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3 years ago

Given that 4704/m = n^2, where m and n are whole numbers and n is as large as possible, find the value of m and of n

Mathematics

1 answer:

Andrews [41]3 years ago

5 0

Answer:

n=28,m=6

Step-by-step explanation:

Rearrange to give 4704 = m * n^2

Express 4704 in its prime factorisation:

2^5 * 3^1 * 7^2

We know that n must be as large as possible and, because it is squared, be the even numbers.

We get n^2 = 2^4 * 7^2 so n = 2^2*7 = 28

m is the leftovers

2^1*3 = 6

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cupoosta [38]

Answer:

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3 years ago

An article reported that for a sample of 52 kitchens with gas cooking appliances monitored during a one-week period, the sample

kolbaska11 [484]

Answer:

a) $654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

b) $n=(\frac{1.960(175)}{25})^2 =188.23 \approx 189$

Step-by-step explanation:

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=52-1=51$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,51)".And we see that $t_{\alpha/2}=2.01$

Replacing we got:

$654.16-2.01\frac{164.55}{\sqrt{52}}=608.29$

$654.16+2.01\frac{164.55}{\sqrt{52}}=700.03$

And we can conclude that we are 95% confident that the true mean of Co2 level is between 608.29 and 700.03 ppm

Part b

The margin of error is given by :

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

The desired margin of error is ME =50/2=25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , and we use an estimator of the population variance the value of 175 replacing into formula (b) we got: