Answer:
The correct option is C). When it was purchased, the coin was worth $6
Step-by-step explanation:
Given function is f(t)=![6\times2^{t}](https://tex.z-dn.net/?f=6%5Ctimes2%5E%7Bt%7D)
Where t is number of years and f(t) is function showing the value of a rare coin.
A figure of f(t) shows that the graph has time t on the x-axis and f(t) on the y-axis.
Also y-intercept at (0,6)
hence, when time t was zero, the value of a rare coin is 6$
f(t)=![6\times2^{t}](https://tex.z-dn.net/?f=6%5Ctimes2%5E%7Bt%7D)
f(0)=![6\times2^{0}](https://tex.z-dn.net/?f=6%5Ctimes2%5E%7B0%7D)
<em>f(0)=6</em>
Thus,
The correct option is C). When it was purchased, the coin was worth $6
The band has sold 1,890,000 copies or 1.89 x 10^6
Answer:
d. interquartile range
Step-by-step explanation:
the iqr measures the first, second, and third quartile, and the median.
3terms = trinomial (7a^2, 4a, -12)
Degree of each term 2nd, 1st, constant
Largest degree: 2nd
Answer: 2nd degree trinomial
Let's call the aces
for hearts, diamonds, clubs and spades. So,
are red and [ted] c, s[/tex] are black.
Since the first card is replaced, the two picks are identical. This means that the sample space is given by all the possible couple
![(x,y)\, x, y \in \{h,d,c,s\}](https://tex.z-dn.net/?f=%20%28x%2Cy%29%5C%2C%20x%2C%20y%20%5Cin%20%5C%7Bh%2Cd%2Cc%2Cs%5C%7D%20)
There are 16 such couples (we have four choices for the first card, and the same four choices for the second card). Now let's compute the odds in our favour to deduce the probability of winning:
If we want a player to draw two card of the same colour, the following couples are good:
![(h,h),\ (h,d),\ (d,h),\ (d,d),\ (c,c),\ (c,s),\ (s,c),\ (s,s)](https://tex.z-dn.net/?f=%20%28h%2Ch%29%2C%5C%20%28h%2Cd%29%2C%5C%20%28d%2Ch%29%2C%5C%20%28d%2Cd%29%2C%5C%20%28c%2Cc%29%2C%5C%20%28c%2Cs%29%2C%5C%20%28s%2Cc%29%2C%5C%20%28s%2Cs%29%20)
so 8 possible couples over 16. This means that the probability that a player draws two cards of the same color is 8/16 = 1/2.
Similarly, the probability of drawing a red ace first and then a black ace is represented by the following couples:
![(h,c),\ (h,s),\ (d,c),\ (d,s)](https://tex.z-dn.net/?f=%20%28h%2Cc%29%2C%5C%20%28h%2Cs%29%2C%5C%20%28d%2Cc%29%2C%5C%20%28d%2Cs%29%20)
which are 4 over the same 16 as above, thus leading to a probability of 4/16 = 1/4.