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san4es73 [151]
3 years ago
9

Find the greatest 3 digit number which when divided by 75,45 and 60 leaves no remainder and the remainder 4 in each case

Mathematics
1 answer:
pishuonlain [190]3 years ago
7 0
LCM of 75,45 and 60 is 900.
The divisor of 75,45 and 60 must be a multiple of their LCM.
75×900=67500 )⤵
45×900=40500 )↪ not a 3 digits no.
60×900=54000 )⤴
Since they are not 3 digits no..the greatest 3 digits no exactly divisible by 75,45,60 must be the LCM itself i.e. 900 >>Ans

Now,the next question is very simple.As the 900
greatest divisor of all given no., add 4 to get remainder 4 when divided by 75,60 and 45.
So, ans=904
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Wires manufactured for use in a computer system are specified to have resistances between 0.11 and 0.13 ohms. The actual measure
Marina CMI [18]

Answer:

a) P(0.11

And we can find this probability with this difference and with the normal standard table or excel:

P(-1.11

b) P(0.11 < \bar X < 0.13)

And we can use the z score defined by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using the limits we got:

z = \frac{0.11-0.12}{\frac{0.009}{\sqrt{4}}}= -2.22

z = \frac{0.13-0.12}{\frac{0.009}{\sqrt{4}}}= 2.22

And we want to find this probability:

P(-2.22< Z

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the resitances of a population, and for this case we know the distribution for X is given by:

X \sim N(0.12,0.009)  

Where \mu=0.12 and \sigma=0.009

We are interested on this probability

P(0.11

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(0.11

And we can find this probability with this difference and with the normal standard table or excel:

P(-1.11

Part b

We select a sample size of n =4. And since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want this probability:

P(0.11 < \bar X < 0.13)

And we can use the z score defined by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using the limits we got:

z = \frac{0.11-0.12}{\frac{0.009}{\sqrt{4}}}= -2.22

z = \frac{0.13-0.12}{\frac{0.009}{\sqrt{4}}}= 2.22

And we want to find this probability:

P(-2.22< Z

4 0
3 years ago
Read 2 more answers
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