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Morgarella [4.7K]
3 years ago
5

What real world problem can be solved with the inequality 30+2x<5x

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
7 0

Answer:

Algebra is user in business/finance management, software development, coding, landscaping plans, and nearly every form of engineering. The integrity of society requires all sorts of intuitive math, and the device you are typing on as well as the coding of this website relied on it was well.

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Sorting through unsolicited e-mail and spam affects the productivity of office workers. An InsightExpress survey monitored offic
nydimaria [60]

Answer:

1. Refer to the explanation for the frequency table.

2. (a) 60%

   (b) 25%

Step-by-step explanation:

Part 1. The question is asking us to group the data according to the classes given and fill in the frequency (F), relative frequency (RF), cumulative frequency (CF) and relative cumulative frequency (RCF).

To compute the frequency for each class count the number of data that lies in the class range and write it. For the first class (1-5) we can see that there are 12 numbers which lie in the range 1 to 5. Those numbers are: 2, 4, 4, 1, 2, 1, 5, 5, 5, 3, 4, 4. Similarly, for the second class, the frequency is 3 because 8, 8, 8 lie in this range from our existing data. For the third class 11-15 there are 2 numbers in the given data which lie in this range and those numbers are 12, 15. Similarly, the rest of the frequencies can be computed.

For the relative frequency, the frequency of each class is divided by the total frequency i.e. 20.  

For the class 1-5, RF = 12/20 = 0.6.  

For 6-10, RF = 3/20 = 0.15

For 11-15, RF = 2/20 = 0.1

For 16-20, RF = 1/20 = 0.05

For 21-25, RF=1/20 = 0.05

For 26-30, RF = 0/20 = 0.00

For 31-34, RF= 1/20 = 0.05.

To compute the cumulative frequency, add the existing frequency of each class with the previous frequencies.

For 1-5, CF = 0+12 = 12

For 6-10, CF = 12+3=15

For 11-15, CF = 12+3+2 = 17

For 16-20, CF=12+3+2+1 = 18

For 21-25, CF = 12+3+2+1+1=19

For 25-30, CF = 12+3+2+1+1+0=19

For 31-34, CF = 12+3+2+1+1+0+1=20

Now, to compute relative cumulative frequency, add the existing relative frequency of each class with the previous relative frequencies.  

For 1-5, RCF = 0+0.6

For 6-10, RCF = 0.6 + 0.15 = 0.75

For 11-15, RCF = 0.6+0.15+0.1 = 0.85

The rest of the relative cumulative frequencies can be computed in the same way.

Class(Minutes)           F       RF     CF     RCF

      1-5                       12     0.60    12     0.60  

      6-10                      3     0.15     15     0.75  

      11-15                      2     0.10     17     0.85  

      16-20                    1      0.05    18     0.90  

      21-25                    1      0.05    19     0.95  

      26-30                   0     0.00    19     0.95  

      31-34                     1     0.05    20       1  

       <u>Total                  20</u>  

Part 2. Now, we are asked to compute the Ogive graph which is also called as the cumulative frequency graph. The cumulative frequency needs to be plotted on the y-axis and the upper limit of each class needs to be plotted on the x-axis. The graph is attached.

(a) From the graph we can see that the number of workers who spend less than 5 minutes on unsolicited e-mail and spam are 12. So the answer for this part is 12 workers.

Percentage = 12/20 x 100 = 60%

(b) From the graph we can see that the number of workers who spend less than 10 minutes on spam e-mail are 15. The question is asking for the number of people who spend more than 10 minutes. For this we need to subtract 15 from the total number of workers.  

Number of workers spending more than 10 minutes = 20-15 = 5 workers.

Percentage = 5/20 x 100 = 25%

4 0
3 years ago
Joshira makes and sells crafts.
Maksim231197 [3]

Answer: 6

Step-by-step explanation:

i think joshira makes 6 item in 8 hours.

i got my answer by multiplying 3/4 with 8

7 0
3 years ago
Read 2 more answers
13.
Artist 52 [7]

Answer:

5 units

Step-by-step explanation:

An isosceles triangle is a triangle with two legs that have the same length. The perimeter of a triangle is the sum of the lengths of all sides of the triangle. Now taking this into account, we know that:

2L + B = 14 units

Where:

L is the measure of one leg

B is the measure of the base

Since two legs are the same and the base is 1 less, this means the measure of each leg would be:

B = L -1

Now we have two equations:

2L + B = 14 units

B = L- 1

We plug one equation into the other and make 1 equation:

2L + (L-1) = 14 units

Get rid of the parentheses:

2L + L - 1 = 14

Combine like terms:

3L - 1 = 14

Add 1 to both sides of the equation:

3L - 1 + 1 = 14 + 1

3L = 15

Divide both sides by 3:

3L/3 = 15/3

L = 5

So the length of a leg is 5 units

Let's check!

B = L - 1

B = 5 - 1

B = 4

Then we use that to solve for the perimeter:

2L + B

2(5) + 4

10 + 4 = 14

7 0
2 years ago
Summer campers were surveyed about their favorite fruit. The resulting table shows the data collected from the survey. Use the i
NemiM [27]

Answer:

Step-by-step explanation: 90

7 0
3 years ago
Read 2 more answers
Find the measures of the numbered angles in the isosceles trapezoid.<br> I need help ASP
Novay_Z [31]

Answer:

  ∠1 = 50°

  ∠2 = ∠3 = 130°

Step-by-step explanation:

In an isosceles trapezoid, such as this one, the angles at either end of a base are congruent:

  ∠1 ≅ 50°

  ∠2 ≅ ∠3

The theorems applicable to transversals and parallel lines also apply to the sides joining the parallel bases. In particular, "consecutive interior angles are supplementary." That is, angles 1 and 2 are supplementary, for example.

  ∠2 = 180° -∠1 = 180° -50° = 130°

We already know angle 3 is congruent to this.

  ∠1 = 50°

  ∠2 = ∠3 = 130°

_____

<em>Additional comment</em>

It can be easier to see the congruence of the base angles if you remove the length of the shorter base from both bases. This collapses the figure to an isosceles triangle and makes it obvious that the base angles are congruent.

Alternatively, you can drop an altitude to the longer base from each end of the shorter base. That will create two congruent right triangles at either end of the figure. Those will have congruent corresponding angles.

5 0
2 years ago
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