The distribution of actual weights of 8-ounce chocolate bars produced by a certain machine is Normal with mean 8.1 ounces and standard deviation 0.1 ounces. Company managers do not want the weight of a chocolate bar to fall below 8 ounces, for fear that consumers will complain. (a) Find the probability that the weight of a randomly selected candy bar is less than 8 ounces Forty candy bars are selected at random and their mean weight is computed. (b) Calculate the mean and standard deviation of the sampling distribution of (c) Find the probability that the mean weight of the forty candy bars is less than 8 ounces. (d) Would your answers to (a), (b), or (c) be affected if the weights of chocolate bars produced by this machine were distinctly non-Normal? Explain.

____________________________

So your answer would be C, [tex]\bold{a=5}[/tex].

____________________________

Answer:

45

Step-by-step explanation:

the correct choice is C.

The equations of the images are after the transformations are

a. y = -3x

b. y = x

c. x = 0

From the question, we have the following parameters that can be used in our computation:

a. y = 3x

b. y = -x

c. x = 0.

The rule of the lines when reflected in the x-axis is

(x, y) = (x, -y)

This means that the functions are negated

So, we have the images to be

a. y = -3x

b. y = x

c. x = 0

Hence, the equations of the images are

a. y = -3x

b. y = x

c. x = 0

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The function f(x) = e^(1+x) has at least one real root. The function f(x) = e^(1+x) cannot have more than one real root.

To show that f(x) has at least one real root, we need to find a value of x for which f(x) equals zero. Let's set f(x) = 0 and solve for x:

e^(1+x) = 0

Since e^(1+x) is always positive for any real value of x, there is no value of x that makes f(x) equal to zero. Hence, f(x) = e^(1+x) does not have any real roots. Therefore, we cannot show that f(x) has at least one real root.

b)

To show that f(x) cannot have more than one real root, we need to demonstrate that there cannot be two distinct real values, say c1 and c2, such that f(c1) = f(c2) = 0. Let's assume that f(x) = 0 at two distinct values, c1 and c2:

e^(1+c1) = e^(1+c2) = 0

However, this equation is not possible since e^(1+c1) and e^(1+c2) are always positive for any real values of c1 and c2. Therefore, f(x) = e^(1+x) cannot have more than one real root.

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P(X < 250) = P(X ≤ 249) = 0 (approximately) Hence, P(X ˃ 310) = 0 and P(X ˂ 250) = 0.

Given a random sample size of n = 900 from a binomial probability distribution with P = 0.30. The probability that the number of successes is greater than 310 and the probability that the number of successes is fewer than 250 are to be found.

Solution: a)We know that P(X > 310) can be found using normal approximation.

We have to check whether np and nq are greater than or equal to 10 or not, where p=0.30, q=0.70 and n=900.

Here, np = 900*0.30 = 270 and nq = 900*0.70 = 630. Since np and nq are greater than or equal to 10, we can use normal approximation for this binomial distribution.

Using the normal approximation formula, z = (X - μ) / σwhere X = 310, μ = np and σ = √(npq), we getz = (310 - 270) / √(900*0.30*0.70)z = 4.25

Using the z-table, the probability of z being greater than 4.25 is almost zero.

Therefore, P(X > 310) = P(X ≥ 311) = 0 (approximately)

b)We know that P(X < 250) can be found using normal approximation. We have to check whether np and nq are greater than or equal to 10 or not, where p=0.30, q=0.70 and n=900.  

Here, np = 900*0.30 = 270 and nq = 900*0.70 = 630.

Since np and nq are greater than or equal to 10, we can use normal approximation for this binomial distribution.

Using the normal approximation formula,z = (X - μ) / σwhere X = 250, μ = np and σ = √(npq), we getz = (250 - 270) / √(900*0.30*0.70)z = -4.25Using the z-table, the probability of z being less than -4.25 is almost zero.

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Given data: n = 900, P = 0.30.

a. The probability that the number of successes is greater than 310 is 0.0000.

b. The probability that the number of successes is fewer than 250 is 0.0174.

a. The formula for finding probability of binomial distribution is:

P(X > x) = 1 - P(X ≤ x)

P(X > 310) = 1 - P(X ≤ 310)

Mean μ = np

= 900 × 0.30

= 270

Variance σ² = npq

= 900 × 0.30 × 0.70

= 189

Standard deviation

σ = √σ²

= √189

z = (x - μ) / σ

z = (310 - 270) / √189

z = 4.32

Using normal approximation,

P(X > 310) = P(Z > 4.32)

= 0.00001673

Using calculator, P(X > 310) = 0.0000(rounded to four decimal places)

b. P(X < 250)

Mean μ = np

= 900 × 0.30

= 270

Variance σ² = npq

= 900 × 0.30 × 0.70

= 189

Standard deviation

σ = √σ²

= √189

z = (x - μ) / σ

z = (250 - 270) / √189

z = -2.12

Using normal approximation, P(X < 250) = P(Z < -2.12) = 0.0174.

Using calculator, P(X < 250) = 0.0174(rounded to four decimal places).

Therefore, the probability that the number of successes is greater than 310 is 0.0000 and the probability that the number of successes is fewer than 250 is 0.0174.

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Answer:

x<28

Step-by-step explanation:

Isolate x

First, subtract 4 on both sides

x/2+<14

Then, multiply both sides by 2 to get x alone

x<28

On a number line, there would be an open circle (not filled in dot) on 28, and the entire left side of the number line would be filled in

Answer:

x<28

Step-by-step explanation:

x/2+4<18

multiply the 2 on both sides to get rid of it

x+8<36

isolate the x

x<28

on the number line, it's an open circle with the arrow pointing to the left.

Answer:

4c + 5b + 11

Step-by-step explanation:

4c + 3 + 5b + 8

4c + 5b + 11

The reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3 is [tex]\dfrac{1}{2}[/tex]

Given the limit of a function expressed as:

[tex]\lim_{x \to 3}\frac{2x^2-x+15}{x^3-5x-12}[/tex]

First, we need to substitute x = 3 into the function to have:

[tex]=\frac{2(3)^2-3+15}{3^3-5(3)-12}\\=\frac{18-3+15}{27-15-12}\\=\frac{0}{0} (indeterminate)[/tex]

Apply l'hospital rule on the function:

[tex]=\lim_{x \to 3}\frac{\frac{d}{dx} (2x^2-x+15)}{\frac{d}{dx} (x^3-5x-12)}\\=\lim_{x \to 3}\frac{4x-1}{3x^2-5}\\[/tex]

Subtitute x = 3 into the result

[tex]=\frac{4(3)-1}{3(3)^2-5}\\=\frac{12-1}{27-5}\\=\frac{11}{22}\\=\frac{1}{2}[/tex]

Hence the reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3 is [tex]\dfrac{1}{2}[/tex]

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Answer:

1/2

Step-by-step explanation:

i am in your walls

Answer: x= 12/103

alternate form x =0.116505

Step-by-step explanation:

Try Photo math! It explains step by step!

Hope this helps!!!!

Answer:

fjekwnkewgnelwnlgnendndj

Step-by-step explanation:

Step-by-step explanation:

Volume of Cylinder =

[tex]500 {cm}^{3} = \pi {r}^{2} h[/tex]

given d = 18

r = 1/2 x d = 9cm,

[tex]\pi( {9}^{2} )h = 500 \\ 81\pi \: h = 500 \\ h = \frac{500}{81\pi} cm[/tex]

I will leave the answer in terms of Pi as I am not sure how you want to leave your answer as.

Answer:

False

Step-by-step explanation:

A P E X

The values of k for which the function f(x, y, z) = kx² - kry + y² - 2yz - 22 has a nondegenerate local maximum at (0, 0, 0) are when k > 0.

To find the critical points of the function, we need to calculate the partial derivatives with respect to each variable:

∂f/∂x = 2kx ∂f/∂y = -kr + 2y - 2z ∂f/∂z = -2y

2kx = 0 => x = 0 (Equation 1) -kr + 2y - 2z = 0 => r = y - z (Equation 2) -2y = 0 => y = 0 (Equation 3)

From Equation 3, we can see that y = 0. Substituting this into Equation 2, we get:

r = 0 - z r = -z (Equation 4)

The Hessian matrix is given by:

H = | ∂²f/∂x² ∂²f/∂x∂y ∂²f/∂x∂z | | ∂²f/∂y∂x ∂²f/∂y² ∂²f/∂y∂z | | ∂²f/∂z∂x ∂²f/∂z∂y ∂²f/∂z² |

Calculating the second-order partial derivatives:

∂²f/∂x² = 2k ∂²f/∂y² = 2 ∂²f/∂z² = 0 ∂²f/∂x∂y = 0 ∂²f/∂y∂z = -2 ∂²f/∂z∂x = 0

Thus, the Hessian matrix becomes:

H = | 2k 0 0 | | 0 2 -2 | | 0 -2 0 |

D = ∂²f/∂x² ∂²f/∂y² ∂²f/∂z² + 2∂²f/∂x∂y ∂²f/∂y∂z ∂²f/∂z∂x - (∂²f/∂x² ∂²f/∂y∂z ∂²f/∂z∂x + ∂²f/∂y² ∂²f/∂z∂x ∂²f/∂x∂y ∂²f/∂z²)

Substituting the partial derivatives we calculated earlier:

D = (2k)(2)(0) + 2(0)(-2)(0) - (2k)(-2)(0) - (2)(0)(0) D = 0

If the determinant D is zero, the second derivative test is inconclusive. In such cases, we need to consider the eigenvalues of the Hessian matrix.

To find the eigenvalues, we solve the characteristic equation:

det(H - λI) = 0

where λ is the eigenvalue and I is the identity matrix. Substituting the values from the Hessian matrix:

| 2k-λ 0 0 | | 0 2-λ -2 | | 0 -2 -λ |

The characteristic equation becomes:

(2k - λ)((2 - λ)(-λ) - (-2)(0)) - (0)((2 - λ)(-2) - (0)(0)) = 0 (2k - λ)(λ² - 2λ) = 0

From this equation, we can see that one eigenvalue is (2k - λ) = 0, which implies λ = 2k.

For our case, we have one eigenvalue (λ = 2k). Thus, the sign of λ depends on the value of k.

When k < 0, the point (0, 0, 0) is a nondegenerate local minimum. When k = 0, the second derivative test is inconclusive, and further analysis would be required to determine the nature of the critical point.

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Answer:

1 < -9

Step-by-step explanation:

A positive number can't be less than a negative

Answer:

$36

Step-by-step explanation:

Basically, every shirt is $6 because 30% of 20 is 6. If he buys 6 shirts, then 6 times 6 is 36 dollars spent, tax not included.

(a) MLE of $\lambda$ is obtained by maximizing the log-likelihood. Suppose that X1,X2,…,XnX1,X2,…,Xn are independent and identically distributed exponential random variables with parameter λ, then the probability density function of XiXi is given by $$f(x_i;\lambda) =\lambda e^ {-\lambda x_i}, \quad x_i\geq0. $$

The log-likelihood function is given by$$\begin{aligned}\ln L(\lambda) &= \ln (\lambda^n e^{-\lambda(x_1+x_2+\cdots+x_n)}) \\&=n\ln \lambda-\lambda(x_1+x_2+\cdots+x_n).\end{aligned}$$

The first derivative of the log-likelihood function with respect to λλ is$$\frac {d\ln L(\lambda)} {d\lambda} = \frac{n}{\lambda}-x_1-x_2-\cdots-x_n.$$

The first derivative is zero when $$\frac{n}{\lambda}-\sum_{i=1} ^{n} x_i=0. $$Hence, the MLE of λλ is $$\hat{\lambda} =\frac{n}{\sum_{i=1} ^{n} x_i}. $$

Substituting the value of $\hat{\lambda} $ gives the maximum value of the log-likelihood. So, the MLE of $\lambda$ is given by $$\boxed{\hat{\lambda} =\frac{n}{\sum_{i=1} ^{n} x_i}}. $$

The MLE of $\lambda$ is $\frac {3} {\sum_{i=1} ^{n} x_i}$.

(b) The median of the exponential distribution is given by$$m = \frac {\ln (2)} {\lambda}. $$

Therefore, the log-likelihood function for median is given by$$\begin{aligned}\ln L(m) &= \sum_{i=1}^{n} \ln f(x_i;\lambda)\\&= \sum_{i=1}^{n} \ln \left(\frac{1}{\lambda}e^{-x_i/\lambda}\right)\\&= -n\ln\lambda-\frac{1}{\lambda}\sum_{i=1}^{n}x_i.\end{aligned}$$

The first derivative of the log-likelihood function with respect to mm is$$\frac {d\ln L(m)} {dm} = \frac {1} {\lambda}-\frac {1} {\lambda^2} \sum_{i=1} ^{n}x_i\ln 2. $$

The first derivative is zero when $$\frac {1} {\lambda} =\frac{1}{\lambda^2}\sum_{i=1}^{n}x_i\ln 2.$$Hence, the MLE of mm is $$\boxed{\hat{m} = \frac{\ln 2}{\bar{x}}}.$$where $\bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_i.$Therefore, the MLE of m is $\frac {\ln 2} {\bar{x}}. $

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:))))))

Step-by-step explanation:

videochamp, picsart

Picsart , Inshot , Gandr , Photo lab and Viva video.

Answer:

I found 2 answers for this one

Step-by-step explanation:

B- 2,5,10

D-12,16,20

The possible length in inches of the stick that will make a right angle triangle are as follows

A right angle triangle has one of its angles as 90 degrees.

The right angle triangle must obeys the Pythagorean theorem.

c² = a² + b²

where

Therefore, the sides that makes a right angle are as follows:

6, 8 , 10 : 6² + 8² = 10²

12, 16, 20 : 12² + 16² = 20²

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Answer:

Radius = 8 inches

Area = [tex]\pi {r}^{2} [/tex]

[tex]area = \pi( {8}^{2})[/tex]

[tex] = 64\pi \: square \: inches[/tex]

A = [tex]201.06 ^{2} \: inches[/tex]

Step-by-step explanation:

Hope it is helpful....

By computing the pooled variance given the following data N_1 = 18, n_2 = 14, s_1 = 7, s_2 = 8, the pooled variance is 436.40.

To compute the pooled variance given N_1 = 18, n_2 = 14, s_1 = 7, s_2 = 8, we can use the formula below;

S_p² = [(n₁ - 1)S₁² + (n₂ - 1)S₂²] / (N - 2),

where S_p² = pooled variance, n₁ = sample size of first group, n₂ = sample size of second group, S₁² = variance of first group, S₂² = variance of second group, and N = total sample size.

To plug in the values, we have: N₁ = 18n₂ = 14S₁ = 7S₂ = 8

Substituting the values into the formula above we get;

S_p² = [(18 - 1)(7²) + (14 - 1)(8²)] / (18 + 14 - 2)S_p² = (17 × 49 + 13 × 64) / 30S_p² = 436.4

Round off to two decimal places to get 436.40.

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Answer:

1

Step-by-step explanation:

Answer:

The answer is 1

Step-by-step explanation:

E is -1 and the absolute value is the posotive of any number. The positive of -1 is 1.

The solution to the system is {eq}(x, y) = (2, 1){/eq}. For the given system, we have the solutions (-3, 0) and (0, 6).

To solve a system of equations, we must find the value of each variable that satisfies both equations. One of the most common methods for solving systems of equations is called substitution.

In substitution, we solve for one variable in one equation and then plug that expression into the other equation.

For example, if we have the system {eq}2x + y = 5,

\quad 4x - y = 7 {/eq},

we can solve for y in the first equation: {eq}y = 5 - 2x {/eq} Then we substitute this expression for y into the second equation:

{eq}4x - (5 - 2x) = 7 {/eq}

Simplifying gives {eq}6x = 12 {/eq}, or {eq}x = 2 {/eq} Once we have a value for one variable, we can substitute it into either equation to find the value of the other variable.

Using {eq}y = 5 - 2x {/eq}, we have {eq}y = 5 - 2(2) = 1 {/eq}These solutions represent the values of x and y that satisfy both equations in the system. To check,

we can substitute each solution into both equations to ensure they are valid. If both equations are satisfied, then we have found the correct solutions.

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Answer:

Second One-

[tex] {5}^{14}. {5}^{ - 2} [/tex]

Fifth One-

[tex] {5}^{6} \: . \: {5}^{6} [/tex]

Sixth One-

[tex] \sqrt{ {5}^{24} } [/tex]

Seventh One-

[tex] {5}^{11} \: . \: 5 [/tex]

Answer:

Don't click the link its a virus

Answer:

a) 2x+(x+36)=90

Step-by-step explanation:

b) A1+A2=90°. (A=angle)

2x+(x+36)=90

2x+x+36=90

3x+36=90

3x=90-36

x=54/3

x=18

then A1=2x=2*18=36°

A2=x+36=18+36=54°

Answer:

4225

Step-by-step explanation:

65x65=4225

Answer: squareroot of 58

You can solve this problem simply by using the distance formula . Using the distance formula we can solve this problem by just placing the numbers and then solving the equation.