Find 6A-8B (picture provided)

IRISSAK [1]

Use point slope formula
$y - y1 = m(x - x1)$
$y - 9 = 2(x - 1)$
$y - 9 = 2x - 2$
$y = 2x + 7$

y=2x+7

5 0

3 years ago

En una prueba de duración para determinar la vida útil de un nuevo tipo de lámpara lanzada al mercado, los datos de la muestra s

diamong [38]

Answer:

no habla espenol :((

Step-by-step explanation:

6 0

3 years ago

Determine the slope of the table

AURORKA [14]

Answer:

The slope is -1/2 or -0.5

Hope this helps!

5 0

3 years ago

Read 2 more answers

Simplify<br> a(cube)-1000b(cube)<br> 64a(cube)-125b(cube)

Elanso [62]

The simplification of a³ - 1000b³ and 64a³ - 125b³ is (a - 10b) × (a² + 10ab + 100b²) and 4a - 5b) • (16a² + 20ab + 25b²) respectively.

<h3>Simplification</h3>

Question 1: a³ - 1000b³

a³ - b³

= (a-b) × (a² +ab +b²)

1000 is the cube of 10
a³ is the cube of a¹
b³ is the cube of b¹

So,

(a - 10b) × (a² + 10ab + 100b²)

Question 2: 64a³ - 125b³

a³ - b³

= (a-b) × (a² +ab +b²)

64 is the cube of 4
125 is the cube of 5
a³ is the cube of a¹
b³ is the cube of b¹

So,

(4a - 5b) • (16a² + 20ab + 25b²)

Learn more about simplification:

brainly.com/question/723406

#SPJ1

8 0

2 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

3 years ago