Solve -3+2ln(x)=0 .......
1 answer:
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The solution is shown in the graph attached.
Explanation:
I am goind to teach you how to get that solution.
1) Restricctions: both x and y cannot be negative, i.e. x ≥ 0 and y ≥ 0
⇒ the solution is on the first quadrant.
2) Using the prices of the tickets, $ 8 for adults, and $ 6 for children, the linear equation for the costs is: cost = 8x + 6y.
3) Since the radio station is willing to spend a maximum of $ 172 you have the final restriction:
⇒ 8x + 6y ≤ 172.
4) Then the solutions that meet the three restrictions (x ≥0, y ≥ 0, and 8x + 6y ≤ 172) is found graphically by drawing the line 8 x + 6y = 172
5) To draw the line 8x + 6y = 172, use the axis intercepts:
x = 0 ⇒ y = 172/6 = 86/3 ⇒ point (0, 86/3)
y = 0 ⇒ x = 172/8 = 86/4 ⇒ point (86/4, 0)
6) Once you have the line you shade the region that is between the line and the two axis. That region contain of the possible solutions.
Explanation:
I am goind to teach you how to get that solution.
1) Restricctions: both x and y cannot be negative, i.e. x ≥ 0 and y ≥ 0
⇒ the solution is on the first quadrant.
2) Using the prices of the tickets, $ 8 for adults, and $ 6 for children, the linear equation for the costs is: cost = 8x + 6y.
3) Since the radio station is willing to spend a maximum of $ 172 you have the final restriction:
⇒ 8x + 6y ≤ 172.
4) Then the solutions that meet the three restrictions (x ≥0, y ≥ 0, and 8x + 6y ≤ 172) is found graphically by drawing the line 8 x + 6y = 172
5) To draw the line 8x + 6y = 172, use the axis intercepts:
x = 0 ⇒ y = 172/6 = 86/3 ⇒ point (0, 86/3)
y = 0 ⇒ x = 172/8 = 86/4 ⇒ point (86/4, 0)
6) Once you have the line you shade the region that is between the line and the two axis. That region contain of the possible solutions.
Answer:
10 square meters will have a standard deviation of 1.897.
Step-by-step explanation:
Standard deviation for n instances of a variable:
If the standard deviation for one instance of a variable is , for n instances of the variable, the standard deviation will be of
The standard deviation of the number of flaws per square yard is 0.6
This means that
For the 10 square yards will a standard deviation of
, so:
10 square meters will have a standard deviation of 1.897.