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love history [14]
4 years ago
13

Two numbers have these properties:

Mathematics
1 answer:
almond37 [142]4 years ago
5 0

Answer:

12\ and\ 30

Step-by-step explanation:

Given both numbers are greater than 6

Their HCF is 6

Their LCM is 60

The product of the HCF and LCM of two numbers is the same as the product of the numbers themselves.

Let us say those number are a and b

a\times b=6\times 60\\a\times b=360

So, the product of those number is 360

Let us factorize 360

360=1\times 2\times 2\times 3\times 5\times 6

It is given that number should be greater than 6

The possible pairs of number are (10,36)\ and\ (12,30)

But only (12,30) has LCM as 60.

So those numbers are 12\ and\ 30

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Step-by-step explanation:

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How far will a runner traveling at 10.6 m/s move in fifteen seconds?
natka813 [3]

Answer: 159 miles because 10.6 times 15 = 159

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A box of ceral contains 10 cups . if each bowl holds 1 1/4 cups of cearal. how many bowls of ceral will you get out of the box
alexandr1967 [171]

Answer:

8 bowls

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
g A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estim
kipiarov [429]

Answer:

a) n= 1045 computers

b) n= 442 computers

c) A. ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

Step-by-step explanation:

Hello!

The variable of interest is

X: Number of computers that use the new operating system.

You need to find the best sample size to take so that the proportion of computers that use the new operating system can be estimated with a 99% CI and a margin of error no greater than 4%.

The confidence interval for the population proportion is:

p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

Z_{1-\alpha /2}= Z_{0.995}= 2.586

a) In this item there is no known value for the sample proportion (p') when something like this happens, you have to assume the "worst-case scenario" that is, that the proportion of success and failure of the trial are the same, i.e. p'=q'=0.5

The margin of error of the interval is:

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

\frac{d}{Z_{1-\alpha /2}} = \sqrt{\frac{p'(1-p)}{n} }

(\frac{d}{Z_{1-\alpha /2}})^2 = \frac{p'(1-p')}{n}

n * (\frac{d}{Z_{1-\alpha /2}})^2 = p'(1-p')

n= [p'(1-p')]*(\frac{Z_{1-\alpha /2}}{d} )^2

n=[0.5(1-0.5)]*(\frac{2.586}{0.04} )^2= 1044.9056

n= 1045 computers

b) This time there is a known value for the sample proportion: p'= 0.88, using the same confidence level and required margin of error:

n= [p'(1-p')]*(\frac{Z_{1-\alpha /2}}{d} )^2

n= [0.88*0.12]*(\frac{{2.586}}{0.04})^2= 441.3681

n= 442 computers

c) The additional information in part b affected the required sample size, it was drastically decreased in comparison with the sample size calculated in a).

I hope it helps!

4 0
4 years ago
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