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2 years ago

Graph the solution set for the inequality x-2/-3 ≤ -5/6

Mathematics

2 answers:

ZanzabumX [31]2 years ago

4 0

Answer: x ≤ 1/6
Explanation: Solve for x
by simplifying both sides of the inequality, then isolating the variable.

zhuklara [117]2 years ago

3 0

Answer:

x $\leq$ - 3/2 (The graph is the image I attached)

Explanation:

1) Determine the sign of the fraction next to the variable, in this case, it's positive, which turns the equation into : x + 2/3 $\leq$ - 5/6

2) Move the constant to the right so that x is only on one side, this then turns the equation into: x $\leq$ - 5/6 - 2/3

3) Subtract the fractions, and the answer is x $\leq$ -3/2

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The midpoint of GH are M(4,3) and D(5,-6). The coordinates of the endpoint are?

Phantasy [73]

Answer:

The endpoint coordinates are (3,12)

Step-by-step explanation: $\frac{5+x^{2} }{2} =4$ $\frac{-6+y^{2} }{2} = 3$

You multiply 4 by 2 which gives you 8 and write the equation $5+x^{2} =8$ then you subtract 5 from 8 to get 3 for the x coordinate.

You multiply 3 by 2 which gives 6 and writhe the equation $-6+y^{2}=6$ then you substitute the -6 for +6 and add +6 and 6 to get 12 for the y coordinate.

To check use the midpoint formula with coordinates D and your endpoint coordinates $\frac{5+3}{2}=\frac{8}{2} =4 \\$ $\frac{-6+12}{2} =\frac{6}{2} =3$ and you get the answer for the midpoint which is (4,3)

6 0

2 years ago

The table shows the different packs of sports cards available for purchase. What is the smallest number of baseball and football

WARRIOR [948]

<span>The least common multiple would be 280 cards of each.

20 Packs of football, 8 packs of Hockey, and 7 packs of Baseball.</span>

5 0

2 years ago

Read 2 more answers

IF TWO ANGLES ARE CONSECUTIVE EXTERIOR ANGLES AND <1=59, WHAT IS THE MEASURE OF <2?

lutik1710 [3]

Answer:

<2 = 121

Step-by-step explanation:

Consecutive exterior angles add up to 180 degrees.

6 0

3 years ago

Ines has saved $5. She doubles the amount she saves each week. Does the function have a constant difference or a constant ratio?

salantis [7]

Answer:

The function have a constant ratio

No, it does not represent an exponential function.

Step-by-step explanation:

As given,

Ines has saved $5. She doubles the amount she saves each week.

⇒ In first week she saved = $5

In second week she saved = 2×$5 = $10

In third week she saved = 2×$10 = 20

In fourth week she saved = 2×$20 = 40

and so on..

∴ we get the series

5, 10, 20, 40 ,.......

If the series is given as a, b, c, d, ..... , then

The common difference is defined as d = b-a = c-b

The common ration is defined as r = $\frac{b}{a} = \frac{c}{b}$

Here, in the given question

The series is - 5, 10, 20, 40

This has no common difference , d = 10 - 5 (= 5 )≠ 20 - 10 (= 10)

As 5 ≠ 10

So the function does not define common difference .

Now,

This has common ratio , r = $\frac{10}{5} (=2) = \frac{20}{10}(=2)$

As 2 = 2

So, the function represent common ratio

Now,

The function does not represent an exponential function because it is increased by common ratio.

7 0

3 years ago

Luda [366]

Answer:

a) H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) $\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

c) $\chi^2_{crit}=5.991$

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

Step-by-step explanation:

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".

Assume the following dataset:

High school Some College Bachelor or higher Total

Public Broadcasting 15 15 10 40

Commercial stations 5 25 10 40

Total 20 40 20 80

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

Part a

H0: There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

The level os significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

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

Part b

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:

$E_{1} =\frac{20*40}{80}=10$

$E_{2} =\frac{40*40}{80}=20$

$E_{3} =\frac{20*40}{80}=10$

$E_{4} =\frac{20*40}{80}=10$

$E_{5} =\frac{40*40}{80}=20$

$E_{6} =\frac{20*40}{80}=10$

And the expected values are given by:

High school Some College Bachelor or higher Total

Public Broadcasting 10 20 10 40

Commercial stations 10 10 20 40

Total 20 30 30 80

Part b

And now we can calculate the statistic:

$\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75$

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

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

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is $\chi^2_{crit}=5.991$

Part d

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}$

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

"=1-CHISQ.DIST(33.75,2,TRUE)"

Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

7 0

3 years ago