There are 4 white and 5 red (indistinguishable) balls in a bag. Suppose you draw one ballout at a time without replacement and stop when you have drawn all the white (4 white) orall the red (5 red) balls. What is the probability that the last ball you drawn was a whiteball?

Answer:

5

Step-by-step explanation:

This is a right triangle, so we can use the Pythagorean theorem

a^2+b^2 = c^2

where a and b are the legs and c is the hypotenuse

QR^2 + 12^2 = 13^2

QR^2 +144 =169

QR^2 = 169-144

QR^2 =25

Take the square root of each side

QR = sqrt(25)

QR =5

Step-by-step explanation:

Any number that can be divided by 2 without having remainder is considered an even number.

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Answer:

0.4060

Step-by-step explanation:

To calculate the sample proportion, phat, we take the ratio of the number of preferred outcome to the total number of trials ;

Phat = number of times coin lands on head (preferred outcome), x / total number of trials (total coin flips), n

x = 406

n = 1000

Phat = x / n = 406/ 1000 = 0.4060

The estimate of the chance that this coin will land on heads is 0.406

If a coin is flipped 1000 times, the total outcomes will 1000

If it landed on the head 406 times, the expected outcome will be 406.

Pr(the coin lands on the head) = 406/1000

Pr(the coin lands on the head) = 0.406

Hence the estimate of the chance that this coin will land on heads is 0.406

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Answer:

[tex]\displaystyle \frac{dy}{dx} = \frac{1}{(x - 2)^2\sqrt{\frac{x}{2 - x}}}[/tex]

General Formulas and Concepts:

Pre-Algebra

Algebra I

Algebra II

Calculus

Differentiation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Derivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Basic Power Rule:

Derivative Rule [Quotient Rule]:                                                                           [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]

Derivative Rule [Chain Rule]:                                                                                 [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Step-by-step explanation:

*Note:

You can simply just use the Quotient and Chain Rule to find the derivative instead of using ln.

Step 1: Define

Identify

[tex]\displaystyle y = \sqrt{\frac{x}{2 - x}}[/tex]

Step 2: Rewrite

Step 3: Differentiate

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Book: College Calculus 10e

Answer:

if I am not mistakedn the answe id e

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

a. The linear equation for the first point (-4,1) is 16A-4B+C=1

b. The linear equation for the second point (-2, 0.94) is 4A-2B+C=0.94

c. The linear equation for the third point (0,1) is 0A+0B+C=1

d. The matrix equation looks like this:

[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]

e. The inverse of the coefficient matrix looks like this:

[tex]A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right][/tex]

f. The equation of the parabola is: [tex]\frac{3}{200}x^{2}+\frac{3}{50}x+1=y[/tex]

a. In order to build a linear equation from the given points, we need to substitute them into the general form of the equation.

Let's take the first point (-4,1). When substituting it into the general form of the quadratic equation we end up with:

[tex](-4)^{2}A+(-4)B+C=1[/tex]

which yields:

[tex]16A-4B+C=1[/tex]

b. Let's take the second point (-2,0.94). When substituting it into the general form of the quadratic equation we end up with:

[tex](-2)^{2}A+(-2)B+C=0.94[/tex]

which yields:

[tex]4A-2B+C=0.94[/tex]

c. Let's take the third point (0,1). When substituting it into the general form of the quadratic equation we end up with:

[tex](0)^{2}A+(0)B+C=1[/tex]

which yields:

[tex]0A+0B+C=1[/tex]

d. A matrix equation consists on three matrices. The first matrix contains the coefficients (this is the numbers on the left side of the linear equations). Make sure to write them in the right order, this is, the numbers next to the A's should go on the first column, the numbers next to the B's should go on the second column and the numbers next to the C's should go on the third column.

The equations are the following:

16A-4B+C=1

4A-2B+C=0.94

0A+0B+C=1

So the coefficient matrix looks like this:

[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right][/tex]

Next we have the matrix that has the variables, in this case our variables are the letters A, B and C. So the matrix looks like this:

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right][/tex]

and finally the matrix with the answers to the equations, in this case 1, 0.94 and 1:

[tex]\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]

so if we put it all together we end up with the following matrix equation:

[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]

e. When inputing the coefficient matrix in our graphing calculator we end up with the following inverse matrix:

[tex]A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right][/tex]

Inputing matrices and calculating their inverses depends on the model of a calculator you are using. You can refer to the user's manual on how to do that.

f. Our matrix equation has the following general form:

AX=B

where:

A=Coefficient matrix

X=Variables matrix

B= Answers matrix

In order to solve this type of equations, we can make use of the inverse of the coefficient matrix to end up with an equation that looks like this:

[tex]X=A^{-1}B[/tex]

Be careful with the order in which you are doing the multiplication, if A and B change places, then the multiplication will not work and you will not get the answer you need. So when solving this equation we get:

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right]*\left[\begin{array}{c}1\\\frac{47}{50}\\1\end{array}\right][/tex]

(Notice that I changed 0.94 for the fraction 47/50 you can get this number by dividing 94/100 and simplifying the fraction)

So, in order to do the multiplication, we need to multiply each row of the coefficient matrix by the answer matrix and add the results. Like this:

[tex]\frac{1}{8}*1+(-\frac{1}{4})(\frac{47}{50})+\frac{1}{8}*1[/tex]

[tex]\frac{1}{8}-\frac{47}{200}+\frac{1}{8}=\frac{3}{200}[/tex]

So the first number for the answer matrix is [tex]\frac{3}{200}[/tex]

[tex]\frac{1}{4}*1+(-1)(\frac{47}{50})+\frac{3}{4}*1[/tex]

[tex]\frac{1}{4}-\frac{47}{50}+\frac{3}{4}=\frac{3}{50}[/tex]

So the second number for the answer matrix is [tex]\frac{3}{50}[/tex]

[tex]0*1+0(\frac{47}{50})+1*1[/tex]

[tex]0+0+1=1[/tex]

So the third number for the answer matrix is 1

In the end, the matrix equation has the following answer.

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}\frac{3}{200}\\\frac{3}{50}\\1\end{array}\right][/tex]

which means that:

[tex]A=\frac{3}{200}[/tex]

[tex]B=\frac{3}{50}[/tex]

and C=1

so, when substituting these answers in the general form of the equation of the parabola we get:

[tex]Ax^{2}+Bx+C=y[/tex]

[tex]\frac{3}{200}x^{2}+\frac{3}{50}x+1=y[/tex]

For further information, you can go to the following link:

https://brainly.com/question/12628757?referrer=searchResults

[tex]7.2\:\text{ft/s}[/tex]

Step-by-step explanation:

We can apply the Pythagorean theorem here:

[tex]26^2 = x^2 + y^2\:\:\:\:\:\:\:\:\:(1)[/tex]

where x is the distance of the ladder base from the wall and y is the distance of the ladder top from the ground. Taking the time derivative of the expression above, we get

[tex]0 = 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt}[/tex]

Solving for [tex]\frac{dx}{dt},[/tex] we get

[tex]\dfrac{dx}{dt} = -\dfrac{y}{x}\dfrac{dy}{dt}[/tex]

We can replace y by rearranging Eqn(1) such that

[tex]y = \sqrt{26^2 - x^2}[/tex]

Therefore,

[tex]\dfrac{dx}{dt} = - \dfrac{\sqrt{26^2 - x^2}}{x}\dfrac{dy}{dt}[/tex]

Since y is decreasing as the ladder is being lowered, we will assign a negative sign to [tex]\frac{dy}{dt}[/tex]. Hence,

[tex]\dfrac{dx}{dt} = - \dfrac{\sqrt{26^2 - (10)^2}}{10}(-3\:\text{ft/min})[/tex]

[tex]\:\:\:\:\:\:\:= 7.2\:\text{ft/min}[/tex]

Step-by-step explanation:

Z will stay the same since the 2 will divide into 1

Answer:

a. 161

b. 181

c. 100

Step-by-step explanation:

a. 162 comes just after 161 (160, 161, 162, 163...)

b. 181 comes just before 182 (180, 181, 182, 183...)

c. 100 is between 99 and 101 (98, 99, 100, 101, 102...)

Answer:

last one

Step-by-step explanation:

they both whole

Answer:

321

Step-by-step explanation:

Answer: y=1x+1

Step-by-step explanation:

y=1x+3

that should be it

Answer:

Step-by-step explanation:

Because then the value on the other side will be unbounded by the infinity sign while expressing the answers on a number line.

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Answer:

The answer is "0.6368 and 0.773".

Step-by-step explanation:

The manufacturer of organic compounds guarantees that its clients have at least 100 lbs. of solvent in every fluid drum they deliver. [tex]X\ is\ N(101.8, 3.76)\\\\P(X>100) =P(Z> \frac{100-101.8}{3.76}=P(Z>-0.47))[/tex]

For point a:

Therefore the Probability =0.6368  

For point b:

[tex]P(Z\geq \frac{100-101.8}{\sigma})=0.99\\\\P(Z\geq \frac{-1.8}{\sigma})=0.99\\\\1-P(Z< \frac{-1.8}{\sigma})=0.99\\\\P(Z< \frac{-1.8}{\sigma})=0.01\\\\z-value =0.01\\\\area=-2.33\\\\ \frac{-1.8}{\sigma}=-2.33\\\\ \sigma= \frac{-1.8}{-2.33}=0.773[/tex]

9514 1404 393

Answer:

  105.0°, 255.0°

Step-by-step explanation:

Many calculators do not have a secant function, so the cosine relation must be used.

  sec(θ) = -3.8637

  1/cos(θ) = -3.8637

  cos(θ) = -1/3.8637

  θ = arccos(-1/3.8637) ≈ 105.000013°

The secant and cosine functions are symmetrical about the line θ = 180°, so the other solution in the desired range is ...

  θ = 360° -105.0° = 255.0°

The angles of interest are θ = 105.0° and θ = 255.0°.

Answer:

Step-by-step explanation:

When you are asked to find the union of sets you find numbers that are present in both sets.

So a number that appears in both the sets of M and N are 2 and 8.

So M U N = { 2,8} where U is the symbol for union.

Answer:

a) forming a bell

b) 5

c) 4.7

d) mean

is the correct answer

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Answer:

a.. rs 7000

Because 15×1000=15000 it is SP when selling 1000units in the rate of Rs 15/unit& 8×1000=8000 this is cp when buying 1000 units in the rate of Rs 8/unit.

So,by formula of profit,

Rs (15000-8000)=Rs7000

Answer:

12; 8

Step-by-step explanation:

length-x

breadth-2/3x

2(x+2/3 x)=40

2×5/3x=40

10/3x=40

x=40÷10/3

x=40×3/10

x=12 (cm) length

2/3×12=8 (cm) breadth

(A)

P(X < 61.25) = P((X - 55.4)/4.1 < (61.25 - 55.4)/4.1)

… ≈ P(Z ≤ 0.1427)

… ≈ 0.5567

(B)

P(X > 46.5) = P((X - 55.4)/4.1 > (46.5 - 55.4)/4.1)

… ≈ P(Z > -2.1707)

… ≈ 1 - P(Z ≤ -2.1707)

… ≈ 0.9850

Answer:

[tex]14[/tex]

Step-by-step explanation:

Given

[tex]\displaystyle \sum_{k=0}^3k^2[/tex]

Let's break down each part. The input at the bottom, in this case [tex]k=0[/tex], is assigning an index [tex]k[/tex] at a value of [tex]0[/tex]. This is the value we should start with when substituting into our equation.

The number at the top, in this case 3, indicates the index we should stop at, inclusive (meaning we finish substituting that index and then stop). The equation on the right, in this case [tex]k^2[/tex], is the equation we will substitute each value in. After we substitute our starting index, we'll continue substituting indexes until we reach the last index, then add up each of the outputs produced.

Since [tex]k=0[/tex] is our starting index, start by substituting this into [tex]k^2[/tex]:

[tex]0^2=0[/tex]

Now continue with [tex]k=1[/tex]:

[tex]1^1=1[/tex]

Repeat until we get to the ending index, [tex]k=3[/tex]. Remember to still use [tex]k=3[/tex] before stopping!

Substituting [tex]k=2[/tex]:

[tex]2^2=4[/tex]

Substituting [tex]k=3[/tex]:

[tex]3^2=9[/tex]

Since 3 is the index we end at, we stop here. Now we will add up each of the outputs:

[tex]0+1+4+9=\boxed{14}[/tex]

Therefore, our answer is:

[tex]\displaystyle \sum_{k=0}^3k^2=0+1+4+9=\boxed{14}[/tex]

Answer:

14

Step-by-step explanation:

∑3/k=0 k^2

Let k=0

0^2 =0

Let k = 1

1^2 =1

Let k =2

2^2 = 4

Let k = 3

3^2 = 9

0+1+4+9 = 14

Answer:

0.966

Step-by-step explanation:

When typed into a calculator, sin75 = -0.3877816354

Upon converting to degrees, the full answer is 0.96592582628

Answer: -n+(-n-1)

Step-by-step explanation:

Expression will be -n + (-1)

Series

-4 +(-5)+(-6)+(-7)+(-8)+(-9)+(-10)+(-11)+(-12)+(-13) and so on

Here number -n has + (-n-1) being added to it

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Answer:

x²+12x+35

Step-by-step explanation:

in factored form it would just be

(x+7)(x+5)=0

expand this

x²+12x+35=0

Complete the square in the denominator.

[tex]s^2 - 4s + 8 = (s^2 - 4s + 4) + 4 = (s-2)^2 + 4[/tex]

Rewrite the given transform as

[tex]\dfrac{3s-14}{s^2-4s+8} = \dfrac{3(s-2) - 8}{(s-2)^2+4} = 3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}[/tex]

Now take the inverse transform:

[tex]L^{-1}_t\left\{3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3L^{-1}_t\left\{\dfrac{s-2}{(s-2)^2+2^2}\right\} - 4L^{-1}_t\left\{\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3e^{2t} L^{-1}_t\left\{\dfrac s{s^2+2^2}\right\} - 4e^{2t} L^{-1}_t\left\{\dfrac{2}{s^2+2^2}\right\} \\\\ \boxed{3e^{2t} \cos(2t) - 4e^{2t} \sin(2t)}[/tex]

Answer:

show us a screenshot or image

or type it out, copy paste

Step-by-step explanation:

Answer:

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Answer:

[tex]\displaystyle p(x) + q(x) = \frac{2x}{(x+1)(x-1)}[/tex]

The result is indeed another rational expression.

Step-by-step explanation:

We are given the two functions:

[tex]\displaystyle p(x) = \frac{1}{x+1}\text{ and } q(x) = \frac{1}{x-1}[/tex]

And we want to perform the operation:

[tex]\displaystyle p(x) + q(x)[/tex]

And show that the result is another rational expression.

Add:

[tex]\displaystyle = \frac{1}{x+1} + \frac{1}{x-1}[/tex]

To combine the fractions, we will need a common denominator. So, we can multiply the first fraction by (x - 1) and the second by (x + 1):

[tex]\displaystyle = \frac{1}{x+1}\left(\frac{x-1}{x-1}\right) + \frac{1}{x-1}\left(\frac{x+1}{x+1}\right)[/tex]

Simplify:

[tex]=\displaystyle \frac{x-1}{(x+1)(x-1)} + \frac{x+1}{(x+1)(x-1)}[/tex]

Add:

[tex]\displaystyle = \frac{(x-1)+(x+1)}{(x+1)(x-1)}[/tex]

Simplify. Hence:

[tex]\displaystyle p(x) + q(x) = \frac{2x}{(x+1)(x-1)}[/tex]

The result is indeed another rational expression.