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Debora [2.8K]
3 years ago
13

At a large local university, 40% of the students live in the dormitories. A random sample of 80 students is selected for a parti

cular study. The probability that the sample proportion (the proportion living in the dormitories) is between 0.30 and 0.50 is what?
Mathematics
1 answer:
likoan [24]3 years ago
7 0

Answer:

0.9325

Step-by-step explanation:

Given  

n = sample size = 80

p = probability = 0.4

q = 1 – p = 0..6

standard deviation for the proportion = √ (p * q) /n = √(0.4*0.6)/80 = 0.0547

for the proportion mean is 0.4

now we can find z and the probability  

P (0.3<mean<0.5) = P((0.3– 0.4)/0.0547 < z < (0.5– 0.4)/0.0547)

P (0.3<mean<0.5) = P(-1.828< z < 1.828)

Using a z table

P (0.3<mean<0.5) = 0.9325

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1. 6(4x+1)=78<br>2. 2(x+10)=60<br><br>Solve for x​
postnew [5]

Answer:

1. x= -3

2. x= 20

Step-by-step explanation:

sorry if this is wrong,

7 0
3 years ago
Please help!!! Will give brainliest to the first correct answer!
NARA [144]

Answer:

a. (-4,8)

Step-by-step explanation:

the two lines intersect at this point

4 0
3 years ago
What is the measure of secant line CD
ZanzabumX [31]

Answer: 45

Step-by-step explanation:

By the tangent secant theorem we have AB^2 = BD(BD + CD)

So...

14^2 = 4(CD +4)

196  = 4CD + 16

4CD = 180

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8 0
3 years ago
Assume that the helium porosity of coal samples taken from any particular seam is Normally distributed with true standard deviat
riadik2000 [5.3K]

Answer:

a) 4.85-2.33\frac{0.75}{\sqrt{20}}=4.46    

4.85+2.33\frac{0.75}{\sqrt{20}}=5.24    

b) 4.56-2.33\frac{0.75}{\sqrt{16}}=4.12    

4.56+2.33\frac{0.75}{\sqrt{16}}=4.99  

c) n=(\frac{1.960(0.75)}{0.2})^2 =54.02 \approx 55

Step-by-step explanation:

Part a

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The Confidence is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that z_{\alpha/2}=2.33

Now we have everything in order to replace into formula (1):

4.85-2.33\frac{0.75}{\sqrt{20}}=4.46    

4.85+2.33\frac{0.75}{\sqrt{20}}=5.24    

Part b

4.56-2.33\frac{0.75}{\sqrt{16}}=4.12    

4.56+2.33\frac{0.75}{\sqrt{16}}=4.99  

Part c  

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =0.4/2 =0.2  we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.960, replacing into formula (b) we got:

n=(\frac{1.960(0.75)}{0.2})^2 =54.02 \approx 55

5 0
3 years ago
What is 38 + 7k = 8(k + 4)
lisov135 [29]

Steps

38 + 7k = 8k + 32

38 + 7k - 38 = 8k + 32 - 38

7k = 8k - 6

7k - 8k = 8k - 6 - 8k

-k = -6

-k/-1 = -6/-1

k = 6

Answer

k = 6

3 0
2 years ago
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