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3 years ago

The t value for a 95% confidence interval estimation with 24 degrees of freedom is

Mathematics

1 answer:

Alborosie3 years ago

4 0

Answer: the value us 2.064

Step-by-step explanation:

To determine the t value, we would need the t distribution table. Since degree of freedom is known as 24, we would determine alpha/2

A 95% confidence level is 95/100 = 0.95

1 - alpha = 0.95

alpha = 1 - 0.95

alpha = 0.05

alpha/2 = 0.05/2 = 0.025

this is the area to the left. The area to the right 1 - 0.025 = 0.975

alpha/2 = 0.975

Looking at the t distribution table, the t value is

2.064

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Find the ratio AP/PB for 100 points ASAP

shtirl [24]

Answer:

$\frac{1}{2}$

Step-by-step explanation:

To find the ratio of AP/PB, find the distance between A and P, then P and B.

Distance between A(-5, 6) and P(1, 4):

$AP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$A(-5, 6) = (x_1, y_1)$

$P(1, 4) = (x_2, y_2)$

$AP = \sqrt{(1 -(-5))^2 + (4 - 6)^2}$

$AP = \sqrt{(6)^2 + (-2)^2}$

$AP = \sqrt{36 + 4} = \sqrt{40}$

Distance between P(-1, 2) and B(1, -2):

$PB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$P(1, 4) = (x_1, y_1)$

$B(13, 0) = (x_2, y_2)$

$PB = \sqrt{(13 - 1)^2 + (0 - 4)^2}$

$PB = \sqrt{(12)^2 + (-4)^2}$

$PB = \sqrt{144 + 16} = \sqrt{160}$

$\frac{AP}{PB} = \frac{\sqrt{40}}{\sqrt{160}}$

$= \frac{\sqrt{40}}{2\sqrt{40}}$

$= \frac{1}{2}$

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MaRussiya [10]

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myrzilka [38]

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Step-by-step explanation:

Check attachment

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mario62 [17]

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