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3 years ago

Can anyone help me find the answer to this question?

Mathematics

1 answer:

galina1969 [7]3 years ago

3 0

Find a 92.9% confidence interval for the difference p1−p2p1−p2 of the population proportions.

Hint: For a level of confidence (1−α)⋅100%(1−α)⋅100%, z∗z∗ is the value that leaves an area of (1−α)/2(1−α)/2 to its right in a standard normal distribution

I hope that helps

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The angle measures of a triangle are 2y+3 degrees 9y-7 and 7y+4 degrees what is the value of y

Roman55 [17]

Answer:

2y+3+9y-7+7y+4=180

2y+9y+7y=180(+7-7)

18y=180

y=180/18

y=10

then

2y+3

2×10+3

20+3

9y-7

9×10-7

7y+4

7×10+4

6 0

3 years ago

Given b = 4 and c = -5, evaluate b 2 - c 2.<br> 41<br> -2<br> -9

ANTONII [103]

The answer to this question -9

5 0

4 years ago

Westkost [7]

Answer: a

Step-by-step explanation:

5 0

3 years ago

Perform the indicated operation. 4 - (7 + 9i) =

Fed [463]

I think that -(3+9i) hope so it will help

8 0

3 years ago

Read 2 more answers

The matrix A = −14 −6 6 28 12 −4 0 0 4 has characteristic polynomial

rosijanka [135]

Answer:

Characteristic equation:

$p(\lambda) = -\lambda^3 + 2\lambda^2 + 8\lambda$

Eigen values:

$\lambda_1 = 0, \lambda_2 = -2, \lambda_3= 4$

Step-by-step explanation:

We are given the matrix:

$\displaystyle\left[\begin{array}{ccc}-14&-6&6\\28&12&-4\\0&0&4\end{array}\right]$

The characteristic equation can be calculated as:

$det(A-\lambda I) = 0\\|A-\lambda I| = 0$

We follow the following steps to calculate characteristic equation:

$=det\Bigg(\displaystyle\left[\begin{array}{ccc}-14&-6&6\\28&12&-4\\0&0&4\end{array}\right]-\lambda\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\Bigg)\\\\= det\Bigg(\displaystyle\left[\begin{array}{ccc}-14-\lambda&-6&6\\28&12-\lambda&-4\\0&0&4-\lambda\end{array}\right]\Bigg)\\\\=(-14-\lambda)[(12-\lambda)(4-\lambda)]+6[28(4-\lambda)]-6[(28)(0)-(12-\lambda)(0)]\\\\= -\lambda^3 + 2\lambda^2 + 8\lambda$

$p(\lambda)= -\lambda^3 + 2\lambda^2 + 8\lambda$

To obtain the eigen values, we equate the characteristic equation to 0:

$p(\lambda) = -\lambda^3 + 2\lambda^2 + 8\lambda = 0\\-\lambda(\lambda^2-2\lambda-8) = 0\\-\lambda(\lambda^2-4\lambda+2\lambda-8) = 0\\-\lambda[(\lambda(\lambda-4)+2(\lambda-4)] = 0\\-\lambda(\lambda+2)(\lambda-4) = 0 \\\lambda_1 = 0, \lambda_2 = -2, \lambda_3= 4$

We can arrange the eigen values as:

$\lambda_1 = 0, \lambda_2 = -2, \lambda_3= 4\\-2 < 0 < 4\\\lambda_2 < \lambda_1 < \lambda_3$

3 0

3 years ago