The length is 10 meters.
150 divided by 15 is 10.
Hope this helps!
Option 3:
m∠ABC = 66°
Solution:
Given
and ABH is a transversal line.
m∠FAB = 48° and m∠ECB = 18°
m∠ECB = m∠HCB = 18°
<u>Property of parallel lines:
</u>
<em>If two parallel lines cut by a transversal, then the alternate interior angles are equal.</em>
m∠FAB = m∠BHC
48° = m∠BHC
m∠BHC = 48°
<u>Exterior angle of a triangle theorem:
</u>
<em>An exterior angle of a triangle is equal to the sum of the opposite interior angles.</em>
m∠ABC = m∠BHC + m∠HCB
m∠ABC = 48° + 18°
m∠ABC = 66°
Option 3 is the correct answer.
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Answer:
1) you're going to have to flip the coins (or fake numbers) for the experimental trials.
2) for the theoretical, there is 1/2 chance for heads or tails with each toss, so you'd expect that out of 10 tosses, 5 heads, 5 tails. out of 100 tosses- 50 heads, 50 tails.
When tossing 2 coins- 1/2×1/2 = 1/4 (25%) chance that 2 heads, 2 tails, or 1 heads & 1 tails. Deviation value comes from after you done your flipping and recorded your data. So if on 100 flips you actually got 50 and 50 (rarely us that exact ;), the deviation from the expected of 50/50 would be 0.00. If however you flipped 100 heads or 100 tails (impossible), then the deviation value would be 1.00.
|(100-50)| ÷ 50 = 50÷50 = 1.00
So usually you may have data like: 47/53 or something a little off than 50/50, making deviation |(47-50)| ÷ 50 = 3÷50 = 0.06.
Now the number of flips is important for the outcome! So if a coin toss if 10 times had 4 heads, 6 tails, the deviation value would be:
|(4-5)| ÷ 5 = 1÷5 = 0.20
So increasing the # flips DECREASES the deviation value!!
Whether it's from 10 to 100, or from 100 to 200. Look at my example of how the 10-flip deviation of 0.20 decreased to 0.06 with 100-flip