Answer: 2.2876792e+13
Step-by-step explanation: calculator
Answer: Our required probability is 0.3387.
Step-by-step explanation:
Since we have given that
Number of red cards = 4
Number of black cards = 5
Number of cards drawn = 5
We need to find the probability of getting exactly three black cards.
Probability of getting a black card = 
Probability of getting a red card = 
So, using "Binomial distribution", let X be the number of black cards:

Hence, our required probability is 0.3387.
Answer:
Point slope intercept form: The equation for line is given by;
......[1] ; where m is the slope and a point
on the line.
Let x represents the number of days and y represents the number of friends.
As per the statement: After day three, he has 25 friends; after day eight, he has 40 friends.
⇒ We have two points i.e,
(3, 25) and (8, 40)
First calculate slope(m);

Substitute the given values we get;
= 3
now, substitute the given values of m=3 and a point (3, 25) in [1] we get;

Using distributive property; 

Add 25 on both sides, we get;

Simplify:
y =3x + 16
if x = 18 days, then;
y = 3(18) + 16 = 54+16 = 70
Therefore, he will have on day 18, if he continues to add the same number of friends each day is, 70 friends.
<em>z</em> = 3<em>i</em> / (-1 - <em>i</em> )
<em>z</em> = 3<em>i</em> / (-1 - <em>i</em> ) × (-1 + <em>i</em> ) / (-1 + <em>i</em> )
<em>z</em> = (3<em>i</em> × (-1 + <em>i</em> )) / ((-1)² - <em>i</em> ²)
<em>z</em> = (-3<em>i</em> + 3<em>i</em> ²) / ((-1)² - <em>i</em> ²)
<em>z</em> = (-3 - 3<em>i </em>) / (1 - (-1))
<em>z</em> = (-3 - 3<em>i </em>) / 2
Note that this number lies in the third quadrant of the complex plane, where both Re(<em>z</em>) and Im(<em>z</em>) are negative. But arctan only returns angles between -<em>π</em>/2 and <em>π</em>/2. So we have
arg(<em>z</em>) = arctan((-3/2)/(-3/2)) - <em>π</em>
arg(<em>z</em>) = arctan(1) - <em>π</em>
arg(<em>z</em>) = <em>π</em>/4 - <em>π</em>
arg(<em>z</em>) = -3<em>π</em>/4
where I'm taking arg(<em>z</em>) to have a range of -<em>π</em> < arg(<em>z</em>) ≤ <em>π</em>.