Answer:
There is not sufficient evidence to support the claim.
Step-by-step explanation:
The claim to be tested is:
The mean respiration rate (in breaths per minute) of students in a large statistics class is less than 32.
To test this claim the hypothesis can be defined as follows:
<em>H₀</em>: The mean respiration rate of students is 32, i.e. <em>μ</em> = 32.
<em>Hₐ</em>: The mean respiration rate of students is less than 32, i.e. <em>μ</em> < 32.
The sample mean respiration rate of students is 31.3.
According to the claim the sample mean is less than 32.
The sample mean value is not unusual if the claim is true, and the sample mean value is also not unusual if the claim is false.
Thus, there is not sufficient evidence to support the claim.
Answer:she will need around 9 weeks and she will have enuogh
Step-by-step explanation:
9
Answer:
y = 3
Step-by-step explanation:
y/9 = 2/6
y/9 = 1/3 (dividing by 2in both numerator and denominator.)
CROSS MULTIPLYING WE GET
3y = 9*1
3y = 9
y = 3
Answer:
Kennan will be from home approximately an hour and 48 minutes.
Step-by-step explanation:
We must know that total time (
) that Keenan will be from home is the sum of run (
), hang out (
) and walk times (
), measured in hours:
If Keenan runs and walks at constant speed, then equation above can be expanded:
Where:
, 
- Run and walk distances, measured in miles.
, 
- Run and walk speeds, measured in miles per hour.
Given that 
, 
, 
and 
, the total time is:
(
)
Kennan will be from home approximately an hour and 48 minutes.
Answer:
We have the equation
![c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]](https://tex.z-dn.net/?f=c_1%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_2%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_3%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_4%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Then, the augmented matrix of the system is
![\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D0%260%260%268%5C%5C0%260%264%264%5C%5C0%263%263%263%5C%5C1%261%261%261%5Cend%7Barray%7D%5Cright%5D)
We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:
![\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%261%261%5C%5C0%263%263%263%5C%5C0%260%264%264%5C%5C0%260%260%268%5Cend%7Barray%7D%5Cright%5D)
This matrix is in echelon form. Then, now we apply backward substitution:
1.
2.
3.
4.
Then the system has unique solution that is 
and this imply that the vectors 
are linear independent.
