What is the distance between the points ‒3 and 19 on a number line?

What would be the new median if 21 was added to the list

-BARSIC- [3]

Answer:

52 is the correct answer

6 0

2 years ago

Solve: 2x ≤ -2/3 (4x + 4)

Sidana [21]

Answer: x ≤ 1

reason:

first, multiply 4x and 4 by -2/3, which leaves you with the equation 2x ≤ -8/3x + 8/3
then, you add 8/3x to both sides, making it 4 2/3x ≤ 8/3
divide both sides by 4 2/3 and you get x ≤ 1

6 0

3 years ago

Read 2 more answers

Find the solution set for the given inequality. -7x + 3 > 45 A) Show your work to solve this inequality (B) Graph the solutio

scoundrel [369]

Answer:

See explanation

Step-by-step explanation:

Given the inequality $-7x+3>45$

A) Add -3 to both sides of inequality:

$-7x+3-3>45-3\\ \\-7x>42$

Divide by -7 (remember, dividing the inequality by negative number changes the sign of inequality):

$x$

B) Plot open circle at -6 on the number line and shade all values of x which go to the left from -6 (see attached diagram).

8 0

3 years ago

An alloy consists of nickel, zinc, and copper in the ratio 2:7:9. How many pounds of nickel have to be used to create alloy that

Hunter-Best [27]

We are given ratio nickel to zinc to copper = 2:7:9.

From this ratio we can see that <em>ratio of nickel to zinc is 2:7</em>.

Let us assume number of pounds of nickel have to be used to create alloy that contains 4.9 lb of zinc is x pounds.

So, we can setup a proportion as :

x : 4.9 = 2:7

Or

$\frac{x}{4.9}=\frac{2}{7}$

On cross multiplication, we get

7x= 4.9× 2

7x = 9.8

Dividing both sides by 7, we get

x = 1.4.

<h3>Therefore, 1.4 pounds of nickel have to be used to create alloy that contains 4.9 lb of zinc.</h3><h3 />

6 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

3 years ago