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3 years ago

Solve x<3 and x≥−1 and write the solution in interval notation. If there is no solution, type ∅ .

Mathematics

1 answer:

bearhunter [10]3 years ago

7 0

Shown in the Figure below

Step-by-step explanation:

In this exercise, we have the following system of inequalities:

$x$

The word and means intersection, so both inequalities must be true at the same time. We can graph this system as indicated in the figure below. Indeed, there is intersection between -1 and 1. Also, the number -1 is included in the solution because $x\geq -1$ and this is represented by the closed circle. On the other hand, the number 3 is not included in the solution because $x$ and this is represented by the open circle.

<h2>Learn more:</h2>

Inequalities: brainly.com/question/12261425

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At the school book fair, Henry brings $10.00. He purchases a book for 8.50, plus a 6% sales tax. How much money in change will H

Irina18 [472]

Answer:

.99 cents

Step-by-step explanation:

8.5 x 1.06= 9.01

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3 years ago

Read 2 more answers

(04.04 LC)

LekaFEV [45]

Reserve Requirement

Took the test got 100

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3 years ago

An office manager recorded the number of reams of paper that were used for a printer each month for six months: 15, 10, 25, 19,

Virty [35]

A)
10, 15, 17, 19, 25 
Min: 10 
Median: 17 
Max: 25 

lower Quartile: (10+15)/2 = 12.5 
upper Quartile: (19+25)/2 = 22 

b) 
http://imageshack.us/photo/my-images/22/...

c) 
Mean: (10 + 15 + 17 + 19 + 25)/5 = 17.2 

[ ( 17.2 - 10 )^2 + ( 17.2 - 15 )^2 + ( 17.2 - 17 )^2 + ( 17.2 - 19 )^2 + ( 17.2 - 25 )^2 ] / 5 = 24.16 

√( 24.16 ) ≈ 4.92 
-------------- 
4) 

[ 4. . .-2 . . . -12 ------> 3R1 - 4R2 ----> R2 
3. . .-1 . . . -3 ] 

[ 4. . .-2 . . . -12 ------> R1 - R2 ----> R1 
0. . .-2 . . . -24 ] 

[ 4. . ..0 . . . .12 ------> R1/4 ----> R1 
0. . .-2 . . . -24 ] R2/-2 ----> R2 

[ 1. . ..0 . . . .3 
0. . .1 . . . 12 ] 

x = 3 & y = 12

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

6 0

3 years ago

Help please. my sister cant figure this out

Elza [17]

Q5 - 6: I'm guessing that it's asking us how many less than right angles, right angles, or more than right angles are present in the shapes given in question 5 and 6.

Keep in mind that right angles are exactly 90 degrees. Less than a right triangle is less than 90 degrees ( x < 90) and more than a right triangle is more than 90 degrees (x > 90) View the attachment for further demonstration and explanation.

Based on this, I think you (or your sister) need to re-check question 5. There are no right angles present there, and the lack of the box and L shaped angles indicate this. You are on the right track for number 6 though as only 2 right triangles are present.

For number 8, Kaylee (lol, feels weird saying that since my name is Kaylee), is incorrect. There are no right angles. You can further explain and elaborate on this using the information I supplied you with.

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3 years ago

In an anthropology class there are 77 anthropology majors and 1414 non-anthropology majors. 55 students are randomly selected to

Nata [24]

Answer:

P = 0.9989

Step-by-step explanation:

In order to do this, I will use the following numbers to make the calculations easier. In this case, We'll say that we have 7 majors and 14 non majors anthopology to present a topic.

This means that in the class we have 21 students.

Now, we choose 5 of them, and we want to know the probability that 1 of them is non major.

First, we need to calculate the number of ways we can select the students in all cases, and then, the probability.

First, we'll use the combination formula, to calculate the number of ways we can select the 5 students out of the 21. We use combination, because it does not matter the order that the students are selected.

C = m! / n!(m - n)!

Where:

m: number of students

n: number of selected students out of m.

With this expression we will calculate first, how many ways we can choose the 5 students out of 21:

C1 = 21! / 5!(21-5)! = 20,349

Now let's calculate the number of ways you can get the all 5 students are non majors:

C2 = 14! / 5!(14 - 5)! = 2002

Now we need to know the number of ways we can get 4 non majors and 1 major:

C3 = C3' * C3''

C3' represents the number of ways we can get 4 non majors and the C3'' represents the number of ways we can get 1 major.

C3' = 14! / 4!(14 - 4)! = 1,001

C3'' = 7! / 1!(7 - 1)! = 7

C3 = 1001 * 7 = 7,007 ways to get 4 non majors and 1 major

Now the way to get 3 non majors and 2 majors, we do the same thing we do to get 4 non majors and 1 major, but changing the numbers. Then the way to get 2 non majors and 3 majors, and finally 1 non major and 4 majors:

3 non majors and 2 majors:

C4 = C4' * C4'' = [14! / 3!(14 - 3)!] * [7! / 2!(7 - 2)!] = 7,644

2 non majors and 3 majors:

C5 = C5' * C5'' = [14! / 2!(14 - 2)!] * [7! / 3!(7 - 3)!] = 3,185

1 non major and 4 majors:

C6 = C6' * C6'' = [14! / 1!(14 - 1)!] * [7! / 4!(7 - 4)!] = 490

Finally to know the probability of getting 1 out of the 5 to be non major, we have to sum all the previous results, and divide them by the ways we can choose the 5 students (C1):

P = 2,002 + 7,007 + 7,644 + 3,185 + 490 / 20,349

P = 0.9989

5 0

3 years ago