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luda_lava [24]
4 years ago
5

Please anyone please help me with this question

Mathematics
1 answer:
marin [14]4 years ago
8 0
Well, It's going to be 8/12 and then reduced it will be 4/6 or 0.667
You might be interested in
The mean of 6 numbers is 30.If one number is excluded the mean is 29.Find the value of the excluded number
MAXImum [283]

Answer:

The value of the excluded number is 35.

Step-by-step explanation:

Write out six letters to represent these six numbers;

a + b + c + d + e + f

---------------------------- = 30.  Note that if "f" is excluded,

              6

a + b + c + d + e

---------------------------- = 29.  

              5.  

Thus, a + b + c + d + e = (5)(29) = 145.  Replacing a + b + c + d + e in the first equation by 145, we get:

145 + f

----------- = 30,  which means that 145 + f = 180.  Solving for f, f = 35.

     6

The value of the excluded number is 35.

3 0
3 years ago
What is 13% of 13%?<br>show ur work
aliina [53]
% = 1/100
of = *
13/100 * 13/100
169 / 10000
0.0169 = 1.69%
6 0
3 years ago
Which equation is true when k = -15?
Inga [223]

Answer:

c) k/3 + 17 = 12

#JUST A TEXT

8 0
3 years ago
Two integers have a sum of 42 and a difference of 22. The greater of the two<br> integers is
DochEvi [55]
Let
x = larger integer
y = smaller integer

The two integers (x and y) have a sum of 42 which means they add to 42

x+y = 42

solve for y to get 

y = 42-x

simply by subtracting x from both sides

---------------------------------------------------

The two integers have a difference of 22. This translates to "subtract the values and the result will be 22", i.e.,

x-y = 22

Plug in y = 42-x. Solve for x

x-y = 22
x-(y) = 22
x - (42-x) = 22
x - 42 + x = 22
2x - 42 = 22
2x - 42+42 = 22+42
2x = 64
2x/2 = 64/2
x = 32

If x = 32, then y is...
y = 42-x
y = 42-32
y = 10

Therefore, 
x = 32
y = 10

The final answer is 10

5 0
4 years ago
Read 2 more answers
On a test victor asked to find the square root of 0.83. Wich of the following is closest to the square root of 0.83?
SVETLANKA909090 [29]

Answer:

I do not know what choices you have, but 0.9110433579144299 is the square root of 0.83

Step-by-step explanation:

Step 1:

Divide the number (0.83) by 2 to get the first guess for the square root .

First guess = 0.83/2 = 0.415.

Step 2:

Divide 0.83 by the previous result. d = 0.83/0.415 = 2.

Average this value (d) with that of step 1: (2 + 0.415)/2 = 1.2075 (new guess).

Error = new guess - previous value = 0.415 - 1.2075 = 0.7925.

0.7925 > 0.001. As error > accuracy, we repeat this step again.

Step 3:

Divide 0.83 by the previous result. d = 0.83/1.2075 = 0.6873706004.

Average this value (d) with that of step 2: (0.6873706004 + 1.2075)/2 = 0.9474353002 (new guess).

Error = new guess - previous value = 1.2075 - 0.9474353002 = 0.2600646998.

0.2600646998 > 0.001. As error > accuracy, we repeat this step again.

Step 4:

Divide 0.83 by the previous result. d = 0.83/0.9474353002 = 0.8760492667.

Average this value (d) with that of step 3: (0.8760492667 + 0.9474353002)/2 = 0.9117422835 (new guess).

Error = new guess - previous value = 0.9474353002 - 0.9117422835 = 0.0356930167.

0.0356930167 > 0.001. As error > accuracy, we repeat this step again.

Step 5:

Divide 0.83 by the previous result. d = 0.83/0.9117422835 = 0.9103449681.

Average this value (d) with that of step 4: (0.9103449681 + 0.9117422835)/2 = 0.9110436258 (new guess).

Error = new guess - previous value = 0.9117422835 - 0.9110436258 = 0.0006986577.

0.0006986577 <= 0.001. As error <= accuracy, we stop the iterations and use 0.9110436258 as the square root.

So, we can say that the square root of 0.83 is 0.911 with an error smaller than 0.001 (in fact the error is 0.0006986577). this means that the first 3 decimal places are correct. Just to compare, the returned value by using the javascript function 'Math.sqrt(0.83)' is 0.9110433579144299.

5 0
4 years ago
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