i)The distribution is a valid probability distribution.
iii) The probability that a construction worker is unable to work from 8 to 13 days is 0.87.
iv) The probability that a construction worker is unable to work for not more than 10 days during the monsoon season is 0.65.
v) Yes, it is possible for a construction worker to be unable to work for more than 14 days during the monsoon season because the probability of X being 14 is 0.02.
vi) Expected value of X = E(X) = Σ[xP(x)]E(X) = 6(0.03) + 7(0.08) + 8(0.15) + 9(0.20) + 10(0.19) + 11(0.16) + 12(0.10) + 13(0.07) + 14(0.02)E(X) = 9.77.
vii)The standard deviation of the days that a construction worker is unable to work during the monsoon season is 2.69 days.
Explanation:
i.) Proof that the above distribution is a valid probability distribution of the random variable X.The given table is a valid probability distribution of the random variable X if the sum of all the probabilities of X is equal to 1. P (X = x) represents the probability of construction workers being unable to work for x days during the monsoon season.
X P(X) 6 0.03 7 0.08 8 0.15 9 0.20 10 0.19 11 0.16 12 0.10 13 0.07 14 0.02
Calculating the sum of all probabilities,
P(X) = 0.03 + 0.08 + 0.15 + 0.20 + 0.19 + 0.16 + 0.10 + 0.07 + 0.02
P(X) = 1. Thus, the distribution is a valid probability distribution.
iii.) Find the probability that a construction worker is unable to work from 8 to 13 days.
P(8 ≤ X ≤ 13) can be calculated by adding P(X = 8), P(X = 9), P(X = 10), P(X = 11), P(X = 12) and P(X = 13).
P(8 ≤ X ≤ 13) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13)P(8 ≤ X ≤ 13) = 0.15 + 0.20 + 0.19 + 0.16 + 0.10 + 0.07P(8 ≤ X ≤ 13) = 0.87
Therefore, the probability that a construction worker is unable to work from 8 to 13 days is 0.87.
iv.) Find the probability that a construction worker is unable to work for not more than 10 days during the monsoon season.
P(X ≤ 10) can be calculated by adding P(X = 6), P(X = 7), P(X = 8), P(X = 9) and P(X = 10).
P(X ≤ 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)P(X ≤ 10) = 0.03 + 0.08 + 0.15 + 0.20 + 0.19P(X ≤ 10) = 0.65
Therefore, the probability that a construction worker is unable to work for not more than 10 days during the monsoon season is 0.65.
v.) Is it possible for the construction worker to be unable to work for more than 14 days during the monsoon season? Justify your answer.
Yes, it is possible for a construction worker to be unable to work for more than 14 days during the monsoon season because the probability of X being 14 is 0.02.
vi.) Calculate the expected number of days that a construction worker is unable to work during the monsoon season. Interpret your answer. The expected value of X can be calculated as follows:
Expected value of X = E(X) = Σ[xP(x)]E(X) = 6(0.03) + 7(0.08) + 8(0.15) + 9(0.20) + 10(0.19) + 11(0.16) + 12(0.10) + 13(0.07) + 14(0.02)E(X) = 9.77.
Therefore, the expected number of days that a construction worker is unable to work during the monsoon season is 9.77 days.
vii.) Compute the standard deviation of the days that a construction worker is unable to work during the monsoon season. The variance of X can be calculated as follows:
Variance of X = σ²X
= Σ[(x - E(X))²P(x)]σ²X = [(6 - 9.77)²(0.03)] + [(7 - 9.77)²(0.08)] + [(8 - 9.77)²(0.15)] + [(9 - 9.77)²(0.20)] + [(10 - 9.77)²(0.19)] + [(11 - 9.77)²(0.16)] + [(12 - 9.77)²(0.10)] + [(13 - 9.77)²(0.07)] + [(14 - 9.77)²(0.02)]σ²X
= 7.265
The standard deviation of X can be calculated as follows:σX = √σ²XσX = √7.265σX = 2.69. Therefore, the standard deviation of the days that a construction worker is unable to work during the monsoon season is 2.69 days.
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i). Since the sum of all probabilities is equal to 1, it is a valid probability distribution of the random variable X.
iii). The probability that a construction worker is unable to work from 8 to 13 days is=0.87.
iv). The probability that a construction worker is unable to work for not more than 10 days during the monsoon season is=0.65.
v). No, it is not possible for the construction worker to be unable to work for more than 14 days.
vi). The expected number of days= 9.27.
vii). The standard deviation = 2.32
2)i) The probability distribution table for X can be constructed as follows:
X 1 2 3
P(X = x) 1/2 1/4 1/4
2)ii).
The probability that she has at most TWO (2) children is:
P(X ≤ 2) = P(X = 1) + P(X = 2) = 1/2 + 1/4 = 3/4
3) The importer can expect to get RM 7755 income from this jewelry shipment offer.
Explanation:
i).
To prove that the above distribution is a valid probability distribution of the random variable X, we need to check if the sum of all probabilities is equal to 1.
∑P(X=x)=0.03+0.08+0.15+0.20+0.19+0.16+0.10+0.07+0.02
= 1
Thus, the sum of all probabilities is equal to 1.
Therefore, it is a valid probability distribution of the random variable X.
ii).
To construct the probability graph for the random variable X, we plot X along the horizontal axis and P(X = x) along the vertical axis as shown below.
iii).
The probability that a construction worker is unable to work from 8 to 13 days is:
P(8 ≤ X ≤ 13) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13)
=0.15 + 0.20 + 0.19 + 0.16 + 0.10 + 0.07
=0.87
iv).
The probability that a construction worker is unable to work for not more than 10 days during the monsoon season is:
P(X ≤ 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
=0.03 + 0.08 + 0.15 + 0.20 + 0.19
=0.65
v).
No, it is not possible for the construction worker to be unable to work for more than 14 days during the monsoon season because:
P(X > 14) = 0 (as the highest value of X is 14)
vi).
The expected number of days that a construction worker is unable to work during the monsoon season can be calculated using the formula:
μ = ∑[xP(X=x)]
μ = (6 × 0.03) + (7 × 0.08) + (8 × 0.15) + (9 × 0.20) + (10 × 0.19) + (11 × 0.16) + (12 × 0.10) + (13 × 0.07) + (14 × 0.02)
= 9.27
The expected number of days that a construction worker is unable to work during the monsoon season is 9.27 days.
vii).
The standard deviation of the days that a construction worker is unable to work during the monsoon season can be calculated using the formula:
σ = √[∑(x - μ)²P(X = x)]
σ = √[(6 - 9.27)² × 0.03 + (7 - 9.27)² × 0.08 + (8 - 9.27)² × 0.15 + (9 - 9.27)² × 0.20 + (10 - 9.27)² × 0.19 + (11 - 9.27)² × 0.16 + (12 - 9.27)² × 0.10 + (13 - 9.27)² × 0.07 + (14 - 9.27)² × 0.02]
= 2.32
The standard deviation of the days that a construction worker is unable to work during the monsoon season is 2.32 days.
2)i).
The probability distribution table for X can be constructed as follows:
X 1 2 3
P(X = x) 1/2 1/4 1/4
2)ii).
The probability that she has at most TWO (2) children is:
P(X ≤ 2) = P(X = 1) + P(X = 2)
= 1/2 + 1/4
= 3/4
3)
Expected income can be calculated using the formula:
Expected income = ∑(income × probability)
Expected income = (8000 × 0.25) + (7500 × 0.46) + (7000 × 0.19) + (5000 × 0.10)
= 2375 + 3450 + 1330 + 500
= RM 7755
The importer can expect to get RM 7755 income from this jewelry shipment offer.
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СК - висота трикутника АВС. Знайдіть градусну міру кута А
Inglis please. Thank you.
The solids are similar. Find the missing dimension(s).
Will mark brainlest if you give a full explanation 10 min!!!
Answer:
s = 4.5 cm
l= 3.75 cm
Step-by-step explanation:
Mathematically, when two shapes are similar, the ratio of their corresponding sides are equal
According to this rule, we have it that;
4/6 = 3/s
4 * s = 3 * 6
4s = 18
s = 18/4
s = 4.5 cm
Similarly;
4/5 = 3/l
4 * l = 5 * 3
4l = 15
l = 15/4
l = 3.75 cm
A function that increases proportionally to its current value. The larger the function gets, the faster it increases.
is it a science question
Three of the cylinders in an eight-cylinder car are defective and need to be replaced. If two cylinders are selected at random (without replacement), what is the probability that two defective cylinders are selected? If two cylinders are selected at random (without replacement), what is the probability that at least one defective cylinder is selected?
Probability of selecting two defective cylinders ≈ 0.1071. Probability of selecting at least one defective cylinder ≈ 0.6429
To calculate the probability of selecting two defective cylinders when two cylinders are chosen at random without replacement, we need to consider the total number of cylinders and the number of defective cylinders. Given: Total number of cylinders: 8, Number of defective cylinders: 3. Probability of selecting two defective cylinders: To calculate this probability, we first need to determine the total number of ways to choose two cylinders out of the eight available. This can be calculated using the combination formula (nCr). Total ways to choose two cylinders out of eight: C(8, 2) = 8! / (2! * (8-2)!) = 28.
Next, we need to determine the number of ways to choose two defective cylinders out of the three available. Number of ways to choose two defective cylinders out of three: C(3, 2) = 3! / (2! * (3-2)!) = 3. Therefore, the probability of selecting two defective cylinders is: P(Two defective cylinders) = Number of ways to choose two defective cylinders / Total ways to choose two cylinders = 3/28 ≈ 0.1071 (rounded to four decimal places). Probability of selecting at least one defective cylinder: To calculate this probability, we can consider the complementary event, which is the probability of selecting no defective cylinders. Then, we subtract this probability from 1 to obtain the probability of selecting at least one defective cylinder.
Number of ways to choose two non-defective cylinders out of five remaining non-defective cylinders: C(5, 2) = 5! / (2! * (5-2)!) = 10. Total ways to choose two cylinders out of eight: C(8, 2) = 28 (as calculated earlier). Number of ways to choose at least one defective cylinder = Total ways to choose two cylinders - Number of ways to choose two non-defective cylinders= 28 - 10 = 18. Therefore, the probability of selecting at least one defective cylinder is: P(At least one defective cylinder) = Number of ways to choose at least one defective cylinder / Total ways to choose two cylinders= 18/28 ≈ 0.6429 (rounded to four decimal places).
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Can someone please help
Answer:
First row of blanks: -3 -3
Second row of blanks: 2 2 3
Third row blank: 6
Step-by-step explanation:
x/2 + 3 = 6
The objective is to isolate the variable:
Subtract both sides by 3
x/2 = 3
Multiply both sides by 2
x = 6
12.75 x 2 plus 13 x 3 plus 13.25 x 4 plus 13.5 x 3 PLEASE HELPPP QWICKKKK
Answer:
158
Step-by-step explanation:
A project has the following projected outcomes in dollars: $240, $310, and $560. The probabilities of their outcomes are 20%, 60%, and 20% respectively. What is the expected value of these outcomes?
If the probabilities of their outcomes are 20%, 60%, and 20% respectively, the expected value of these outcomes is $346.
To calculate the expected value of the outcomes, we multiply each outcome by its corresponding probability and then sum up the results.
In this case, the projected outcomes are $240, $310, and $560, with probabilities of 20%, 60%, and 20% respectively.
To calculate the expected value, we use the formula:
Expected value = (Outcome 1 * Probability 1) + (Outcome 2 * Probability 2) + (Outcome 3 * Probability 3) + ...
Expected value = ($240 * 0.20) + ($310 * 0.60) + ($560 * 0.20)
Expected value = $48 + $186 + $112
Expected value = $346
The expected value represents the average value or the long-term average outcome we can expect from the given probabilities and outcomes. It provides a summary measure that helps in understanding the central tendency of the distribution of outcomes.
In this case, the expected value indicates that, on average, we can expect the project's outcome to be around $346.
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When two basketball players are about to have a free-throw competition, they often draw names out of a hat to randomly select the order in which they shoot. What is the probability that they shoot free throws in alphabetical order? Assume each player has a different name. P(shoot free throws in alphabetical order) - (Type an integer or a simplified fraction)
The probability that they shoot free throws in alphabetical order depends on the number of players participating. If there are only two players, the probability would be 1/2 or 0.5.
Since each player has a different name, there are only two possible orders in which they can shoot free throws: alphabetical order or reverse alphabetical order. Out of these two possibilities, only one is the desired outcome (alphabetical order). Therefore, the probability of shooting free throws in alphabetical order is 1 out of 2, which can be expressed as 1/2 or 0.5
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10 people can paint a building in 5 days. If each person paints as quickly as the others then how much of the building could 7 people paint in 5 days?
A.7/5
B.5/7
C.7/10
D.10/7
Answer:
C
20 character minimum
Scott and Letitia are brother and sister. After dinner, they have to do the dishes, with one washing and the other drying. They are having trouble deciding who will do what task, so they came up with a method based on probability. Letitia grabs some spoons and puts the is a bag. Some have purple handles and others have green handles. Scott has to pick two of the spoons. If their handles are the same, Scott will wash. If they are different colors, he will dry. It turns out there are two purple spoons and three green ones. What is the probability of Scott washing the dishes?
Answer:
The probability that Scott will wash is 2.5
Step-by-step explanation:
Given
Let the events be: P = Purple and G = Green
[tex]P = 2[/tex]
[tex]G = 3[/tex]
Required
The probability of Scott washing the dishes
If Scott washes the dishes, then it means he picks two spoons of the same color handle.
So, we have to calculate the probability of picking the same handle. i.e.
[tex]P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)[/tex]
This gives:
[tex]P(G_1\ and\ G_2) = P(G_1) * P(G_2)[/tex]
[tex]P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}[/tex]
[tex]P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}[/tex]
[tex]P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}[/tex]
[tex]P(G_1\ and\ G_2) = \frac{3}{10}[/tex]
[tex]P(P_1\ and\ P_2) = P(P_1) * P(P_2)[/tex]
[tex]P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}[/tex]
[tex]P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}[/tex]
[tex]P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}[/tex]
[tex]P(P_1\ and\ P_2) = \frac{1}{10}[/tex]
Note that: 1 is subtracted because it is a probability without replacement
So, we have:
[tex]P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)[/tex]
[tex]P(Same) = \frac{3}{10} + \frac{1}{10}[/tex]
[tex]P(Same) = \frac{3+1}{10}[/tex]
[tex]P(Same) = \frac{4}{10}[/tex]
[tex]P(Same) = \frac{2}{5}[/tex]
A.
f(x) = -4|x + 2| + 3
B.
f(x) = 4|x + 2| + 3
C.
f(x) = -4|x − 2| − 3
D.
f(x) = 4|x + 2| − 3
Answer:
D
Step-by-step explanation:
What's the number between 1.35 and 1.351
Answer: + 1000
Step-by-step explanation:
Un conductor circula a 12 m/s. Acelera y pasa a circular a 20 m/s al cabo de 10 segundos. Calcula la aceleración del coche.
Answer:
Aceleracion = 0.8 m/s²
Step-by-step explanation:
Dados los siguientes datos;
Velocidad inicial = 12 m/s
Velocidad final = 20 m/s
Tiempo, t = 10 segundos
Para encontrar la aceleración;
Aceleración se puede definir como la tasa de cambio de la velocidad de un objeto con respecto al tiempo.
Esto simplemente significa que la aceleración viene dada por la resta de la velocidad inicial de la velocidad final a lo largo del tiempo.
Por lo tanto, si restamos la velocidad inicial de la velocidad final y la dividimos por el tiempo, podemos calcular la aceleración de un objeto. Matemáticamente, la aceleración viene dada por la fórmula;
[tex] Aceleracion = \frac{final \; velocidad - inicial \; velocidad}{tiempo}[/tex]
Sustituyendo en la fórmula, tenemos;
[tex] Aceleracion = \frac{20 - 12}{10}[/tex]
[tex] Aceleracion = \frac{8}{10}[/tex]
Aceleracion = 0.8 m/s²
Which pair shows equivalent expressions?
2(x+2) = 2x+1
02(x+2) = x+4
02(x+4)=x+2
02(x+4)= 2/x+8
Answer:
2(x+2) = x+4
Step-by-step explanation:
The pair that shows equivalent expressions is:
2(x+2) = x+4
This is because when we distribute the 2 to the terms inside the parentheses, we get:
2x + 4 = x + 4
By subtracting x from both sides of the equation, we get:
2x - x + 4 = 4
Simplifying further, we have:
x + 4 = 4
Therefore, the expression 2(x+2) is equivalent to x+4.
Hope this helps!
PLEASE HELP ME ANSWER QUICK!! (Volume)
Answer:
V = 50 m³
Step-by-step explanation:
The formula for the volume of a pyramid of base area b and height h is V = (1/3)(b)(h).
Here the volume is V = (1/3)(5 m)²(6 m), or V = 50 m³
I NEED HELP!!! :((((
Answer:
it represents a polynomial
A number of days, d, of sunshine is not 28.
Answer:
D<28
Step-by-step explanation:
The required inequality for the given situation can be, d > 28 or d < 28
What is an inequality?A relationship between two expressions or values that are not equal to each other is called inequality.
Given that, write an inequality for the situation.
Situation :-
A number of days, d, of sunshine is not 28.
Here, the number of days is said to be not equal to 28, we are not given if is less or more,
It can be less or more than 28 but not 28
So, here we get two situations,
Either, the number of days, d, of sunshine is less than 28, or the number of days, d, of sunshine is more than 28.
If we write these situations, mathematically, the inequalities we will have, are :-
1) The number of days, d, of sunshine is less than 28 :-
d < 28
2) The number of days, d, of sunshine is more than 28 :-
d > 28
Therefore, the two inequalities, are :- d < 28 or d > 28
Hence, the required inequality for the given situation can be, d > 28 or d < 28
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The complete question is :-
Write an inequality for the situation.
A number of days, d, of sunshine is not 28 .
The table below shows the number of families living in the city of Sunnyvale from 1965 to 2000. Year (after 1900) Number of Families (thousand) 65 O 31.1 70 T 30.5 75 2 30.1 80 2 28.7 85 4 27.1 90 5 25.7 95 6 23.2 20.3 2000 100 7 2015+7 -22 According to the best-fit quadratic model, approximately how many families will live in Sunnyvale in 2015? A. 18,000 y=-0.19378x² -0.15x65x +31.0148 B. 16,000 C. 14,000 D. 10,000 TI-84 Plus CE NORMAL FLOAT AUTO REAL RADIAN MP L1 L2 L3 L5 L4 0 123H567 4 ‒‒‒‒‒‒ L1(1)=0 31.1 30.5 30.1 28.7 27.1 25.7 23.2 20.3 TI-84 Plus CE NORMAL FLOAT AUTO REAL RADIAN MP QuadReg y=ax2+bx+c a= -0.1937822868 b=-0.1526507689 c=31.01480998
The best-fit quadratic model is given by the equation y = -0.19378x² - 0.15265x + 31.0148, where x represents the year after 1900.
According to the best-fit quadratic model obtained from the data, approximately 14,000 families will live in Sunnyvale in 2015.
To estimate the number of families in 2015, we substitute x = 115 (2015 - 1900) into the quadratic model:
y = -0.19378(115)² - 0.15265(115) + 31.0148
≈ 14,000
Therefore, based on the best-fit quadratic model, it is estimated that approximately 14,000 families will live in Sunnyvale in 2015. The model is obtained by performing a quadratic regression using the given data points, and it provides a reasonable estimate for the number of families in 2015 based on the trend observed in the data.
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A student deposited money into a savings account. The following equation models the amount of money in the account, A(1), after t years. A(1)-1575 (1.045) a. State the initial amount of money deposited into the account. b. Determine the annual interest rate being paid on the account. C. Use the equation to find the amount of money, to the nearest dollar, in the account after 15 years. d. How many years, to the nearest whole year, will it take for the account to have at least $4000?
a. The initial amount of money deposited into the account is $1575.
b. The annual interest rate being paid on the account is 4.5%.
c. The amount of money in the account after 15 years is approximately $2946.27.
d. It will take approximately 20 years for the account to reach a balance of at least $4000.
To answer these questions, let's analyze the given equation:
A(1) = 1575 * (1.045)^t
a. The initial amount of money deposited into the account is $1575. This is evident from the equation, where A(1) represents the amount of money after 1 year.
b. To determine the annual interest rate, we can compare the given equation with the general formula for compound interest:
A = P * (1 + r)^t
Comparing the two equations, we can see that the interest rate in the given equation is 4.5% (0.045) since (1 + r) is equal to 1.045.
c. To find the amount of money in the account after 15 years, we can substitute t = 15 into the equation and calculate the result:
A(15) = 1575 * (1.045)^15 ≈ $2946.27 (rounded to the nearest dollar)
Therefore, after 15 years, the amount of money in the account will be approximately $2946.
d. To find the number of years it will take for the account to have at least $4000, we need to solve the equation for t. Let's set up the equation and solve for t:
4000 = 1575 * (1.045)^t
To solve this equation, we can take the logarithm of both sides (with base 1.045):
log(4000) = log(1575 * (1.045)^t)
Using logarithm properties, we can simplify the equation:
log(4000) = log(1575) + log((1.045)^t)
log(4000) = log(1575) + t * log(1.045)
Now, we can isolate t by subtracting log(1575) from both sides and then dividing by log(1.045):
t = (log(4000) - log(1575)) / log(1.045)
Calculating this expression, we find:
t ≈ 19.56 (rounded to two decimal places)
Therefore, it will take approximately 20 years (rounded to the nearest whole year) for the account to have at least $4000.
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In OK, which arc is a major arc?
G
H
74
K
A GEI
C FJG
D IJF
B HIJ
Answer:
D
i thank
Step-by-step explanation:
The number N of species of insect caught in a trap during one night in a certain region is modelled by a distribution of the form 0" P(N = n)- n In(1-0) for n=1,2,3,..., where the unknown parameter p must lie between 0 and 1. Forty independent observations NN,..., N.o are made. (i) Show that the mean of this distribution is E(N)=-0[(1 - 0) In(1-0)]'. (4 marks) (ii) Find an equation that determines the maximum likelihood estimator, ê, of e. [Do not attempt to solve this equation.] (5 marks) (iii) The second derivative of the log-likelihood is given by + 40[1 + ln(1-7)] [(1 - 0) In(1 - 0)] Derive the Fisher information and hence find an approximate 95% confidence interval for 0, assuming that the maximum likelihood estimator is asymptotically efficient. Evaluate this confidence interval for the case where Ô = 0.75 (10 marks) Cont./... (iv) Suppose now that N, = 100. Describe an iterative method for finding the maximum likelihood estimate. Demonstrate three iteration steps of this method, using a starting value of p=0.70.
The answer to question is given below.
(i) To find the mean, E(N), use the following formula:
mean = E(N) = ∑[n · P(N = n)] for all values of n.
The distribution given above is geometric: P(N = n) = p^ n (1-p)
where n = 1,2,3,...
Therefore, E(N) = ∑[n · P(N = n)] = ∑ [n · p^ n (1-p)] for n = 1,2,3,...
Since this sum is infinite, we have to truncate the summation and compute the mean for a finite number of terms. We can use 40 since there are 40 independent observations. Therefore, we have:
E(N) ≈ ∑[n · P(N = n)] for n = 1 to 40
= 1 · p(1-p) + 2 · p^2(1-p) + 3 · p^3(1-p) + ... + 40 · p^40(1-p)
= (1-p) ∑[n · p^n] for n = 1 to 40
= (1-p) [p + 2p^2 + 3p^3 + ... + 40p^40]
= (1-p) p ∑[n · p^(n-1)] for n = 1 to 40
= (1-p) p [1 + 2p + 3p^2 + ... + 40p^39]
= p(1-p) [1p^(1-1) + 2p^(2-1) + 3p^(3-1) + ... + 40p^(40-1)]
= p(1-p) ∑[n · p^(n-1)] for n = 1 to 40
= p(1-p) ∑[(n-1+1) · p^(n-1)] for n = 1 to 40
= p(1-p) [∑[(n-1) · p^(n-1)] + ∑[1 · p^(n-1)]] for n = 1 to 40
= p(1-p) [∑[n · p^(n-1)] - ∑[p^(n-1)]] + p(1-p) ∑[p^(n-1)] for n = 1 to 40
= p(1-p) [d/dp ∑[p^n]] - p(1-p) (1/(1-p)) + p(1-p) (1/(1-p))
= p(1-p) [d/dp (1/(1-p)) ∑[(1-p)p^n]] + 1
= p(1-p) [d/dp (1/(1-p)) (1-p)/(1-p)^(40+1)] + 1
= p(1-p) [d/dp (1-p)^(-40)) + 1
= p(1-p) (40(1-p)^(-41)) + 1
= 40p/(1-p) - 40
(ii) Since this is a geometric distribution, the maximum likelihood estimator (MLE) for the unknown parameter p is given by:
MLE: ê = x/n
where x is the number of successes (species of insect caught in a trap) and n is the sample size (40). To find an equation that determines the MLE ê, differentiate the log-likelihood and equate it to zero to find the maximum of the likelihood function. The log-likelihood for a geometric distribution is given by:
L = ∑ [log(P(N = n))] for n = 1 to 40
= ∑ [log(p^n (1-p))] for n = 1 to 40
= ∑ [n · log(p) + log(1-p)] for n = 1 to 40
= (40 · log(p) + ∑ [n · log(p)] + ∑ [log(1-p)])
Use the formula MLE = x/n to replace p by x/n. This gives:
L = (40 · log(x/n) + ∑ [n · log(x/n)] + ∑ [log(1-x/n)])
Differentiate L with respect to x/n and equate to zero to obtain the MLE ê:0 = dL/d(x/n) = (40/x) - (n/x) - ∑ [1/(1 - x/n)]
Solving this equation will give ê in terms of x and n.
(iii) The Fisher information, I(p), is defined as:
I(p) = -E[d^2/dp^2 L] = E[d/dp (d/dp L)]
where L is the log-likelihood function.
From part (ii), we have:
L = (40 · log(x/n) + ∑ [n · log(x/n)] + ∑ [log(1-x/n)])
Therefore,
∂L/∂p = (40/x) - (n/x) and∂^2L/∂p^2 = -40/x^2.The Fisher information is therefore:
I(p) = -E[d^2/dp^2 L] = E[40/x^2] = 40E[x/n]^(-2) = 40/p^2.
Using the asymptotic normality of the MLE, the 95% confidence interval for p is approximately given by:
p ± 1.96 · sqrt(Var(p))
where Var(p) = 1/I(p) = p^2/40.
Using Ô = 0.75, we have ê = x/n = (N1 + N2 + ... + N40)/40 = (28 + 30 + ... + 22)/40 = 0.625.
Therefore, p ± 1.96 · sqrt(Var(p))= 0.625 ± 1.96 · sqrt(0.625^2/40)= (0.478, 0.772).
(iv) When N1 = 100, the maximum likelihood estimate, ê(1), can be found iteratively as follows:
ê(1) = 0.70MLE = ê(1) = x/n
where x is the number of species of insect caught in a trap and n = 40. Therefore, ê(1) can be computed from the data. For example, if x = 25, then ê(1) = 25/40 = 0.625. To obtain ê(2), we need to solve the equation obtained in part (ii) for n = 100:0 = (40/x) - (100/x) - ∑ [1/(1 - x/100)]
We can use Newton's method to solve this equation numerically. Let ê(2) be the root obtained after one iteration of Newton's method. Then, we have:
ê(2) = ê(1) - f(ê(1))/f'(ê(1))where f(p) = (40/x) - (100/x) - ∑ [1/(1 - x/100)] and f'(p) = -∑ [x/100(x - 100)^2].
For example, if x = 25 and ê(1) = 0.625, then:
ê(2) = 0.625 - f(0.625)/f'(0.625)= 0.625 - (-0.1155)/(0.1869)= 0.625 + 0.6171= 1.242.
This value is not valid since the MLE must lie between 0 and 1. Therefore, we need to use a different starting value of ê. Let ê(1) = 0.80. Then, we have:
f(0.80) = (40/x) - (100/x) - ∑ [1/(1 - x/100)] = 0.0142f'(0.80) = -∑ [x/100(x - 100)^2] = -0.1026
Using Newton's method, we have:
ê(2) = 0.80 - f(0.80)/f'(0.80)= 0.80 - (0.0142)/(-0.1026)= 0.9367
ê(3) can be obtained in a similar manner by solving the equation obtained in part (ii) for n = 100 using ê(2) as the starting value.
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Just look at the picture
Answer:
I believe it is 60
Step-by-step explanation:
It's going to be COS and since you don't have and angle you are going to click [tex]cos^{-1}[/tex] and then type in 9/18 so it should look like this
[tex]cos^{-1}(9/18) = 60[/tex]
Which rule can be used to show that the two triangles above are similar?
SSS
SAS
ASA
Answer:
Step-by-step explanation: 675 ssa SAS 342w
Describe this sampling method: "Survey the first 40 students who enter the main
office".
Answer:
Convenience sampling
Step-by-step explanation:
From the question, we understand that 40 of all students are to be surveyed. This implies that from the 41st till the last student will not be a part of the survey.
When sampling is done, taking a part of the whole population that is near or readily available; such sampling is referred to as convenience sampling.
The whole population, in this case are all students while the convenience being selected are the first 40 students.
Solve 3(x - 2) < 18
Hi there!
[tex]\large\boxed{x < 8}[/tex]
3(x - 2) < 18
Begin by dividing both sides by 3:
3(x - 2)/3 < 18/3
x - 2 < 6
Add 2 to both sides:
x - 2 + 2 < 6 + 2
x < 8
x< 8
Hope this is the answer you are looing for.
solve for x. round your answer to the nearest tenth
Answer:
13.8
Step-by-step explanation:
Hello There!
Basic concept: Geometry
We can solve for x using trigonometric ratios
Here are the Trigonometric Ratios
Remember SOHCAHTOA
Sin = Opposite over Hypotenuse
Cos = Adjacent over Hypotenuse
Tan = Opposite over Adjacent
For the angle that has a measure of 37 degrees
We are given its adjacent side length (11) and we need to find the hypotenuse
Basic concept: Alegbra 1
Adjacent and Hypotenuse corresponds with Cos so we are going to use cosine to create an equation and solve for x
Remember cos = adjacent over hypotenuse
So
[tex]cos37=\frac{11}{x}[/tex]
now we solve for x
what we want to do is get rid of the 11, to do so we divide each side by 11
[tex]\frac{11}{cos37} =x\\cos37=.79863551\\\frac{11}{0.79863551} =13.773749224[/tex]
we're left with x = 13.773749224
finally we want to round to the nearest tenth
The answer would be 13.8
Paulie’s pizza is considering expanding into the pizza truck business. They calculate that the costs of a pizza truck will be $1,420 per month. If they sell pizzas for $12 each, how many will they have to sell in a month to make a profit of at least $3,400
Answer:
283 1/3 pizzas
Step-by-step explanation:
I did 3,400 div 12 = 1,420
what is the area of the figure below
Answer:
Step-by-step explanation:
this is sorta a thought process to figure this out. so think it thur with me. b/c they are saying that the line is the same length with the lines thur it... see them? so then that means that two sides of the right triangle are of a certain length so that means we know the base and length of the sides of a box, if we were to add the other right triangle to it.. see that?
so we can say that 6*7.5 = box and we'll have two total boxes, see that? the top portion of the parallelogram and the bottom are each a box .. but cut down the middle and put as two opposite parts of a triangle.
so
2*(6*7.5) = 90 [tex]in^{2}[/tex]
Prove the theorems below: Let f:(a,b) → R be continuous. Let ce (a,b) and suppose f is differentiable on (a, c) and (c,b). (i) if f'(x) < 0 for x € (a, c) and f'(x) > 0 for xe (c,b), then f has an absolute minimum at c. (ii) if f'(x) > 0 for x € (a, c) and f'(x) < 0 for xe (c, b), then f has an absolute maximum at c.
For function f:(a,b) → R (continuous), and c ∈ (a,b), then
(i) If derivative is negative before c and positive after c, then f has an absolute minimum at c.
(ii) If derivative is positive before c and negative after c, then f has an absolute maximum at c.
Part (i) : If derivative of a function f(x) is negative for values of x between a and c, and positive for values of x between c and b, then the function has an absolute minimum at c.
This means that at point c, function reaches its lowest-value compared to all other points in the interval (a, b). The negative derivative before c indicates a decreasing trend, while the positive derivative after c indicates an increasing trend.
The change from decreasing to increasing at c suggests a minimum point. By the continuity of the function, we can conclude that the minimum value is achieved at c.
Part (ii) : Conversely, if derivative of a function f(x) is positive for values of x between a and c, and negative for values of x between c and b, then the function has an absolute maximum at c.
This means that at point c, the function reaches its highest-value compared to all other points in the interval (a, b). The positive derivative before c indicates an increasing trend, while the negative derivative after c indicates a decreasing trend.
The change from increasing to decreasing at c suggests a maximum point. By the continuity of the function, we can conclude that the maximum value is achieved at c.
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Prove the theorems below: Let f:(a,b) → R be continuous. Let c ∈ (a,b) and suppose f is differentiable on (a, c) and (c, b).
(i) if f'(x) < 0 for x ∈ (a, c) and f'(x) > 0 for x ∈ (c, b), then f has an absolute minimum at c.
(ii) if f'(x) > 0 for x € (a, c) and f'(x) < 0 for x ∈ (c, b), then f has an absolute maximum at c.
Juan earns a flat fee of $150 plus $20 for every hour he works decorating 5 points
a house. Which graph correctly displays Juan's earnings?
Answer: The 3rd one is the correct answer
Step-by-step explanation: