Write an equation of the circle with center (9, -3) and radius 8.

Answer:

about 29,748.3

Step-by-step explanation:

4000040000 - 4018040180 = 18000180

(11 gallons were used there^)

4018040180 - 4048440484 = 30400304

1616+11 = 1627 total gallons

18000180 + 30400304 = 48400484

48400484/ 1627 = 29,748.3

Answer:

a) 40, 50, 55, 57.5

b) [tex]S_n=h_0+2h_0\sum_{n=1}^{\infty}r^n[/tex]

Step-by-step explanation:

h₀ = Initial height of the ball =20

r = Rebound fraction = 0.5

a) The series of bouncing balls is given by

Sₙ=h₀+2h₀(r¹+r²+r³+r⁴.........rⁿ)

S₁=h₀+2h₀r¹=20+2×20×0.5=40

S₂=h₀+2h₀(r¹+r²)=20+2×20×(0.5+0.5²)=50

S₃=h₀+2h₀(r¹+r²+r³)=20+2×20×(0.5+0.5²+0.5³)=55

S₄=h₀+2h₀(r¹+r²+r³+r⁴)=20+2×20×(0.5+0.5²+0.5³+0.5⁴)=57.5

b) General expression for the nth term of the sequence

[tex]S_n=h_0+2h_0\sum_{n=1}^{\infty}r^n[/tex]

Final answer:

The first 4 heights after each bounce form a geometric sequence with the first term h₀ being 20 meters and subsequent terms being 10, 5, and 2.5 meters. The general expression for the nth term (hn) is given by the formula hₙ = 20 * (0.5)ⁿ

Explanation:

Given an initial height h₀ of 20 meters and a rebound fraction r of 0.5, the sequence of heights after each bounce forms a geometric sequence.

The first height is h₀ which is 20 meters. Subsequent heights can be found by multiplying the previous height by the rebound fraction r.

Third term : h2 = h= 10 * 0.5 = 5 meters

General Expression for the nth Term (b)

The nth term (hₙ) of the sequence can be found using the formula for the nth term of a geometric sequence:

hₙ = h₀ * rⁿ

For this particular sequence:

hₙ = 20 * (0.5)ⁿ

The standard form of the hyperbola is derived using trigonometric identities and substituted values. The given hyperbola has vertices (0,±2) and foci (0,±5), which yields a = 2, and c = 5. Using these, the standard form of the hyperbola would be y²/4 - x²/21 = 1, and the parametric equations are x = sqrt(21) tan(θ), y = 2 sec(θ).

To eliminate the parameter and obtain the standard form of the rectangular equation for a hyperbola, use the properties of trigonometric identities and apply the Pythagorean identity tan²(θ) + 1 = sec²(θ). Now, express tan(θ) and sec(θ) in terms of x and y, and substitute these into the Pythagorean identity to obtain the equation of the hyperbola.

In this case, tan(θ) = (x - h) / b and sec(θ) = (y - k) / a. Substitute these into the Pythagorean identity to get ((x - h) / b)² + 1 = ((y - k) / a)². Rearrange to obtain ({(x - h)²}/{b²}) - ({(y - k)²}/{a²}) = 1. This is the standard form of the hyperbola equation centered at (h, k).

For the specific hyperbola given with vertices (0,±2) and foci (0,±5), you can determine that a = 2, and c = 5. Using the relationship c² = a² + b² (for hyperbolas), you can find b = sqrt(c² - a²) = sqrt((5)² - (2)²) = sqrt(21).

So, the standard form of the equation would be y²/4 - x²/21 = 1. The parametric equations revert back to the original equation with specific values, i.e., x = sqrt(21) tan(θ) and y = 2 sec(θ).

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Answer:  Yes, the given set of vectors is a basis of R³.

Step-by-step explanation:  We are given to determine whether the following set of three vectors in R³ is a basis of R³ or not :

B = {(-1, 1,-1), (1, 0, 2), (1, 1, 0)} .

For a set to be a basis of R³, the following two conditions must be fulfilled :

(i) The set should contain three vectors, equal to the dimension of R³

and

(ii) the three vectors must be linearly independent.

The first condition is already fulfilled since we have three vectors in set B.

Now, to check the independence, we will find the determinant formed by theses three vectors as rows.

If the value of the determinant is non zero, then the vectors are linearly independent.

The value of the determinant can be found as follows :

[tex]D\\\\\\=\begin{vmatrix} -1& 1 & -1\\ 1 & 0 & 2\\ 1 & 1 & 0\end{vmatrix}\\\\\\=-1(0\times0-2\times1)+1(2\times1-1\times0)-1(1\times1-0\times1)\\\\=(-1)\times(-2)+1\times2-1\times1\\\\=2+2-1\\\\=3\neq 0.[/tex]

Therefore, the determinant is not equal to 0 and so the given set of vectors is linearly independent.

Thus, the given set is a basis of R³.

To find the conditional probabilities, you need to use the definition of conditional probability. Given that a card is black, the probability that it is a spade is 1/2. Given that a card is a spade, the probability that it is black is 2. Given that a card is black, the probability that it is a seven is 1/13. Given that a card is a face card, the probability that it is a king is 1/3.

To find these conditional probabilities, we need to use the definition of conditional probability:

P(A|B) = P(A and B) / P(B)

a) The card is a spade, given that it is black:

In a standard deck of cards, there are 26 black cards and 13 spades. So, P(S|B) = P(S and B) / P(B) = 13/26 / 26/52 = 1/2

b) The card is black, given that it is a spade:

P(B|S) = P(B and S) / P(S) = 26/52 / 13/52 = 26/13 = 2

c) The card is a seven, given that it is black:

In a standard deck of cards, there are 4 black sevens and 26 black cards. So, P(7|B) = P(7 and B) / P(B) = 4/26 / 26/52 = 1/13

d) The card is a king, given that it is a face card:

In a standard deck of cards, there are 4 kings and 12 face cards. So, P(K|F) = P(K and F) / P(F) = 4/52 / 12/52 = 1/3

Answer:

The correct option is D.

Step-by-step explanation:

The slope of a line is the change in y with respect to x.

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

If the slope of a line is undefined it means it is a vertical line and a vertical line  can not passes through three quadrants. So, option A is incorrect.

If the slope of a line is 0 it means it is a horizontal line and a horizontal line  can not passes through three quadrants. So, option B is incorrect.

If the slope of a line is positive it means the value of y increases as x increases.

Since it is an increasing line, therefore after a certain period both x and y will positive. It means the line will passes through first quadrant. So, option C is incorrect.

If the slope of a line is negative it means the value of y decreases as x increases. It can passes through each of Quadrants II, III, and IV.

Therefore the correct option is D.

Answer:

As given in the problem statement

mean score was less than 106

H0:mean score<106

so There is sufficient evidence to reject the claim μ < 106. is correct option.

Answer:

The coefficient is 15120.  

Step-by-step explanation:

Since, by the binomial expansion formula,

[tex](x+y)^n=\sum_{r=0}^n^nC_r x^{n-r} y^r[/tex]

Where, [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

Thus, we can write,

[tex](3x+2y)^7 = \sum_{r=0}^n ^7C_r (3x)^{7-r} (2y)^r[/tex]

For finding the coefficient of [tex]x^3y^4[/tex],

r = 4,

So, the term that contains [tex]x^3y^4[/tex] = [tex]^7C_4 (3x)^3 (2y)^4[/tex]

[tex]=35 (27x^3) (16y^4)[/tex]

[tex]=15120 x^3 y^4[/tex]

Hence, the coefficient of [tex]x^3y^4[/tex] is 15120.

Answer:[tex][/tex]

Coefficient of  [tex]x^3y^4[/tex] in [tex](3x+2y)^7[/tex] is 15120

Step-by-step explanation:

We know that [tex](x+y)^{n}[/tex]) can be expanded in (n+1) terms by using binomial theorem and each term is given as

[tex]n_C_{r}x^{n-r}y^{r}[/tex]

Here value of r is taken from n to 0

we have to determine the coefficient of [tex]x^3y^4[/tex] in [tex](3x+2y)^7[/tex]

in this problem we have given n=7

We have to determine the coefficient of [tex]x^3y^4[/tex]

it means in the expansion we have to find the the 3rd power of x and therefore

r=n-3

here n=7

therefore, r=7-3=4

Hence the coefficient of [tex]x^3y^4[/tex]  can be determine by using formula

[tex]n_C_{r}x^{n-r}y^{r}[/tex]

here n=7, r=4

[tex]7_C_{4}x^{7-4}y^{4}[/tex]

=[tex]\frac{7\times 6\times 5\times 4}{1\times 2\times 3\times 4} (3x)^3(2y)^4[/tex]

=[tex]15120x^3y^4[/tex]

Therefore the coefficient of  [tex]x^3y^4[/tex] in [tex](3x+2y)^7[/tex] is 15120

Answer:

In year 2013 annual maintenance done = $ 3,000,000

Out of this 85% is expected to renew in 2014 = [tex]0.85\times3000000[/tex] = $2,550,000

Now Sales in 2013 = $3,000,000

60% of these sales = [tex]0.60\times3000000[/tex] = $1,800,000

Now out of this 60%, 20% annual maintenance was expected to be paid

In 2014 = [tex]0.20\times1800000[/tex] = $360000

So, Total annual maintenance in 2014 = [tex]2550000+360000[/tex]

= $2,910,000

Answer:

Your Winnings are $ -1

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Algebraic Equations.

Since we are not told what two balls were actually drawn, then the expected winnings would be of the greatest probability occurring. To figure this out we need to calculate the probability of getting two red balls consecutively and the probability of getting two different colors consecutively. Since there are a total of 10 balls the probabilities of getting one ball would be the following,

Red: [tex]\frac{4}{10} =  40%[/tex]

Green: [tex]\frac{6}{10} = 60%[/tex]

Now we need to know the probability of us getting 2 reds together, two greens, or one of each.

2 Reds: [tex]\frac{4}{10} *\frac{4}{10} = \frac{16}{100}  = 16%[/tex]

2 Greens:  [tex]\frac{6}{10} *\frac{6}{10} = \frac{36}{100}  = 36%[/tex]

1 Each:  [tex]\frac{4}{10} *\frac{6}{10} = \frac{24}{100}  = 24%[/tex]

So we can see that the highest probability is getting 2 Greens in  a row. Meaning you would have to pay $1

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

The question is about calculating the conditional probability density functions of two random variables X and Y given their joint density function. However, without the marginal density functions for X and Y, it is not possible to provide exact numerical answers. General formulas involve dividing the joint density function by the respective marginal density function.

The question relates to the field of probability and statistics, specifically pertaining to joint density functions of two random variables (X and Y). We have the joint density function f(x, y) = 6xe^(−x(y+6)) where x > 0 and y > 0.

For the conditional probability density function, the general formula for fX|Y(x, y) is the joint density function divided by the marginal density function of Y. Similarly, for fY|X(x, y), it is the joint density function divided by the marginal density function of X.

However, the marginal density functions for X and Y are not specified in the question. Typically, to find these, one would integrate the joint density function with respect to the other variable. Due to this missing information, it's not possible to provide an exact numerical answer.

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Answer:

[tex]y=2e^{sin(x)}[/tex]

Step-by-step explanation:

Given equation can be  re written as

[tex]\frac{\mathrm{d} y}{\mathrm{d} x}-ycos(x)=0\\\frac{\mathrm{d} y}{\mathrm{d} x}=ycos(x)\\\\=> \frac{dy}{y}=cox(x)dx\\\\Integrating  \\ \int \frac{dy}{y}=\int cos(x)dx \\\\ln(y)=sin(x)+c[/tex]............(i)

Now it is given that y(π/2) = 2e

Applying value in (i) we get

ln(2e) = sin(π/2) + c

=> ln(2) + ln(e) = 1+c

=> ln(2) + 1 = 1 + c

=> c = ln(2)

Thus equation (i) becomes

ln(y) = sin(x) + ln(2)

ln(y) - ln(2) = sin(x)

ln(y/2) = sin(x)

[tex]y= 2e^{sinx}[/tex]

If [tex]f(x)=4x-3[/tex]:

[tex]\displaystyle\lim_{\Delta x\to0}\frac{(4(x+\Delta x)-3)-(4x-3)}{\Delta x}=\lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4[/tex]

If [tex]f(x)=4x^{-3}[/tex]:

[tex]\displaystyle\lim_{\Delta x\to0}\frac{\frac4{(x+\Delta x)^3}-\frac4{x^3}}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac{4x^3-4(x+\Delta x)^3}{x^3(x+\Delta x)^3}}{\Delta x}[/tex]

[tex]\displaystyle=\lim_{\Delta x\to0}\frac{4x^3-4(x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3)}{x^3\Delta x(x+\Delta x)^3}[/tex]

[tex]\displaystyle=\lim_{\Delta x\to0}\frac{-12x^2\Delta x-12x(\Delta x)^2-4(\Delta x)^3}{x^3\Delta x(x+\Delta x)^3}=-\frac{12}{x^4}[/tex]

Answer: 0.9834

Step-by-step explanation:

Given : The test scores are normally distributed with

Mean : [tex]\mu=\ 0[/tex]

Standard deviation :[tex]\sigma= 1[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x = -2.13

[tex]z=\dfrac{-2.13-0}{1}=-2.13[/tex]

For x = 3.88

[tex]z=\dfrac{3.88-0}{1}=3.88[/tex]

The p-value = [tex]P(-2.13<z<3.88)=P(z<3.88)-P(z<-2.13)[/tex]

[tex]0.9999477-0.0165858=0.9833619\approx0.9834[/tex]

Hence, the probability that a given score is between negative 2.13 and 3.88 = 0.9834

Answer:

b) 4,000,000

Step-by-step explanation:

Let the population is measured since 1995,

Given,

The initial population, P = 3,942,000,

Annual rate of growing, r = 0.7% = 0.007,

If y represents the population after t years

So, the population after t years would be,

[tex]y=Pe^{rt}[/tex]

[tex]y=3942000(2.7)^{0.007x}[/tex]

Therefore, the population after 5 years,

[tex]y=3942000(2.7)^{0.007\times 5}=3942000(2.7)^{0.035}=4081448.78924\approx 4000000[/tex]

Hence, the population in 2000 would be approximately 40,00,000.

Option 'b' is correct.

By using the accurate population growth formula P = P0(1 + r)^n and substitifying the given values, the population of the Asian country in 2000 would be approximately 4,160,000 residents.

The question involves the concept of exponential growth, specifically applied to the population growth of an Asian country. Now, in given formula y = 3,942,000(2.7)^0.0074, we should use the correct growth rate formula to solve the problem which should be P = P0(1 + r)^n since the population growth is annually and continuous. Here, P0 is the initial population (3,942,000), r is the growth rate (0.7% or 0.007) and n is the number of years (2000-1995 = 5 years).

Using this formula, if you substitute these values in, you should get:

P = 3,942,000(1 + 0.007)^5

If you calculate this out, you reach a population of approximately 4,160,000 residents at the end of year 2000. Thus, answer 'c' is the correct option. Please remember exponential growth is a concept necessary for understanding population dynamics across multiple fields like demography, biology and mathematical modeling.

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The correct answers are:

(a) [tex]P(0\leq Z\leq 2.38)=0.4913[/tex]

(b)  [tex]P(0\leq Z\leq 1)=0.3413[/tex]

(c)  [tex]P(-2.70\leq Z\leq 0)=0.4965[/tex]

(d)  [tex]P(-2.70\leq Z\leq 2.70)=0.9926[/tex]

(e)  [tex]P(Z\leq 1.62)=0.9474[/tex]

(f)  [tex]P(-1.55\leq Z)=0.9394[/tex]

(g)  [tex]P(-1.70\leq Z\leq 2.00)=0.9326[/tex]

(h)  [tex]P(1.62\leq Z\leq 2.50)=0.0457[/tex]

(i) [tex]P(1.70\leq Z)=0.0446[/tex]

(j) [tex]P(|Z|\leq 2.50)=1.9862[/tex]

Let's calculate these probabilities step by step using the standard normal distribution table (also known as the z-table).

For reference, the standard normal distribution table provides the probabilities associated with the standard normal random variable (Z), which has a mean of [tex]0[/tex] and a standard deviation of [tex]1[/tex].

We'll use the standard normal distribution table to find the probabilities corresponding to the given Z-values.

(a [tex]P(0\leq Z\leq 2.38)[/tex] From the z-table,  [tex]P(Z\leq 2.38)=0.9913 P(0\leq Z\leq 2.38)=0.9913-0.5=0.4913[/tex] (subtracting the cumulative probability up to [tex]0[/tex] from the cumulative probability up to [tex]2.38[/tex])

(b) [tex]P(0\leq Z\leq 1)[/tex] From the z-table,  [tex]P(Z\leq 1)=0.8413 P(0\leq Z\leq 1)=0.8413-0.5=0.3413[/tex]

(c) [tex]P(-2.70\leq Z\leq 0)[/tex] From the z-table,  [tex]P(Z\leq 0)=0.5[/tex] and  [tex]P(Z\leq -2.70)=0.0035 P(-2.70\leq Z\leq 0)=0.5-0.0035=0.4965[/tex]

(d)  [tex]P(-2.70\leq Z\leq 2.70)[/tex] From the z-table,  [tex]P(Z\leq 2.70)=0.9961[/tex] and  [tex]P(Z\leq -2.70)=0.0035 P(-2.70\leq Z\leq 2.70)=0.9961-0.0035=0.9926[/tex]

(e)  [tex]P(Z\leq 1.62)[/tex] From the z-table,  [tex]P(Z\leq 1.62)=0.9474[/tex]

(f)  [tex]P(-1.55\leq Z)[/tex] From the z-table,  [tex]P(Z\leq -1.55)=0.0606 P(-1.55\leq Z)=1-0.0606=0.9394[/tex] (subtracting the cumulative probability up to [tex]-1.55[/tex] from [tex]1[/tex])

(g [tex]P(-1.70\leq Z\leq 2.00)[/tex] From the z-table,  [tex]P(Z\leq 2.00)=0.9772[/tex] and  [tex]P(Z\leq -1.70)=0.0446 P(-1.70\leq Z\leq 2.00)=0.9772-0.0446=0.9326[/tex]

(h)   [tex]P(1.62\leq Z\leq 2.50)[/tex] From the z-table,   [tex]P(Z2.50)=0.9931[/tex] and  [tex]P(Z\leq 1.62)=0.9474 P(1.62\leq Z\leq 2.50)=0.9931-0.9474=0.0457[/tex]

(i) [tex]P(1.70\leq Z)[/tex][tex]From the z-table, P(Z\leq 1.70)=0.9554 , P(1.70\leq Z)=1-0.9554=0.0446[/tex] (subtracting the cumulative probability up to [tex]1.70[/tex] from [tex]1[/tex])

(j) [tex]P(|Z|\leq 2.50)[/tex] Since the standard normal distribution is symmetric,   [tex]P(|Z|\leq 2.50)=2*P(0\leq Z\leq 2.50)=2*0.9931=1.9862[/tex] (multiply by [tex]2[/tex] because the probability of Z being between [tex]-2.50[/tex] and [tex]2.50[/tex] is twice the probability of Z being between [tex]0[/tex] and [tex]2.50[/tex] )

Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Round your answers to four decimal places.)

(a) [tex]P(0 \leq Z \leq 2.38) .4913[/tex]

(b) [tex]P(0 \leq Z\leq 1) .3413[/tex]

(c) [tex]P(-2.70 \leq Z \leq 0) .4965[/tex]

(d) [tex]P(-2.70 \leq Z \leq 2.70) .9931[/tex]

(e) [tex]P(Z \leq 1.62) .9474[/tex]

(f) [tex]P(-1.55 \leq Z) .9394[/tex]

(g) [tex]P(-1.70 \leq Z \leq 2.00) .9327[/tex]

(h) [tex]P(1.62 \leq Z \leq 2.50) .0464[/tex]

(i) [tex]P(1.70 \leq Z) .0445[/tex]

(j) [tex]P(|Z| \leq 2.50) .9876[/tex]

Answer with explanation:

The given non Homogeneous linear differential equation is:

   y" +4 y'=3 Sin 2 x-------(1)

Put , u=y'

Differentiating once

u'=y"

Substituting the value of , y' and y" in equation (1)

⇒u' +4u =3 Sin 2x

This is a type of linear differential equation.

Integrating factor [tex]=e^{4t}[/tex]

Multiplying both sides of equation by Integrating factor

[tex]e^{4 x}(u'+4u)=e^{4x}3 \sin 2x\\\\ \text{Integrating both sides}\\\\ue^{4x}=\int {3 \sin 2x \times e^{4x}} \, dx \\\\ue^{4x}=\frac{3e^{4x}}{2^2+4^2}\times (4\sin 2x -2 \cos 2x)\\\\ue^{4x}=\frac{3e^{4x}}{20}\times (4\sin 2x -2 \cos 2x)+C_{1}\\\\ \text{Using the formula of}\\\\\int{e^{ax}\sin bx } \, dx=\frac{e^{ax}}{a^2+b^2}\times (a \sin bx-b \cos bx)+C[/tex]

where C and [tex]C_{1}[/tex] are constant of integration.

Replacing , u by , y' in above equation we get the solution of above non homogeneous differential equation

  [tex]y'(x)=\frac{3}{20}\times (4\sin 2x -2 \cos 2x)+C_{1}e^{-4 x}[/tex]

Answer:

1,084 tickets were sold that cost $10

1,291 tickets were sold that cost $20

767 tickets were sold that cost $30

Step-by-step explanation:

Let

x ----> the number of tickets  that cost $10 sold

y ----> the number of tickets  that cost $20 sold

z ----> the number of tickets  that cost $30 sold

we know that

x+y+z=3,142 -----> equation A

10x+20y+30z=59,670 ----> equation B

y=x+207 ----> equation C

substitute equation C in equation A and equation B

x+(x+207)+z=3,142 ----> 2x+z=2,935 ----> equation D

10x+20(x+207)+30z=59,670 ---> 30x+30z=55,530 ----> equation E

Solve the system of equations D and E by graphing

The solution is the intersection point both graphs

The solution is the point (1,084,767)

so

x=1,084, z=767

see the attached figure

Find the value of y

y=x+207 ----> y=1,084+207=1,291

therefore

1,084 tickets were sold that cost $10

1,291 tickets were sold that cost $20

767 tickets were sold that cost $30

Answer:

5.13%.

Step-by-step explanation:

Amount  accumulated in 1 year

= 1600(1 + 0.05/360)^360

= $1682.03

Account's effective annual yield

= 82.03 * 100  / 1600 %

= 5.13%.

The account's effective annual yield (EAY) for an investment of $1600 with a 5% interest rate compounded daily (assuming a 360-day year) is approximately 5.12% when rounded to two decimal places.

The student has invested $1600 in an account that offers 5% interest compounded daily with the assumption of a 360-day year. To find the effective annual yield, we use the formula for compound interest and the definition of effective annual yield (EAY), which accounts for the compounding effect:

EAY = (1 + r/n)n - 1

Where:

In this case:

Now, substituting the values, we get:

EAY = (1 + 0.05/360)360 - 1

Calculating this out:

EAY = (1 + 0.0001388888889)360 - 1

EAY

to find the EAY:

EAY = ((1 + (0.05/360))^360) - 1

After calculating the above expression, the approximate effective annual yield comes out to be:

EAY = 0.05116 or 5.116%

Therefore, after rounding to two decimal places as required, the effective annual yield of the account is 5.12%.

Answer:

The number of cases prior to the increase is 50.

Step-by-step explanation:

It is given that the number of measles cases increased by 13.6% and the number of cases after increase is 57.

We need to find the number of cases prior to the increase.

Let x be the number of cases prior to the increase.

x + 13.6% of x = 57

[tex]x+x\times \frac{13.6}{100}=57[/tex]

[tex]x+0.136x=57[/tex]

[tex]1.136x=57[/tex]

Divide both the sides by 1.136.

[tex]\frac{1.136x}{1.136}=\frac{57}{1.136}[/tex]

[tex]x=50.176[/tex]

[tex]x\approx 50[/tex]

Therefore the number of cases prior to the increase is 50.

You can use

[tex]x=u\cos v[/tex]

[tex]y=u\sin v[/tex]

[tex]z=u\cos v+3[/tex]

with [tex]0\le u\le3[/tex] and [tex]0\le v\le2\pi[/tex].

The parametric equations for the part of the plane z = x + 3 that lies inside the cylinder x² + y² = 9 can be written as x = 3cos(θ), y = 3sin(θ), and z = 3cos(θ) + 3.

The parametric representation of a surface can be found by expressing the variables x, y, and z in terms of parameters. Given the cylinder equation x² + y² = 9, we can express x and y in terms of a single parameter θ as follows:

Here we've used the parametric equations for a circle of radius 3. Moving further with the given plane equation z = x+3, we substitute x from our parametric equations above:

So, the parametric representation for the given surface is:

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Answer:

Step-by-step explanation:

13, 15, 16, 16, 19, 20, 20, 21, 22, 22, 25, 25, 25, 25, 30, 33, 33, 35, 35, 35, 35, 36, 40, 45, 46, 52, 70.

a) Smoothing by bin means

Each bin has depth of 3

Dividing data into bins

so, Bin 1= 13, 15, 16

Bin 2=  16, 19, 20

Bin 3=  20, 21, 22

Bin 4= 22, 25, 25

Bin 5= 25, 25, 30

Bin 6= 33, 33, 35

Bin 7 = 35, 35, 35

Bin 8= 36, 40, 45

Bin 9=  46, 52, 70

Now, smoothing data by bin mean

so, Bin 1= 13, 15, 16 = (13+15+16)/3 = 15 Bin 1 = 15,15,15

Bin 2=  16, 19, 20 = (16+19+20)/3 = 18 Bin 2 = 18,18,18

Bin 3=  20, 21, 22= (20+21+22)/3 = 21 Bin 3 = 21,21,21

Bin 4= 22, 25, 25 = (22+25+25)/3 = 24 Bin 4 = 24,24,24

Bin 5= 25, 25, 30 =(25+25+30)/3 = 27 Bin 5 = 27,27,27

Bin 6= 33, 33, 35 = (33+33+35)/3 = 34 Bin 6 = 34,34,34

Bin 7 = 35, 35, 35 = (35+35+35)/3 = 35 Bin 7 = 35,35,35

Bin 8= 36, 40, 45 = (36+40+45)/3 = 40 Bin 8 = 40,40,40

Bin 9=  46, 52, 70=(46+52+70)/3 = 56 Bin 9 = 56,56,56

This technique is used to smooth the data. Data may have noise, using binning techniques we can remove noise from the data. It helps in providing more accurate results

b) How might you determine outliers in the data?

Outliers are the data that are abnormal to other data points. Outliers can be found by Box and whisker chart (box plot). Inter Quartile range can also be used to identify outliers

c)  What other methods are there for data smoothing?

Other methods of smoothing data are

a) binning by boundaries

b) Exponential smoothing

c) Random walk

The correct statement will be that the method of binning is extensively used for smoothening the data and cancels the noise a data contains. Outliers of a data can be determined using the quartile range.

There are different methods of data smoothening, such as binning with the help of boundaries, binning with the help of outliers, exponential binning, etc.

The bins will be created using the depth of 3 as,

Bin 1- 13,15,16 ; Bin 2- 16,19,20 ; Bin 3 20, 21, 22 ; Bin 4- 22,25,25 ; Bin 5- 25, 25, 30 ; Bin 6- 33,33,35 ; Bin 7- 35,35,35 ; Bin 8- 36,40,45 ; Bin 9- 46,52,70.

The vales of bins will be averaged and rounded off to the nearest whole numbers, considering bins as b in the following way.

[tex]b1= \dfrac {13+15+16}{3}\\\\b1= 14.67[/tex]

Continuing further in similar ways, we will find the values of remaining bins as

The outliers in the data can be found by using the box plot method and quartile range functions. Normally, such outliers are referred to as the mismatching data in a bin.

There are various different methods for smoothing the data. Some ways of smoothing the data are exponential smoothing, random walk smoothing, boundary binning method, etc.

Hence, the data is smoothened using the binning method and the values obtained are as above.

Learn more about bin data here:

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Answer:

We are given that The dependent variable is production; that is, it is assumed that different levels of production result from a different number of employees.

Number of Assemblers(x)  One-Hour Production(y) (units)

2                                                           11

4                                                           18

1                                                            7

5                                                           29

3                                                           20

a. Draw a scatter diagram.

Solution : Refer the attached figure  

b. Based on the scatter diagram, does there appear to be any relationship between the number of assemblers and production? Explain.

Solution: The equation that shows the relationship between the number of assemblers and production is [tex]y=5.1x+1.7[/tex]

Where y is One-Hour Production (units)  and x is the Number of Assemblers

c.Compute the correlation coefficient.

Solution:

Formula of correlation coefficient:[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{[n \sum x^2 -(\sum x)^2][n \sum y^2 -(\sum y)^2]}[/tex]

            x        y         xy       [tex]x^2[/tex]   [tex]y^2[/tex]

           2        11        22              4                   121

           4        18       72               16                 324

           1          7         7                 1                    49

           5         29      145             25                  841

           3         20      60               9                    400            

Sum:   15     85       306           55                  1735

n=5

Substitute the values in the formula :

[tex]r=\frac{5(306)-(15)(85)}{[5 (55) -(15)^2][5 (1735) -(85)^2]}[/tex]

[tex]r=0.00351[/tex]

The correlation coefficient is 0.00351

Answer:

Part 1:

Let the nickels be = n

Let the dimes be = d

Let the quarters be = q

Plato has 36 coins in nickels, dimes, and quarters. So, equation forms:

[tex]n+d+q=36[/tex]    .....(1)

The number of nickels is three less than twice the number of dimes.

[tex]n=2d-3[/tex]    ....(2)

The total value of the coins is $5.20.

[tex]0.10d+0.05n+0.25q=5.20[/tex]   .... (3)

Substituting n=2d-3 in (1) and (3)

[tex]2d-3+d+q=36[/tex]

=> [tex]3d+q=39[/tex]    ....(4)

[tex]0.10d+0.05(2d-3)+0.25q=5.20[/tex]

=> [tex]0.10d+0.10d-0.15+0.25q=5.20[/tex]

=> [tex]0.20d+0.25q=5.35[/tex]    ...(5)

Multiplying (4) by 0.25 and subtracting (5) from (4)

[tex]0.75d+0.25q=9.75[/tex] now subtracting (5) from this we get;

[tex]0.55d=4.4[/tex]

=> d = 8

Substituting d = 8 in [tex]3d+q=39[/tex]

[tex]3(8)+q=39[/tex]

[tex]24+q=39[/tex]

=> q = 15

Substituting values of d and q in [tex]n+d+q=36[/tex], we get n

[tex]n+8+15=36[/tex]

[tex]n=36-23[/tex]

=> n = 13

Therefore Plato has 13 nickels, 15 quarters and 8 dimes.

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

Part 2:

Let the original price of the ski hat be = x

Original price was marked down by 35% means value was lowered by 35%.

So, we can calculate as:

[tex]x-\frac{35x}{100}=15.60[/tex]

=> [tex]\frac{65x}{100}=15.60[/tex]

=> [tex]65x=1560[/tex]

x = 24

Hence, the original price was $24 but after 35% marking down, it was available for $15.60.

Answer:

p(0) = 800

p(8) = 997

Step-by-step explanation:

p(t) = 800 * (1.028)^t

The current price is when t=0

p(0) = 800 * (1.028)^0

       = 800(1)

       = 800

The price in 8 years

p(8) = 800 * (1.028)^8

       =997.7802522414861936754688

To the nearest dollar

        = 998

Answer:

[tex]p(0)=\$\ 800[/tex]

[tex]p(8)=\$\ 998[/tex]

Step-by-step explanation:

The function that the mode in the price is a function of exponential growth

[tex]p(t)=800(1.028)^t[/tex]

If t represents time in years, then to find the current price we do [tex]t = 0[/tex]

Then:

[tex]p(t=0)=800(1.028)^0[/tex]

[tex]p(0)=800(1)[/tex]

[tex]p(0)=\$\ 800[/tex]

To find the price after 8 years substitute t = 8 in the equation

[tex]p(t=8)=800(1.028)^8[/tex]

[tex]p(8)=\$\ 998[/tex]

Answer:

25%

Step-by-step explanation:

Great question, since a regular coin has two sides one heads and one tails. That gives us a 50% probability of it landing on either side of the coin. Since we would like to know the probability of getting 2 heads in a row, we would need to multiply the probability of the first toss landing on heads with the second toss landing on heads, like so...

[tex]\frac{1}{2} *\frac{1}{2} =\frac{1}{4}[/tex]

So we can see that the probability of us getting two heads in a row is that of \frac{1}{4}[/tex] or 25%.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

The probability of flipping at least two heads in four tosses of a fair coin is calculated using the binomial distribution, and the total probability is found to be 0.6875 or 68.75%.

To calculate the probability of flipping at least two heads in a series of four coin tosses with a fair coin, we need to consider all the possible outcomes in which we can get at least two heads. The different numbers of heads that can be obtained are 0, 1, 2, 3, or 4. To find the probability of each specific event, we use the binomial distribution formula, which for flipping two heads is:

P(2 heads) = (4 choose 2) × (0.5)² × (0.5)² = 6 × 0.25 × 0.25 = 0.375.

We can also find the probabilities of obtaining three and four heads:

P(3 heads) = (4 choose 3) × (0.5)³ × (0.5)¹ = 4 × 0.125 × 0.5 = 0.25,

P(4 heads) = (4 choose 4) × (0.5)⁴ = 1 × 0.0625 = 0.0625.

Next, we add these probabilities together to get the total probability of flipping at least two heads:

Total probability = P(2 heads) + P(3 heads) + P(4 heads) = 0.375 + 0.25 + 0.0625 = 0.6875.

Therefore, the probability of flipping at least two heads in four tosses of a fair coin is 0.6875 or 68.75%.

Answer:

  (d)  10

Step-by-step explanation:

Multiply by x and divide by its coefficient:

  (43.65)(8.79) = (0.4365)(87.9)x

  (43.65)(8.79)/((0.4365)(87.9)) = x

At this point, any calculator can give you the answer. It is, perhaps, more satisfying to work out the answer without a calculator.

  x = (43.65)/(0.4365) × (8.79)/(87.9)

In the first quotient, the numerator is 100 times the denominator; in the second, the denominator is 10 times the numerator.

  x = (100) × (1/10) = 100/10

  x = 10

_____

Moving the decimal point to the right 1 place multiplies the numerical value by 10.

The value of x in the given equation that satisfies the condition is 10.

In this question, we're given a mathematical expression in which the value of x is unknown. We're looking for the value of x that satisfies the equation:

( 43.65 ) ( 8.79 ) / x = ( 0.4365 ) ( 87.9 )

To solve this equation for x, we can start by noting the similarity between the left and right sides. We have larger numbers on the left side that appear, in reduced form, on the right side.

Follow these steps:

So, the correct option would be (d) 10.

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The rate at which the water level is rising when the water level is 6 cm is 8.33 cm/s.

We can find the rate at which the water level is rising by using similar triangles. Let the height of the water level be h (in cm). Since the pyramid is inverted, the volume of water inside the pyramid is given by V = (6-h)^2 * h. Taking the derivative of both sides with respect to time, we get dV/dt = 50. Solving for dh/dt, we find that the rate at which the water level is rising is dh/dt = 50 / (12 - h).

When the water level is 6 cm, we substitute h = 6 into the equation to find the rate at which the water level is rising. dh/dt = 50 / (12 - 6) = 50 / 6 = 8.33 cm/s. Therefore, when the water level is 6 cm, the rate at which the water level is rising is 8.33 cm/s.

Answer:

[tex]P(5)=0.238[/tex]

[tex]P(x\geq 6)=0.478[/tex]

[tex]P(x<4)=0.114[/tex]

Step-by-step explanation:

In this case we can calculate the probability using the binomial probability formula

[tex]P(X=x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}[/tex]

Where p is the probability of obtaining a "favorable outcome " x is the number of desired "favorable outcome " and n is the number of times the experiment is repeated. In this case n = 10 and p = 0.54.

(a) exactly​ five

This is:

[tex]x=5,\ n=10,\ p=0.54.[/tex]

So:

[tex]P(X=5)=\frac{10!}{5!(10-5)!}*0.54^x*(1-0.54)^{10-5}[/tex]

[tex]P(5)=0.238[/tex]

(b) at least​ six

This is: [tex]x\geq 6,\ n=10,\ p=0.54.[/tex]

[tex]P(x\geq 6)=P(6) + P(7)+P(8)+P(9) + P(10)[/tex]

[tex]P(x\geq 6)=0.478[/tex]

(c) less than four

This is: [tex]x< 4,\ n=10,\ p=0.54.[/tex]

[tex]P(x<4)=P(3) + P(2)+P(1)+P(0)[/tex]

[tex]P(x<4)=0.114[/tex]

This question is based on the probability. Therefore, the required probabilities  are :  (a) [tex]P(5) = 0.238[/tex],  (b)[tex]P(x \geq 6) = 0.478[/tex]  and (c) [tex]P(x <4) = 0.114[/tex].

Given:

54​% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults.

We have to find the probability that the number of U.S. adults who have very little confidence in newspapers is ​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four.

According to the question,

[tex]P(5) = 0.238\\P(x \geq 6) = 0.478\\P(x <4) = 0.114[/tex]

In this we have to calculate the probability using the binomial probability formula,

[tex]P(X=x) = \dfrac{n!}{x!(n-x)!} \times p^{x} \times (1-p)^{n-x}[/tex]

Where, p is the probability of obtaining a "favorable outcome ", x is the number of desired "favorable outcome " and n is the number of times the experiment is repeated. In this case n = 10 and p = 0.54.

(a) exactly​ five  

x=5, n= 10, p = 0.54

[tex]P(X=5)= \dfrac{10!}{5!(10-5)!} \times 0.5^{x} \times (1-0.54)^{10-5}[/tex]

P(X=5) = 0.238

(b) at least​ six

[tex]x\geq 6, n=10, p=0.54\\P(x\geq 6) = P(6) + P(7) + P(8) + P(9)+P(10)\\P(x\geq 6) = 0.478[/tex]

(c) less than four

[tex]x< 6, n=10, p=0.54\\P(x< 4) = P(3) + P(2) + P(1) + P(0)\\P(x< 4) = 0.114[/tex]

Therefore, the answers are :  (a) [tex]P(5) = 0.238[/tex],  (b)[tex]P(x \geq 6) = 0.478[/tex]

and (c) [tex]P(x <4) = 0.114[/tex].

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