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almond37 [142]
3 years ago
13

Can two positive integers have their H.C.F and L.C.M as 12 and 512 respectively? justify.

Mathematics
2 answers:
DaniilM [7]3 years ago
7 0

Here is your answer

NO

Reason:

H.C.F. divides the L.C.M. completely.

Here 512 is not completely divided by 12

So, two numbers can't have their H.C.F and L.C.M as 12 and 512 respectively.

HOPE IT IS USEFUL

MAXImum [283]3 years ago
4 0

Answer:

No.

Step-by-step explanation:

The  answer is no because 12 has a factor of 3 but 512 hasn't:

12 = 2*2*3

512 = 2^9.

The LCM  would have to have 3 as one of its factors.

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3 years ago
GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.24 mg of mercury. A sample of 25 bulbs shows a m
amm1812

Answer:

Null hypothesis:\mu \leq 3.24  

Alternative hypothesis:\mu > 3.24  

Step-by-step explanation:

1) Previous concepts  and data given

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".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

A right tailed test (sometimes called an upper test) is when the alternative hypothesis statement contains a greater than (>) symbol.

\bar X=3.29 represent the sample mean  

s represent the sample standard deviation

n represent the sample selected

\alpha significance level  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean for fluorescent bulbs is no more than 3.24 mg of mercury, the system of hypothesis would be:  

Null hypothesis:\mu \leq 3.24  

Alternative hypothesis:\mu > 3.24  

IIf we know the population deviation we can apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

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3 years ago
From our lesson, what is your difficulty in dealing with exponential expressions?
aleksandrvk [35]

Answer:

From the said lesson, the difficulty that I have been trough in dealing over the exponential expressions is the confusion that frequently occurs across my system whenever there's a thing that I haven't fully understand. It's not that I did not actually understand what the topic was, but it is just somewhat confusing and such. Also, upon working with exponential expressions — indeed, I have to remember the rules that pertain to dealing with exponents and frequently, I will just found myself unconsciously forgetting what those rule were — rules which is a big deal or a big thing in the said lesson because it is obviously necessary/needed over that matter. Surely, it is also a big help for me to deal with exponential expressions since it's so much necessary — it's so much necessary but I keep fogetting it.. hence, that's why I call it a difficulty. That's what my difficulty. And in order to overcome that difficulty, I will do my best to remember and understand well the said rules as soon as possible.

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