234÷72 divided with compatible numbers and estimate it

Please help thanks a(n)=−6+3(n−1)

erica [24]

Answer:

a=-1

Step-by-step explanation:

a(n) = -6+3(n-1)

a(n) = -6+3n-3

a(n) -n=-6+3n-n-3

a=-6+3-3

+3 +3

a3=-3

---- ----

3 3

a=-1

5 0

3 years ago

The following table summarises the data from a survey on the ownership of iPods among families with different levels of income.

AURORKA [14]

The two nominal variables are related.

The following table summarises the data from a survey on the ownership of iPods among families with different levels of income.

Ownership C1 C2 C3

No 40 32 48

Yes 30 48 52

The first thing to do in order to determine if they are related or not is to state our null and alternative hypothesis

Null hypothesis

$\mathbf{H_o: Two \nomial \ variables \ are \ related}$

Alternative hypothesis

$\mathbf{H_a: Two \nomial \ variables \ are \ not \ related}$

Using the Chi-square test statistics which can be expressed by using the formula

$X ^2 = \sum \dfrac{(O-E)^2}{E}$

Ownership C1 C2 C3 Total

No 40 32 48 120

Yes 30 48 52 130

Total 70 80 100 250

The expected values are calculated as:

$\mathsf{E_{a,b} = \dfrac{(row \ total \times column \ total )}{grand \ total }}$

$\mathsf{E_{1,1} = \dfrac{(70 \times120 )}{250 }}$

$\mathsf{E_{1,1} = 33.6}$

$\mathsf{E_{1,2} = \dfrac{(70 \times130 )}{250 }}$

$\mathsf{E_{1,2} = 36.4}$

$\mathsf{E_{2,1} = \dfrac{(80 \times120 )}{250 }}$

$\mathsf{E_{2,1} = 38.4}$

$\mathsf{E_{2,2} = \dfrac{(80 \times 130 )}{250 }}$

$\mathsf{E_{2,2} = 41.6}$

$\mathsf{E_{3,1} = \dfrac{(100 \times120 )}{250 }}$

$\mathsf{E_{3,1} = 48}$

$\mathsf{E_{3,2} = \dfrac{(70 \times130 )}{250 }}$

$\mathsf{E_{3,2} = 52}$

∴ Using the Chi-square test statistics, we have:

$X ^2 = \sum \dfrac{(O-E)^2}{E}$

$X ^2 = \Bigg( \dfrac{(40-33.6)^2}{33.6}+ \dfrac{(30-36.4)^2}{36.4}+ \dfrac{(32-38.4)^2}{38.4}+ \dfrac{(48-41.6)^2}{41.6} + \dfrac{(48-48)^2}{48}+ \dfrac{(52-52)^2}{52} \Bigg)$

$X ^2 = \Bigg( \dfrac{40.96}{33.6}+ \dfrac{40.96}{36.4}+ \dfrac{40.96}{38.4}+ \dfrac{40.96}{41.6} + \dfrac{0}{48}+ \dfrac{0}{52} \Bigg)$

$X ^2 = \Bigg( 1.2190+ 1.1253+ 1.0667+ 0.9846+0+ 0 \Bigg)$

$\mathbf{X ^2 =4.3956}$

The degree of freedom df = ((r - 1) × (c - 1))

= (3 - 1) (2 -1 )

= 2 × 1

= 2

∴

Assuming the level of significance = 5%

The p-value of the Chi-square test statistics at df of 2 is:

= $\mathbf{P(X^2 > 4.3956) \implies 0.111}$

Therefore, we can conclude that since the p-value (0.111) is greater than the level of significance (0.05), we fail to reject the null hypothesis.

Hence, the two nominal variables are related.

Learn more about Chi-square test statistics here:

brainly.com/question/2365682?referrer=searchResults

7 0

3 years ago

at store A, pencils are sold on individually. The cost y of x pencils is represented by the equation y = 0.55x. The costs of pen

Helga [31]

Answer:

The pencils at store A cost more

Step-by-step explanation:

The complete question in the attached figure

store A

we have

$y=0.55x$

This is a proportional relationship between the total cost (y) and the number of pencils (x)

The constant of proportionality is equal to the slope

The slope or unit rate is equal to

$m=\$0.55\ per\ pencil$

Store B

This is a proportional relationship between the total cost (y) and the number of pencils (x)

The constant of proportionality or slope is equal to

$k=\frac{y}{x}$

For (6,3.06)

substitute

$k=\frac{3.06}{6}$

$m=\$0.51\ per\ pencil$

The linear equation at store B is

$y=0.51x$

Verify each statement

case 1) The pencils at store A cost more

The statement is true

Because, one pencil at store A cost $0.55 and one pencil at store B cost $0.51

case 2) The pencils at store A cost $0.27 each

The statement is false

Because, one pencil at store A cost $0.55

case 3) The pencils at store B cost $0.30 each

The statement is false

Because, one pencil at store B cost $0.51

case 4) The pencils at store B cost more

The statement is false

Because, pencils at store A cost more

3 0

3 years ago

The vertex of a parabola is ( 3, -1). One point on the parabola is

Rudik [331]

Answer:

Step-by-step explanation:

If you plot the vertex and the point, you see that the point is above the vertex. Therefore, this is a positive parabola with the work form of

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

We have values for x, y, h, and k. Let's write the equation of the parabola, put it into function notation, then find another x value at which to evaluate it.

$8=a(6-3)^2-1$ and

$8=a(3)^2-1$ and

8 = 9a - 1 and

9 = 9a so

a = 1. The equation of the parabola in function notation is

$f(x)=(x-3)^2-1$

Since the vertex is at (3, -1) it would make sense to evaluate the function at x values close to the vertex. Let's evaluate the function at an x value of 4:

$f(4)=(4-3)^2-1$ and

$f(4)=(1)^1-1$ and

f(4) = 0. That means that another point on this parabola will be (4, 0).

8 0

4 years ago

three tickets to attend an off-broadway show cost $81 , 4 tickets cost $108, and 5 ticckets cost $135. A. show that the relation

alex41 [277]

1 ticket costs $27, 2 tickets cost $54, 3 cost $81, 4 cost $108, and 5 cost $135.

There is a $27 difference between each ticket above or below

4 0

3 years ago

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