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mezya [45]
3 years ago
6

you are a plumber installing a pipe the ratio of the outside diameter to the inside diameter is 1.5 to 1. you use a pipe with a

0.5 inch inside diameter what is the smallest diameter that you will need to drill to allow the pipes to pass through the wall
Mathematics
1 answer:
yanalaym [24]3 years ago
7 0
Since the outside is 1.5 to 1, we take half of the value of the inside diameter and add it to itself to get the smallest value needed. 0.5+0.25=0.75

So the smallest diameter possible would be 0.75 inches.
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y=-3x+3

Step-by-step explanation:

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Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and
sergij07 [2.7K]

Answer:

a) sample size is 40

Step-by-step explanation:

(a)

\sigma=4 minutes

Margin of error, E is 75 seconds E=75/60= 1.25 minutes.

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For 95% confidence interval z_{\frac{\alpha}{2}}=1.96 from standard deviation table

Sample size required, n

n=\left (\frac{z_{\frac{\alpha}{2}}\cdot \sigma}{E} \right )^{2}=\left (\frac{1.96\cdot4}{75/60} \right )^{2}=39.337984

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3 0
3 years ago
Use the exponential decay​ model, Upper A equals Upper A 0 e Superscript kt​, to solve the following. The​ half-life of a certai
Akimi4 [234]

Answer:

It will take 7 years ( approx )

Step-by-step explanation:

Given equation that shows the amount of the substance after t years,

A=A_0 e^{kt}

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If the half life of the substance is 19 years,

Then if t = 19, amount of the substance = \frac{A_0}{2},

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\frac{A_0}{2}=A_0 e^{19k}

\frac{1}{2} = e^{19k}

0.5 = e^{19k}

Taking ln both sides,

\ln(0.5) = \ln(e^{19k})

\ln(0.5) = 19k

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0.78 A_0=A_0 e^{-0.03648t}

0.78 = e^{-0.03648t}

Again taking ln both sides,

\ln(0.78) = -0.03648t

-0.24846=-0.03648t

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Hence, approximately the substance would be 78% of its initial value after 7 years.

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