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Firlakuza [10]
2 years ago
13

For a certain population of sea urchins, 16% are longer than 8 inches. A random sample of 75 sea urchins will be selected. What

is the standard deviation of the sampling distribution of the sample proportion of sea urchins longer than 8 inches for samples of size 75?
Mathematics
1 answer:
Alona [7]2 years ago
4 0

Answer:

0.042

Step-by-step explanation:

Given :

p = 16% = 0.16

q = 1 - p = 1 - 0.16 = 0.84

Sample size, n = 75

The standard deviation of sampling distribution :

√(pq)/n

√(0.16 * 0.84) / 75

√0.1344 / 75

√0.001792

Standard deviation of sampling distribution = 0.0423320

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Answer:

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Step-by-step explanation:

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8 0
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How many solutions exist for the system of equations graphed below?
Nat2105 [25]

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no solutions

Step-by-step explanation:

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3 0
3 years ago
What is the domain of the given function?
lana [24]

Answer:

The domain of the given function is

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6 0
3 years ago
PLEASE HELP ME ANSWER! im so desperate bro !!
sergey [27]

a)13

b) skewed

c) IQR

d)Sam's observation is correct

e)6.17

Step-by-step explanation:

Step 1 :

From the histogram we can infer that

0 to 2 texts has been sent by 1 student

2 to 4 texts has been sent by 3 student

4 to 6 texts has been sent by 4 student

6 to 8 texts has been sent by 4 student

8 to 10 texts has been sent by 5 student

Step 2:

a)

From step 1 , we can see that there are total of 17 students represented by the histogram

Step 3:

b)

This is a skewed histogram because we have all large values of students on the right of the histogram and smaller values on the left of the histogram.

In symmetrical histogram, there would be large values on the center and the smaller values on both sides.

Step 4:

c)

IQR would be better recommendation for this as this is a skewed histogram

Standard deviation would be recommended for a symmetrical histogram

Step 5:

d)

Total number of students is 17

Number of students who text between 8 to 10 messages is 5

Hence the students who text between 8 and 10 messages is 5/17

which is nearly 1/3 rd of the total students.

Hence Sam's observation is correct

Step 6:

e)

Mean amount of the texts = (1 * 3 + 3* 3 + 5*4 +7*4 + 9* 5) /17 = 6.17

6 0
3 years ago
What is the area of the figure use pi
Inessa05 [86]
Area of triangle = 1/2 (6)(5) = 15
Area of semicircle = 1/2(3.14)(3^2) = 14.13

Area of figure = 15 + 14.13 = 29.13

Answer
29.13 square units
6 0
3 years ago
Read 2 more answers
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