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tankabanditka [31]

3 years ago

8

Which set of ordered pairs does NOT represent a function? A) (0, 1), (2, 2), (4, 8), (-2, 7), (5, 8) B) (0, 1), (2, 2), (4, 8),

(2, 7), (5, 8), (7, 9) C) (-3, 6), (2, 7), (0, 5), (1, 5), (4, 9), (5, 4) D) (-4, 2), (-3, 2), (-2, 2), (-1, 2), (0, 2), (1, 2) Submit

1 answer:

Viktor [21]3 years ago

8 0

That is B
Because there are duplicate x-values in the ordered pairs, ( the (2,2) and the (2,7).

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In a list of households, own homes and do not own homes. Five households are randomly selected from these households. Find the p

hichkok12 [17]

Answer:

The probability of success for this case would be:

$p =\frac{9}{15}= 0.6$ representing the proportion of homes that are own homes

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=15, p=0.6)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want this probability:

$P(X=3)$

And uing the probability mass function we got:

$P(X=3)= 15C3 (0.6)^3 (1-0.6)^{15-3}= 0.00165$

Step-by-step explanation:

Adduming the following info: In a list of 15 households, 9 own homes and 6 do not own homes. Five households are randomly selected from these 15 households. Find the probability that the number of households in these 5 own homes is exactly 3.

Round your answer to four decimal places

P (exactly 3)=

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

The probability of success for this case would be:

$p =\frac{9}{15}= 0.6$ representing the proportion of homes that are own homes

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=15, p=0.6)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want this probability:

$P(X=3)$

And uing the probability mass function we got:

$P(X=3)= 15C3 (0.6)^3 (1-0.6)^{15-3}= 0.00165$

4 0

3 years ago

Please can you simplify -x+2z+x+2y

slavikrds [6]

Answer:

2y + 2z

Step-by-step explanation:

-x + 2z + x + 2y

= (-1 + 1)x + 2y + 2z

= 2y + 2z

Hope this helped!

3 0

3 years ago

olga2289 [7]

For the number before x you would look for the rise over the run. in this case you can see that there is an intersection at (-2,0) and (0,1) you find the difference ox each so you have 2/1 or just 2 you can tell that the line is going up so it is positive. for the addition part look at where the y intercept is in this case it is at positive one. this makes your equation y=2x+1

8 0

3 years ago

Read 2 more answers

Divide. Reduce the answer to lowest terms. 6 ÷ 4/5

Butoxors [25]

Answer:

I think its 7.5 but I might be wrong

Step-by-step explanation:

7 0

3 years ago

How do I solve question 6 through 8?<br> Solve for me

rewona [7]

The equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2

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

A quadratic function is represented as:

y = a(x - h)^2 + k

<u>Question #6</u>

The vertex of the graph is

(h, k) = (-1, 2)

So, we have:

y = a(x + 1)^2 + 2

The graph pass through the f(0) = -2

So, we have:

-2 = a(0 + 1)^2 + 2

Evaluate the like terms

a = -4

Substitute a = -4 in y = a(x + 1)^2 + 2

y = -4(x + 1)^2 + 2

<u>Question #7</u>

The vertex of the graph is

(h, k) = (2, 1)

So, we have:

y = a(x - 2)^2 + 1

The graph pass through (1, 3)

So, we have:

3 = a(1 - 2)^2 + 1

Evaluate the like terms

a = 2

Substitute a = 2 in y = a(x - 2)^2 + 1

y = 2(x - 2)^2 + 1

<u>Question #8</u>

The vertex of the graph is

(h, k) = (1, -2)

So, we have:

y = a(x - 1)^2 - 2

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So, we have:

-3 = a(0 - 1)^2 - 2

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a = -1

Substitute a = -1 in y = a(x - 1)^2 - 2

y = -(x - 1)^2 - 2

Hence, the equations of the functions are y = -4(x + 1)^2 + 2, y = 2(x - 2)^2 + 1 and y = -(x - 1)^2 - 2

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5 0

1 year ago