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BaLLatris [955]
2 years ago
6

Students in a statistics class are conducting a survey to estimate the mean number of units students at their college are enroll

ed in. The students collect a random sample of 47 students. The mean of the sample is 12.3 units. The sample has a standard deviation of 1.9 units.What is the 95% confidence interval for the average number of units that students in their college are enrolled in? Assume that the distribution of individual student enrollment units at this college is approximately normal.(____,______)Your answer should be rounded to 2 decimal places.
Mathematics
1 answer:
GaryK [48]2 years ago
3 0

Answer:

(11.76, 12.84)

Step-by-step explanation:

Given that we can assume that the distribution of individual student enrollment units at this college is approximately normal.

Sample mean =12.3

sample std dev s = 1.9 units

Sample size = 47

Std error = \frac{1.9}{\sqrt{47} } \\=0.277

Z critical value for 95% = 1.96

Margin of error = 1.96*0.277=0.543

Confidence interval =

(12.3-0.543, 12.3+0.543)\\= (11.757, 12.843)

=(11.76, 12.84)

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tino4ka555 [31]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards  

Step-by-step explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

y = 1.6 x - 0.032x^2

we know that maximum height occurs is given as

x = -\frac{b}{2a}

y = - \frac{1.6}{2( - 0.032)} = 25

and maximum height is

y = 1.6(25) - 0.032 (25)^2

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards  

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2 years ago
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adelina 88 [10]

Step-by-step explanation:

Only Q4 has clear details

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then,  £12.60 = 12.60×1.28 =$16.128

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2 years ago
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3 0
2 years ago
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Answer: no Tony is not correct It is less than 5 hundred dollars.

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A college basketball coach has 12 players on his team. eight players are receiving scholarships, and four are not. the coach dec
EleoNora [17]

Solution: The given problem is a binomial distribution.

The probability that a team member receives a scholarship is \frac{8}{12} = 0.67

Therefore, the given problem follows binomial with n = 5 and p = 0.6667

Now the probability that 4 of 5 players selected are on scholarship is:

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Therefore, the probability that 4 of 5 players selected are on scholarship is 0.329

                 

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