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EastWind [94]
3 years ago
13

For which sample size (n) and sample proportion (p) can a normal curve be

Mathematics
1 answer:
Nonamiya [84]3 years ago
7 0

Answer:

Option c.

Step-by-step explanation:

Using the normal curve to approximate a sampling distribution:

For a sample size n and a proportion n, the normal curve can be used if:

np \geq 10 and n(1-p) \geq 10

Option a:

np = 35*0.8 = 28 > 10

n(1-p) = 35*0.2 = 7 < 10

So option a cannot be used.

Option b:

np = 65*0.9 = 58.5 > 10

n(1-p) = 65*0.1 = 6.5 < 10

So option b cannot be used.

Option c:

np = 65*0.8 = 52 > 10

n(1-p) = 65*0.2 = 13 > 10

So option c can be used, and is the answer

Option d:

np = 35*0.9 = 31.5 > 10

n(1-p) = 35*0.1 = 3.5 < 10

So option d cannot be used.

The answer is given by option c.

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➷ Area = length x width

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2 years ago
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dmitriy555 [2]

Answer:

84+28w

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Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
WILL GIVE BRAINLIEST THANKS AND 5 STARS FOR CORRECT ANSWER Desmond ran a lemonade stand in front of his house for the fans walki
Fiesta28 [93]

Answer:

56

Step-by-step explanation:

First, convert the 70% so that we can use it with the 80 cups. To do this, convert the percent to a decimal by moving the decimal 2 spaces to the left:

70% → 70.00 → 0.70 → 0.7

70% percent can be seen as 0.7. Multiply this by 80:

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So, 70% of 80 is 56. Desmond sold 56 cups on the second Saturday.

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7 0
3 years ago
The 10th term of an a.p is -27 and the 5th term is -12 .find the first term?​
horsena [70]

Step-by-step explanation:

a_{10} = - 27... (Given) \\\\\therefore a + 9d = - 27...... (1)\\\\a_{5} = - 12... (Given) \\\\\therefore a + 4d = - 12.......(2)\\\\

Subtracting equation (2) from equation (1)

a + 9d-( a + 4d )= - 27-(-12)\\\\\therefore a +9d - a-4d = - 27+12\\\\\therefore 5d = - 15\\\\\therefore d = \frac {- 15}{5}\\\\\huge \orange {\boxed {\therefore d = -3}} \\\\

Substituting d = - 3 in equation (1), we find:

a + 9\times (-3)= - 27\\\\\therefore a - 27 = - 27\\\\\therefore a  = 27- 27\\\\\huge \red {\boxed {\therefore a  = 0}}\\\\

Hence, first term is zero.

3 0
3 years ago
14. Let R^2 have inner product defined by ((x1,x2), (y,, y2)) 4x1y1 +9x2y2 A. Determine the norm of (-1,2) in this space B. Dete
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The norm of a vector \vec x is equal to the square root of the inner product of \vec x with itself.

a. \|(-1,2)\|=\sqrt{\langle(-1,2),(-1,2)\rangle}=\sqrt{4(-1)^2+9(2)^2}=\sqrt{40}=2\sqrt{10}

b. \|(3,2)\|=\sqrt{\langle(3,2),(3,2)\rangle}=\sqrt{4(3)^2+9(2)^2}=\sqrt{72}=6\sqrt2

7 0
3 years ago
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