It has been proven that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
<h3>How to prove a Line Segment?</h3>
We know that in a triangle if one angle is 90 degrees, then the other angles have to be acute.
Let us take a line l and from point P as shown in the attached file, that is, not on line l, draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In ΔPNM, ∠N = 90°
∠P + ∠N + ∠M = 180° (Angle sum property of a triangle)
∠P + ∠M = 90°
Clearly, ∠M is an acute angle.
Thus; ∠M < ∠N
PN < PM (The side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to all of them. Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
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There are ⌊ 999 6 ⌋ − ⌊ 99 6 ⌋ numbers between 100 and 999 (i.e. three digit numbers) which are divisible by six.
Answer:
98%
Step-by-step explanation:
If the probability of winning a school raffle is 2%, the probability of not winning the raffle would be 98%.
Answer:
Step-by-step explanation:
In a study of the relationship between pet ownership and physical activity in older adults, 586 subjects reported that they owned a pet, while 1931 reported that they did not. Give a 98% confidence interval for the proportion of older adults in this population who are pet owners. Lower limit Incorrect: Your answer is incorrect. Upper limit
p ± z × (1 - p/n)
z = 98%
p = 586/1931
= 0.3034697048
n = 1931
Approximately = 0.3035
Confidence interval =
0.3035 + 2.326 × ( 1 - 0.3035/ 1931)