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3 years ago

Find the geometric proof

Mathematics

2 answers:

GrogVix [38]3 years ago

6 0

Answer:

See Explanation

Step-by-step explanation:

Line l is bisector of MO... (given)

$\therefore MN = NO.... (1)$

$\because MN + NO = MO..(M-N-O)$

$\therefore NO + NO = MO..[From \:(1)]$

$\therefore 2NO = MO$

$\therefore NO = \frac{1}{2} MO$

$\therefore NO = \frac{1}{2}\times 24..(\because MO=24)$

$\therefore NO =12$

Thus Proved

Crazy boy [7]3 years ago

5 0

Answer:

67 the dhjguuugf

Step-by-step explanation:

Ducati I am so

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Item 12

katovenus [111]

Y=25(1.025)^x

Rate: r=1.025-1
r=0.025>0 (positive), then the function is increasing
r=0.025*100%
r=2.5%

Answer: Second option: Increasing by 2.5%

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3 years ago

40 is what percent of 50?

maks197457 [2]

40/50 is equal to 80 percent, a calculator can provide this result

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3 years ago

Read 2 more answers

What is the approximate value for the modal daily sales?

Aleksandr [31]

Answer:

Step-by-step explanation:

Hello!

The table shows the daily sales (in $1000) of shopping mall for some randomly selected days 

Sales 1.1-1.5 1.6-2.0 2.1-2.5 2.6-3.0 3.1-3.5 3.6-4.0 4.1-4.5 

Days 18 27 31 40 56 55 23 

Use it to answer questions 13 and 14. 

13. What is the approximate value for the modal daily sales? 

To determine the Mode of a data set arranged in a frequency table you have to identify the modal interval first, this is, the class interval in which the Mode is included. Remember, the Mode is the value with most observed frequency, so logically, the modal interval will be the one that has more absolute frequency. (in this example it will be the sales values that were observed for most days)

The modal interval is [3.1-3.5]

Now using the following formula you can calculate the Mode:

$Md= Li + c[\frac{(f_{max}-f_{prev})}{(f_{max}-f_{prev})(f_{max}-f_{post})} ]$

Li= Lower limit of the modal interval.

c= amplitude of modal interval.

fmax: absolute frequency of modal interval.

fprev: absolute frequency of the previous interval to the modal interval.

fpost: absolute frequency of the posterior interval to the modal interval.

$Md= 3,100 + 400[\frac{(56-40)}{(56-40)+(56-55)} ]= 3,476.47$

A. $3,129.41 B. $2,629.41 C. $3,079.41 D. $3,123.53 

Of all options the closest one to the estimated mode is A.

14. The approximate median daily sales is … 

To calculate the median you have to identify its position first:

For even samples: PosMe= n/2= 250/2= 125

Now, by looking at the cumulative absolute frequencies of the intervals you identify which one contains the observation 125.

F(1)= 18

F(2)= 18+27= 45

F(3)= 45 + 31= 76

F(4)= 76 + 40= 116

F(5)= 116 + 56= 172 ⇒ The 125th observation is in the fifth interval [3.1-3.5]

$Me= Li + c[\frac{PosMe-F_{i-1}}{f_i} ]$

Li: Lower limit of the median interval.

c: Amplitude of the interval

PosMe: position of the median

F(i-1)= accumulated absolute frequency until the previous interval

fi= simple absolute frequency of the median interval.

$Me= 3,100+400[\frac{125-116}{56} ]= 3164.29$

A. $3,130.36 B. $2,680.36 C. $3,180.36 D. $2,664

Of all options the closest one to the estimated mode is C.

5 0

3 years ago

Original price of a pen $1.99 tax:6%

egoroff_w [7]

Answer:

The final price would be 2.11$

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3 years ago

3-*= x3 + x to the nearest tenth.

Vesna [10]

Answer:

− 1.52

Step-by-step explanation:

3 - 8 = x^3 + x

3 0

3 years ago