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katovenus [111]
2 years ago
15

Help solve 100 points!!!!!

Mathematics
1 answer:
patriot [66]2 years ago
4 0
85 + w(10+8)
85 + (4)(10+8)
b= 157
So your answer is 4 weeks
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PROBLEM SOLVING Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during
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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  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean numbers of seals = 73

            \sigma = standard deviation = 14.1

Now, the probability that at most 50 seals were observed during a randomly chosen survey is given by = P(X \leq 50 seals)

            P(X \leq 50) = P( \frac{X-\mu}{\sigma} \leq \frac{50-73}{14.1} ) = P(Z \leq -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.

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Step-by-step explanation:

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Anything inside a square foot can be multiplied together. In this case, it forms :
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