A tunnel is in the shape of a parabola. The maximum height is 50 m and it is 10 m wide at the base as shown below. What is the v
ertical clearance 2 m from the edge of the tunnel?
1 answer:
Refer to the figure shown below.
Because the maximum height of the parabola is 50 m, its equation is of the form
y = ax² + 50
This equation places the vertex at (0,50). The constant a should be negative for the vertex to be the maximum of y.
The base of the parabola is 10 m wide. Therefore the x-intercepts are (5,0) and (-5,0).
Set x=5 and y=0 to obtain
a(5²) + 50 = 0
25a = -50
a = -2
The equation of the parabola is
y = - 2x² + 50
At 2 m from the edge of the tunnel, x = 5 - 2 = 3 m.
Therefore the height of the tunnel (vertical clearance) at x = 3 m is
h = y(3)
= -2(3²) + 50
= - 18 + 50
= 32 m
Answer: 32 m
Because the maximum height of the parabola is 50 m, its equation is of the form
y = ax² + 50
This equation places the vertex at (0,50). The constant a should be negative for the vertex to be the maximum of y.
The base of the parabola is 10 m wide. Therefore the x-intercepts are (5,0) and (-5,0).
Set x=5 and y=0 to obtain
a(5²) + 50 = 0
25a = -50
a = -2
The equation of the parabola is
y = - 2x² + 50
At 2 m from the edge of the tunnel, x = 5 - 2 = 3 m.
Therefore the height of the tunnel (vertical clearance) at x = 3 m is
h = y(3)
= -2(3²) + 50
= - 18 + 50
= 32 m
Answer: 32 m
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Answer:
One Angle = 110°
Other Angle = 70°
Step-by-step explanation:
A linear pair means that two angles are in a straight line (or, a straight angle).
A straight line is 180 degrees.
THey are supplementary.
We can say one angle is "a" and another one is "b".
<em>One angle is 10 MORE THAN 2/3rds of the other, we can write:</em>
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<em>Also, since they are supplementary (add up to 180), we can write:</em>
<em>a + b = 180</em>
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We can now substitute 1st equation in this one and find b:
Since a + b = 180, we can write:
a + 110 = 180
so,
a = 180 - 110
a = 70
Thus,
One Angle = 110°
Other Angle = 70°
Answer:
The correct option is (A).
Step-by-step explanation:
Let <em>X</em> = number of orange milk chocolate M&M’s.
The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.
The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.
The event of an milk chocolate M&M being orange is independent of the other candies.
The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.
The expected value of a Binomial random variable is:
Compute the expected number of orange milk chocolate M&M’s in a bag of 55 candies as follows:
It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.
This proportion is not surprising.
This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.
All unusual events have a very low probability, i.e. less than 0.05.
Compute the probability of P (X ≥ 14) as follows:
The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.
This probability is quite larger than 0.05.
Thus, the correct option is (A).