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1 year ago

Which of the following points could be added to the graph f(x)=-|x+3| to keep the graph a function?

Mathematics

1 answer:

sasho [114]1 year ago

6 0

None of the points could be added to the graph f(x)=-|x+3| to keep the graph a function

<h3>How to determine the point?</h3>

The equation of the function is given as:

f(x) = - |x + 3|

The points are given as:

(0, 3) and (-3, -6)

When x = 0, we have:

f(0) = - |0 + 3|

f(0) = -3 --- different y value from (0, 3)

When x = -3, we have:

f(-3) = - |-3 + 3|

f(-3) = 0 --- different y value from (-3, -6)

This means that the x values point to different y values (this does not represent a function)

Hence, none of the points could be added to the graph f(x)=-|x+3| to keep the graph a function

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You might be interested in

These box plots show the basketball scores for two teams.

Arisa [49]

The most appropriate statement is the interquartile range for the Wolverines, 30 is less than the IQR for the panthers, 40.

<h3>What is the correct statement?</h3>

The box plot is used to show the distribution of data. The box plot can be used to determine the range, interquartile range and median of the data set.

The range is the difference between the two ends of the whiskers.

Range for the Wolverines = 96 - 35 = 61

Range for the Panthers = 107 - 33 = 74

The interquartile range is the difference between the first and third lines on the box

IQR for the Wolverines = 85 - 55 = 30

Range for the Panthers = 90 - 50 = 40

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6 0

1 year ago

There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=

yulyashka [42]

Answer:

S = {0,2,3,4}

P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001

Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

P(X = 0) = $\frac{12P4}{12^{4} }$

P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = $\frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }$

P(X=2) = 0.401

P(X=3) = $\frac{4C3 * 12P2}{12^{4} }$

P(x=3) = 0.025

P(X=4) = $\frac{12}{12^{4} }$

P(X=4) = 0.001

Mean, $\mu = \sum xP(x)$

$\mu = (0*0.573) + (2*0.401) + (3*0.025) + (4*0.001)\\\mu = 0.879$

Standard deviation, $SD = \sqrt{\sum x^{2} P(x) - \mu^{2}} \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2} * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}$