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Kaylis [27]
3 years ago
14

A statistics professor drew a random sample of 81 observations and found that x with bar on top equals 62 s equals 15. Estimate

the LCL of the population mean with 90% confidence. Report your answer to two decimal places.
Mathematics
1 answer:
ratelena [41]3 years ago
3 0

Answer:

LCL = 59.26 to two decimal places

Step-by-step explanation:

Here, we want to estimate the LCL of the population mean with 90% confidence

We proceed as follows;

Given alpha = 0.1, then Z(0.05)=1.645 (from standard normal table), s = 15

Mathematically;

LCL =x_bar -Z*s/√( n)= 62 - (1.645 * 15)/√81

LCL = 62- (24.675)/9 = 59.2583

LCL = 59.26 to two decimal places

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Write an equation for the line in the given form.
nexus9112 [7]

Answer:

  5x +y = 34

Step-by-step explanation:

Since a point and slope are given, it is convenient to start with point-slope form, then rearrange the equation to put it into standard form.

For slope m through point (h, k), the point-slope form of the equation of a line is usually written ...

  y -k = m(x -h)

For m=-5 and (h, k) = (6, 4), the equation is ...

  y -4 = -5(x -6)

  y -4 = -5x +30 . . . . . eliminate parentheses

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Standard form looks like ...

  ax + by = c

where a > 0 and GCD(a, b, c) = 1. Here's how we can get there.

  5x +y -4 = 30 . . . . . . add 5x

  5x +y = 34 . . . . . . . . . add 4 to get standard form.

3 0
3 years ago
What Ln(z) is answer????!!
dimaraw [331]

Write <em>z</em> in exponential form:

z=1-i=\sqrt2 e^{-i\frac\pi4}

Then taking the logarithm, we get

\mathrm{Ln}(z)=\ln(\sqrt2) + \ln e^{-i\frac\pi4} = \boxed{\ln(\sqrt2)-\dfrac\pi4i}

so a is the correct answer.

3 0
3 years ago
Let Y1 and Y2 have the joint probability density function given by:
Ann [662]

Answer:

a) k=6

b) P(Y1 ≤ 3/4, Y2 ≥ 1/2) =  9/16

Step-by-step explanation:

a) if

f (y1, y2) = k(1 − y2), 0 ≤ y1 ≤ y2 ≤ 1,  0, elsewhere

for f to be a probability density function , has to comply with the requirement that the sum of the probability of all the posible states is 1 , then

P(all possible values) = ∫∫f (y1, y2) dy1*dy2 = 1

then integrated between

y1 ≤ y2 ≤ 1 and 0 ≤ y1 ≤ 1

∫∫f (y1, y2) dy1*dy2 =  ∫∫k(1 − y2) dy1*dy2 = k  ∫ [(1-1²/2)- (y1-y1²/2)] dy1 = k  ∫ (1/2-y1+y1²/2) dy1) = k[ (1/2* 1 - 1²/2 +1/2*1³/3)-  (1/2* 0 - 0²/2 +1/2*0³/3)] = k*(1/6)

then

k/6 = 1 → k=6

b)

P(Y1 ≤ 3/4, Y2 ≥ 1/2) = P (0 ≤Y1 ≤ 3/4, 1/2 ≤Y2 ≤ 1) = p

then

p = ∫∫f (y1, y2) dy1*dy2 = 6*∫∫(1 − y2) dy1*dy2 = 6*∫(1 − y2) *dy2 ∫dy1 =

6*[(1-1²/2)-((1/2) - (1/2)²/2)]*[3/4-0] = 6*(1/8)*(3/4)=  9/16

therefore

P(Y1 ≤ 3/4, Y2 ≥ 1/2) =  9/16

8 0
3 years ago
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