This expression can't be factored with rational numbers.
Answer:
Yes bro
Step-by-step explanation:
To solve a compound inequality, first separate it into two inequalities. Determine whether the answer should be a union of sets ("or") or an intersection of sets ("and"). Then, solve both inequalities and graph.
Answer:




Step-by-step explanation:
The probability mass function P(X = x) is the probability that X happens x times.
When n trials happen, for each
, the probability mass function is given by:

In which p is the probability that the event happens.
is the permutation of n elements with x repetitions(when there are multiple events happening(like one passes and two not passing)). It can be calculated by the following formula:

The sum of all P(X=x) must be 1.
In this problem
We have 3 trials, so 
The probability that a wafer pass a test is 0.7, so 
Determine the probability mass function of the number of wafers from a lot that pass the test.




7x / 8 = 21
7x = 8*21
x = 24
Answer:
7707 tickets
Step-by-step explanation:
From the question we know that:
the number of tickets sold at the counter on
First day = 1094 tickets
Second day = 1812 tickets
Third day = 2050 tickets
Fourth day = 2751 tickets
The total number of tickets sold on all the four days is calculated as:
(1094 + 1812 + 2050 + 2751)tickets
= 7707 tickets