Answer:
Probability that at most 50 seals were observed during a randomly chosen survey is 0.0516.
Step-by-step explanation:
We are given that Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during each survey.
The numbers of seals observed were normally distributed with a mean of 73 seals and a standard deviation of 14.1 seals.
Let X = <u><em>numbers of seals observed</em></u>
The z score probability distribution for normal distribution is given by;
Z =
~ N(0,1)
where,
= population mean numbers of seals = 73
= standard deviation = 14.1
Now, the probability that at most 50 seals were observed during a randomly chosen survey is given by = P(X
50 seals)
P(X
50) = P(
) = P(Z
-1.63) = 1 - P(Z < 1.63)
= 1 - 0.94845 = <u>0.0516</u>
The above probability is calculated by looking at the value of x = 1.63 in the z table which has an area of 0.94845.
Answer:
95
Step-by-step explanation:
Let 2x be how many red/green marbles there are,
let 5x be how many blue marbles there are, and
let 7x be how many white marbles there are.
2x + 2x + 5x + 7x = 304
16x = 304
x = 19
As 5x is defined as how many blue marbles there are, we could just plug x for 19:
5*19 = 95
I hope this helped, and if it did please consider giving me brainliest :)
First, make it easy for yourself by putting like terms together
-7x+8x+5>3
then, combine like terms
-7x+8x=1x
now you have this
1x+5>3
hope this helps!
the answer is 0.605 if you need help message me