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schepotkina [342]

4 years ago

5

Which set of ordered pairs represents a function? {(2, –2), (1, 5), (–2, 2), (1, –3), (8, –1)} {(3, –1), (7, 1), (–6, –1), (9, 1

), (2, –1)} {(6, 8), (5, 2), (–2, –5), (1, –3), (–2, 9)} {(–3, 1), (6, 3), (–3, 2), (–3, –3), (1, –1)}

2 answers:

ra1l [238]4 years ago

7 0

A function can't have x repeating any of the same number twice for example the first one (2, -2), (1, 5), (-2, 2), (1,-3), (8,-1) you have two 1's (1,5) and (1,-3) the x is the first number. Now a function can have the same y value. So your answer is (3, -1), (7,1), (-6,-1), (9,1), and (2,-1) you have to have all different x values in order for it to be a function. Hope that helps.

Aneli [31]4 years ago

7 0

Answer:

B

Step-by-step explanation:

BECAUSE I GOT AN 100

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Given the function, f(x)=lx+1l+2 , choose the correct transformation.

Wewaii [24]

moves left 1 unit and up tow units any thing on the inside of the brackets means it will move left or right + menas left - means right on the outside it will move up or down - menas down + means up

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3 years ago

What are the possible solutions to the equation: x-12 = sqrt x+44

Ainat [17]

possible solutions for equation $x-12 = \sqrt{x+44}$ is $x=5 , x=20$

<u>Step-by-step explanation:</u>

Here we have to find the possible solutions to the equation: x-12 = sqrt x+44 .Let's find out:

⇒ $x-12 = \sqrt{x+44}$

⇒ $(x-12)^2 = (\sqrt{x+44})^2$

⇒ $(x^2+144-24x) =x+44$

⇒ $(x^2+144-24x) -(x+44) =0$

⇒ $(x^2+100-25x) =0$

⇒ $x^2-25x+100 =0$

⇒ $x^2-20x-5x+100 =0$

⇒ $x(x-20-5(x-20) =0$

⇒ $(x-5)(x-20)=0$

⇒ $x-5=0 , x-20=0\\$

⇒ $x=5 , x=20$

Therefore , possible solutions for equation $x-12 = \sqrt{x+44}$ is $x=5 , x=20$

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3 years ago

A grocery store sells a bag of 8 oranges for $6.24. If Chase spent $8.58 on oranges, how many did he buy?

sammy [17]

Answer:

11 oranges

Step-by-step explanation:

divide the price by the number which is 0.78

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5 0

3 years ago

Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa

xxMikexx [17]

Answer:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins. Yes No Total

Small 32 18 50

Medium 68 7 75

Large 89 11 100

_____________________________________

Total 189 36 225

We need to conduct a chi square test in order to check the following hypothesis:

H0: independence between heath insurance coverage and size of the company

H1: NO independence between heath insurance coverage and size of the company

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{50*189}{225}=42$

$E_{2} =\frac{50*36}{225}=8$

$E_{3} =\frac{75*189}{225}=63$

$E_{4} =\frac{75*36}{225}=12$

$E_{5} =\frac{100*189}{225}=84$

$E_{6} =\frac{100*36}{225}=16$

And the expected values are given by:

Size Company/ Heal. Ins. Yes No Total

Small 42 8 50

Medium 63 12 75

Large 84 16 100

_____________________________________

Total 189 36 225

And now we can calculate the statistic:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

3 0

4 years ago

Given P(A)=0.34P(A)=0.34, P(B)=0.45P(B)=0.45 and P(A\text{ or }B)=0.517P(A or B)=0.517, find the value of P(A\text{ and }B)P(A a

maria [59]

Easy this answer is going to be 0.517 you can read and round

8 0

3 years ago