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.
0.00037 = 3.7×10^-4
37,000 = 3.7×10^4
_____
The exponent of 10 is the place-value multiplier that the most-significant digit has.
In the first number, the most-significant digit (3) is multiplied by 0.0001 to find its place value. Of course 0.0001 = 10^-4.
In the second number, the most significant digit (3) is multiplied by 10,000 to find its place value. You know that 10,000 = 10^4.
Pretty sure it’s the second one
Answer:
hello the answer is G because I don’t know why