Given that <span>Line m is parallel to line n.
We prove that 1 is supplementary to 3 as follows:
![\begin{tabular} {|c|c|} Statement&Reason\\[1ex] Line m is parallel to line n&Given\\ \angle1\cong\angle2&Corresponding angles\\ m\angle1=m\angle2&Deifinition of Congruent angles\\ \angle2\ and\ \angle3\ form\ a\ linear\ pair&Adjacent angles on a straight line\\ \angle2\ is\ supplementary\ to\ \angle3&Deifinition of linear pair\\ m\angle2+m\angle3=180^o&Deifinition of supplementary \angle s\\ m\angle1+m\angle3=180^o&Substitution Property \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7C%7D%0AStatement%26Reason%5C%5C%5B1ex%5D%0ALine%20m%20is%20parallel%20to%20line%20n%26Given%5C%5C%0A%5Cangle1%5Ccong%5Cangle2%26Corresponding%20angles%5C%5C%0Am%5Cangle1%3Dm%5Cangle2%26Deifinition%20of%20Congruent%20angles%5C%5C%0A%5Cangle2%5C%20and%5C%20%5Cangle3%5C%20form%5C%20a%5C%20linear%5C%20pair%26Adjacent%20angles%20on%20a%20straight%20line%5C%5C%0A%5Cangle2%5C%20is%5C%20supplementary%5C%20to%5C%20%5Cangle3%26Deifinition%20of%20linear%20pair%5C%5C%0Am%5Cangle2%2Bm%5Cangle3%3D180%5Eo%26Deifinition%20of%20supplementary%20%5Cangle%20s%5C%5C%0Am%5Cangle1%2Bm%5Cangle3%3D180%5Eo%26Substitution%20Property%0A%5Cend%7Btabular%7D)
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We prove that 1 is supplementary to 3 as follows:
An electrician earns 100 dollars for a project installing light fixtures. She must pay for the light fixtures herself and they c
2 answers:
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Answer:
12 feet per second.
Step-by-step explanation:
Please consider the complete question.
Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds after she jumps for the bone can be represented.
What is Spot's average rate of ascent, in feet per second, from the time she jumps into the air to the time she catches the bone at t=1/2?
We will use average rate of change formula to solve our given problem.
Therefore, Spot's average rate of ascent is 12 feet per second.
Answer:
a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.
b) 0.996 is the probability that more than half of the vehicles carry just one person.
Step-by-step explanation:
We are given the following information:
A) Binomial distribution
We treat vehicle on road with one passenger as a success.
P(success) = 64% = 0.64
Then the number of vehicles follows a binomial distribution, where
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 10
We have to evaluate:
0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.
B) By normal approximation
Sample size, n = 92
p = 0.64
We have to evaluate the probability that more than 47 cars carry just one person.
After continuity correction, we will evaluate
Calculation the value from standard normal z table, we have,
0.996 is the probability that more than half out of 92 vehicles carry just one person.