Given that slope = -1 and a point (3,-1), write an equation in slope-
intercept form of the line,
у
X +

Answer:

5f

Step-by-step explanation:

Answer:

Step-by-step explanation:

Find the attachment in the diagram below,

In order to get the right match, we will use the conversion

180° = πrad

- For angle 315°;

If 180° = πrad

315° = a

Cross multiplying

180a = 315π rad

a = 315π rad/180

a = 63π/36 rad

a = 7π/4 rad

Hence 315° = 7π/4 rad

- For angle 80°;

If 180° = πrad

80° = b

Cross multiplying

180b = 80π rad

b = 80π rad/180

b = 8π/18 rad

b = 4π/9 rad

Hence, 80° = 4π/9 rad

- For angle 225°;

If 180° = πrad

225° = c  

Cross multiplying

180c = 225π rad

c = 225π rad/180

c = 15π/12 rad

c = 5π/4 rad

Hence, 225° = 5π/4 rad

- For angle 324°;

If 180° = πrad

324° = d

Cross multiplying

180d = 8324π rad

d = 324π rad/180

d = 27π/15 rad

d = 9π/5 rad

Hence, 324° = 9π/5 rad

Answer:

1 hour and 45 minutes or 105 minutes

Step-by-step explanation:

if 2 miles is equal to 30 minutes

4 miles would be 60 minutes/1 hour

6 miles would be 90 minutes/1 hour 30 minutes

7 miles would be 105 minutes/1 hour 45 minutes

Given a cyclic quadrilateral

As shown:

The measure of the arc FG = 97

The measure of the arc GH = 117

The measure of the arc EHG = 164

The measure of the arc is two times the measure of the inscribed angle opposite to the arc.

So, the measure of the angle E = 1/2 the measure of the arc FGH =

The measure of the angle F = 1/2 the measure of the arc EHG =

For the cyclic quadrilateral, every two opposite angles are supplementary.

So,

And:

So, the answer will be:

The value of P(10) is 0.11 to ensure that the given distribution is a valid probability distribution.

To determine the value of P(10) so that the given distribution is valid, we need to ensure that the sum of all the probabilities equals 1.

Given:

X: 5, 10, 20, 30, 40, 50

P(X): 0.13, P(10), 0.11, 0.14, 0.21, 0.3

We know that the sum of the probabilities should be equal to 1.

Sum of probabilities = 0.13 + P(10) + 0.11 + 0.14 + 0.21 + 0.3

To find the value of P(10), we can set up the equation:

0.13 + P(10) + 0.11 + 0.14 + 0.21 + 0.3 = 1

Simplifying the equation:

P(10) + 0.89 = 1

Subtracting 0.89 from both sides:

P(10) = 1 - 0.89

P(10) = 0.11

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we can conclude that f'(0) does not exist because the limit of the difference quotient as x approaches 0 does not exist.

To determine whether the derivative of f(x), denoted as f'(0), exists at x = 0, we need to check if the limit of the difference quotient exists as x approaches 0.

The difference quotient formula for the derivative is:

f'(a) = lim (h->0) [(f(a + h) - f(a))/h]

In this case, we need to evaluate f'(0), so our formula becomes:

f'(0) = lim (h->0) [(f(0 + h) - f(0))/h]

Substituting the definition of f(x) into the formula, we have:

f'(0) = lim (h->0) [(0 + h * sin(1/(0 + h)) - 0)/h]

Simplifying the expression further, we get:

f'(0) = lim (h->0) [h * sin(1/h)/h]

Canceling out h in the numerator and denominator, we have:

f'(0) = lim (h->0) [sin(1/h)]

Now, as h approaches 0, the value 1/h approaches infinity. As sin(1/h) oscillates between -1 and 1 for all non-zero values of 1/h, the limit of sin(1/h) as h approaches 0 does not exist.

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Answer:

Slope =-3, y intercept =4

Step-by-step explanation:

y2-y1/x2-x1 = slope

4-(-2)/0-2=6/-2 or -3

y intercept is where line crosses y axis so it is 4.

Answer:

Slope: -3, y-intercept: 4

Explanation:

=> In order to find slope when a graph is given, the formula would be: \(\frac{rise}{run}\)(pronounced as "rise over run")

Rise = change in y-value = -3

Run = change in x-value = 1

Slope = \(\frac{rise}{run}\) = \(\frac{-3}{1}\) also could be written as -3

=> Now, to find the y-intercept, find a value on the graph where the line touches the y-axis, that's where x = 0, which would be your y-intercept;

Here the line touches the y-axis at "y = 4", this is also where x = 0;

therefore y-intercept = 4

Hope this helps!

The Fees Receivable average account is used to record amounts due from Wilson Company for services provided by Tsang Company.

The Fees Receivable account is used to track amounts due from Wilson Company to Tsang Company for services that have been provided but not yet invoiced. The account is created when Tsang Company begins a contract to provide services to Wilson Company for six months at a total cost of $21,600. This amount is paid at the end of the six-month period, with equal services being provided each month. The Fees Receivable account is used to record the amount due each month, with the total balance of the account being equal to the total cost of the services provided. It is important to keep track of amounts due from customers so that the business can accurately track its accounts receivable and ensure that it is receiving payment for the services it has provided.

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Given the function f(x)=sin(2πx), with x = 0.3, f(x) = 0.951057. The objective is to approximate the value of f(0.2) using the first two terms in the Taylor series and Vx=0.1.

We know that the Taylor series for a function f(x) can be written as:f(x)=f(a)+f′(a)(x−a)+f′′(a)2(x−a)2+…+f(n)(a)n!(x−a)n+…The first two terms of the Taylor series are given by:f(x)=f(a)+f′(a)(x−a)The first derivative of f(x) is given by:f′(x)=2πcos(2πx)On substituting x = a = 0.1, we get:f′(0.1) = 2πcos(2π * 0.1) = 5.03118603447The value of f(x) at a=0.1 is given by:f(0.1) = sin(2π * 0.1) = 0.587785252292With a=0.1, the first two terms of the Taylor series become:f(x)=0.587785252292+5.03118603447(x−0.1) = 0.587785252292+0.503118603447x−0.503118603447×0.1Using x=0.2 and substituting the values of a and f(a) in the equation above, we get:f(0.2)=0.587785252292+0.503118603447*0.2−0.503118603447×0.1=0.712261After approximating the value of f(0.2) using the first two terms in the Taylor series,

we can conclude that the value of f(0.2) = 0.712261 with a = 0.1, with an error of approximately 0.012796.

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Answer:

should be a hope it helps

Step-by-step explanation:

Answer:

c

Step-by-step explanation:

90/1=90

140/2=70

190/3=63.3333333

240/4=60

so the first hour cost more than the others, hope this helps!

Answer:

Step-by-step explanation:

The yield of stock at yesterday's closing price is 3.43%.

Annual dividend is the profit of the stock which is divided among the shareholders, both common and preferred shareholders annually.

Given,

Annual dividend = $1.20

Today's closing price = $34.50

Net change = -$0.50

Let x be yesterday's closing price.

Net change = Today's closing price - yesterday's closing price

⇒ -0.50 = 34.50 - x

⇒ x = 34.50 + 0.50

⇒ x = 35

Yield = \(\frac{Annual dividend}{Price}\) × 100

         = \(\frac{1.2}{35}\) × 100

         = 3.4286 % ≈ 3.43 %

Hence the stock's yield at yesterday's closing price of preferred stock is 3.43 %.

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I guess you're asking about the probability density for the random variable \(|A-B|\) where \(A,B\) are independent and identically distributed uniformly on the interval (0, 15). The PDF of e.g. \(A\) is

\(\mathrm{Pr}(A=a) = \begin{cases}\dfrac1{15} & \text{if } 0 < a < 15 \\\\ 0 & \text{otherwise}\end{cases}\)

It's easy to see that the support of \(|A-B|\) is the same interval, (0, 15), since \(|x|\ge0\), and

• at most, if \(A=15\) and \(B=0\), or vice versa, then \(|A-B|=15\)

• at least, if \(A=B\), then \(|A-B|=0\)

Compute the CDF of \(C=|A-B|\) :

\(\mathrm{Pr}(C\le c) = \mathrm{Pr}(|A - B| \le c) = \mathrm{Pr}(-c \le A - B \le c)\)

This probability corresponds to the integral of the joint density of \(A,B\) over a subset of a square with side length 15 (see attached). Since \(A,B\) are independent, their joint density is

\(\mathrm{Pr}(A=a,B=b) = \begin{cases}\dfrac1{15^2} & \text{if } (a,b) \in (0,15) \times (0,15) \\ 0 &\text{otherwise}\end{cases}\)

The easiest way to compute this probability is by using the complementary region. The triangular corners are much easier to parameterize.

\(\displaystyle \mathrm{Pr}(|A-B|\le c) = 1 - \mathrm{Pr}(|A-B| > c) \\\\ ~~~~~~~~ = 1 - \int_0^{15-c} \int_{a+c}^{15} \frac{db\,da}{15^2} - \int_c^{15} \int_0^{a-c} \frac{db\,da}{15^2} \\\\ ~~~~~~~~ = 1 - \frac1{225} \left(\int_0^{15-c} (15 - a - c) \, da + \int_c^{15} (a - c) \, da\right)\)

In the second integral, substitute \(a=15-a'\) and \(da=-da'\), so that

\(\displaystyle \int_c^{15} (a-c) \, da = \int_{15-c}^0 (15-a'-c) (-da') = \int_0^{15-c} (15 - a' - c) \, da'\)

which is the same as the first integral. This tells us the joint density is symmetric over the two triangular regions.

Then the CDF is

\(\displaystyle \mathrm{Pr}(|A-B|\le c) = 1 - \frac2{225} \int_0^{15-c} (15 - a - c) \, da \\\\ ~~~~~~~~ = 1 - \frac2{225} \left((15-c) a - \frac12 a^2\right) \bigg|_{a=0}^{a=15-c} \\\\ ~~~~~~~~ = \begin{cases}0 & \text{if } c < 0 \\\\ 1 - \dfrac{(15-c)^2}{225} = \dfrac{2c}{15} - \dfrac{c^2}{225} & \text{if } 0 \le c < 15 \\\\ 1 & \text{if } c \ge 15\end{cases}\)

We recover the PDF by differentiating with respect to \(c\).

\(\mathrm{Pr}(|A-B| = c) = \begin{cases}\dfrac2{15} - \dfrac{2c}{225} & \text{if } 0 < c < 15 \\\\ 0 & \text{otherwise}\end{cases}\)

Expected gain/loss with survey = (p * (probability of success given favorable survey * gain from success + probability of failure given favorable survey * loss from failure)) + ((1-p) * (probability of success given unfavorable survey * gain from success + probability of failure given unfavorable survey * loss from failure))
= (0.4 * (0.6 * $540,000 + 0.1 * -$450,000)) + (0.6 * (0.2 * $540,000 + 0.7 * -$450,000))

he company is considering whether to market a new product. They assess that the probabilities of the outcomes - success or failure - are p and 1-p, respectively. If the product is marketed and it fails, the company will have a net loss of $450,000. If the product is marketed and it succeeds, the company will have a net gain of $540,000. If the company decides not to market the product, there is no gain or loss.
The company can conduct a survey of prospective buyers to determine the market potential. The survey results can be classified as favorable, neutral, or unfavorable. Based on similar surveys for previous products, the company assesses the probabilities of favorable, neutral, and unfavorable survey results to be 0.6, 0.3, and 0.1 for a product that will evntually be a success, and 0.1, 0.2, and 0.7 for a product that will eventually be a failure.
The total cost of conducting the survey is c dollars. The question asks for which values of c, if any, would the company choose to conduct the survey, given that p=0.4.
To determine whether to conduct the survey, we need to compare the expected gains/losses from marketing the product with and without the survey.
If the company markets the product without the survey, the expected gain/loss can be calculated as follows:
Expected gain/loss without survey = (p * gain from success) + ((1-p) * loss from failure)
= (0.4 * $540,000) + (0.6 * -$450,000)
If the company markets the product with the survey, the expected gain/loss can be calculated as follows:
Expected gain/loss with survey = (p * (probability of success given favorable survey * gain from success + probability of failure given favorable survey * loss from failure)) + ((1-p) * (probability of success given unfavorable survey * gain from success + probability of failure given unfavorable survey * loss from failure))
= (0.4 * (0.6 * $540,000 + 0.1 * -$450,000)) + (0.6 * (0.2 * $540,000 + 0.7 * -$450,000))

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Area depends on the product of sides,

so if the sides are shortened by a factor of 2, area will reduce by a factor of 4. (2×2)

new area = 10/4=2.5 cm²

The approximate diameter of the circular plate with circumference of 68.5 centimetre is 21.82 centimetres.

The circumference of a circle is the perimeter of a circle. Therefore, the formula of a circle can be represented as follows:

circumference = 2πr

where

Therefore,

circumference = 68.5 cm

68.5 = 2 × 3.14 × r

r = 68.5 / 6.28

r = 10.9076433121

radius = 10.907

Diameter = 2(radius)

Diameter  = 2 × 10.9076433121

Diameter = 21.8152866242

Diameter = 21.82 cm

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The number of children and adults enrolled is 53 and 179 respectively

let

x + y = 232 (1)

27x + 38y = 8,233 (2)

from (1)

x = 232 - y

27x + 38y = 8,233

27(232 - y) + 38y = 8233

6264 - 27y + 38y = 8233

- 27y + 38y = 8233 - 6264

11y = 1969

y = 1969/11

y = 179

x + y = 232

x + 179 = 232

x = 232 - 179

x = 53

Therefore, the number of children and adults enrolled is 53 and 179 respectively

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Answer:

x= 40

Step-by-step explanation:

the way to figure out a cale factor is to divide a new side by an old side, in this case you could divide 24 by 15 to get 1.6

multiply the side congruent to x by 1.6, so 25 and 1.6

you should get 40

As per the information provided, a = 4√3/3 + 4, b = 4 - 4√3/3 the answer can be calculated with optimization method. it will be as follows:

Sum of a and b is 8, we get

a+b=8

b=8−a

Now, let x=a−b

Then we get,

\(x=a−(8−a) \\ x=2a−8 \\ x+8=2 \\ 1 \div 2x+4=a

\)

we use this to answer to solve for b

\(b=8−a \\ =8−(1 \div 2x+4) \\ =4−1 \div 2x

\)

Now we use the product of two numbers and its difference. This can be expressed as:

\(a⋅b⋅x=(1 \div 2x+4)(4−1 \div 2x) \\ x=2 {x}^{2} − \frac{1}{4} {x}^{3} +16x−2x^{2} \\ =−14x^{3} +16x

\)

Thus, this expression that we need to maximize. Take the derivative, set it equal to zero, and solve for x

\(−3 \div 4x ^{2} +16=0 \\ 16 =3 \div 4 x ^{2} \\ 643=x \\ 28√3 \div 3=x\)

Now for us to check that this is a maximum, we have to note that the second derivative is

\(−3 \div 2x

\\ At \\

x=8√3 \div 3

\)

the second derivative is −4√3. Since this number is negative, the point is a maximum.

Now we must find the values of a and b for this x. We have to use the relationship

\(a=1 \div 2x+4\)

\(a=1 \div 2 \times 8√3 \div 3+4 \\ =4√3 \div 3+4\)

now we use the relationship b=8−a

\(b=8−(4√3 \div 3+4) \\ =4−4√3 \div 3\)

The first step in determining a function's maximum or minimum value is differentiating it. Then, set this derivative to zero and conduct the computation.

x. This will reveal the location of a function's maximum or minimum, but it won't reveal which.

Take the second derivative to get more details. A local maximum occurs when both the first and second derivatives are negative. You have a local minimum when both the first derivative and the second derivative are zero.

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Answer:

5

Step-by-step explanation:

Dont ask why i think its 5 mph

Answer:

Step-by-step explanation:

Initial value of house is $478000

Change in value is 8% greater:

We need 8% of Value

\(\\ \sf\longmapsto 8\%\:of\;478000\)

\(\\ \sf\longmapsto 0.08(478000)\)

\(\\ \sf\longmapsto 38240\$\)

Current value:-

\(\\ \sf\longmapsto 478000+38240=516240\)

The probability of a student doing well on an exam given that they spend time reading is approximately 0.983 or 98.3%.

To calculate the probability of a student doing well on an exam given that the student spends time reading, we need to use conditional probability.

Let's denote:

P(R) as the probability of a student spending time reading (P(R) = 0.59),

P(E) as the probability of a student doing well on an exam (P(E)),

P(E|R) as the probability of a student doing well on an exam given that they spend time reading (P(E|R) = 0.58).

The formula for conditional probability is:

P(E|R) = P(E and R) / P(R).

Given that P(E and R) = 0.58 (the probability of a student doing well on an exam and spending time reading) and P(R) = 0.59 (the probability of a student spending time reading), we can substitute these values into the formula:

P(E|R) = 0.58 / 0.59 = 0.983.

Therefore, the probability of a student doing well on an exam given that the student spends time reading is approximately 0.983 or 98.3%.

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Answer:

Step-by-step explanation:

\(11-13x-7-2x=7\\\\\mathrm{Group\:like\:terms}\\-13x-2x+11-7=7\\\\\mathrm{Add\:similar\:elements:}\:-13x-2x=-15x\\-15x+11-7=7\\\\\mathrm{Add/Subtract\:the\:numbers:}\:11-7=4\\-15x+4=7\\\\\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}\\-15x+4-4=7-4\\\\Simplify\\-15x=3\\\\\mathrm{Divide\:both\:sides\:by\:}-15\\\frac{-15x}{-15}=\frac{3}{-15}\\\\Simplify\\x=-\frac{1}{5}\)

Step-by-step explanation:

Hey, there!!

Let's solve it simply.

Given,

11 - 13x - 7 - 2x = 7

Simplify the like terms.

11 - 7 - 13x - 2x = 7

4 - 15x = 7.

shifting 4 in right side.

-15x =7-4

-15x = 3

Divide 3 by -15.

\(x = \frac{3 }{ - 15} \)

Therefore, x = (-3/15) or (-1/5).

Hope it helps....

Solving the given equation,

Substituting the value of x in the original equation,

Thus, x=8 is an extraneours solution.

Answer:

x < -3

Step-by-step explanation:

Step 1: Subtract 6 from both sides.

Step 2: Divide both sides by 4.

Therefore, the answer is x < -3.

Answer:

x = ca + ba

Step-by-step explanation:

x/a -b=c

x/a = c+b

x = ca + ba

Answer:

Step-by-step explanation:

(a) To find the probability that it is raining when you arrive in Oneirabad on a random day, we need to use the law of total probability.

Let A be the event that it is raining, and B be the event that it is the Wet season.

P(A) = P(A|B)P(B) + P(A|B')P(B')

Given that the Wet season lasts for 1/3 of the year, we have P(B) = 1/3. The probability that it is raining during the Wet season is 3/4, so P(A|B) = 3/4.

The Dry season lasts for 2/3 of the year, so P(B') = 2/3. The probability that it is raining during the Dry season is 1/6, so P(A|B') = 1/6.

Now we can calculate the probability that it is raining when you arrive:

P(A) = (3/4)(1/3) + (1/6)(2/3)

= 1/4 + 1/9

= 9/36 + 4/36

= 13/36

Therefore, the probability that it is raining when you arrive in Oneirabad on a random day is 13/36.

(b) Given that it is raining when you arrive, we can use Bayes' theorem to calculate the probability that your visit is during the Wet season.

Let C be the event that your visit is during the Wet season.

P(C|A) = (P(A|C)P(C)) / P(A)

We already know that P(A) = 13/36. The probability that it is raining during the Wet season is 3/4, so P(A|C) = 3/4. The Wet season lasts for 1/3 of the year, so P(C) = 1/3.

Now we can calculate the probability that your visit is during the Wet season:

P(C|A) = (3/4)(1/3) / (13/36)

= 1/4 / (13/36)

= 9/52

Therefore, given that it is raining when you arrive, the probability that your visit is during the Wet season is 9/52.

(c) Given that it is raining when you arrive, the probability that it will be raining when you return to Oneirabad in a year's time depends on the season. If you arrived during the Wet season, the probability of rain will be different from if you arrived during the Dry season.

Let D be the event that it is raining when you return.

If you arrived during the Wet season, the probability of rain when you return is the same as the probability of rain during the Wet season, which is 3/4.

If you arrived during the Dry season, the probability of rain when you return is the same as the probability of rain during the Dry season, which is 1/6.

Since the season you arrived in is independent of the weather when you return, we need to consider the probabilities based on the season you arrived.

Let C' be the event that your visit is during the Dry season.

P(D) = P(D|C)P(C) + P(D|C')P(C')

Since P(C) = 1/3 and P(C') = 2/3, we can calculate:

P(D) = (3/4)(1/3) + (1/6)(2/3)

= 1/4 + 1/9

= 9/36 + 4/36

= 13/36

Therefore, the probability that it will be raining when you return to Oneirabad in a year's time, given that it is raining when you arrive, is 13/36.

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Part c: -3x + 5 + 5x - 1 = 0;  x = 2 is incorrect.

Part d: -(x -1) = 4x -5 - 3x ; x = 3 is correct.

For the given equations,

Part c: -3x + 5 + 5x - 1 = 0.

Simplify the equation.

-3x + 5 + 5x - 1 = 0

2x + 4 = 0

x = -4/2

x = -2

Thus, the given value x = 2 is incorrect.

Part d: -(x -1) = 4x -5 - 3x

Simplify the equation.

-(x -1) = 4x -5 - 3x

-x + 1 = x - 5

2x = 6

x= 3

Thus, the given value x = 3 is correct.

Thus, the given conditions for the equation is found.

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