I don’t understand this can someone help me

The cross product of  three vectors is not associative in nature.

Considering three vectors such  as : A=(2,1,3), B=(1,0,2) and  C=(1,1,0)

we  know that the  associative law depicts that :

A x (B x C) = (A x B) x C ,while doing the operation for the  considered vectors :

A x (B x C) =(2,1,3)x ((1,0,2)x (1,1,0))

                  =(2,1,3)x (-2,2,1)

     =(-5,-8,6)

(A x B) x C=  ((2,1,3)x (1,0,2))x (1,1,0)

            = (2,-1,1) x (1,1,0)

  =(-1,1,3)

So  for this vector operation, A x (B x C)  is not  equal to  (A x B) x C, so you can sum up that cross product of three vectors can't be associative every possible time.

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Answer: false

Step-by-step explanation:

linear equations do not have x's with exponents

Answer: The probability is 100.

Step-by-step explanation:

Maine has a cold climate in winter which means that we can expect cold weather when winter season reaches Maine.

January is a winter month so the probability that the temperature in Maine will drop below 32° in January is a certain occurrence which means that the probability is a 100%.

The sum of angle in a triangle is 180°

A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted \triangle ABC.

The sum of angle A, B and C is 180 i.e A+B +C = 180°

Also a triangle is a 3 sided polygon. The sum of of angle in a polygon is( n-2)180

How do we prove that the sum of angle in a triangle is 180°?

Since triangle is 3 sided, n= 3, because n denote the number if sides

therefore the sum of angle = (n-2) 180 = (3-2)×180

= 180°

therefore the sum of angle In a triangle is 180°

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Answer:

x=2

Step-by-step explanation:

Substitute y=0    0=0.5(x-2) 21

multiply the numbers     0=10.5(x-2)        *(21x0.5)

remove the parenthesis     0=10.5x-21

move the variable      -10.5x=-21

divide both sides by   -10.5

solution     x=2

The coin is flipped at least 52 times in order to obtain a 99% confidence interval of width of at most . 18 for the probability of flipping a head.

Hence, Option F is correct answer.

In a sample with a number n of people surveyed with a probability of a success of π , and a confidence level of (1-α), we have the following confidence interval of proportions.

\(\pi\) ± z\(\sqrt{\frac{\pi \p(1-\pi )}{n} }\)

z is the z-score that has a pvalue of \(1-\frac{\alpha }{2}\)

Since, the coin is fair, so \(\pi =0.5\).

The margin of error is:

\(M=z\sqrt{(\frac{\pi (1-\pi )}{n} )}\)

99% confidence level:

So \(\alpha =0.01\) ,z is the value of Z that has a p value of 1-\(\frac{0.01}{2}\)=0.995,

so Z= 2.575

We have to flip the coin in order to obtain a 99% confidence interval of width of at most 18 for the probability of flipping a head at least n times.

n is found when M=0.18 .

So

M=\(z\sqrt{\frac{\pi (1-\pi )}{n} }\)

\(0.18=2.575\sqrt{\frac{0.5*0.5}{n} }\)

\(0.18\sqrt{n} =2.575*0.5\)

\(\sqrt{n}=\frac{2.575*0.5}{0.18}\)

\(\sqrt{n} ^{2} =(\frac{2.575*0.5}{0.18} )^{2}\)

n = 51.16

Thus, we have to flip the coin at least 52 times.

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Answer:

Step-by-step explanation:

The first answer choice is correct i ma 100% sure

Answer:

(-1.2,0.3)

Step-by-step explanation:

does this help you

Laura has 15 red sheets after using the final blue sheet and 3 red sheets from the given ratio. There were originally 19 sheets of construction paper in the bundle.

B/R = 2/7

The day before

=18 red + 1 blue

=19 sheets.

Given that she still possesses 15 red and that the ratio of B/R=2/7, she must have started with:

The original batch of paper contained 135 sheets, or 15 x 7 = 105 red sheets and 2/7 x 105 = 30 blue sheets.

4 sheets every day, during a period of 135 - 19 = 116 sheets, were used: 116/4 =29 days.

18 red sheets remain for the final day (105 - 29*3).

30 - 29*1 = 1 blue sheet remaining for the final day.

The correct question is : In a pack of construction paper, the numbers of blue and red sheets are originally in the ratio 2:7. Each day, Laura uses 1 blue sheet and 3 red sheets. One day, she uses 3 red sheets and the last blue sheet, leaving her with 15 red sheets. How many sheets of construction paper were in the pack originally?

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Answer:

.?

I don’t understand:/

The probability density function (pdf) of the largest of the five waiting times is given by: f(x) = 4/10^5 * x^4, where x is a real number between 0 and 10. The expected value of the largest of the five waiting times is 8.33 minutes.

The pdf of the largest of the five waiting times can be found by considering the order statistics of the waiting times. The order statistics are the values of the waiting times sorted from smallest to largest.

In this case, the order statistics are X1, X2, X3, X4, and X5. The largest of the five waiting times is X5.

The pdf of X5 can be found by considering the cumulative distribution function (cdf) of X5. The cdf of X5 is given by: F(x) = (x/10)^5

where x is a real number between 0 and 10. The pdf of X5 can be found by differentiating the cdf of X5. This gives: f(x) = 4/10^5 * x^4

The expected value of X5 can be found by integrating the pdf of X5 from 0 to 10. This gives: E[X5] = ∫_0^10 4/10^5 * x^4 dx = 8.33

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Answer:The top answer is 15

Step-by-step explanation: good luck, kid

The correct answer is option B. No value of k will make AB = BA

To check whether AB = BA, we need to multiply the matrices AB and BA and see if they are equal.

AB = [ 2 3-1 1] [ 2 9-3 k ]

[ 1 0 2 ] [ 0 0 1 ]

= [ 4 + 3(0) - 2k 18 - 9(0) - 3k ]

[ 0 + 0 + 2 0 + 0 + 1 ]

= [ 4 - 2k 18 - 3k ]

[ 2 1 ]

BA = [ 2 9-3 k ] [ 2 3-1 1]

[ 0 0 1 ] [ 1 0 2 ]

= [ 4 + 27 - 6k 6 - 3k - 3 ]

[ 1 0 ]

= [ 31 - 6k 3 - 3k ]

[ 1 0 ]

For AB = BA, we need to have

4 - 2k = 31 - 6k and 18 - 3k = 3 - 3k

Solving the first equation gives k = 9, and substituting k = 9 into the second equation gives 15 = 0, which is not true. Therefore, there is no value of k that makes AB = BA.

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I'm not quite sure what the question is, but I think that John's age would be the independent variable, and John's height would be the dependent variable.

The unit rate of the cost of the fluid per ounce is $0.11

How to calculate the cost of the fluid per ounce ?

Chee buys two pint bottle of juice

The two pints costs $3.52

The unit rate of the cost of the juice per fluid ounce can be calculated as follows

1 pint is equals to 16 fluid ounce

2 pints= 2 × 16

= 32

The cost per fluid ounce is

= 3.52/32

= 0.11

Hence the unit rate of the cost of the juice per fluid ounce is $0.11

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Answer:

Step-by-step explanation:

When the diver hits the water the height, h(t), becomes zero.  We set h(t) equal to zero and solve the resulting equation for t:

h(t) = -4t^2+ 11t + 3 = 0

The coefficients of this quadratic equation are {-4, 11, 3}, and so the discriminant (needed in the quadratic formula) is -11 - 4(-4)(3) = 47.

                                            -11 ± √37               11 + √37            11 - √37

Therefore the roots are t = --------------, or t = ------------- or t = ---------------

                                                  -8                          8                        8

Both of these results are positive.  Picture the graph as one opening down and intersecting the x-axis in two places.  The diver hits the water (and h becomes zero) at the later time:

       11 + √37

t = ---------------- sec (a little over 2 sec)

              8

The best description about the sampling distribution of means for the student's sample is mu x = 45, sigma x = 3.75, shape of distribution unknown.

Sampling distribution is defined as the probability distribution of a statistical measure which is got by the repeated sampling of a specific population.

Given that,

μ = 45 minutes and σ = 15 minutes

Sampling distribution of means for the student's sample is :

μₓ = μ = 45

σₓ = σ / √n, where n is the sample size.

σₓ = 15 / √(16) = 15 / 4 = 3.75

Now since the sample size 16 which is less than 30, shape of the distribution is unknown.

Hence the sampling distribution is, μₓ = 45, σₓ = 3.75, and the shape of distribution is unknown.

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Answer:

increase

Step-by-step explanation:

Answer:

20

Step-by-step explanation:

Answer:

0.05

Step-by-step explanation:

Calc it

The discrete-time representation of the system at T = 0.1 is given by the ratio of two polynomials, where the numerator and denominator coefficients are determined using the bilinear transform. The BIBO stability and stability nature (asymptotic or marginal) cannot be determined without the pole locations in the z-plane.

To obtain the discrete-time representation of the system at T = 0.1, we can use the bilinear transform method. The bilinear transform maps the s-domain to the z-domain using the equation: \(s = (2/T)((z-1)/(z+1))\).

First, let's apply the bilinear transform to the given transfer function G(s):

\(G(z) = G(s)|s=(2/T)((z-1)/(z+1))\)

Substituting the transfer function \(G(s) = (6s^3 + s^2 + 3s - 20)/(2s*4 + 7s^3 + 15s^2 + 16s + 10)\), we have:

\(G(z) = [(6((2/T)((z-1)/(z+1)))^3 + ((2/T)((z-1)/(z+1)))^2 + 3((2/T)((z-1)/(z+1))) - 20)] / [2((2/T)((z-1)/(z+1)))^4 + 7((2/T)((z-1)/(z+1)))^3 + 15((2/T)((z-1)/(z+1)))^2 + 16((2/T)((z-1)/(z+1))) + 10]\)

Simplifying the expression, we have:

\(G(z) = [(6(2/T)^3((z-1)^3/(z+1)^3) + (2/T)^2((z-1)^2/(z+1)^2) + 3(2/T)((z-1)/(z+1)) - 20] / [2(2/T)^4((z-1)^4/(z+1)^4) + 7(2/T)^4((z-1)^3/(z+1)^3) + 15(2/T)^2((z-1)^2/(z+1)^2) + 16(2/T)((z-1)/(z+1)) + 10]\)

Simplifying further, we obtain the discrete-time representation of the system:

\(G(z) = [c_0z^4+ c_1z^3+ c_2z^2 + c_3z + c_4] / [d_0z^4+ d_1z^3 + d_2z^2 + d_3z + d_4]\)

where:

\(c_0 = (6(2/T)^3 + 2(2/T)^2 + 3(2/T) - 20)\\c_1 = (-3(6(2/T)^3+ 2(2/T)^2) + 16(2/T))\\c_2 = (3(6(2/T)^3) - 15(2/T)^2)\\c_3 = (-6(2/T)^3 + 7(2/T)^3)\\c_4 = 2(2/T)^4\\d_0 = 2(2/T)^4\\d_1 = 7(2/T)^4 + 16(2/T)\\d_2 = 15(2/T)^2\\d_3 = -6(2/T)^2\\d4 = 10\)

Now, let's analyze the properties of the discrete system:

(i) BIBO Stability:

To determine if the discrete-time system is BIBO stable, we need to check if all the poles of the transfer function G(z) lie inside the unit circle in the z-plane. If all poles are inside the unit circle, the system is BIBO stable.

(ii) Asymptotic or Marginally Stable:

By analyzing the pole locations of the discrete system, we can determine if it is asymptotically stable (all poles inside the unit circle) or marginally stable (some poles on the unit circle).

(iii) Controllability:

The controllability of the discrete system can be assessed by examining the controllability matrix based on the system's state-space representation. Unfortunately, without the state-space representation, it is not possible to determine controllability.

(iv) Observability:

The observability of the discrete system can be determined by examining the observability matrix based on the system's state-space representation. However, without the state-space representation, it is not possible to determine observability.

To fully analyze the discrete system, we would need the state-space representation or further information. However, we can still determine the BIBO stability and stability nature (asymptotic or marginal) based on the pole locations in the z-plane.

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There are 420 ways to select a committee of 4 people with at least 2 women.

To solve these problems, we can use the formula for combinations. The number of ways to choose k items from a set of n items is given by:

C(n, k) = n! / (k! * (n - k)!)

where "!" denotes the factorial function.

To select a committee of 4 people from a pool of 12 people, we can use the formula:

C(12, 4) = 12! / (4! * 8!) = 495

So there are 495 ways to select a committee of 4 people without any additional criteria.

To select a committee of 4 people with 2 men and 2 women, we can first count the number of ways to choose 2 men from the 5 men and 2 women from the 7 women. This gives us:

C(5, 2) * C(7, 2) = (5! / (2! * 3!)) * (7! / (2! * 5!)) = 10 * 21 = 210

So there are 210 ways to select a committee of 4 people with 2 men and 2 women.

To select a committee of 4 people with at least 2 women, we can count the number of committees with exactly 2 women, 3 women, or 4 women, and add them together.

The number of committees with exactly 2 women is:

C(7, 2) * C(5, 2) = 21 * 10 = 210

The number of committees with exactly 3 women is:

C(7, 3) * C(5, 1) = 35 * 5 = 175

The number of committees with all 4 women is:

C(7, 4) = 35

So the total number of committees with at least 2 women is:

210 + 175 + 35 = 420

Therefore, there are 420 ways to select a committee of 4 people with at least 2 women.

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For the given numerical example in a classification problem, the logistic loss function is always less than or equal to zero, indicating that a solution that minimizes the objective function cannot be found, possibly due to non-linearly separable data.

Since this is a classification problem with binary labels, we can use the logistic loss function given by:

L(y, h(x)) = log(1 + exp(-y * h(x)))

where y is the true label (either 1 or -1), h(x) is the predicted value, and exp is the exponential function.

In this case, λ = 0.5, y = 1, x = [10], and θ^ = [θ1^,θ2^]. We want to minimize the objective function:

f(θ^) = λ/2 * ||θ^||^2 + L(y, θ^ · x)

Substituting in the values, we get:

f(θ^) = 0.25 * (θ1^2 + θ2^2) + log(1 + exp(-θ1^ * 10))

To solve for θ1^ and θ2^, we need to find the partial derivatives of f(θ^) with respect to θ1^ and θ2^ and set them equal to zero:

∂f(θ^)/∂θ1^ = 0.1 * exp(-θ1^ * 10) * (1 / (1 + exp(-θ1^ * 10))) + 0.5 * θ1^ = 0

∂f(θ^)/∂θ2^ = 0.5 * θ2^ = 0

The second equation gives θ2^ = 0. Setting the first equation to zero and simplifying, we get:

exp(-θ1^ * 10) = -5

Since the exponential function is always positive, there are no solutions to this equation. However, we can use the hint given in the problem to show that the loss function is always less than or equal to zero for this example:

L(y, θ^ · x) = log(1 + exp(-10 * θ1^))

= log(1 + exp(-10 * (θ1^ - log(5))))

Since exp(-10 * (θ1^ - log(5))) is always less than or equal to 1, log(1 + exp(-10 * (θ1^ - log(5)))) is always less than or equal to 0. Therefore, the loss function is always less than or equal to 0.

So, we cannot find a solution that minimizes the objective function for this example. This suggests that the data is Non -linearly separable.

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_____The given question is incomplete, the complete question is given below:

Consider minimizing the above objective fuction for the following numerical example:

λ=0.5,y=1,x=[10]

Note that this is a classification problem where points lie on a two dimensional space. Hence θ^ would be a two dimensional vector.

Let θ^=[θ1^,θ2^], where θ1^,θ2^ are the first and second components of θ^ respectively.

Solve for θ1^,θ2^.

Hint: For the above example, show that Lossh(y(θ^⋅x))≤0

Answer:

ok. its  8/24. but if you got to simplify its 1/3

Step-by-step explanatin

you do 5*4 (the numerator) and 6*4(the denomerator).

this is the lcd(least comon )denominator.

repaet the steps for 2/4 but multiply by 6 on each number

once    

thats

done

its

your

answer

Answer:

5/6 – 2/4

= (10–6)/12

= 4/12

= 1/3

89 /3

frogs = 29.6

fish =  29.6

dragonflies =  29.6

all of them equal 89

The expected value of the winnings on a $6 bet placed on red in European roulette is -$0.81.

In European roulette, the probability of winning a bet on red is 18/37, and the probability of losing is 19/37. If the bet wins, the payout is even money, which means the player receives $6 in addition to getting their original $6 bet back. If the bet loses, the player loses their $6 bet. Therefore, the expected value of the winnings can be calculated as:

Expected value = (probability of winning * payout if win) - (probability of losing * amount bet)

Expected value = (18/37 * $12) - (19/37 * $6)

Expected value = -$0.81

This means that on average, the player can expect to lose $0.81 for every $6 bet placed on red in European roulette.

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square: a quadrilateral with four right angles and four congrugent sides.

rectangle: a quadrilateral with four right angles

quadrilateral: 4-sided closed figure

rhombus: a quadrilateral with four congrugent sides

parallelogram: a quadrilateral in which opposite sides are parallel

The cycle efficiency, also known as the operating cycle or cash conversion cycle, is a measure of how efficiently a company manages its working capital.

In this case, with 23 days' sales in accounts receivable, 80 days' sales in inventory, and 43 days' payable outstanding, the cycle efficiency can be calculated.

The cycle efficiency measures the time it takes for a company to convert its resources into cash flow. It is calculated by adding the days' sales in inventory (DSI) and the days' sales in accounts receivable (DSAR), and then subtracting the days' payable outstanding (DPO).

In this case, the DSI is 80 days, which indicates that it takes 80 days for the company to sell its inventory. The DSAR is 23 days, which means it takes 23 days for the company to collect payment from its customers after a sale. The DPO is 43 days, indicating that the company takes 43 days to pay its suppliers.

To calculate the cycle efficiency, we add the DSI and DSAR and then subtract the DPO:

Cycle Efficiency = DSI + DSAR - DPO

= 80 + 23 - 43

= 60 days

Therefore, the cycle efficiency for the company is 60 days. This means that it takes the company 60 days, on average, to convert its resources (inventory and accounts receivable) into cash flow while managing its payable outstanding. A lower cycle efficiency indicates a more efficient management of working capital, as it implies a shorter cash conversion cycle.

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The 90% confidence interval for the mean length of stay for patients with abdominal surgery is [5.64, 5.86].

The formula for the confidence interval is given by: \($CI=\bar{X}\pm Z_{\alpha/2} \times \frac{S}{\sqrt{n}}$\)

Given data:Sample size, n = 2121,Sample mean, \($\bar{X}$\) = 5.75,Sample standard deviation, S = 1.11 Confidence interval, CI = 90%

Now, we need to find the value of Zα/2 from the Z-table.

The value of α/2 is given by 1-90/100 = 0.1.

Therefore, α/2 = 0.05.From the Z-table, we can find the Z-value corresponding to 0.05, which is 1.645.

Substituting all these values in the formula, we get:\(\[\begin{aligned}CI&=\bar{X}\pm Z_{\alpha/2} \times \frac{S}{\sqrt{n}}\\&=5.75\pm 1.645 \times \frac{1.11}{\sqrt{2121}}\\&=5.75\pm 0.106\\&=[5.64, 5.86]\end{aligned}\]\)

Therefore, the 90% confidence interval for the mean length of stay for patients with abdominal surgery is [5.64, 5.86].

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We can conclude that gcd(a, b) divides 15 if and only if there exist integers s and t such that as + bt = 15.

If there are integers s and t such that as + bt = 15, then we can conclude that gcd(a, b) divides 15. This is known as Bézout's identity, which states that for any two integers a and b, there exist integers s and t such that as + bt = gcd(a, b).

To see why this is true, consider the set of all linear combinations of a and b, that is, the set {ax + by : x, y are integers}. This set contains all multiples of gcd(a, b) since gcd(a, b) divides both a and b.

Therefore, gcd(a, b) is the smallest positive integer that can be expressed as a linear combination of a and b.

Now, if as + bt = 15, then 15 is a linear combination of a and b, which means that gcd(a, b) divides 15.

Conversely, if gcd(a, b) divides 15, then we can find integers s and t such that as + bt = gcd(a, b), and we can scale this equation to obtain as' + bt' = 15, where s' = (15/gcd(a, b))s and t' = (15/gcd(a, b))t.

Therefore, we can conclude that gcd(a, b) divides 15 if and only if there exist integers s and t such that as + bt = 15.

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