Answer:
x^2y^5
Step-by-step explanation:
Simplify the expressions.
Answer:
168
Step-by-step explanation:
Answer:
So then the best option would be:
a. 1/25
Step-by-step explanation:
For this case we assume that the sample space for the numbers is :
![S_1= [A,B,C,D,E]](https://tex.z-dn.net/?f=%20S_1%3D%20%5BA%2CB%2CC%2CD%2CE%5D)
And the sample space for the numbers is:
![S_2 =[1,2,3,4,5]](https://tex.z-dn.net/?f=%20S_2%20%3D%5B1%2C2%2C3%2C4%2C5%5D)
Both sampling spaces with a size of 5.
We define the following events:
A="We select a 2 from the numbers"
B= "We select a E from the letters"
We can find the individual probabilities for each event like this:
And assuming independence we can find the probability required like this:
The last probability is the probability of obtain obtain a 2 AND an E
So then the best option would be:
a. 1/25
Answer:
7x+54
Step-by-step explanation:
because you distribute the 7 to the x and the 8 first then you bring that down which becomes 7x+56 then you to the addition so you do 56+-2 which would give you 54.
Answer:
Step-by-step explanation:
On a normal distribution table with z scores of 0 as the mean and the standard deviations going to the right and to the left, 1.9 on the normal curve where 3 is the mean falls at -.275 on the normal curve where 0 is the mean. I used the normal distribution z table for negative values to get that .39165 lies to the left of -.275, which means that 1 - .39165 of the data lies to the right.
The probability that the value is greater than 1.9 is .60835, or as a percentage, 60.835%.
