All categories

3 years ago

Refer to the equation 3x − 2y = 12.

Mathematics

1 answer:

-Dominant- [34]3 years ago

4 0

Answer: don't know sorry what kind if grade is this

Step-by-step explanation: YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET !!!!!!!!!!!!!!!!!!!!!!!!!!

You might be interested in

Construct a 95% confidence interval for the population mean, mu. Assume the population has a normal distribution. A random sam

aniked [119]

Given Information:

Number of lithium batteries = n = 16

Mean life of lithium batteries = μ = 645 hours

Standard deviation of lithium batteries = σ = 31 hours

Confidence level = 95%

Required Information:

Confidence Interval = ?

Answer:

$CI = 628.5 \: to \: 661.5 \: hours$

Step-by-step explanation:

The confidence interval is given by

$CI = \mu \pm t_{\alpha/2} (\frac{\sigma}{\sqrt{n}})$

Where μ is the mean life of lithium batteries, σ is the standard deviation, n is number of lithium batteries selected, and t is the critical value from the t-table with significance level of

tα/2 = (1 - 0.95) = 0.05/2 = 0.025

and the degree of freedom is

DoF = n - 1 = 16 - 1 = 15

The critical value (tα/2) at 15 DoF is equal to 2.131 (from the t-table)

$CI = 645 \pm 2.131(\frac{31}{\sqrt{16}})$

$CI = 645 \pm 2.131(7.75})$

$CI = 645 \pm 16.515$

$CI = 645 - 16.515 \: and \: 645 + 16.515$

$CI = 628.5 \: to \: 661.5 \: hours$

Therefore, the 95% confidence interval is 628.5 to 661.5 hours

What does it mean?

It means that we are 95% confident that the mean life of 16 lithium batteries is within the interval of (628.5 to 661.5 hours)

6 0

3 years ago

What is the perimeter of a rectangle with length 16.8 units and the width 9.6 units

Oxana [17]

Hey there! :) In order to find the perimeter, you must add all 4 sides. You first do 16.8 + 16.8 + 9.6 + 9.6 which is 52.8.

Hope it helps!

5 0

3 years ago

Any know how do these problems?

Ksju [112]

13. (6-9)^-3= -1/27 or -0.04(rounded to nearest hundredth.
15. 4^-1+(5-7)^-2= 1/2 or .5

3 0

3 years ago

(1 point) The matrix A=⎡⎣⎢−4−4−40−8−4084⎤⎦⎥A=[−400−4−88−4−44] has two real eigenvalues, one of multiplicity 11 and one of multip

serious [3.7K]

Answer:

We have the matrix $A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]$

To find the eigenvalues of A we need find the zeros of the polynomial characteristic $p(\lambda)=det(A-\lambda I_3)$

Then

$p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda$

Now, we fin the zeros of $p(\lambda)$ .

$p(\lambda)=-\lambda^3-8\lambda^2-16\lambda=0\\\lambda(-\lambda^2-8\lambda-16)=0\\\lambda_{1}=0\; o \; \lambda_{2,3}=\frac{8\pm\sqrt{8^2-4(-1)(-16)}}{-2}=\frac{8}{-2}=-4$

Then, the eigenvalues of A are $\lambda_{1}=0$ of multiplicity 1 and $\lambda{2}=-4$ of multiplicity 2.

Let's find the eigenspaces of A. For $\lambda_{1}=0$ : $E_0 = Null(A- 0I_3)=Null(A)$ .Then, we use row operations to find the echelon form of the matrix