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2 years ago

How many 3 digit numbers are possible when a) the leading digit cannot be zero and the number must be a multiple of 4?

Mathematics

1 answer:

guajiro [1.7K]2 years ago

6 0

Step-by-step explanation:

I assume the digits can be repeated.

so, e.g. 555 is a valid number for this problem, right ?

that means we start with permutations with repetition :

n^r

n = the total number of items to pick from.

r = the number of items being picked per result.

we have 10 digits (0,1,2,3,4,5,6,7,8,9), and we pick 3 of them.

that gives us (with very little surprise, I hope)

10³ = 1000 different possible numbers from 000 to 999.

from these numbers we eliminate all with leading 0.

as we handled all digits the same way and with the same priority, there is the same amount of numbers for every digit in the leading position.

that means 1/10 of the total amount of numbers has a leading 0, or a leading 1, or a leading 2, ...

so, we need to subtract 1/10 × 1000 from 1000 :

1000 - 1000×1/10 = 1000 - 100 = 900

that would be the numbers 100 to 999.

and we have one more condition : the number must be a multiple of 4.

how many are there ?

well, that's the funny thing about numbers : from all numbers 1/2 of them are multiples of 2 (or divisible by 2), 1/3 of them are multiples of 3 (or divisible by 3), and ... you guessed it, 1/4 of them are multiples of 4 (or divisible by 4). and so on.

and so, 1/4 of our 900 numbers are multiples of 4 :

1/4 × 900 = 225

so, there are 225 possible 3-digit numbers that are multiples of 4 and do not start with a 0.

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My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the

AVprozaik [17]

Answer:

a) $n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

b) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

Step-by-step explanation:

Previous concepts

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

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

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.64$

The margin of error for the proportion interval is given by this formula:

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

And on this case we have that $ME =\pm 0.02$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07$

And rounded up we have that n=1476

Part b

For this case since we don't have a prior estimate we can use $\hat p =0.5$

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681$

And rounded up we have that n=1681

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Answer:

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<em>P.S. Hope it makes sense. If you have any questions, feel free to share them in the comment senction below. I'll be happy to help. Have a wonderful day!</em>

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Step-by-step explanation:

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