Answer:
<em>H</em>₀: <em>μ</em> = 4 vs. <em>H
ₐ</em>: <em>μ </em>> 4
Step-by-step explanation:
A null hypothesis is a sort of hypothesis used in statistics that intends that no statistical significance exists in a set of given observations.
It is a hypothesis of no difference.
It is typically the hypothesis a scientist or experimenter will attempt to refute or discard. It is denoted by H₀.
Whereas, the alternate hypothesis is the contradicting statement to the null hypothesis.
The alternate hypothesis describes direction of the hypothesis test, i.e. if the test is left tailed, right tailed or two tailed.
It is also known as the research hypothesis and is denoted by H
ₐ.
In this case we need to test whether the amount is paid after the grace period, on average, more than 4 times in 2018.
The hypothesis can be defined as follows:
<em>H</em>₀: <em>μ</em> = 4 vs. <em>H
ₐ</em>: <em>μ </em>> 4
9514 1404 393
Answer:
$20.01
Step-by-step explanation:
In 2004–2012, the interest rate is 0.002%. In 2013, it is 0.004%. In 2014–2021, the interest rate is 0.002%. That is, in the 18 years between 2004 and 2021 (inclusive), the interest rate is 0.002% for 17 of them. The effective account multiplier is ...
(1.00002^17)(1.00004^1) = 1.00038006801
Then the account balance is ...
$20 × 1.00038006801 ≈ $20.01
_____
<em>Additional comment</em>
The annual interest earned on $20.00 is $0.0004. If the account balance is rounded to the nearest cent annually, at the end of the 18 years, the balance will still be $20.00. Not enough interest is earned in one year to increase the balance above $20. At the end of the 18 years, the amount of interest earned is 0.76¢ (a fraction of a penny) <em>only if there is no rounding in intervening years</em>.
18+6x?= 168
try taking away the 18 and 6 then dividing it
Answer:
Verified
Step-by-step explanation:
Let the 2x2 matrix A be in the form of:
![\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
Where det(A) = ad - bc # 0 so A is nonsingular:
Then the transposed version of A is
![A^T = \left[\begin{array}{cc}a&c\\b&d\end{array}\right]](https://tex.z-dn.net/?f=A%5ET%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26c%5C%5Cb%26d%5Cend%7Barray%7D%5Cright%5D)
Then the inverted version of transposed A is
![(A^T)^{-1} = \frac{1}{ad - cb} \left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]](https://tex.z-dn.net/?f=%28A%5ET%29%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20cb%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-c%5C%5C-b%26d%5Cend%7Barray%7D%5Cright%5D)
The inverted version of A is:
![A^{-1} = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-b\\-c&d\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-b%5C%5C-c%26d%5Cend%7Barray%7D%5Cright%5D)
The transposed version of inverted A is:
![(A^{-1})^T = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]](https://tex.z-dn.net/?f=%28A%5E%7B-1%7D%29%5ET%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-c%5C%5C-b%26d%5Cend%7Barray%7D%5Cright%5D)
We can see that
So this theorem is true for 2 x 2 matrices
