If the letters are considered distinct, then the number of permutations is
.
If we count either C as the same character, then we would be double-counting - to correct this, we would simply divided by the number of ways we can choose C from the available characters, or
.
Answer:
you can get 80 1/2 oz servings.
Step-by-step explanation:
Answer:
(a) B
(b) $2
Step-by-step explanation:
(a) Let's say the cost of a ticket is t and the cost of popcorn is p. Then we can write the two equations from the table:
12t + 8p = 184
9t + 6p = 138
We need to solve this, so let's use elimination. Multiply the first equation by 3 and the second equation by 4:
3 * (12t + 8p = 184)
4 * (9t + 6p = 138)
We get:
36t + 24p = 552
36t + 24p = 552
Subtract the second from the first:
36t + 24p = 552
- 36t + 24p = 552
________________
0 = 0
Since we get down to 0 = 0, which is always true, we know that we cannot determine the cost of each ticket because there is more than one solution (infinitely many, actually). The answer is B.
(b) Our equation from this, if we still use t and p, is:
5t + 4p = 82
Now, just choose any of the two equations from above. Let's just pick 9t + 6p = 138. Now, we have the system:
5t + 4p = 82
9t + 6p = 138
To solve, let's use elimination again. Multiply the first equation by 6 and the second one by 4:
6 * (5t + 4p = 82)
4 * (9t + 6p = 138)
We get:
30t + 24p = 492
36t + 24p = 552
Subtract the second from the first:
36t + 24p = 552
- 30t + 24p = 492
________________
6t + 0p = 60
So, t = 60/6 = $10. Plug this back into any of the equations to solve for p:
5t + 4p = 82
5 * 10 + 4p = 82
50 + 4p = 82
4p = 32
p = 32/4 = $8
So the ticket costs 10 - 8 = $2 more dollars than the popcorn.
Answer:
Let's define the cost of the cheaper game as X, and the cost of the pricer game as Y.
The total cost of both games is:
X + Y
We know that both games cost just above AED 80
Then:
X + Y > AED 80
From this, we want to prove that at least one of the games costed more than AED 40.
Now let's play with the possible prices of X, there are two possible cases:
X is larger than AED 40
X is equal to or smaller than AED 40.
If X is more than AED 40, then we have a game that costed more than AED 40.
If X is less than or equal to AED 40, then:
X ≥ AED 40
Now let's take the maximum value of X in this scenario, this is:
X = AED 40
Replacing this in the first inequality, we get:
X + Y > AED 80
Replacing the value of X we get:
AED 40 + Y > AED 80
Y > AED 80 - AED 40
Y > AED 40
So when X is equal or smaller than AED 40, the value of Y is larger than AED 40.
So we proven that in all the possible cases, at least one of the two games costs more than AED 40.
Answer:
<em>343 metres per second
</em>
<em>In dry air at 20 °C (68 °F), the speed of sound is 343 metres per second (about 767 mph, 1234 km/h or 1,125 ft/s).</em>
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