sale price
x - 0.45x = $715
0.55x = 715
x = 715/0.55
x = $1,300
which is the original price