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lapo4ka [179]
3 years ago
9

The population that you're interested in has a mean of 60 and a standard deviation of 72 Step 2 of 2: If you calculate a(n) 93.1

2 % Confidence Interval on the Mean, what will your Margin of Error be? (Round to 2 decimal places.)
Mathematics
1 answer:
aivan3 [116]3 years ago
5 0

Answer:

Z = 8.41

Step-by-step explanation:

Given that standard deviation = 72

Mean = 60

Z = marginal error

Z = 72/√60

Z = 72/7.746

Z = (9.036)(0.9312)

Z = 8.41

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Answer:

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Step-by-step explanation:

We have been given that Trevor has a bag of jelly beans. There are 10 red, 15 green, 5 orange, and 5 purple jelly beans in the bag.

The probability of choosing an orange jelly will be total number of orange jellies over all jellies.

\text{P(Orange)}=\frac{5}{10+15+5+5}

\text{P(Orange)}=\frac{5}{35}

Now, he again chooses an orange jelly and eats it. Since he already ate 1 orange jelly, so orange jelly left is 4 and total jellies would be 34.

\text{P(Orange)}=\frac{4}{34}

Using multiplication rule of probability, we will get:

P(A\cap B)=P(A)\cdot P(B|A)

P(\text{Orange}\cap\text{Orange})= \frac{5}{35}\times \frac{4}{34}

P(\text{Orange}\cap\text{Orange})=\frac{1}{7}\times \frac{2}{17}

P(\text{Orange}\cap\text{Orange})= \frac{2}{119}

Therefore, the probability that Trevor randomly chooses 1 orange jellybean, eats it, and then chooses another orange jellybean and eats it would be \frac{2}{119}.

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