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2 years ago

Please please me I need help

Mathematics

1 answer:

sergiy2304 [10]2 years ago

4 0

As the guy on top of me said,

1) A
2) C
3) A
4) A
5) D
6) D

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Please answer this question only if you know it!

irinina [24]

C would be chur answer i hoped i helped

7 0

3 years ago

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Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 1

Dahasolnce [82]

Answer:

<h3>Option c) $H_a:\mu$ is correct</h3><h3>The alternate hypothesis for the significance test is $H_a:\mu$ </h3>

Step-by-step explanation:

The alternate hypothesis use the sumbol for the population value

Let p be the proportion , Mean be $\mu$ and stadard deviation be $\sigma$

The null hypothesis states that population value is equal to the value mentioned in the given claim.

$H_0:\mu=18$ (by given).

The alternate hypothesis for the significance test states that the opposite of the null hypothesis(based on the claim )

∴ $H_a:\mu$

The symbol "<" because we want to test if the maze is completed faster and thus if the time has decreased.

<h3>∴ option c) $H_a:\mu$ is correct.</h3>

3 0

3 years ago

Is it ASA , SSS, AAS, SAS!! Help

Sonja [21]

Answer:

490

Step-by-step explanation:

5 0

3 years ago

What else would need to be congruent to show that ???

Fed [463]

D. Side bc is congruent to side ef
It then creates a SAS proof (side angle side)

5 0

3 years ago

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What eigen value for this matix <br> (1 -2)<br> (-2 0)

natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

Compute the matrix $A' = A-\lambda I$ , where I is the identity matrix (1s on the diagonal, 0s elsewhere)
Compute the determinant of A'
Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$

The determinant of this matrix is

$\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4$

Finally, we have

$\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}$

So, the two eigenvalues are

$\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}$

5 0

3 years ago

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