(c) how large a sample size is necessary if the width of the 95% interval is to be 0.45? (round your answer up to the nearest whole number.)

Answer: con que

Step-by-step explanation:

Answer:

The prime factorization of 21 is;

Explanation:

We want to find the prime factorization of 21;

The number 21 is a product of 3 and 7;

since 3 and 7 are both prime number, then the prime factorization of 21 is;

To solve for k, you can start by isolating k on one side of the equation.

10 + 3k = 22

Subtract 10 from both sides:

3k = 12

Finally, divide both sides by 3:

k = 4

So the solution is k = 4.

Answer:

x = -2

Step-by-step explanation:

Step 1: Write equation

5(x + 6) + 11 = 25 - 3x

Step 2: Solve for x

Step 3: Check

Plug in x to verify it's a solution.

5(-2 + 6) + 11 = 25 - 3(-2)

5(4) + 11 = 25 + 6

20 + 11 = 25 + 6

31 = 31

Answer:

x=-8

Step-by-step explanation:

5x+30+11=

5x +41= 25-3x

5x+ 41-25= 25-25-3x

5x+16=3x

5x-5x= 3x-5x

16=-2x

16/-2=x

-8=x

Answer:

To be honest with you it would more likely be more easier to divide first

The reason to that is because dividing is harder then subtracting and if your subtract second it would just be easier

Answer:

To reduce bias when estimating mean commuting time

Step-by-step explanation:

Given that the selection of adults is based on the three strata found in the region; urban, suburban, and rural areas where the population and proportion of adults commuting to work is progressively decreasing, using a uniform number or sample size will not increase sample error as there are more commuters in the urban areas than in the suburban or rural areas, likewise, there are more commuters in the suburban area than in the rural areas, and as such the proportion of those sampled in the urban  areas should be more than those surveyed in the suburban while those surveyed in the suburban areas should be more than those surveyed in the rural areas to reduce bias when estimating the mean as there will be more general representation.

For a Poisson process with rate λ, the interarrival times between events are exponentially distributed with parameter μ = 1/λ. So, the time between the (n-1)-th and n-th event, denoted as Tn, follows an exponential distribution with parameter μ = 1/3 minutes.

Since Sn is the sum of the first n interarrival times, we have:

Sn = T1 + T2 + ... + Tn

The sum of n exponential random variables with parameter μ is a gamma random variable with shape parameter n and scale parameter μ. Therefore, Sn follows a gamma distribution with shape parameter n and scale parameter μ.

In this case, n = 10 and μ = 1/3. So, E[S10] can be calculated as:

E[S10] = n * μ = 10 * (1/3)

= 10/3 minutes.

Therefore, E[S10] = 10/3 minutes.

b. E[S4|N(1) = 3]:

Given that N(1) = 3, we know that there are 3 events in the first minute. Therefore, the time of the 4th event, S4, will be the sum of the first 3 interarrival times plus the time between the 3rd and 4th event.

Using the same reasoning as in part a, we know that the sum of the first 3 interarrival times follows a gamma distribution with shape parameter 3 and scale parameter 1/3. The time between the 3rd and 4th event, denoted as T4, follows an exponential distribution with parameter 1/3.

So, S4 = T1 + T2 + T3 + T4.

Since T1, T2, T3 are independent of T4, we can calculate E[S4|N(1) = 3] as:

E[S4|N(1) = 3] = E[T1 + T2 + T3 + T4]

= E[T1 + T2 + T3] + E[T4]

= (3/3) + (1/3)

= 4/3 minutes.

Therefore, E[S4|N(1) = 3] = 4/3 minutes.

c. Var(S10):

The variance of Sn, Var(Sn), for a Poisson process with rate λ, is given by:

Var(Sn) = n * σ^2,

where σ^2 is the variance of the interarrival times.

In this case, n = 10 and the interarrival times are exponentially distributed with parameter μ = 1/3. The variance of an exponential distribution is \(\mu^2\)So, \(\sigma^2 = \left(\frac{1}{3}\right)^2\)

= 1/9.

Substituting the values into the formula, we have:

Var(S10) = 10 * (1/9)

= 10/9.

Therefore, Var(S10) = 10/9.

d. E[N(4) − N(2)|N(1) = 3]:

Given that N(1) = 3, we know that there are 3 events in the first minute. Therefore, at time t = 2 minutes, there will be 3 - 1 = 2 events that have already occurred.

Now, we need to find the expected value of the difference in the number of events between time t = 4 minutes and t = 2 minutes, given that there were 3 events at t = 1 minute.

Since the number of events in a Poisson process follows a Poisson distribution with rate λt, where t is

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The transition probability matrix is:

\(\left[\begin{array}{cccc}0.5&0.1&0.2&0.2\\0.75&0&0.6 &0.3\\0.1&0.5&0 &0.1\\0.05&0.1&0.25&0\end{array}\right]\)

What is matrix?

A matrix is a rectangular array of numbers or other mathematical objects for which operations such as addition, subtraction, multiplication, and scalar multiplication are defined.

The Markov chain for housing movements can be formulated as follows:

State 1: Apartment

State 2: Condominium

State 3: Own House

State 4: Rented House

The transition probability matrix P is given by:

\(\left[\begin{array}{cccc}P_{11}&P_{12}&P_{13} &P_{14}\\P_{21}&P_{22}&P_{23} &P_{24}\\P_{31}&P_{32}&P_{33} &P_{34}\\P_{41}&P_{42}&P_{43} &P_{44}\end{array}\right]\)

where \($P_{ij}$\) is the probability of moving from state i to state j. To calculate these probabilities, we need to use the data from the survey.

For example, \($P_{12}$\) is the probability of moving from an apartment to a condominium. From the survey data, we can see that out of 200 apartment dwellers, 100 moved to another apartment, 20 moved to a condominium, 40 moved to their own house, and 40 moved to a rented house. Therefore, \($P_{12} = 20/200 = 0.1$\).

Similarly, we can calculate the other transition probabilities:

\(P_{13} = 40/200 = 0.2$\\$P_{14} = 40/200 = 0.2$\\$P_{21} = 150/200 = 0.75$\\$P_{23} = 120/200 = 0.6$\\$P_{24} = 60/200 = 0.3$\\$P_{31} = 20/200 = 0.1$\\$P_{32} = 100/200 = 0.5$\\$P_{34} = 20/200 = 0.1$\\$P_{41} = 10/200 = 0.05$\\$P_{42} = 20/200 = 0.1$\\$P_{43} = 50/200 = 0.25$\)

Therefore, the transition probability matrix is:

\(\left[\begin{array}{cccc}0.5&0.1&0.2&0.2\\0.75&0&0.6 &0.3\\0.1&0.5&0 &0.1\\0.05&0.1&0.25&0\end{array}\right]\)

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The total cost of Kayla's purchase of 0.1 pounds of medium binder clips and 0.2 pounds of jumbo binder clips is $1.20.

To calculate the total cost, we need to determine the cost of each type of binder clip separately and then sum them up.

The cost of medium binder clips is $5 per pound, and Kayla purchased 0.1 pounds. Therefore, the cost of medium binder clips is 0.1 pounds * $5 per pound = $0.50.

The cost of jumbo binder clips is $6 per pound, and Kayla purchased 0.2 pounds. Thus, the cost of jumbo binder clips is 0.2 pounds * $6 per pound = $1.20.

Finally, to find the total cost, we add the individual costs together: $0.50 + $1.20 = $1.70.

Therefore, the total cost of Kayla's purchase of 0.1 pounds of medium binder clips and 0.2 pounds of jumbo binder clips is $1.70.

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Hello !

_______

\(\huge {\boxed {\underline{Explanation :}}}\)

\(\sf{5 \: x \: \frac{3}{4}\)

_______

\(\sf{Apply \: the \: fraction \: rule :} \: a \: ^. \: \frac{b}{c} = \frac{a \: ^. \: b}{c}\)

\(\sf{\frac{5x \: ^. \: 3}{4}\)

_______

\(\sf{Multiply \: the \: numbers:} \: 5 \: ^. \: 3 = 15\)

\(\sf{= \frac{15x}{x}\)

Hopefully This Helps ! ~

#LearnWithBrainly

\(\underline{Answer :}\)

Dollixna ~

Answer: 75% increase

Step-by-step explanation:

First, you subtract the two numbers to find out how much increased.

483-276= 207

Divide by original number

207/276= \(\frac{3}4}\)

Multiply by 100

\(\frac{3}{4}*100=\) 75% increase

Answer:

16 is definitely a

17 is c

Step-by-step explanation:

an app that I use that helps a lot is desmos u should try it out

hope this helps :)

can i get brainliest?

The number of heads tossed on four distinct coins is a discrete random variable.

A discrete random variable can be a count or a finite set of values. Out of the options given in the question, the random variable that is discrete is the number of heads tossed on four distinct coins.

The correct option is: The number of heads tossed on four distinct coins is a discrete random variable.

The time spent waiting for a bus at the bus stop is a continuous random variable because time can take on any value in a given range. The amount of water traveling over a waterfall in one minute is also a continuous random variable because the water can flow at any rate.

The mass of a test cylinder of concrete is also a continuous random variable because the mass can take on any value within a certain range.

The number of heads tossed on four distinct coins, on the other hand, is a discrete random variable because it can only take on certain values: 0, 1, 2, 3, or 4 heads.

Hence, the number of heads tossed on four distinct coins is a discrete random variable.

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Answer:

5. 3\(x^{4}\)

6. id k this one

The measure of the mid - segment is 53 .

Given :

the measures of the bases of a trapezoid are 35 and 71.

What is trapezoid ?

A trapezoid is a four-sided shape which has one pair of sides as parallel. It is basically a two-dimensional shape or figure similar to a square etc .

Mid segment :

A mid segment of a trapezoid is the line segment connecting the midpoints of the two non-parallel sides of a trapezoid.

A trapezoid mid segment is parallel to the set of parallel lines in a trapezoid and is equal to the average of the lengths of the bases.

Measure of mid segment = sum of two bases / 2

= 35 + 71 / 2

= 106 / 2

= 53

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The first option (n − 1)s/ s1^2/s2^2 has an F distribution.

In statistics, the F-distribution is a probability distribution that is used to compare the variances of two independent samples. The F-distribution is defined as the ratio of two independent chi-square distributions, each divided by its degrees of freedom. The query lists three options: (n − 1)s/ s12/s22, (n − 1)s1/s2, and s1/s2, and asks which of these has an F-distribution. The correct option is (n − 1)s1/s2, which has an F-distribution with (n − 1) numerator degrees of freedom and (n − 1) denominator degrees of freedom [1]. The other options are not F-distributions.

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Answer:

A. 232 acres

Step-by-step explanation:

Got it right

The number of acres of land that were harvested is given by A = 232 acres

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

Substituting the values in the equation , we get

Let the total amount of harvest of bushels be = 9,396 bushels

And , the amount of bushels per acre = 40.5

So , number of acres of land that were harvested is A = amount of harvest of bushels / amount of bushels per acre

On simplifying the equation , we get

The number of acres of land that were harvested is A = 9,396 / 40.5

The number of acres of land that were harvested is A = 232 acres

Hence , the number of acres is 232 acres

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Based on the given values, we can compute the expected returns and expected standard deviations for different weightings of the stocks in the portfolio. The results are as follows:

a. When w1 = 1.00, the expected return of the two-stock portfolio is 0.13, and the expected standard deviation is 0.03.

b. When w1 = 0.65, the expected return of the two-stock portfolio is 0.1095, and the expected standard deviation is 0.0214.

c. When w1 = 0.60, the expected return of the two-stock portfolio is 0.104, and the expected standard deviation is 0.0222.

d. When w1 = 0.30, the expected return of the two-stock portfolio is 0.074, and the expected standard deviation is 0.0262.

e. When w1 = 0.10, the expected return of the two-stock portfolio is 0.038, and the expected standard deviation is 0.0324.

To calculate the expected return of the two-stock portfolio, we use the weighted average of the individual expected returns based on the given weights. For example, in part (a), where w1 = 1.00, the expected return is simply equal to E(R1) = 0.13.

To calculate the expected standard deviation of the two-stock portfolio, we use the formula:

σ = √(w1^2 * E(a1)^2 + w2^2 * E(q2)^2 + 2 * w1 * w2 * E(a1) * E(q2) * ρ)

where E(a1) is the expected standard deviation of stock 1, E(q2) is the expected standard deviation of stock 2, and ρ is the correlation coefficient.

Regarding the risk-return graph, without the specific details of the graph options provided, it is not possible to determine which graph is correct for the given weightings and correlation coefficient. The graph would typically depict the risk-return tradeoff for different weightings and correlation coefficients, showing the relationship between expected return and expected standard deviation of the portfolio.

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Answer:

2

Step-by-step explanation:

To evaluate f(x) when x = - 4, substitute x = - 4 into f(x), that is

f(- 4) = 3(- 4) + 14 = - 12 + 14 = 2

Answer:

Step-by-step explanation:

Answer:

4,060,000,000

Step-by-step explanation:

Take 4.06 remove the decimal then add a zero for the number beside ten. So 10^7 add 7 zeros behind 406.

Answer:

$4400

Step-by-step explanation:

I=PRT

I=(2000)(8%)(15)

I=4400

Hicksian demand functions are:x1** = 2P₁x₂ ; x₂** = P₂²

Utility function: u(x1+x2) = .5ln(x1) + .25ln(x₂) .The Marshallian demand functions are: x1* = - and x₂* = P₂.

The indirect utility function is found by substituting Marshallian demand functions into the utility function and solving for v(P₁, P₂, Y).u(x1*,x2*) = v(P₁,P₂,Y) ⇒ u(-, P₂) = v(P₁,P₂,Y) ⇒ .5ln(-) + .25ln(P₂) = v(P₁,P₂,Y) ⇒ v(P₁,P₂,Y) = - ∞ (as ln(-) is not defined)

Thus the indirect utility function is undefined.

Minimum expenditure function can be derived from the Marshallian demand function and prices of goods:

Exp = P₁x1* + P₂x2* = P₁(-) + P₂P₂ = -P₁ + P₂²

Minimum expenditure function is thus:

Exp = P₁(-) + P₂²

Hicksian demand functions can be derived from the utility function and prices of goods:

H1(x1, P1, P2, U) = x1*H2(x2, P1, P2, U) = x2*

Hicksian demand functions are:

x1** = 2P₁x₂

x₂** = P₂²

If there are no restrictions on the amount of money the consumer can spend, the Hicksian demand functions for x1 and x2 coincide with Marshallian demand functions.

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Answer:

11/14

Step-by-step explanation:

11/4 / 3 1/2

3 1/2= 3*2+1= 6+1=7/2

11/4 * 2/7= 22/28

22/2=11

28/2= 14

11/14

11/14 is the answer! (Quotient in simplest form)

Answer:

90% are good to eat !

Step-by-step explanation:

Answer:

90%

Step-by-step explanation:

Well using a fraction

1/10 had mold or in this case 10 percent had mold on it.

The other 90percent are still safe

If the columns of matrix B are linearly dependent, then the columns of AB are also linearly dependent.

Suppose the columns of matrix B are linearly dependent, which means there exist constants c₁, c₂, ..., cₙ (not all zero) such that c₁b₁ + c₂b₂ + ... + cₙbₙ = 0, where b₁, b₂, ..., bₙ are the columns of B.

Now, let's consider the product AB. Suppose A is an m × n matrix.

The jth column of AB is given by the product AB(:, j) = A(b₁j) + A(b₂j) + ... + A(bₙj), where b₁j, b₂j, ..., bₙj are the jth columns of B.

Substituting the linear dependence relation for the columns of B into the expression for AB(:, j), we get:

AB(:, j) = A(c₁b₁ + c₂b₂ + ... + cₙbₙ)

         = c₁A(b₁) + c₂A(b₂) + ... + cₙA(bₙ)

Since matrix multiplication distributes over column vectors, we have:

AB(:, j) = c₁A(b₁) + c₂A(b₂) + ... + cₙA(bₙ)

This shows that the jth column of AB can be expressed as a linear combination of the columns of A, multiplied by the corresponding coefficients c₁, c₂, ..., cₙ. Therefore, if the columns of B are linearly dependent, the columns of AB are also linearly dependent.

In conclusion, if the columns of matrix B are linearly dependent, then so are the columns of the product AB.

Complete Question:

Show that if the columns of B are linearly dependent, then so are the columns of AB.

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Answer:

37 boxes

Step-by-step explanation:

At first, Dymin had some boxes. Let x be the starting number of boxes.

Then, she bought 7 more boxes. Let y be the new number of boxes that she had.

y = x + 7

Next, her little brother destroyed half of them. It means half of y. So, the number of boxes left is

Number of boxes = y/2

The number of the boxes left is 22 boxes. Then, we can use the equation above.

Number of boxes = 22

y/2 = 22

Substitute y = x + 7 into the equation.

(x + 7)/2 = 22

x + 7 = 44

x = 44 - 7

x = 37

Thus, Dymin has 37 boxes at first.

Answer:

0.25

Step-by-step explanation:

:)

Answer: the absolute value of 1/4 is 1

Step-by-step explanation:

Answer:

x = 6

Step-by-step explanation:

Coach ticket = $180

first class ticket = $1020

people = 8

Let x be number of coach tickets

Then number of first class tickets = (8 - x)

x * 180 + (8 - x) * 1020 = 3120

180 X + 8 * 1020 - 1020 x = 3120

180 x + 8160 - 1020 x = 3120

180 x - 1020 x = 3120 -8160

- 840 x = - 5040

840 x = 5040

x = 6

So 6 coach class tickets and 2 first class tickets

Answer:

She bought 8 coach tickets

And 5 first class tickets

Step-by-step explanation:

390 x 8 = 3,120

1000 x 5 = 5,000

3,120 + 5,000 = 8,120

a) The distance of the boat from its starting point is 20 km.

b. The nearest degree  is 63°

c. The bearing g that the boat should travel on to return to the starting point is 360° - 63° = 297°.

To find the distance of the boat from its starting point, we can use the Pythagorean theorem.

x^2 + y^2 = d^2

Substituting the values for x and y, we get:

12^2 + 16^2 = d^2

d^2 = 144 + 256

d^2 = 400

d = 20

Hence, the distance of the boat from its starting point is 20 km.

b) To find the bearing that the boat should travel on to return to the starting point, we must calculate the angle between the two legs of the journey.

b = arctan (y / x) = arctan (16 / 12) = 63.43°

To the nearest degree  is 63°.

c) The bearing that the boat should travel on to return to the starting point is the opposite direction of the angle formed by the two legs of the journey. Therefore, the bearing is 360° - 63° = 297°.

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