This is an interesting question. I chose to tackle it using the Law of Cosines.
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355
<h3>
Answer: 13/28</h3>
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Reason:
The table shows he got 26 heads out of 26+30 = 56 coin flips.
26/56 = (2*13)/(2*28) = 13/28 is the empirical or experimetnal probablity of getting heads.
Side note: 13/28 = 0.4643 = 46.43% approximately which is fairly close to 50%
2 tens, and 8 ones. I say this because 20 = 10 + 10 so, two tens. 8= 1+1+1+1+1+1+1+1 so, eight ones.
I cannot see it all is their something you could do to help me help you?
Answer:
5
Step-by-step explanation: