Sin44 is the same as Cos46. Hope this helped.
Answer:
Step-by-step explanation:
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:
The price of
1 adult ticket = $15
1 student ticket = $9
Step-by-step explanation:
Let
The price of adult tickets be represented by a
The price of student tickets be represented by s
Therefore:
On the first day of ticket sales the school sold 4 adult tickets and 10 student tickets for a total of $150.
4a + 10s = $150.... Equation 1
The school took in $105 on the second day by selling 1 adult ticket and 10 student tickets.
a + 10s = $105.... Equation 2
a = $105 - 10s
Therefore, we substitute : $105 - 10s = a in Equation 1
4a + 10s = $150.... Equation 1
4($105 - 10s) + 10s = $150
$420 - 40s + 10s = $150
Collect like terms
- 40s + 10s = $150 - $420
-30s = -$270
Divide both sides by -30
-30s/-30 = -$270/-30
s = $9
We find a
a = $105 - 10s
a = $105 - 10($9)
a = $105 - $90
a = $15
Therefore, the price of
1 adult ticket = $15
1 student ticket = $9