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Brrunno [24]
3 years ago
7

At Appalachian State University it is known that 52% of undergraduates are female. If a sample of 150 undergraduate students was

taken, which of the following would accurately describe the sampling distribution?
Select all that apply.
Select one or more:

a. The sampling distribution will be approximately normal.
b. The sampling distribution will be skewed right.
c. The sampling distribution will be skewed left. Incorrect
d. The mean of the sampling distribution will be close to 52% Correct
e. The mean of the sampling distribution will be close to 50%
f. We can not determine the mean of the sampling distribution from the given information.
g. The standard deviation of the sampling distribution will be 0.0408
h. The standard deviation of the sampling distribution will be 0.0017 Incorrect
i. The standard deviation of the sampling distribution will be 0.4996.
j. We can not determine the standard deviation of the sampling distribution from the given information.
Mathematics
1 answer:
In-s [12.5K]3 years ago
3 0

Answer:

a. The sampling distribution will be approximately normal.

d. The mean of the sampling distribution will be close to 52%

g. The standard deviation of the sampling distribution will be 0.0408

Step-by-step explanation:

For this problem the sample size is large enough (n>30), and then the sampling distribution \hat p would be approximately normal. The mean of the sampling distributions is given by p=0.52

The expected value for the sampling distribution would be 0.52 since E(\hat p) = p

And for the standard deviation we know that is given by:

Sd= \sqrt{\frac{p (1-p)}{n}}=\sqrt{\frac{0.52(1-0.52)}{150}}=0.0408

So the correct answers on this case are:

a. The sampling distribution will be approximately normal.

d. The mean of the sampling distribution will be close to 52%

g. The standard deviation of the sampling distribution will be 0.0408

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Answer:

1 g/cm³

Step-by-step explanation:

Volume of the model:

V=1/3bh= 1/3*100*6= 300 cm³

Density= weight/volume= 300 g/300 cm³= 1 g/cm³

The lowest density is 1 g/cm³

7 0
3 years ago
How many times greater is the value of the 3 in 4,367 than the value of the 3 in 39?​
Lesechka [4]

Answer: Roughly 111.97

Step-by-step explanation:

1. 39 divided by 3 is 13.  

2. 4367 divided by 3 is not evenly distributed. Rounding, it would work out to be roughly 1455.66 

3. 1455.66 divided by 13 is roughly 111.97 (Again, this does not distribute evenly)

4. To check your work, multiply 111.97 times 13 and you should get somewhere around 1455.66 which is (4367 divided by 3).

Hopefully this helps! Feel free to mark brainliest! :)

8 0
3 years ago
In the diagram, point O is the center of the circle. If<br> m∠ABC = 70°, what is m∠AOC?
marysya [2.9K]
I can’t see the picture I’m sorry if I can’t help you
6 0
3 years ago
A lighthouse is built on an exposed reef, 4.5 miles off-shore. The shoreline is perfectly straight, and a town is located 6 mile
Alisiya [41]

Answer: 3.94 hours

Step-by-step explanation:

From the question, the following parameters are given:

Distance offshore = 4.5 miles

Distance downshore = 6 miles

Speed when running = 3.4 mph

Speed with boat = 1.8 mph

distance of boat rowing = sqrt(4.5^2 + x^2)

Where

Speed = distance/time

Time = distance/speed

Time = distance of boat rowing/1.8

distance of running = 6 - x

Time = (6-x)/3.4

total travel time

t = sqrt(4.5^2 + x^2)/1.8 + (6 - x)/3.4

dt/dx = x/(1.8×sqrt(4.5^2 + x^2) - 1/3.4

d^2t/dx^2 = +ve at any x

x is at its minimum when dt/dx=0

x/(1.8×sqrt(4.5^2 + x^2) = 1/3.4

x = 6/sqrt(253) × 4.5 = 1.697 miles

Substitute x in

t = sqrt(4.5^2 + x^2)/1.8 + (6 - x)/3.4

We obtaine

t = sqrt(4.5^2 + 1.697^2)/1.8 + (6 - 1.69)/3.4

t = 2.672+ 1.266 = 3.94 hours

7 0
3 years ago
Write an algebraic expression for each word expression. Then evaluate the expression for these values of the variable : 1/2,4,an
Yuki888 [10]

Answer:

1)p/4,2,1,1.625   2)x+1,3/2,5,7.5

Step-by-step explanation:

1) p/4

1/2 divided by 4=2

4/4=1

6.5/4=1.625

2) x+1

1/2+1=3/2

4+1=5

6.5+1=7.5


7 0
3 years ago
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