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3 years ago

Is the sum of .75 and -2.25 a rational number or irrational

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

3 0

The sum of 0.75 and -2.25 is -1.5. -1.5 is rational

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Compare 117.93 and 117.859

Licemer1 [7]

117.93 is greater than 117.859 by .035.

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3 years ago

Evaluate the triple integral_E x^6e^y dV where E is bounded by the parabolic cylinder z = 25 - y^2 and the planes z = 0, x = 5,

Nimfa-mama [501]

Nothing crazy here, just a matter of figuring out the limits of integration.

$\displaystyle\iiint_Ex^6e^y\,\mathrm dV=\int_{-5}^5\int_{-5}^5\int_0^{25-y^2}x^6e^y\,\mathrm dz\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_{-5}^{-5}\int_{-5}^5x^6e^y(25-y^2)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\left(\int_{-5}^5x^6\,\mathrm dx\right)\left(\int_{-5}^5e^y(25-y^2)\,\mathrm dy\right)=\boxed{\frac{625,000(3+2e^{10})}{7e^5}}$

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3 years ago

A student is solving the problem x^2 + 10x+ ___ = 16 + ___ by the method of completing the square. What number should the studen

kvv77 [185]

Answer:

I believe it could be any number because:

let y=any number

x^2+10x+y=16+y

x^2+10x+y-y=16+y-y

x^2+10x=16

you can add any number to both side and not change the solution.

5 0

3 years ago

PLEASE HURRY IT DO THE NEXT DAY PLZ HELP ME

liberstina [14]

The Bulldogs are least likely to win.

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3 years ago

Read 2 more answers

A political candidate has asked you to conduct a poll to determine what percentage of people support him. If the candidate only

Nadusha1986 [10]

Answer:

The sample size required is, n = 502.

Step-by-step explanation:

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}$

The margin of error is:

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

Assume that 50% of the people would support this political candidate.

The margin of error is, MOE = 0.05.

The critical value of z for 97.5% confidence level is:

z = 2.24

Compute the sample size as follows:

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

$n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)}}{MOE}]^{2}$

$=[\frac{2.24\times \sqrt{0.50(1-0.50)}}{0.05}]^{2}\\\\=501.76\\\\\approx 502$

Thus, the sample size required is, n = 502.

4 0

3 years ago