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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.

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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Write a paragraph comparing the two classes' semester grades. Be sure to compare the extremes, the quartiles, the medians, and t

NISA [10]

Answer:

Class second have higher score and have greater spread.

Step-by-step explanation:

For first box plot

$\text{Minimum value }= 53$

$Q_1=62$

$Median=80$

$Q_3=86$

$\text{Maximum value }= 89$

$Range=Maximum-Minimum=89-53=36$

$IQR=Q_3-Q-1=86-62=24$

For second box plot

$\text{Minimum value }= 56$

$Q_1=62$

$Median=74$

$Q_3=89$

$\text{Maximum value }= 96$

$Range=Maximum-Minimum=96-56=40$

$IQR=Q_3-Q-1=89-62=27$

First class has greater minimum value, it means first class has lower grades.

First quartile of both classes are same, it means equal number students in both classes have less than 62 marks.

First class has greater median.

second class has greater third quartile.

Second class has greater Maximum value. It means second class have higher score than first class.

Second class has greater range. It means the data of second class has greater spread.

Second class has greater inter quartile range. It means the data of second class has greater spread.

Therefore, the class second have higher score and have greater spread.

5 0

3 years ago

Find the interval(s) on which the function is increasing.

Tatiana [17]

Answer:

II (THE Second one )

Step-by-step explanation:

Because when you substitute X by (-2)

you will get :

y=(-2)^2 - 6(-2) - 16 = Zero

So, -2 is not including in the interval .

4 0

3 years ago

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mr. brown class and mrs. farmers class collected cans for the school recycling project. mrs. farmers class collected 3 times as

babunello [35]

159-Mrs. Farmers class
53-Mr. Browns class

4 0

3 years ago

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If you add Natalie's age and Fred's age, the result is 39 If you add Fred's age to 3 times Natalie's age, the result is 69 W

vaieri [72.5K]

Answer:

Natalie is 15

Fred is 24

Step-by-step explanation:

You would first create equations to represent the problem. You would then solve for a variable (I chose to solve for y). I subtracted each equation from each other which helped me isolate y, which equaled 15. I then substituted y for 15 which allowed me to isolate x, which gave me 24.

(the equations)

x + y = 39

x + 3y = 69

(solving for y)

x + 3y = 69 - x + y = 39

x-x +3y -y = 69 - 39

2y = 30

y = 15

(solving for x)

x + 15 = 39

x +15 -15 = 39 -15

x = 24

3 0

3 years ago

Jack bought a new car in 2014 for 28,000. If the value of the car decreases by 14% each year, write an exponential model for the

elena55 [62]

Answer:

In 2026 car will have a value of $5,000.

Step-by-step explanation:

We have been given that Jack bought a new car in 2014 for 28,000. If the value of the car decreases by 14% each year.

Since we know that an exponential function is in form: $y=a*b^x$ , where,

a = Initial value,

b = For decay or decrease b is in form (1-r), where r represents decay rate in decimal form.

Let us convert our given decay rate in decimal form.

$14\%=\frac{14}{100}=0.14$

Upon substituting a =28,000 and r=0.14 in exponential decay function we will get,

$y=28,000(1-0.14)^x$ , where x represents number of years after 2014.

Therefore, the function $y=28,000(0.86)^x$ represents the value of car x years after 2014.

To find the number of years it will take to car have the value of $5,000, we will substitute y=5,000 in our function.

$5,000=28,000(0.86)^x$

Let us divide both sides of our equation by 28,000.

$\frac{5,000}{28,000}=\frac{28,000(0.86)^x}{28,000}$

$0.1785714285714286=(0.86)^x$

Let us take natural log of both sides of our equation.

$ln(0.1785714285714286)=ln((0.86)^x)$

Using natural log property $ln(a^b)=b*ln(a)$ we will get,

$ln(0.1785714285714286)=x*ln(0.86)$

$\frac{ln(0.1785714285714286)}{ln(0.86)}=\frac{x*ln(0.86)}{ln(0.86)}$

$\frac{-1.7227665977411033893}{-0.1508228897345836}=x$

$x=11.422447\approx 12$

As in the 12th year after 2014 car will have a value of $5,000, so we will add 12 to 2014 to find the year.

$2014+12=2026$

Therefore, in 2026 car will have a value of $5,000.

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2 years ago

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