<img src="https://tex.z-dn.net/?f=5%5Cfrac%7B42%7D%7B170%7D%20%20-%203%5Cfrac%7B80%7D%7B170%7D%20" id="TexFormula1" title="5\fra

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yaroslaw [1]

3 years ago

c{42}{170} - 3\frac{80}{170} " alt="5\frac{42}{170} - 3\frac{80}{170} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Reika [66]3 years ago

7 0

$5\dfrac{42}{170}-3\dfrac{80}{170}=(5-3)+\left(\dfrac{42}{170}-\dfrac{80}{170}\right)\\\\=2+\left(-\dfrac{38}{170}\right)=2-\dfrac{32}{170}=1\dfrac{170}{170}-\dfrac{38}{170}\\\\=1\dfrac{170-38}{170}=1\dfrac{132}{170}=1\dfrac{132:2}{170:2}=1\dfrac{66}{85}$

$5\dfrac{42}{170}-3\dfrac{80}{170}=4\dfrac{170+42}{170}-3\dfrac{80}{170}\\\\=(4-3)+\left(\dfrac{212}{170}-\dfrac{80}{170}\right)=1+\dfrac{212-80}{170}\\\\=1+\dfrac{132}{170}=1\dfrac{132:2}{170:2}=1\dfrac{66}{85}$

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diamong [38]

1. Convert the mixed numbers into improper numbers (-3 2/5 = -17/5 and 8 1/5 = 41/5). Then add 17/5 to both sides when means b=58/5

2. Multiply both sides by the reciprocal of -7/3 (which is -3/7). So n=18

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5 yrs ago, Nuri was thrice as old as Sonu. 10 yrs later, Nuri will be twice as old Sonu. How old are Nuri n Sonu?

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Answer:

Answer will be 50

Step-by-step explanation:

Let us suppose, present age of Nuri be ‘x’ years and present age of Sonu be ‘y’ years.

Now, it is given that five years ago, Nuri was thrice old as Sonu. Hence,

Five years ago,

Nuri’s age = x-5 years

Sonu’s age = y-5 years

And relation between ages can be given as

Nuri’s age = 3×sonu’s age or

x-5 = 3(y-5)

x-5 = 3y-15

x-3y+10 = 0 ………..(i)

Another relation is given in the problem that ten years later, Nuri is twice as old as Sonu.

So, ten years ago,

Nuri’s Age = x+10

Sonu’s Age = y+10

And relation between ages can be written as

x+10 = 2(y+10)

x+10 = 2y+20

x-2y-10 = 0 …………..(ii)

Now we can solve the equation (i) and (ii) to get values of x and ‘y’ or present ages of Nuri and Sonu.

Value of ‘x’ from equation (i) be

x = 3y-10 ……….(iii)

Putting value of ‘x’ from equation (iii) in equation (ii) we get,

3y-10-2y-10 = 0

y = 20

Now, from equation (iii) value of ’x’ can be given as,

x= 3(20)-10

x = 50

Hence, the present ages of Nuri and Sonu are 50 years and 20 years respectively.

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2 years ago

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Answer:

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= 13.8564064606

Step-by-step explanation:

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3 years ago

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A probability experiment is conducted in which the sample space of the experiment is upper s equals startset 4 comma 5 comma 6 c

olchik [2.2K]

The correct question statement is:

A probability experiment is conducted in which the sample space of the experiment is S = {4,5,6,7,8,9,10,11,12,13,14,15}. Let event E={7,8,9,10,11,12,13,14,15}. Assume each outcome is equally likely. List the outcomes in $E^{c}$ . Find $P(E^{c})$ .

Solution:

Part 1:

$E^{c}$ means compliment of the set E. A compliment of a set can be obtained by finding the difference of the set from the universal set. The universal set is the set which contains all the possible outcomes of the events which is S in this case.

So, compliment of E will be equal to S - E. S - E will result in all those elements of S which are not present in E. So, we can write:

$E^{c}=S-E \\ \\ 
E^{c}=(4,5,6,7,8,9,10,11,12,13,14,15)-(7,8,9,10,11,12,13,14,15) \\ \\ 
E^{c}=(4,5,6)$

Thus the set compliment of E will contain the elements {4,5,6}.So

$E^{c}$ = {4,5,6}

Part 2)

$P(E^{c})$ means probability that if we select any number from the Sample Space S, it will belong the set E compliment.

$P(E^{c})$ = (Number of Elements in $E^{c}$ )/Number of elements in S

Number of elements in set S = n(S) = 12
Number of elements in set $E^{c}$ = $n(E^{c})$ =3

So,

$P(E^{c})= \frac{n(E^{c}) }{n(S)} \\ \\ 
P(E^{c})= \frac{3}{12} \\ \\ 
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3 years ago