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8x + 5y = 9.....(1)3x + 2y = 4......(2)

Mutiply eq(1) by 2 and eq(2) by 5 and then subtract

We get,16x + 10y = 1815x + 10y = 20

x = - 2

Put value of x = - 2 in eq(1)8*-2 + 5y = 9-16 + 5y = 95y = 25y = 25/5 = 5

Therefore x = - 2, y = 5

As 450 is the nearest number divisible by 15Hence =450/15=30 rows Trees left over is 458-450=8 trees.

Number of rows = 53

Number of equal saplings on each row:

Number on each row = 15,019 ÷53 = 283 remaining 20

Find the number of saplings he need to buy to make another row:

Number of saplings = 53 - 20 = 33

Answer: He should bring in 33 more saplings to have 284 equal rows.

A deterministic mathematical model is meant to yield a single solution describing the outcome of some "experiment" given appropriate inputs. A probabilistic model is, instead, meant to give a distribution of possible outcomes (i.e. it describes all outcomes and gives some measure of how likely each is to occur).It should be noted that a probabilistic model can be quite useful even for a person who believes the entire universe to be deterministic. This utility arises because even a deterministic process may have so many variables that any model that attempts to account for them all is too cumbersome to work with. For example, a coin toss might be deterministic if one could precisely measure everything about the flip, the coin, the floor, the air currents, the tides, the precise location on earth, etc. In practice, this level of deterministic modeling is impossible, so stochastic models are used instead. On the other hand, if one takes quantum mechanics seriously, everything has some level of non-deterministic behavior.

Similarly, deterministic models can be used to great effect even in real-world process that is clearly stochastic. For example, the heat equation works great in many situations despite the fact that it ignores the "random" motion of the atoms involved. Usually, in these scenarios, the distribution of possible final answers is so sharply peaked (i.e. has such a small variance) that there is no need to complicate the model by forcing it to calculate the distribution rather than just a single value.

x+4/x-4 is the answer

x+4/x-4 is the right answer

6y+2x=3xy/6=xy/2;; 3y+2x=xy;//2y=1; y=1/2; 3/x+2/2=0; 3=X

6y+2x=3xy/6=xy/2;; 3y+2x=xy;//2y=1;y=1/2;3/x+2/2=0;3=x

3x + 5x + x - 8x = 0

8x + x - 8x = 0

x = 0

3x+5x+x-8x=0 9x -8x=0 x=0

4x^2+8x=4x(x+2) is the answer

i) x+x+ 70 = 1802x = 1802x = 110x = 55

ii) 2y + 90 = 180 2y = 180 2y = 90 y = 90/ 2= 45

AC = BDdeduct BC from bothso we getAC - BC= BD - BCAB = CD

Number of rows = 53

Number of equal saplings on each row:

Number on each row = 15,019 ÷53 = 283 remaining 20

Find the number of saplings he need to buy to make another row:

Number of saplings = 53 - 20 = 33

Answer: He should bring in 33 more saplings to have 284 equal rows.

the right answer. is 33more saplings to have 284 equalrows

Sin8x + Sin2x/Cos2x - Cos8x>(2sin(x+y/2)cos(x-y/2))/(-2sin(x+y/2)sin(x-y/2)

>2sin5xcos3x/-2sin3x(-sin5x)>2sin5xcos3x/2sin3xsin5x>Cos2x/sin3x>Cot3x

please jaldi

Please mention the figure!

I am trying your questions answering

p=(-4,4) Q=(3,3) R=(4,2) S=(-5,-3)

∆ 3/4 ∆ 3/4 ∆ 3/4 ∆so, 3/4×3=9/4

3/4, 3/4, 3/4 then 3/4×3= 9/4

9/4 is the correct answer of the given question.

We knowProduct of numbers a and b= LCM (a, b) * HCF(a, b)

(1) Let other number be x Then, 9*x = 9*27 x = 27 Other number is 27

24 and 26

thanks

24² and 26² are the correct answer

Thanks

The sample space S is the following:

S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5), (5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Now favorable event = (5,5)(5,6)(6,5)(6,6)total event= 36probability = 4/36= 1/9

1.

{As the formula of the picture, = 1/2 x 2 x cos (7x - x) - cos (7x + x )}= cos 6x - cos 4x

24/6 = 4 এই প্রশ্নের উত্তর

We have4events :HH,HT,TH,TT. We have success, ifHH or,HT, orTH appears. Since all events have the same probability, the probability of success is 3/4.

Maximum one tailH, HH, TT, HT, Tboth head and tail will come two times soprobability = 2/4 = 1/2

12 level question

1- 0.2422-0.7083-.44-28.755- 0.666676-0.69444447- 3.148- 1.222222

1) 0.242, 2) -0.708; 3)-4; 4)-28.75

This kind of questions follows the same pattern:

Let the speed of the boat in still water = xkm/hr.

Let the speed of the stream = ykm/hr.

Speed upstream = x - y.

Speed Downstream = x + y.

Now,

Given that boat can travel 30km upstream and 28km downstream in 7 hours.

30/x-y + 28/x+y = 7

Let 1/x - y = a and 1/x + y = b

30a + 28b = 7 ---------------------------- (1).

Also, Given that it can travel 21 km upstream and return in 5 hours.

21/x - y + 21/x + y = 5

Let 1/x - y = a and 1/x + y = b

21a + 21b = 5 ------------------------ (2)

On solving (1) * 21 & (2) * 28, we get

630a + 588b = 147

588a + 588b = 140-----------------------------42a = 7

a = 1/6.

Substitute a = 6 in (1), we get

30a + 28b = 7

30(1/6) + 28b = 7

5 + 28b = 7

28b = 7 - 5

28b =2

b = 2/28

b = 1/14.

We know that,

a = 1/x - y

1/6 = 1/x - y

x - y = 6 ----------- (3)

We know that,

b = 1/x + y

1/14 = 1/x + y

x + y = 14 ------------ (4).

On solving (3) & (4), we get

x + y = 14

x - y = 6

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

2x = 20

x = 10

Substitute x = 10 in (4), we get

x + y = 14

10 + y = 14

y = 14 - 10

y = 4.

Therefore the speed of the boat in still water = 10km/hr.

Therefore the speed of the stream = 4km/hr.

It is not possible to predict the outcome of coin before the result.

6.4 is the mode of the data

Thanks