Answer:
"A translation of 7 units to the left followed by a translation of 1 unit down".
Step-by-step explanation:
There are multiple transformations that map one point into another, here is one example that works particularly for translations, which are the simplest (and usually the most used) transformations.
Suppose that we have the point (a, b) which is transformed into (a', b')
Then we have a horizontal translation of (a' - a) units followed by a vertical translation of (b' - b) units.
(the order of the translations does not matter, is the same having first the vertical translation and then the horizontal one).
Here we have the point A (3, 4) transformed into (-4, 3)
Then we have a horizontal translation of ((-4) - 3) = -7 units followed by a vertical translation of (3 - 4) = -1 units.
Where a horizontal translation of -7 units is a translation of 7 units to the left, and a vertical translation of -1 unit is a translation of 1 unit down.
Then we can write this transformation as:
"A translation of 7 units to the left followed by a translation of 1 unit down".
Answer:
32 days
Step-by-step explanation:
Assume that the jar is full initially, containing 24 oz. After how many withdrawals of 3/4 oz will the jar be empty?
24 - (3/4)n = 0, or
24 = (3/4)n, or 96 = 3n. Dividing both sides by 3, we get n = 32.
The jar will be empty after 32 days.
Answer:
It’s 14
Step-by-step explanation:
Just add
As the variation is direct we have:
We must find the value of k.
For this, we use the following data:
y = 200 when x = 5
Substituting values we have:
Clearing k:
Then, the function is:
We evaluate the function for x = -3
Answer:
the value of y when x = -3 is:
c.
72
Answer:
(a) E(X) = -2p² + 2p + 2; d²/dp² E(X) at p = 1/2 is less than 0
(b) 6p⁴ - 12p³ + 3p² + 3p + 3; d²/dp² E(X) at p = 1/2 is less than 0
Step-by-step explanation:
(a) when i = 2, the expected number of played games will be:
E(X) = 2[p² + (1-p)²] + 3[2p² (1-p) + 2p(1-p)²] = 2[p²+1-2p+p²] + 3[2p²-2p³+2p(1-2p+p²)] = 2[2p²-2p+1] + 3[2p² - 2p³+2p-4p²+2p³] = 4p²-4p+2-6p²+6p = -2p²+2p+2.
If p = 1/2, then:
d²/dp² E(X) = d/dp (-4p + 2) = -4 which is less than 0. Therefore, the E(X) is maximized.
(b) when i = 3;
E(X) = 3[p³ + (1-p)³] + 4[3p³(1-p) + 3p(1-p)³] + 5[6p³(1-p)² + 6p²(1-p)³]
Simplification and rearrangement lead to:
E(X) = 6p⁴-12p³+3p²+3p+3
if p = 1/2, then:
d²/dp² E(X) at p = 1/2 = d/dp (24p³-36p²+6p+3) = 72p²-72p+6 = 72(1/2)² - 72(1/2) +6 = 18 - 36 +8 = -10
Therefore, E(X) is maximized.
