I don’t understand what to do

The probability that the selected individual has children in public schools AND favors the school tax is 0.32

The probability that the selected individual favors the school tax AND has children in public schools is 0.32.

The probability that the selected individual favors the school tax AND does NOT have children in public schools is 0.2.

The probability that the selected individual favors the school tax AND is over 56 years old is 0.15.

The probability that the selected individual favors the school tax AND is 18-25 years old is 0.45.

Based on the given information, the probability of event F (the selected individual favors the school tax) is 0.54, as 54% of the respondents are in favor of the tax. The probability of event C (the selected individual has children in public schools) is 0.59, as the approval rating rises to 59% for those with children in public schools. The probability of event O (the selected individual is over 56 years old) is 0.37, as only 37% of those over age 56 said they would vote for the tax. The probability of event Y (the selected individual is 18-25 years old) is 0.71, as 71% of 18- to 25-year-olds said they would vote for the tax.

Using these probabilities, we can estimate the values of the following probabilities:

(1) P(CF) is the probability that the selected individual has children in public schools AND favors the school tax. Based on the given information, we can multiply the probabilities of events C and F: P(CF) = 0.59 * 0.54 = 0.318, or approximately 0.32.

(ii) P(FIC) is the probability that the selected individual favors the school tax AND has children in public schools. This is the same as P(CF), so P(FIC) = 0.32.

(iii) P(FIN) is the probability that the selected individual favors the school tax AND does NOT have children in public schools. To calculate this, we can use the fact that the approval rating falls to 44% for those with no children in public schools. So, P(FIN) = 0.44 * (1 - 0.59) = 0.18, or approximately 0.2.

(iv) P(FTO) is the probability that the selected individual favors the school tax AND is over 56 years old. To calculate this, we can use the fact that the approval rating for those over 56 years old is only 37%. So, P(FTO) = 0.37 * (1 - 0.59) = 0.1523, or approximately 0.15.

(v) P(FY) is the probability that the selected individual favors the school tax AND is 18-25 years old. To calculate this, we can use the fact that the approval rating for those 18-25 years old is 71%. So, P(FY) = 0.71 * (1 - 0.37) = 0.4477, or approximately 0.45.

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Answer:

Yield per share = 7.68% (Approx.)

Step-by-step explanation:

Given:

Dividend paid = $4.15

Price per dividend = $54

Find:

Yield per share

Computation:

Yield per share = [Dividend paid / Price per dividend]100

Yield per share = [4.15 / 54]100

Yield per share = [0.0768]100

Yield per share = 7.68% (Approx.)

Probability of ultimate extinction: To get the probability of ultimate extinction in a branching process, we need to calculate the value of o(q) = q where q is the probability of ultimate extinction.

Now, o(s) = (1/6) + (5/6)s²

Therefore, for the probability of ultimate extinction,

o(q) = q = (1/6) + (5/6)q²=> 6q² - 5q + 1 = 0

On solving the quadratic equation,

we get q = 1 or q = 1/6.

Thus, the probability of ultimate extinction is 1/6. Mean time to extinction:

To calculate the mean time to extinction, we can use the formula, E(T) = (1/o'(1))

where o'(1) is the first derivative of the generating function o(s) evaluated at s = 1.

Now, o(s) = (5/6) + (1/6)s=> o'(s) = (1/6)

On substituting s = 1, o'(1) = (1/6).

Thus, E(T) = (1/o'(1))= (1/ (1/6))= 6

The mean time to extinction is 6.

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X/2= 87.2

to find X:

87.2 X 2= 174.4

therefore X is 174.4

Answer:

YOU DO IT X 2

Step-by-step explanation:

Answer:

Dominos is the better deal.

The student is implicitly using the commutative property and the associative property of addition.

i. (9-1) + 9: The student uses the commutative property, which states that the order of adding numbers does not affect the sum. They rearrange the terms to (9-1) + 9, recognizing that adding 9 after subtracting 1 is the same as adding 9 before subtracting 1.

ii. (9-1) + 9 = 9 + (9-1): Here, the student demonstrates the associative property. The associative property states that the grouping of numbers being added does not affect the sum. They regroup the terms to show that adding (9-1) first and then adding 9 is equivalent to adding 9 first and then subtracting 1.

iii. 9 + (9-1): The student does not explicitly demonstrate a property here. They simply perform the calculation of adding 9 and (9-1), recognizing that 9 minus 1 is 8.

iv. (9+9)-1-18-1: In this step, the student performs the subtraction calculations but does not demonstrate any particular property.

v. 18-1: Finally, the student calculates the subtraction and arrives at the answer 17.

So, the student implicitly uses the commutative property and the associative property of addition in their mental calculation.

Answer:

150 cm squared

Step-by-step explanation:

I guess that's the answer if I'm wrong you can tell me right away so that I can try another method thank you.

Answer:

it's 1,-1

Step-by-step explanation:

hoped I helped. I'm good at these.

Answer:

She saved 1.20

Step-by-step explanation:

Purchase Price:

$12

Discount:

(12 x 10)/100 = $1.20

Final Price:

12 - 1.20 = $10.80

Yes, there is a uniformly most powerful (UMP) test at level a for testing for the given density function.

The test is based on the likelihood ratio, where the critical region is (θ_o, ∞) and the test statistic is (nθ_o^3)/Y(n), where Y(n) is the largest observation in the sample.

To obtain the UMP test at level a for testing H_0: θ ≤ θ_o vs. H_1: θ > θ_o, we need to find the likelihood ratio test with the largest power for all possible alternatives. The likelihood ratio test is constructed as the ratio of the likelihood function under H_0 to the likelihood function under H_1. By algebraic manipulation, we obtain the likelihood ratio test statistic as (nθ_o^3)/Y(n), where Y(n) is the largest observation in the sample.

Under H_0, this test statistic has a chi-squared distribution with one degree of freedom. Therefore, the critical region for rejecting H_0 at level a is the right tail of the chi-squared distribution with one degree of freedom, which is (θ_o, ∞).

This test is UMP because it has the highest power for all possible alternatives. This is because the distribution of Y(n) is stochastically increasing in θ, which means that for a given sample size n, the probability of obtaining an observation larger than a threshold value increases as θ increases.

Therefore, the likelihood ratio test statistic decreases as θ increases, which means that the rejection region (θ_o, ∞) has the highest power for all possible alternatives. Hence, the test based on the likelihood ratio is the UMP test at level a for testing H_0: θ ≤ θ_o vs. H_1: θ > θ_o for the given density function

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Answer:  Complementary     x=15

Step-by-step explanation:

Complementary angles add up to 90°, supplementary angles add up to 180°.

We know they add up to 90 so...

3x+45=90

3x=45

x=15

Main answer: An orthogonal basis for the given span H is {x1, x2-x1, x3 - (x1 + x2 - x1)} which simplifies to {x1, x2-x1, x3-x2}.

Supporting explanation: Given, x₁ = 0, x₂ = 1, x₃ = √3The span of H is the set of all linear combinations of x1, x2 and x3.So, we have to find an orthogonal basis for H using the Gram-Schmidt process. Let's start with the first vector x1 = [0, 0, 0]The second vector x2 is the projection of x2 onto the subspace perpendicular to x1. x2 is already perpendicular to x1 so x2-x1 = x2. So, the second vector is x2 = [0, 1, 0].The third vector x3 is the projection of x3 onto the subspace perpendicular to x1 and x2. x3 is not perpendicular to x1 and x2, so we subtract the projections of x3 onto x1 and x2 from x3. Projection of x3 onto x1:projx₁(x₃) = x₁ [(x₁ . x₃)/(x₁ . x₁)] = [0, 0, 0]Projection of x3 onto x2:projx₂(x₃) = x₂ [(x₂ . x₃)/(x₂ . x₂)] = [0, √3/3, 0]Therefore, x3 - projx₁(x₃) - projx₂(x₃) = [0, √3/3, √3]So, the orthogonal basis for H is {x1, x2-x1, x3 - (x1 + x2 - x1)} which simplifies to {x1, x2-x1, x3-x2}.

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Question:

A population consists  1, 2, 4, 5, 8. Draw all possible samples of size 2  without replacement from this population.

Verify that the sample mean is an unbiased estimate of the population mean.  

Answer:

[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]

[tex]\hat p = \frac{3}{5}[/tex] --- proportion of evens

The sample mean is an unbiased estimate of the population mean.

Step-by-step explanation:

Given

[tex]Numbers: 1, 2, 4, 5, 8[/tex]

Solving (a): All possible samples of 2 (W.O.R)

W.O.R means without replacement

So, we have:

[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]

Solving (b): The sampling distribution of the proportion of even numbers

This is calculated as:

[tex]\hat p = \frac{n(Even)}{Total}[/tex]

The even samples are:

[tex]Even = \{2,4,8\}[/tex]

[tex]n(Even) = 3[/tex]

So, we have:

[tex]\hat p = \frac{3}{5}[/tex]

Solving (c): To verify

[tex]Samples: \{(1,2),(1,4),(1,5),(1,8),(2,4),(2,5),(2,8),(4,5),(4,8),(5,8)\}[/tex]

Calculate the mean of each samples

[tex]Sample\ means = \{1.5,2.5,3,4.5,3,3.5,5,4.5,6,6.5\}[/tex]

Calculate the mean of the sample means

[tex]\bar x = \frac{1.5 + 2.5 +3 + 4.5 + 4 + 3.5 + 5 + 4.5 + 6 + 6.5}{10}[/tex]

[tex]\bar x = \frac{40}{10}[/tex]

[tex]\bar x = 4[/tex]

Calculate the population mean:

[tex]Numbers: 1, 2, 4, 5, 8[/tex]

[tex]\mu = \frac{1 +2+4+5+8}{5}[/tex]

[tex]\mu = \frac{20}{5}[/tex]

[tex]\mu = 4[/tex]

[tex]\bar x = \mu = 4[/tex]

This implies that [tex]\bar x[/tex] is an unbiased estimate of the [tex]\mu[/tex]

After 100 years, the long-term distribution of population in the city and suburbs of New York state can be calculated based on the given migration rates. The population in the city and suburbs will stabilize at approximately 3.96 million and 16.04 million, respectively. The population distribution can be visualized using a graph that shows the population of the city and suburbs over the 100-year period.

To calculate the long-term population distribution, we can use the concept of equilibrium. Let C represent the population in the city and S represent the population in the suburbs. The equilibrium equations can be written as follows:

C = C - 0.035C + 0.017S

S = S + 0.035C - 0.017S

Simplifying these equations, we have:

C = 0.965C + 0.017S

S = 0.035C + 0.983S

Solving these equations simultaneously, we find that C stabilizes at approximately 3.96 million and S stabilizes at approximately 16.04 million.

To plot the population of the city and suburbs over the 100-year period, you can use the following MATLAB code:

Copy code

years = 0:100;

C = zeros(1, 101);

S = zeros(1, 101);

C(1) = 8.4;

S(1) = 20 - C(1);

for i = 2:101

   C(i) = 0.965*C(i-1) + 0.017*S(i-1);

   S(i) = 0.035*C(i-1) + 0.983*S(i-1);

end

plot(years, C, 'b', 'LineWidth', 2);

hold on;

plot(years, S, 'r', 'LineWidth', 2);

xlabel('Years');

ylabel('Population');

legend('City', 'Suburbs');

title('Population of City and Suburbs Over 100 Years');

This MATLAB code calculates and plots the population of the city (in blue) and suburbs (in red) over the 100-year period.

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Answer:

x = 37 mm

Step-by-step explanation:

using the Pythagorean theorem:

m² = 13² - 5² = 169 - 25 = 144

m = √144 = 12 mm

x² = 12² + 35² = 144 + 1225 = 1369

x = √1369 = 37 mm

Answer:

x=37.3

Step-by-step explanation:

Since we're looking for "x" we need to use "the Pythagorean theorem".

13mm will be our "A" and 35mm will be our "B" will "x" being the hypotenuse.

Since x is the hypotenuse, it needs to be a number greater than 13 and 35.

c=√(a² + b²)

Making it c=√(13² + 35²)

Making C=37.336 or 37.3

Therefore making, x=37.3

When the total number of units produced over the five-year period is 125,000 units, the depreciation will be $1.20 pee unit.

The total number of units produced over the five-year period is 125,000 units. Therefore, the depreciation expense for each year would be calculated as follows:

Depreciation expense = (Cost of asset - Residual value) / Total number of units produced

Depreciation expense = ($150,000 - $12,000) / 125,000 units

Depreciation expense = $1.20 per unit

There are several factors that a company should consider when selecting a depreciation method. These factors include:

The type of asset being depreciated is an important factor to consider because different assets have different useful lives and expected patterns of depreciation. For example, a machine that is expected to be used for five years would be depreciated using a different method than a building that is expected to be used for 30 years.

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Step-by-step explanation:

D. In 2 _ In

maaf kalo salah

Answer: You want to calculate the interest on $380 at 5.3% interest per year after .5 year(s).

The formula we'll use for this is the simple interest formula, or:

Where:

P is the principal amount, $380.00.

r is the interest rate, 5.3% per year, or in decimal form, 5.3/100=0.053.

t is the time involved, 0.5....year(s) time periods.

So, t is 0.5....year time periods.

To find the simple interest, we multiply 380 × 0.053 × 0.5 to get your answer.

Step-by-step explanation:

Answer:

none of these

Step-by-step explanation:

There are 3 boys walking

There are a total of 20 people

3/20 = 0.15

That is 15 percent, therefore none of these answers.

Step-by-step explanation:

any has at least one mode

Answer:

 Forgot picture?

Step-by-step explanation:

Answer:

You can get your grade up by studying, getting a tutor, paying attention in class, taking good notes, asking questions, and cheating (i don't recommend this one :/)

Answer:

1/5

Step-by-step explanation:

i searched it up yikessssssss

The value of the given term is 1296.

A fraction consisting of a quotient and remainder is a mixed fraction. we can convert the mixed fraction to improper fraction by first dividing the numerator by denominator and then taking the quotient as whole number and remainder as the numerator of proper fraction keeping the denominator same.

We are given the number as [tex]\dfrac{1}{6} ^{-4}[/tex]

Here we can see that the power is negative so we can just reciprocal the term.

Then we get;

6^4

this means 6 x6x6x6 = 1296

The correct value is 1296.

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Answer:

$30.99

Step-by-step explanation:

The formula for simple interest is I = PRT where I = interest earned, P = principal amount borrowed/deposited, R = rate as a decimal, and T = time in years.

I = (30)(0.033)(1)

I = 0.99

Then add that to the amount deposited ($30) and you're done.

30 + 0.99 = $30.99

Please let me know if you have questions.

The answer is $29.01

Answer:

25

Step-by-step explanation:

First, plug -5 in for a, -(-5)^2. We treat the negative on the outside of the paranthese as a -1 so we do -1 times -5 and we get 5. Then we square 5 and get 25.

Answer:

here,hope this helps : )

Step-by-step explanation:

Answer: A= 32

a (Base) 6

b (Base) 10

h (Height) 4

Step-by-step explanation: A=a+b

2h=6+10

2·4=32    I really hoped this helped

The reformulated problem includes the original variables x₁, x₂, x₃, and the additional slack variables s₁, s₂, and s₃. The objective function and constraints remain the same, but the problem has been transformed into an equality form suitable for applying the big-M method or other linear programming techniques.

To reformulate the given linear programming problem using the big-M method, we introduce additional binary variables and big-M constants. The reformulated problem is as follows:

maximize 2x₁ + x₂ + x₃

subject to:

x₂ + x₃ + s₁ = 5

x₁ - x₂ + s₂ = 1

x₁ + x₂ + 3x₃ + s₃ = 25

x₁, x₂, x₃, s₁, s₂, s₃ ≥ 0

where s₁, s₂, and s₃ are slack variables, and we choose a large positive constant M to represent infinity.

Introducing slack variable s₁ for the inequality constraint x₂ + x₃ ≥ 5:

We add s₁ to the left-hand side of the inequality to transform it into an equation: x₂ + x₃ + s₁ = 5.

Introducing slack variable s₂ for the inequality constraint x₁ - x₂ ≥ 1:

We subtract s₂ from both sides of the inequality to transform it into an equation: x₁ - x₂ + s₂ = 1.

Introducing slack variable s₃ for the inequality constraint x₁ + x₂ + 3x₃ ≤ 25: We subtract s₃ from the left-hand side of the inequality to transform it into an equation: x₁ + x₂ + 3x₃ + s₃ = 25.

Since all variables in the original problem are non-negative (x₁, x₂, x₃ ≥ 0), we explicitly state this in the reformulation.

Therefore, the reformulated problem includes the original variables x₁, x₂, x₃, and the additional slack variables s₁, s₂, and s₃. The objective function and constraints remain the same, but the problem has been transformed into an equality form suitable for applying the big-M method or other linear programming techniques.

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Incomplete question:

Reformulate the following LP by the big-M method (only the reformulation is required, no need to solve the LP).

maxₓ 2x₁ + x₂ + x₃

s.t.

x₂+x₃ ≥ 5

x₁ - x₂ ≥ 1

x₁ + x₂+ 3x₃ ≤ 25

x₁, x₂, x₃  ≥0

Answer:

the graph has two vertical asymptotes, line q intersects the line at -8 and is the more important one.

Step-by-step explanation:

This is visible based off of the picture.

The probability that in the next 18 samples, exactly 2 contain the pollutant is

P(x = 2) =[tex]C(18, 2) (0.1)^2 (0.9)^16[/tex]

= 0.252.

The given is Each sample of water has a 10% chance of containing a particular organic pollutant.

Assume that the samples are independent with regard to the presence of the pollutant.

We need to find the probability that in the next 18 samples, exactly 2 contain the pollutant.

We can use the Binomial distribution formula to solve this problem.

The binomial distribution formula is given by:

P(x) = [tex]C(n, x) px qn-x[/tex]

Where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure.

Since there is a 10% chance of containing a particular organic pollutant, the probability of success is:

p = 0.1

The probability of failure is:

q = 1 - p

= 1 - 0.1

= 0.9

Number of trials (samples) is n = 18

(a) To find P(x≥≥4)

We need to find the probability that 2 or more samples contain the pollutant.

We can do this by subtracting the probability that less than 2 samples contain the pollutant from 1.

P(x ≥ 2) = 1 - P(x < 2)P(x < 2)

= P(x = 0) + P(x = 1)P(x = 0)

= C(18, 0) (0.1)0 (0.9)18

= [tex]0.9^18[/tex]

P(x = 1) = [tex]C(18, 1) (0.1)^1 (0.9)^17[/tex]

= 0.33

P(x < 2) = [tex]0.9^18[/tex]+ 0.33

= 0.335

P(x ≥ 2) = 1 - 0.335

= 0.665

(b) To find P(x≤≤4)

We need to find the probability that 4 or less samples contain the pollutant.

P(x ≤ 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)P(x = 0)

= C(18, 0) (0.1)0 (0.9)18

= [tex]0.9^18[/tex]

P(x = 1) = [tex]C(18, 1) (0.1)^1 (0.9)^17[/tex]

= 0.33

P(x = 2) = [tex]C(18, 2) (0.1)^2 (0.9)^16[/tex]

= 0.053

P(x = 3) = [tex]C(18, 3) (0.1)^3 (0.9)^15[/tex]

= 0.0047

P(x = 4) =[tex]C(18, 4) (0.1)^4 (0.9)^14[/tex]

= 0.0003

P(x ≤ 4) = 0.9^18 + 0.33 + 0.053 + 0.0047 + 0.0003

= 0.288

(d) To find P(3≤≤x≤≤7)

We need to find the probability that between 3 and 7 (inclusive) samples contain the pollutant.

P(3 ≤ x ≤ 7) = P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7)P(x = 3)

= [tex]C(18, 3) (0.1)^3 (0.9)^15[/tex]

= 0.0047

P(x = 4) = C(18, 4) (0.1)^4 (0.9)^14

= 0.0003

P(x = 5) = [tex]C(18, 5) (0.1)^5 (0.9)^13[/tex]

= 0.00001

P(x = 6) = [tex]C(18, 6) (0.1)^6 (0.9)^12[/tex]

= 2.15e-07

P(x = 7) = [tex]C(18, 7) (0.1)^7 (0.9)^11[/tex]

= 4.7e-10

P(3 ≤ x ≤ 7) = 0.0047 + 0.0003 + 0.00001 + 2.15e-07 + 4.7e-10

= 0.00471

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