Answer:
- 18x + 15y ≤ 300
- x ≥ 0; y ≥ 6
Step-by-step explanation:
Variables are defined in the problem statement.
18x +15y ≤ 300 . . . . total budget
y ≥ 6 . . . . . . . . . . . . minimum number of annuals
x ≥ 0 . . . . . number of perennials cannot be negative
This system of inequalities describes the situation.
Answer: x = 61.81925228
Step-by-step explanation:
Answer:
<h2>F. 12</h2>
Step-by-step explanation:
The distance between A and M i.e JM = M - J = 5-(-19)
JM = 5+19
JM = 24
If JK:KL:KM = 2:1:3
The total ratio = 2+1+3
Total ratio = 6
Dividing each length based on ratio
length JK = 2/6 * JM
JK = 1/3 * 24
JK = 8
length KL = 1/6 * JM
KL = 1/6 * 24
KL = 4
length KM = 3/6 * JM
KM = 1/2 * 24
KM = 12
<em>Hence the value of KM is 12</em>
Answer: ![(23.3,\ 29.7)](https://tex.z-dn.net/?f=%2823.3%2C%5C%2029.7%29)
Step-by-step explanation:
The confidence interval for population mean is given by :-
, where
is the sample mean and ME is the margin of error .
Given : The sample mean : ![\overline{x}=26.5](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D26.5)
Margin of error : ![ME=3.2](https://tex.z-dn.net/?f=ME%3D3.2)
Then , the range of values (confidence interval) likely to contain the true value of the population parameter will be :-
![26.5\pm 3.2=(26.5-3.2,\ 26.5+3.2)=(23.3,\ 29.7)](https://tex.z-dn.net/?f=26.5%5Cpm%203.2%3D%2826.5-3.2%2C%5C%2026.5%2B3.2%29%3D%2823.3%2C%5C%2029.7%29)
Hence, the range of values likely to contain the true value of the population parameter = ![(23.3,\ 29.7)](https://tex.z-dn.net/?f=%2823.3%2C%5C%2029.7%29)