Answer:
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
<u>Algebra II</u>
- Distance Formula:
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>
Point (21, 13)
Point (3, 13)
<u>Step 2: Find distance </u><em><u>d</u></em>
Simply plug in the 2 coordinates into the distance formula to find distance <em>d</em>
- Substitute in points [Distance Formula]:
- [√Radical] (Parenthesis) Subtract:
- [√Radical] Evaluate exponents:
- [√Radical] Add:
- [√Radical] Evaluate:
Answer:
Please check the explanation.
Step-by-step explanation:
<u>Calculating the area of the outer rectangle:</u>
Given
- The length outer rectangle = l = 3x - 1
- The width of outer rectangle = w = 5x + 2
Thus,
The area of the outer rectangle:
<u>Calculating the area of the inner rectangle:</u>
Given
- The length inner rectangle = l = x + 7
- The width of inner rectangle = w = x
Thus,
The area of the outer rectangle:
A = wl
= x(x+7)
= x² + 7
<u>Calculating the area of the shaded region:</u>
As
The area of the outer rectangle = 15x² + x - 2
The area of the inner rectangle = x² + 7
- The area of the shaded region can be determined by subtracting the area of the inner rectangle from the area of the outer rectangle.
Thus,
shaded region Area = Outer Rectangle Area - Inner Rectangle Area
= 15x² + x - 2 - (x² + 7)
= 15x² + x - 2 - x² - 7
= 14x² + x - 9
Therefore, the Area of the shaded region is: 14x² + x - 9
It will most likely just rain on Monday
Answer:
The answer is "9 and 7".
Step-by-step explanation:
Given:
![A=\left[\begin{array}{cc}7&9\\0&9\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D)
Using formula:
![\to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]-\lambda \left[\begin{array}{cc}1&0\\0&1\end{array}\right] |=0\\\\\\ \to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]- \left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right] |=0\\\\\\\to|\left[\begin{array}{cc}7-\lambda &9\\0&9-\lambda\end{array}\right]|=0\\\\\\\to|(7-\lambda)(9-\lambda)|=0\\\\\to (7-\lambda)(9-\lambda)=0\\\\\to 7-\lambda=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9-\lambda=0\\\\](https://tex.z-dn.net/?f=%5Cto%20%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D-%5Clambda%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%7C%3D0%5C%5C%5C%5C%5C%5C%20%5Cto%20%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7%269%5C%5C0%269%5Cend%7Barray%7D%5Cright%5D-%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%7C%3D0%5C%5C%5C%5C%5C%5C%5Cto%7C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D7-%5Clambda%20%269%5C%5C0%269-%5Clambda%5Cend%7Barray%7D%5Cright%5D%7C%3D0%5C%5C%5C%5C%5C%5C%5Cto%7C%287-%5Clambda%29%289-%5Clambda%29%7C%3D0%5C%5C%5C%5C%5Cto%20%287-%5Clambda%29%289-%5Clambda%29%3D0%5C%5C%5C%5C%5Cto%207-%5Clambda%3D0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%209-%5Clambda%3D0%5C%5C%5C%5C)
Answer:
The range of the <em>p</em>-value is: 0.050 < <em>p</em>-value < 0.100.
Step-by-step explanation:
For checking the equivalence of two population variances of independent samples, we use the <em>f</em>-test.
The test statistic is given by:
It is provided that the hypothesis test is one-tailed.
The computed value of the test statistic is:
<em>F</em> = 4.23.
The degrees of freedom of the numerator and denominator are:
Use MS-Excel to compute the <em>p</em>-value as follows:
Step 1: Select function fX → F.DIST.RT.
Step 2: A dialog box will open. Enter the values of f-statistic and the two degrees of freedom.
*See the attachment below.
Step 3: Press OK.
The <em>p</em>-value is, 0.0728.
The range of the <em>p</em>-value is:
0.050 < <em>p</em>-value < 0.100
