Use the second method in the preceding segment to find the probability that the 6th toss out of a total of 10 tosses is Heads, g
1 answer:
Answer:
<h2>The answer is 0.008789.</h2>Step-by-step explanation:
The probability of getting a head or tail in a toss is .
It is given that in the 6th toss, the outcome should be head.
Rather than the 6th toss, there are 9 tosses. From these 9 tosses 1 toss can be selected in 9 ways.
Hence, the probability is = 0.008789.
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Answer:
8. c. (-1, -1)
9. a. (-6, -1)
b. True
Step-by-step Explanation:
8. Given the midpoint M(2, 4), and one endpoint D(5, 7) of segment CD, the coordinate pair of the other endpoint C, can be calculated as follows:
let
Rewrite the equation to find the coordinates of C
and
Solve for each:
Coordinates of endpoint C is (-1, 1)
9. a.Given segment AB, with midpoint M(-4, -5), and endpoint A(-2, -9), find endpoint B as follows:
let
and
Solve for each:
Coordinates of endpoint B is (-6, -1)
b. The midpoint of a segment, is the middle of the segment. It divides the segment into two equal parts. The answer is TRUE.
9514 1404 393
Answer:
C. Obtuse
Step-by-step explanation:
The "form factor" I use for this is ...
a^2 + b^2 - c^2 = 6^2 + 8^2 - 11^2 = 36 +64 -121 = -21
The value is negative, indicating an OBTUSE triangle.
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<em>Additional comment</em>
In this expression, 'a' and 'b' are the two shortest sides (in no particular order) and 'c' is the longest side. The interpretation is ...
negative — obtuse
zero — right
positive — acute
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If you're familiar with Pythagorean triples, you know that the 3-4-5 right triangle triple can be doubled to give 6-8-10. The long side of 11 is longer than the hypotenuse for the right triangle, so would correspond to a largest angle greater than 90°. The 6-8-11 triangle would be OBTUSE.
