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3 years ago

How to rearrange this equation to get v by itself

Mathematics

1 answer:

s344n2d4d5 [400]3 years ago

8 0

V^2/(1-v^2/c^2)=R
v^2=R(1-v^2/c^2)
v^2=R-Rv^2/c^2
v^2-Rv^2/c^2=R
v^2(1-R/c^2)=R
v=sqrt(R/(1-R/c^2))
where R was original right side, dont forget plus minus

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3 years ago

If x^2-x-6 and x^2+3x-18 hacpve a common factor (x-a) then find the value of a.

DedPeter [7]

Answer:

a = 3

Step-by-step explanation:

Factor both expressions

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3 0

3 years ago

Read 2 more answers

in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru

melisa1 [442]

Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

The sample size is

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Since both np ≥ 10 and n(1-p) ≥ 10

We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant difference.

The approximation would be to a Normal curve with this parameters:

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The 90% confidence interval for the proportion would be then

$\bf [0.77-z^*0.0298, 0.77+z^*0.0298]$

where $\bf z^*$ is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval $\bf [-z^*,z^*]$ is 10%=0.1

We can either use a table, a calculator or a spreadsheet to get this value.

In Excel or OpenOffice Calc we use the function

NORMSINV(0.95) and we get a value of 1.645

The 90% confidence interval for the proportion is then

$\bf [0.77-1.645^*0.0298, 0.77+1.645^*0.0298]=[0.7210,0.8189]$

This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%

If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Yes, I do.

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3 years ago

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Answer:

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3 years ago

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