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pantera1 [17]
1 year ago
15

Determine the margin of error, m , of a 99% confidence interval for the mean IQ score of all students with the disorder. Assume

that the standard deviation IQ score among the population of all students with the disorder is the same as the standard deviation of IQ score for the general population, σ = 15 points.
Mathematics
1 answer:
madreJ [45]1 year ago
7 0

Answer:

E= 6.45

standard deviation is σ = 15,

The critical value is z(α/2) = 2.58.

Step-by-step explanation:

Margin error is the value that is lie above and below the sample.It gives percentage of numbers.Its is the product of critical value standard deviation and standard error of statistic.

General formula for the margin of error is

Margin of error = critical value  ×   standard error of statistic

                        = z \alpha /2  ×   σ √n

z-value from two tailed is listed below:

From the table of standard normal distribution, probability value of 0.10.

row and column values gives the area to the two tail of z.

The positive z value is 2.58.

standard deviation is σ = 15,

The critical value is z(α/2) = 2.58.

after putting these vales we obtain the margin of error value that is

E= 6.45

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Answer:  \bold{9)\ \sin \theta=\dfrac{1}{3}\qquad 10)\ \sin \theta = \dfrac{4}{5}\qquad 11)\ \cos \theta = \dfrac{\sqrt{11}}{6}\qquad 12)\ \tan \theta = \dfrac{17\sqrt2}{26}}

<u>Step-by-step explanation:</u>

Pythagorean Theorem is: a² + b² = c²   , <em>where "c" is the hypotenuse</em>

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9)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{4}{12}\quad \rightarrow \large\boxed{\dfrac{1}{3}}

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10)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{16}{20}\quad \rightarrow \large\boxed{\dfrac{4}{5}}

Note: 12² + opposite² = 20²   →   opposite = 16

11)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{\sqrt{11}}{6}\quad =\large\boxed{\dfrac{\sqrt{11}}{6}}

Note: adjacent² + 5² = 6²   →   adjacent = √11

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Answer:

For First Solution: y_1(t)=e^t

y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:y_2(t)=cosht

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Step-by-step explanation:

For First Solution: y_1(t)=e^t

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_1(t)=e^t

First order derivative:

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In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

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First order derivative:

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Put Them in equation y''-y=0

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