10.
1 answer:
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Answer: option c.
Step-by-step explanation:
To solve the given exercise and write the expression as a single logarithm, you must keep on mind the following properties:
Therefore, by applying the properties shown above, you can rewrite the expression given, as following:
Then as you can see, the answer is the option c.
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.