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Zina [86]
3 years ago
10

How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not 12?

Mathematics
1 answer:
siniylev [52]3 years ago
7 0

Count the number of multiples of 3, 4, and 12 in the range 1-2005:

⌊2005/3⌋ ≈ ⌊668.333⌋ = 668

⌊2005/4⌋ = ⌊501.25⌋ = 501

⌊2005/12⌋ ≈ ⌊167.083⌋ = 167

(⌊<em>x</em>⌋ means the "floor" of <em>x</em>, i.e. the largest integer smaller than <em>x</em>, so ⌊<em>a</em>/<em>b</em>⌋ is what you get when you divide <em>a</em> by <em>b</em> and ignore the remainder)

Then using the inclusion/exclusion principle, there are

668 + 501 - 2•167 = 835

numbers that are multiples of 3 or 4 but not 12. We subtract the number multiples of 12 twice because the sets of multiples of 3 and 4 both contain multiples of 12. Subtracting once removes the multiples of 3 <em>and</em> 4 that occur twice. Subtracting again removes them altogether.

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Assume that the random variable X is normally distributed with mean u = 45 and standard deviation o = 14.
larisa86 [58]

Answer:

P(57 < X < 69) = 0.1513

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 45, \sigma = 14

Find P(57 < X < 69):

This is the pvalue of Z when X = 69 subtracted by the pvalue of Z when X = 57. So

X = 69

Z = \frac{X - \mu}{\sigma}

Z = \frac{69 - 45}{14}

Z = 1.71

Z = 1.71 has a pvalue of 0.9564

X = 57

Z = \frac{X - \mu}{\sigma}

Z = \frac{57 - 45}{14}

Z = 0.86

Z = 0.86 has a pvalue of 0.8051

0.9564 - 0.8051 = 0.1513

P(57 < X < 69) = 0.1513

4 0
3 years ago
Which is the steepest? Explanation that includes the graph. (The graph is on the picture)
fomenos

Answer:

D

Step-by-step explanation:

3 0
3 years ago
A coin-operated machine sells plastic rings. It contains 6 yellow rings, 11 blue rings, 15 green rings, and 3 black rings. Suzy
aleksandrvk [35]
1) you have to add all the rings together to find the denominator

6 + 11 + 15 + +3 = 35

2) the numerator is the amount of rings of the certain color you are trying to find the probability of

15

3) find the theoretical probability! 15/35

4) simplify

15/35 (divide both by 5) = 3/7
hope it helped :)


8 0
3 years ago
2+2=?<br> What does it equal
Oxana [17]

Answer:

4

Step-by-step explanation:

2 + 2 = 4

4 0
3 years ago
Read 2 more answers
Find the limit, if it exists. (If an answer does not exist, enter DNE.)
gavmur [86]

Answer:

\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{x^2+y^2+49}+7\right)=14

Step-by-step explanation:

to find the limit:

\lim\limits_{(x,y)\rightarrow(0,0)}\left(\dfrac{x^2+y^2}{\sqrt{x^2+y^2+49}-7}\right)

we need to first rationalize our expression.

\dfrac{x^2+y^2}{\sqrt{x^2+y^2+49}-7}\left(\dfrac{\sqrt{x^2+y^2+49}+7}{\sqrt{x^2+y^2+49}+7}\right)

\dfrac{(x^2+y^2)(\sqrt{x^2+y^2+49}+7)}{(\sqrt{x^2+y^2+49}\,)^2-7^2}

\dfrac{(x^2+y^2)(\sqrt{x^2+y^2+49}+7)}{(x^2+y^2)}

\sqrt{x^2+y^2+49}+7

Now this is our simplified expression, we can use our limit now.

\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{x^2+y^2+49}+7\right)\\\sqrt{0^2+0^2+49+7}\\7+7\\14

Limit exists and it is 14 at (0,0)

4 0
3 years ago
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