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3 years ago

When you roll a die, what kind of probability are you using

1 answer:

7 0

It is a compound event with independent probability, which means that it can fall into any one of them (this happens if the die has not been tampered with). This means that one outcome does not affect the next outcome in anyway

hope this helps

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URGENT! Describe step by step how to find the following equations.

yuradex [85]

Answer:

1. x = 45°

2. x = 330°

3. x = 105°

4. x = 30°

5. x = 60°

6. x = 30° or x = -45° (315°) or x = 45°

Step-by-step explanation:

1. For

cos x = sin x

∴ sin x/(cos x) = 1 = tan x

Hence, x = tan⁻¹1 = 45°

2. For

csc(x) + 3 = 1

∴ csc(x) = 1 - 3 = -2

Which gives sin x = -1/2

∴ x = sin⁻¹(-1/2) = -30° = 360 +(-30) = 330 °

3. For

cot(3x) = -1

∵ cot(3x) = 1/(tan(3x))

Hence, 1/(tan(3x)) = -1

∴ tan(3x) = -1

3·x = tan⁻¹(-1) = -45° = 360 + (-45) = 315°

Which gives, x = -45/3 = -15° or x = 105°

4. For

2·sin²(x) + 3·sin(x) = 2

We put sin(x) = y to get

2·y² + 3·y = 2 or 2·y² + 3·y - 2 =0

Factorizing gives

(2·y -1)(y+2) =0

∴ y = 1/2 or y = -2

That is, sin(x) = 1/2 or sin(x) = -2

Hence, x = sin⁻¹(1/2) = 30° or x = sin⁻¹(-2) = (-π/2 + 1.3·i)

∴ x = 30°

5. For

4cos²(x) = 3 we have;

cos²(x) = 3/4

cos(x) = √(3/4) = (√3)/2

∴ x = cos⁻¹((√3)/2) = 60°

6. For

4·sin³(x) + 1 = 2·sin²(x) + 2·sin(x)

We put sin(x) = y to get;

4·y³ + 1 = 2·y² + 2·y which gives;

4·y³ + 1 - (2·y² + 2·y) = 0 or 4·y³ -2·y² - 2·y + 1 = 0

Factorizing gives;

$\frac{(2x-1)(2x+\sqrt{2}) )(2x-\sqrt{2})}{2} =0$

Therefore, x = 1/2 or x = -(√2)/2 or (√2)/2

Therefore, sin(x) = 1/2 or -(√2)/2 or (√2)/2

That is x = sin⁻¹(1/2) = 30 or sin⁻¹(-(√2)/2) = -45 or sin⁻¹((√2)/2) = 45.

8 0

3 years ago

How do you solve this problem In geometry?

JulsSmile [24]

The 2 smaller triangles are similar. Thus, y is to z as x is to 4. You need 3 such ratios to find the 3 unknowns. Could you try your hand at writing 1 more ratio based upon the illustration?

6 0

3 years ago

Please answer this correctly please

Phantasy [73]

If we have 2 more blue pens than black pens, our blue pens can be rewritten as blue = 2 + black. Now we can set up an equation. Originally this equation would involve both blue and black, but since we only have 1 equation to set up, we can only have 1 unknown. That's why we base the number of blue pens on the number of black pens and do a substitution. So instead of blue + black = 94, we have (black + 2) + black = 94. That simplifies to 2 black + 2 = 94, and 2 black = 92. Now if we divide by 2, we get that the number of black pens is 46. If we have 2 more blue than black, the number of blue pens we have is 48. 46 + 48 = 94, so there you go!

7 0

3 years ago

Solve w^2 + 7w - 18 = 0<br><br>with the steps pls

Bess [88]

Answer:

w=2 w = -9

Step-by-step explanation:

w^2 + 7w - 18 = 0

We can factor this equation

What 2 numbers multiply to -18 and add to 7

9*-2 = -18

9+-2 = 7

(w-2) (w+9) = 0

Using the zero product property

w-2 = 0 w+9 =0

w=2 w = -9

4 0

2 years ago

Read 2 more answers

Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

2 years ago