Answer:
Alabama, Alaska, American Samoa, Arizona, Arkansas.
California, Colorado, Connecticut.
Delaware, District of Columbia.
Florida.
Georgia, Guam.
Hawaii.
Idaho, Illinois, Indiana, Iowa.
Kansas, Kentucky.
Answer:
57.8125% or approx. 57.8%
Step-by-step explanation:
There is a 1/4, or 25%, or 0.25 chance that an egg has salmonella.
Thus, there is a 75%, or 0.75 chance that an egg DOESN'T contain salmonella.
Let's find the probability that all 3 of Larry's eggs are free from salmonella. Larry would have to hit that 75% chance 3 times in a row. The chance of that happening is:
0.75 * 0.75 * 0.75 =
= 0.421875
From this, we can deduce that if there is a 0.421875 (42.1875%) chance that all eggs are safe to eat, there must be a...
1 - 0.421875 = 0.578125
...0.578125 (57.8125%) chance that 1 or more of Larry's eggs do have salmonella.
Answer: approx. 57.8% or 57.8125%
If Houa bought 11 chicken wings for $23.10.The cost of each wing will be $2.310.
<h3>What is an arithmetic operation?</h3>
It is defined as the operation in which we do the addition of numbers, subtraction, multiplication, and division. It has basic four operators that are +, -, ×, and ÷.
It is given that, Houa bought 11 chicken wings for $23.10.
We'll apply division by Splitting into equal halves or groups, the division may be used to decide how to distribute a dish of cookies evenly among a group. Doing division is simple it is a type of operation in which we will distribute the items.
Suppose the cost of each wing is x.
x = $23.10 / 10
x = $2.310
Thus, if Houa bought 11 chicken wings for $23.10.The cost of each wing will be $2.310.
Learn more about the arithmetic operation here:
brainly.com/question/20595275
#SPJ1
Answer:
a. P(x = 0 | λ = 1.2) = 0.301
b. P(x ≥ 8 | λ = 1.2) = 0.000
c. P(x > 5 | λ = 1.2) = 0.002
Step-by-step explanation:
If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

a. What is the probability of selecting a carton and finding no defective pens?
This happens for k=0, so the probability is:

b. What is the probability of finding eight or more defective pens in a carton?
This can be calculated as one minus the probablity of having 7 or less defective pens.



c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?
We can calculate this as we did the previous question, but for k=5.
