Answer:
μ = 5.068 oz
Step-by-step explanation:
Normal distribution formula to use the table attached
Z = (x - μ)/σ
where μ is mean, σ is standard deviation, Z is on x-axis and x is a desired point.
98% of 6-oz. cups will not overflow means that the area below the curve is equal to 0.49; note that the curve is symmetrical respect zero, so, 98% of the cases relied between the interval (μ - some value) and (μ + some value)].
From table attached, area = 0.49 when Z = 2.33. From data, σ = 0.4 oz and x = 6 oz (maximum capacity of the cup). Isolating x from the formula gives
Z = (x - μ)/σ
2.33 = (6 - μ)/0.4
μ = 6 - 2.33*0.4
μ = 5.068
This means that with a mean of 5 oz and a standard deviation of 0.4 oz, the machine will discharge a maximum of 6 oz in the 98% of the cases.
Answer:
3.6666 (I didn't round)
Repeating decimal
Step-by-step explanation:
We know that
2/3
is the same as
2÷3
Therefore:
3 2/3 = 3+(2÷3) = 3 + 0.6666 =3.6666
A repeating decimal is one that keeps going and repeats a pattern. This is decimal keeps going and repeats a pattern so it is a repeating decimal.
A terminating decimal is one that terminates or ends. This decimal does not terminate or end. It keeps going.
Hope this helped.
First you will divide 525 by 100. Then you multiply your answer by 4. That will equal 21
Answer:
The set of solutions is ![\{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}12\\-7-r\\r\end{array}\right]: \text{r is a real number} \}](https://tex.z-dn.net/?f=%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12%5C%5C-7-r%5C%5Cr%5Cend%7Barray%7D%5Cright%5D%3A%20%5Ctext%7Br%20is%20a%20real%20number%7D%20%20%5C%7D)
Step-by-step explanation:
The augmented matrix of the system is ![\left[\begin{array}{ccccc}3&6&6&-9\\-2&-3&-3&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D3%266%266%26-9%5C%5C-2%26-3%26-3%263%5Cend%7Barray%7D%5Cright%5D)
.
We will use rows operations for find the echelon form of the matrix.
- In row 2 we subtract
from row 1. (R2- 2/3R1) and we obtain the matrix ![\left[\begin{array}{cccc}3&6&6&-9\\0&1&1&-7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%266%266%26-9%5C%5C0%261%261%26-7%5Cend%7Barray%7D%5Cright%5D)
- We multiply the row 1 by
.
Now we solve for the unknown variables:
The system has a free variable, the the system has infinite solutions and the set of solutions is
