49. From 3 coin tosses, there are 8 possible outcomes:
... TTT, TTH, THT, THH, HTT, HTH, HHT, HHH
All but the first have at least one head, so 7/8 of the possibilities have at least one head. That's 87.5% (not among your choices).
Likewise, all but the last listed outcome have at least one tail. The problem is symmetrical that way when the coin is fair. 87.5% of outcomes have at least one tail.
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Perhaps you can tell I read your question as having two parts. If your question is the probability of getting at least one head AND at least one tail, you can see that condition includes 6 of the 8 outcomes, or 75%, matching selection d.
50. See for yourself: the calculator says 66.82%. Your best choice is selection d.
Answer:
Step-by-step explanation:
Given:
∠2 ≅ ∠7 ≅ ∠19
m∠2 = 125°
Since, m∠1 + m∠2 = 180° [Linear pair of angles]
m∠1 + 125° = 180°
m∠1 = 180 - 125 = 55°
m∠1 = m∠3 = 55° [Vertically opposite angles]
m∠2 = m∠4 = 125° [Vertically opposite angles]
m∠5 = m∠7 = 125°
m∠5 + m∠6 = 180° [Linear pair of angles]
125° + m∠6 = 180°
m∠6 = 180 - 125 = 55°
m∠7 = m∠5 = 125°
m∠8 = m∠6 = 55°
m∠19 = m∠17 = 125° [Vertically opposite angles]
m∠19 + m∠18 = 180° [linear pair of angles]
125° + m∠18 = 180°
m∠18 = 55°
m∠18 = m∠20 = 55° [Vertically opposite angles]
m∠13 = m∠20 = 55° [Corresponding angles]
m∠16 = m∠17 = 125° [Corresponding angles]
m∠15 = m∠18 = 55° [Corresponding angles]
m∠14 = m∠19 = 125° [Corresponding angles]
m∠13 + m∠10 + m∠8 = 180° [Sum of angles in a triangle]
55° + m∠10 + 55° = 180°
m∠10 = 180 - 110 = 70°
m∠10 = m∠12 = 70° [Vertically opposite angles]
m∠9 + m∠10 = 180° [Linear pair of angles]
m∠9 = 180 - 70 = 110°
m∠11 = m∠9 = 110° [Vertically opposite angles]
Answer:
7t +5
Step-by-step explanation:
6t +7 -2 +t
Combine like terms
6t+t +7-2
7t +5
Step-by-step explanation:
The ability to influence another is a function of dependency hahahaja
Answer:
4 of each
Step-by-step explanation:
1.50 x 4 =6
.50 x 4 =2
he bought 4 folders and 4
