Answer:
the answer might be either one of the four
Answer:
$8.5 per hour
Step-by-step explanation:
Number of overtime hours = 4 hours
Since overtime starts after 40 hours, then the number of hours worked = 40 + 4 = 44 hours
Total amount earned = $391
Let Hourly pay = x
Overtime pay = 1 1/2x = 3/2x
Therefore ;
40x + (4*3/2x) = 391
40x + 6x = 391
46x = 391
x = 391 / 46
x = 8.5
Hourly pay = $8.5 per hour
15$ I believe, as we can disregard the two original books (they have been payed for) leaving her to buy 1 book and having seven dollars, however if she needs $8 more you simply need to add 8+7.
Answer:
Solve for x by simplifying both sides of the equation, then isolating the variable.
x = −
6
Step-by-step explanation:
I hope this helps! Have a nice day!
Answer:
a

b

Step-by-step explanation:
From the question we are told that
The number of students in the class is N = 20 (This is the population )
The number of student that will cheat is k = 3
The number of students that he is focused on is n = 4
Generally the probability distribution that defines this question is the Hyper geometrically distributed because four students are focused on without replacing them in the class (i.e in the generally population) and population contains exactly three student that will cheat.
Generally probability mass function is mathematically represented as

Here C stands for combination , hence we will be making use of the combination functionality in our calculators
Generally the that he finds at least one of the students cheating when he focus his attention on four randomly chosen students during the exam is mathematically represented as

Here




Hence


Generally the that he finds at least one of the students cheating when he focus his attention on six randomly chosen students during the exam is mathematically represented as

![P(X \ge 1) =1- [ \frac{^{k}C_x * ^{N-k}C_{n-x}}{^{N}C_n}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B%5E%7Bk%7DC_x%20%2A%20%5E%7BN-k%7DC_%7Bn-x%7D%7D%7B%5E%7BN%7DC_n%7D%5D%20)
Here n = 6
So
![P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{20 -3}C_{6-0}}{^{20}C_6}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B%5E%7B3%7DC_0%20%2A%20%5E%7B20%20-3%7DC_%7B6-0%7D%7D%7B%5E%7B20%7DC_6%7D%5D%20)
![P(X \ge 1) =1- [ \frac{^{3}C_0 * ^{17}C_{6}}{^{20}C_6}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B%5E%7B3%7DC_0%20%2A%20%5E%7B17%7DC_%7B6%7D%7D%7B%5E%7B20%7DC_6%7D%5D%20)
![P(X \ge 1) =1- [ \frac{1 * 12376}{38760}]](https://tex.z-dn.net/?f=P%28X%20%20%5Cge%201%29%20%3D1-%20%5B%20%20%5Cfrac%7B1%20%20%2A%20%2012376%7D%7B38760%7D%5D%20)

