Answer:
$2598.80
Step-by-step explanation:
49 at $12.80 per ticket = $627.20
106 at $18.60 per ticket = $1971.60
Total = $2598.80
Answer:
- as written, c ≈ 0.000979 or c = 4
- alternate interpretation: c = 0
Step-by-step explanation:
<em>As written</em>, you have an equation that cannot be solved algebraically.
(32^2)c = 8^c
1024c = 8^c
1024c -8^c = 0 . . . . . . rewrite as an expression compared to zero
A graphical solution shows two values for c: {0.000978551672551, 4}. We presume you're interested in c = 4.
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If you mean ...
32^(2c) = 8^c
(2^5)^(2c) = (2^3)^c . . . . rewriting as powers of 2
2^(10c) = 2^(3c) . . . . . . . simplify
10c = 3c . . . . . . . . . . . . . .log base 2
7c = 0 . . . . . . . . . . . . . . . subtract 3c
c = 0 . . . . . . . . . . . . . . . . divide by 7
The correct order is:
the probability of heads on 2nd and 4th toss only;
the probability of at least 3 tails in a row;
the probability of 3 or more heads;
the probability of consecutive tails; and
the probability of at least 2 tails.
There are 16 outcomes in the sample space.
There are 8 ways of having consecutive tails; this gives the probability 8/16.
There is 1 way of having heads on the 2nd and 4th toss only, making the probability 1/16.
There are 3 ways to have at least 3 tails in a row, making the probability 3/16.
There are 11 ways to have at least 2 tails, making the probability 11/16.
There are 5 ways to have at least 3 heads, making the probability 5/16.
Answer:
B.
Step-by-step explanation:
B should be (2,3) and D should be (2, -4) because its just a reflection of the two points. :)
