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4 years ago

Classify each substance as either a conductor or insulator.

Mathematics

2 answers:

IceJOKER [234]4 years ago

8 0

A sample of mercury-conductor

A piece of glass-insulator

A rubber hose-insulator

A negatively ionized lithium paste-conductor

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

Conductors are substances that allow the flow electricity through them whereas insulators do not permit the flow of electricity through them.

Mercury is a liquid metal and due to its metallic nature, it is a fair conductor of electricity because metals have free electrons that carry the electric charge through the material.

Rubber and glass are insulators whereas a negatively charged lithium paste is a conductor.

lukranit [14]4 years ago

4 0

Answer:

Mercury=Conductor

Glass=Insulator

Rubber Hose=Insulator

Negatively Ionized Lithium Paste=Conductor

Step-by-step explanation:

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3 years ago

Determine the sample size needed to construct a 90% confidence interval to estimate the population proportion when p = 0.65 and

WINSTONCH [101]

Answer:

n=170

Step-by-step explanation:

1) Notation and definitions

$n$ random sample taken (variable of interest)

$\hat p=0.65$ estimated proportion.

$p$ true population

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.64$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.06$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.65(1-0.65)}{(\frac{0.06}{1.64})^2}=169.968$

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4 years ago

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3 years ago

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