P(at least 2 students have the same birthday)= 1- P(no 2 students have the same birthday)
Because P(A)=1-P(A'), where A is an event, and A' the complement of that event.
P(no 2 students have the same birthday)=
think of the problem as follows. We have an urn of balls, numbered from 1 to 365 (the number of the days of the year.
What is the probability of picking 56 different numbered balls, with replacements?
The first one can be any of the 365
the second any of 364 (since one selection has already been made)
the third any of the 363
.
.
and so on
the 56th selection is one of 310 left
Answer:
Answer:
graph (b)
Explanation:
The farm started with 40 sheep and the number doubled each year.
So, the equation modeling this situation would be:
N(x) = 40*(2)^2
By getting the equation, we can notice that we already excluded the third and fourth choices as they have the wrong equations.
Now, we will need to select the right graph which is either graph a or graph b.
To do this, we will get the value of N(x) for different values of x as follows:
At x = 0: N(0) = 40*(2)^0 = 40
At x = 1: N(1) = 40*(2)^1 = 80
At x = 1: N(2) = 40*(2)^2 = 160
Now, taking a look at graph a, we will find that:
N(1) = 120
N(2) = 360
These numbers are not the same as the numbers we calculated, therefore, this graph is wrong
Taking a look at graph b, we will find that:
N(1) = 80
N(2) = 160
These numbers are exactly the same as the ones calculated, therefore, this graph is the correct one.
Hope this helps :)
Answer:
93,000,000
Step-by-step explanation:
9.3x10^7
since the 10 has the ^7 on it, you move the decimal over 7 spots.
Answer:
"and motion, and touch—a tingling
sensation pervading my frame. Then the mere consciousness
of existence, without thought—a condition which lasted
long. Then, very suddenly, thought, and shuddering terror,"
