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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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The product of -3 and -7 is positive.<br><br><br> True False
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Which equation describes the circle having center point (3,7) and the radius r=4 in Standard form
Annette [7]

The equation of circle in standard form is (x - 3)^2 + (y - 7)^2 = 16

<h3><u>Solution:</u></h3>

Given that circle having center point (3,7) and the radius r = 4

To find: equation of circle in standard form

<em><u>The equation of circle is given as:</u></em>

(x - h)^2 + (y - k)^2 = r^2

Where center (h,k) and radius r units

Given that center point (h , k) = (3, 7) and radius r = 4 units

Substituting the values in above equation of circle,

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6 0
3 years ago
The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)
Lemur [1.5K]

Answer:

Cardiac output:F=0.055 L\s

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output:F=\frac{A}{\int\limits^T_0 {c(t)} \, dt} ---1

A = 3 mg

\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt

Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}

Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

Hence Cardiac output:F=0.055 L\s

4 0
3 years ago
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