Recall that for

, i.e. a random variable

following a binomial distribution over

trials and with probability parameter

,

So you have




The expected value of

is simply

, while the standard deviation is

. In this case, they are

and

, respectively.
<h3>
Answer: B) 2/3</h3>
====================================
Explanation:
The order is important for ABC and XYZ so we can see what letters pair up.
- A pairs with X since they are the first letters
- B pairs with Y since they are the second letters
- C pairs with Z since they are the third letters
Based on that, we can see that AB pairs with XY as they are the first two letters of ABC and XYZ respectively.
Divide the length XY over AB and reduce
XY/AB = 10/15 = 2/3
We could also divide XZ over AC
XZ/AC = 14/21 = 2/3
and we get the same ratio
It will expand. Or it will grow.
47 is 0.47 and 2/3 is 0.66
Answer:
Therefore the company have to print either 47 or 12 t-shirt for its production cost to be $40 or less.
Step-by-step explanation:
Given that,
The function that models the production cost is

To find the number of t-shirt, we put f(t)=40 in given function.
∴40= 0.1t²-6t+100
⇒ 0.1t²-6t+100-40=0
⇒0.1 t²- 6t +60=0
⇒t²-60t+600=0
Applying quadratic formula
, a=1, b= - 60 and c=600

= 47.32,12.67
≈47, 12
Therefore the company have to print either 47 or 12 t-shirt for its production cost to be $40 or less.