Can you help me solve this?

Y-intercept= 4, goes through the point (2, 3)

Please solve using y=Mx+b

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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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:

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....

Answer: The correct answer is A or B

`

Step-by-step explanation:

Answer:

Don't click the link its a virus

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 logistic differential equation for the hyena population is given by:

dP/dt = r * P * (1 - P/K)

where P(t) is the hyena population at time t, r is the growth rate, and K is the carrying capacity.

We are given that:

P(t) = 40 + k * e^(-0.57t)

K = 200

To determine the value of k, we can plug in these values into the logistic differential equation and solve for k:

dP/dt = r * P * (1 - P/K)

dP/dt = r * P * (1 - P/200)

dP/dt = r/200 * (200P - P^2)

dP/(200P - P^2) = r dt

Integrating both sides, we get:

-1/200 ln|200P - P^2| = rt + C

where C is a constant of integration.

Using the initial condition P(0) = 40 + k, we can solve for C:

-1/200 ln|200(40+k)-(40+k)^2| = 0 + C

C = -1/200 ln|8000-480k|

Plugging in this value of C and simplifying, we get:

-1/200 ln|200P - P^2| = rt - 1/200 ln|8000-480k|

ln|200P - P^2| = -200rt + ln|8000-480k|

|200P - P^2| = e^(-200rt) * |8000-480k|

200P - P^2 = ± e^(-200rt) * (8000-480k)

Since the population is increasing, we choose the positive sign:

200P - P^2 = e^(-200rt) * (8000-480k)

Using the initial condition P(0) = 40 + k, we get:

200(40+k) - (40+k)^2 = (8000-480k)

8000 + 160k - 2400 - 80k - k^2 = 8000 - 480k

k^2 + 560k - 2400 = 0

(k + 60)(k - 40) = 0

Thus, k = -60 or k = 40. Since k represents a growth rate, it should be positive, so we choose k = 40. Therefore, the value of the constant k is option (A) 4.

For the second part of the question, the logistic equation that could model the growth in the graph is option (B) dM/dt = (20-M)*(M-50). This is because the carrying capacity is between 20 and 50, and the population growth rate is zero at both of these values (i.e. the population does not increase or decrease when it is at the carrying capacity).

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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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The standard deviation of the number of people over the age of 65 in a sample of 12 coronary bypass patients is 1.6487.

Given that a healthcare research agency reported that 41% of people who had coronary bypass surgery in 2008 were over the age of 65 and twelve coronary bypass patients are sampled.

To determine the mean number of people over the age of 65 in a sample of 12 coronary bypass patients, we use the formula below:

Mean = np

Where n = 12 and p = 0.41.

Mean = 12(0.41)

Mean = 4.92

Therefore, the mean number of people over the age of 65 in a sample of 12 coronary bypass patients is 4.92.

To determine the standard deviation of the number of people over the age of 65 in a sample of 12 coronary bypass patients, we use the formula below:

Standard deviation, σ = √(n p q)

Where n = 12, p = 0.41, and q = 1 - p.

Standard deviation, σ = √(12 × 0.41 × 0.59)

Standard deviation, σ = √2.71948

Standard deviation, σ = 1.6487 (rounded to four decimal places).

Therefore, the standard deviation of the number of people over the age of 65 in a sample of 12 coronary bypass patients is 1.6487.

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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.

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:

$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.

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:

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:

1 < -9

Step-by-step explanation:

A positive number can't be less than a negative

(a) To prove that the set N(R) = {x ∈ R: x is nilpotent} is an ideal of the commutative ring R with 1.

We need to show that it satisfies the two conditions of being an ideal: closure under addition and closure under multiplication by elements of R.

To demonstrate closure under addition, let x and y be nilpotent elements in N(R). This means that there exist positive integers m and n such that xm = 0 and yn = 0.

We want to show that x + y is also nilpotent. By expanding (x + y)^k using the binomial theorem, we can see that each term involves a product of powers of x and y. Since both x and y are nilpotent, their product is also nilpotent.

Therefore, the sum (x + y) raised to a sufficiently high power will result in zero, showing that x + y is indeed nilpotent. Hence, N(R) is closed under addition.

To prove closure under multiplication by elements of R, let x be a nilpotent element in N(R) and r be any element in R. We aim to show that rx is nilpotent. Since x is nilpotent, there exists a positive integer m such that xm = 0.

When we raise rx to a sufficiently high power, (rx)^k, it can be expanded as r^k * x^k. Since x^k is zero due to x being nilpotent, the product r^k * x^k is also zero. Therefore, rx is nilpotent, and N(R) is closed under multiplication by elements of R.

Hence, N(R) satisfies both conditions of being an ideal, and thus, it is an ideal of the commutative ring R.

(b) To prove that N(R/N(R)) = 0, we want to show that every element in R/N(R) is not nilpotent.

Let [x] be an element in R/N(R), where [x] represents the equivalence class of x modulo N(R). Our goal is to demonstrate that [x] is not nilpotent, meaning it is not equal to the zero element in R/N(R).

Suppose, for contradiction, that [x] = 0 in R/N(R). This would imply that x belongs to N(R), the set of nilpotent elements in R. However, if x is an element of N(R), it means that x is nilpotent, and by definition, there exists some positive integer n such that xn = 0. This contradicts our assumption that [x] = 0, since it would imply that x is not nilpotent.

Therefore, our assumption that [x] = 0 leads to a contradiction, and we conclude that every element in R/N(R) is not nilpotent.

Consequently, N(R/N(R)) = 0, indicating that the set of nilpotent elements in the quotient ring R/N(R) is empty.

In summary, we have shown that N(R/N(R)) = 0 and established that N(R) is an ideal of the commutative ring R.

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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.

a. To take a sample of the student population that would not represent the population well, Susan could use a biased sampling method.

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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:

fjekwnkewgnelwnlgnendndj

Step-by-step explanation:

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:

45

Step-by-step explanation:

the correct choice is C.

Answer:

False

Step-by-step explanation:

A P E X

:))))))

Step-by-step explanation:

videochamp, picsart

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

(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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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°