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3 years ago

Simplify: 2m – 6 ≤ 18.

Mathematics

2 answers:

inessss [21]3 years ago

4 0

Answer:

D. m≥12

Step-by-step explanation:

You first need to transfer the -6 over to the 18. The negative turns into a positive, therefore you add. 18+6 is 24. You now have 2m≤24. Bring the 2 over. You are dividing. 24/2 is 12.

MrRissso [65]3 years ago

3 0

Answer:

The answer is D.

Step-by-step explanation:

~Hoped this helped~

You might be interested in

What are the approximate values of the minimum and maximum points of f(x) = x5 − 10x3 + 9x on [-3,3]?

nika2105 [10]

Answer:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

Step-by-step explanation:

Given:

$f(x)=x^5-10x^3+9x$ ; [-3,3]

Explanation:

In order to find minimum/maximum of a function, we need to find the first derivative of the function and then set it equal to 0 to get critical points.

Therefore,

$f'(x)=5x^4-30x^2+9$

Setting derivative equal to 0, we get

$5x^4-30x^2+9=0$

On applying quadratic formula, we get

x=2.4, -2.4, -0.7, 0.7.

So, those are critical points of the given function.

Plugging the values x=2.4, -2.4, -0.7, 0.7, -3 and 3 in above function, we get

f(2.4)=(2.4)^5-10(2.4)^3+9(2.4)= -37.01376 : Minimum.

f(-2.4)=(-2.4)^5-10(-2.4)^3+9(-2.4)= 37.01376 : Maximum.

f(0.7)=(0.7)^5-10(0.7)^3+9(0.7) = 3.03807

f(-0.7)=(-0.7)^5-10(-0.7)^3+9(-0.7) = -3.03807

f(-3)=(-3)^5-10(-3)^3+9(-3) =0

f(3)=(3)^5-10(3)^3+9(3) =0

Therefore the approximate values of the minimum and maximum points of f(x) = $x^5- 10x^3+ 9x$ on [-3,3] are:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

7 0

4 years ago

How many different ways can you write a fraction that has a numerator of 2 as a sum of fractions

Rzqust [24]

At least you can write it as the sum of half of it twice, that is:
2/x = 1/x + 1/x
whatever x number you have, for example:
2/5 = 1/5 + 1/5

But then, you can express a fraction in many different ways, for example again:
2/5 = <span>1/5 + 1/5 = 2/10 + 2/10
</span>= 1/10 + 1/10 + 1/10 + 1/10
so there are an infinite ways of expressing such a fraction.

7 0

4 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

4 years ago

Select the number line that represents all solutions of e>-0.5. (Answer FAST!!!)

Pie

Well the answer couldn’t be C because the arrow has to go to the right not left because for the > in the equation

So the answer has to be D because the dot it on -.5 and the arrow is going right

4 0

3 years ago

Annie has $18.00 to spend at the grocery store. She is going to buy bags of nuts that cost $5.70 each and one box of cookies tha

Tcecarenko [31]

Answer: She can buy up to 2 bags of nuts

Step-by-step explanation:

Hi, to answer this question we have to write an inequality:

The product of the number of bags of nuts bought (n) and the price per bag (5.70); plus the price of one box of cookies (4.25) must be less or equal to Annie's money (18).

5.70n +4.25 ≤18

Solving for x:

5.70n ≤18 -4.25

5.70n ≤13.75

n ≤13.75/5.70

n ≤2.41

n ≤2 (rounded)

She can buy up to 2 bags of nuts

8 0

3 years ago

Read 2 more answers