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2 years ago

4,8,12,16,20 what is the pattern

Mathematics

2 answers:

lilavasa [31]2 years ago

7 0

The pattern is that you multiply the term number by 4 or
add four to the last/previous number in the sequence

Serggg [28]2 years ago

5 0

The pattern is that every number in the pattern increases by 4

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Use a table of numerical values of f(x,y) for (x,y) near the origin to make a conjecture about the value of the limit of f(x,y)

grigory [225]

Seems to be that the limit to compute is

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+2y^2}$

Consider an arbitrary line through the origin $y=mx$ , so that we rewrite the above as

$\displaystyle\lim_{x\to0}\frac{mx^2}{x^2+2m^2x^2}=\lim_{x\to0}\frac m{1+2m^2}=\frac m{1+2m^2}$

The value of the limit then depends on the slope $m$ of the line chosen, which means the limit is path-dependent and thus does not exist.

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Below is the five Number summary for a 136 hikers who recently completed the John muir trail JMT the variable is the amount of t

Helga [31]

Answer:

$IQR = Q_3 -Q_1 = 28-18 = 10$

Now we can find the limits in order to determine outliers like this:

$Left = Q_1 -1.5 IQR = 18 -1.5*10=3$

$Right = Q_1 +1.5 IQR = 18 +1.5*10=33$

So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier

b. 3

Step-by-step explanation:

For this case we have the following summary:

$Minimum = 9$ represent the minimum value

$Q_1 = 18$ represent the first quartile

$Median =Q_2= 21$ represent the median

$Q_3 = 28$ represent the third quartil

$Maximum=56$ represent the maximum

If we use the 1.5 IQR we need to find first the interquartile range defined as:

$IQR = Q_3 -Q_1 = 28-18 = 10$

Now we can find the limits in order to determine outliers like this:

$Left = Q_1 -1.5 IQR = 18 -1.5*10=3$

$Right = Q_1 +1.5 IQR = 18 +1.5*10=33$

So for this case the left boundary would be 3, if a value is lower than 3 we consider this observation as an outlier

b. 3

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2 years ago

Sections of prefabricated fencing are each 4 1/3 feet long. how long are 6 1/2 sections placed end to end

zalisa [80]

we know that

$4\frac{1}{3}= \frac{(4*3+1)}{3} =\frac{13}{3}$

$6\frac{1}{2}= \frac{(6*2+1)}{2}=\frac{13}{2}$

To find how long are $6\frac{1}{2}$ sections placed end to end

Multiply the number of sections by the length of one section

$\frac{13}{2} *\frac{13}{3} =\frac{169}{6}\ ft$

$\frac{169}{6}\ ft =28\frac{1}{6}\ ft$

therefore

<u>the answer is</u>

$28\frac{1}{6} \ ft$

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They are comparing the number of cats in teh shelter to the number of dogs in the shelter.

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Let the side of one square be x.    Area = x*x = x²

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