The additional information which would be sufficient to conclude that LMNO is a parallelogram is; ML ∥ NO, LO ≅ MN, and ML ≅ LO.
<h3>What information renders LMNO a parallelogram?</h3>
The condition for a quadrilateral to be a parallelogram is that; the opposite pairs must be parallel and consequently opposite pairs are congruent as they have equal length measures.
On this note, it can be concluded that the additional information which would be sufficient are; ML ∥ NO, LO ≅ MN, and ML ≅ LO.
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Answer: 4.27% of adults in the USA have stage 2 high blood pressure.
Step-by-step explanation:
Let x be a random variable that denotes a person with high blood pressure .
Given: Average blood pressure: 
Standard deviation: 
Someone qualifies as having Stage 2 high blood pressure if their systolic blood pressure is 160 or higher.
The probability that an adult in the USA have stage 2 high blood pressure:
![P(x\geq160)=P(\dfrac{x-\mu}{\sigma}}\geq\dfrac{160-122}{22})\\\\=P(z\geq1.72)\ \ \ [z=\dfrac{x-\mu}{\sigma}]\\\\=1-P(z](https://tex.z-dn.net/?f=P%28x%5Cgeq160%29%3DP%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%7D%5Cgeq%5Cdfrac%7B160-122%7D%7B22%7D%29%5C%5C%5C%5C%3DP%28z%5Cgeq1.72%29%5C%20%5C%20%5C%20%5Bz%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-P%28z%3C1.72%29%5C%5C%5C%5C%3D1-0.9573%5C%20%5C%20%5BBy%5C%20p-value%5C%20table%5D%5C%5C%5C%5C%3D0.0427%3D4.27%5C%25)
Hence, 4.27% of adults in the USA have stage 2 high blood pressure.
Max value of x is infinite and min value of x is 0
It represents the ounces of water that the bottle can still hold.
Answer:
1) The solve by graphing will the preferred choice when the equation is complex to be easily solved by the other means
Example;
y = x⁵ + 4·x⁴ + 3·x³ + 2·x² + x + 3
2) Solving by substitution is suitable where we have two or more variables in two or more (equal number) of equations
2x + 6y = 16
x + y = 6
We can substitute the value of x = 6 - y, into the first equation and solve from there
3) Solving an equation be Elimination, is suitable when there are two or more equations with coefficients of the form, 2·x + 6·y = 23 and x + y = 16
Multiplying the second equation by 2 and subtracting it from the first equation as follows
2·x + 6·y - 2×(x + y) = 23 - 2 × 16
2·x - 2·x + 6·y - 2·y = 23 - 32
0 + 4·y = -9
4) An example of a linear system that can be solved by all three methods is given as follows;
2·x + 6·y = 23
x + y = 16
Step-by-step explanation:
