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Anni [7]
3 years ago
6

Find the value of x round to the nearest tenth

Mathematics
1 answer:
Alla [95]3 years ago
3 0

Answer:

Approximately 18.0.

Step-by-step explanation:

To find <em>x</em>, we can use the Law of Cosines.

The Law of Cosines is the following:

c^2=a^2+b^2-2ab\cos(C)

Now that the <em>c</em> is the side we're trying to find. Thus, the <em>c</em> is <em>x</em>. Also, Angle C is the angle opposite to <em>c</em> (or <em>x</em>). <em>a </em>and <em>b</em> are the two sides next to Angle C.

Plug in <em>x </em>for <em>c</em>, 16 for <em>a</em>, 30 for <em>b</em>, and 30 for Angle C:

x^2=(16)^2+(30)^2-2(16)(30)\cos(30)\\x^2=1156-960\cos(30)\\x=\sqrt{1156-960\cos(30)} \\x\approx18.0171

(Use a calculator. Make sure you're in degree mode!)

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Determine la razón de la siguiente progresión geométrica: 81,27,9,3,1,....
worty [1.4K]

Answer:

BẠN BỊ ĐIÊN À

Step-by-step explanation:

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5 0
3 years ago
Reena bought a mobile phone for 16,240 that included a GST of 12%. Find the price of the mobile phone before the GST was added.​
NNADVOKAT [17]

Answer: 14500

Step-by-step explanation:

Let the price of the mobile phone before the GST was added be represented by x.

Therefore,

x + (12% × x) = 16240

x + 0.12x = 16240

1.12x = 16240

x = 16240/1.12

x = 14500

The price of the mobile phone before the GST was added was 14500

4 0
3 years ago
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
Solve for v.<br> 19v – 18 = 16 + 17v
GarryVolchara [31]

Answer:

v= 17

Step-by-step explanation:

Start with the given equation.

19v – 18 = 16 + 17v

Move -18 to the other side, and swap it's sign.

19v = 16 + 18 + 17v

Move 17v to the other side and swap it's sign.

19v - 17v = 16 + 18

Combine like terms.

2v = 34

Divide by 2v.

v = 17

8 0
3 years ago
Read 2 more answers
Find the equation of the line that has slope -5 and passes through the point (1,-7)
RUDIKE [14]

Answer: y=-5x-2

Step-by-step explanation:

5 0
3 years ago
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