Decide whether the given expression is a polynomial and tell why or why not.





5. 3x2 – 5x + 2

Answer: The correct answer is A or B

`

Step-by-step explanation:

Answer:

fjekwnkewgnelwnlgnendndj

Step-by-step explanation:

Answer:

87 ft

Step-by-step explanation:

SohCahToa is your best friend here.

You have two values you need to pay attention to:

The length that is adjacent to the 74°C, 25 ft. And the length opposite of the 74°C, the height of how high the rocket traveled.

So adjacent and opposite, O & A. "Toa", find the tangent of 74°C.

tan(74) = [tex]\frac{x}{25}[/tex]

x = (tan(74))(25)

x = 87 ft

Answer:

$400

Step-by-step explanation:

all you do is add 320+540+230+450+280+580/6 and the asnwe comes out to 400

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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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:

45

Step-by-step explanation:

the correct choice is C.

The value of the lawn ornament set in the year 2009 was $51.375. The value of the lawn ornament set in the year 2021 was $558.181. The instantaneous rate of change of the value of the lawn ornament set in the year 2013 was $230.986. The instantaneous rate of change of the value of the lawn ornament set in the year 2021 was $351.076.  The estimated value of the lawn ornament set in 2022 was $909.257.

a)

To find the value of the lawn ornament set in the year 2009, we have to plug in t = 3, as t = 0 corresponds to the year 2006.

V(3) = 0.06(3)4 – 1.05(3)3 + 3.47(3) – 8.896 + 269.95

V(3) = 51.375

So, the value of the lawn ornament set in the year 2009 was $51.375.

b)

To find the value of the lawn ornament set in the year 2021, we have to plug in t = 15, as t = 0 corresponds to the year 2006.

V(15) = 0.06(15)4 – 1.05(15)3 + 3.47(15) – 8.896 + 269.95

V(15) = $558.181

So, the value of the lawn ornament set in the year 2021 is $558.181.

c)

To find the instantaneous rate of change of the value of the lawn ornament set in the year 2013, we have to find V'(7), where V(t) is the given function.

V(t) = 0.06t4 – 1.05t3 + 3.47t – 8.896 +269.95 for Osts 15

V'(t) = 0.24t3 – 3.15t2 + 10.41t + 269.95

V'(7) = 0.24(7)3 – 3.15(7)2 + 10.41(7) + 269.95

V'(7) = $230.986

So, the instantaneous rate of change of the value of the lawn ornament set in the year 2013 was $230.986.

d) To find the instantaneous rate of change of the value of the lawn ornament set in the year 2021, we have to find V'(15), where V(t) is the given function.

V(t) = 0.06t4 – 1.05t3 + 3.47t – 8.896 +269.95 for Osts

15V'(t) = 0.24t3 – 3.15t2 + 10.41t + 269.95

V'(15) = 0.24(15)3 – 3.15(15)2 + 10.41(15) + 269.95

V'(15) = $351.076

So, the instantaneous rate of change of the value of the lawn ornament set in the year 2021 is $351.076.

e)

To ESTIMATE the value of the lawn ornament set in 2022, we can use the formula

V(t) ≈ V(a) + V'(a)(t – a),

where a is the year 2021.

V(a) = V(15) = $558.181

V'(a) = V'(15) = $351.076t = 16 (as we need to estimate the value of the lawn ornament set in 2022)

V(t) ≈ V(a) + V'(a)(t – a)

V(t) ≈ 558.181 + 351.076(16 – 15)

V(t) ≈ $909.257

So, the estimated value of the lawn ornament set in 2022 is $909.257.

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:))))))

Step-by-step explanation:

videochamp, picsart

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

Answer: Mean = 2.36 Median = 4 Range = 0

Step-by-step explanation:

Mean - the sum of the data values divided by the number of data values

Median - the middle number in an ordered set of data

Range - the difference between the greatest and least numbers in a data set

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.

Answer:

5 drinks will be americanos

Step-by-step explanation:

2:8  (2/8)

simplify

1:4  (1/4)

divide 20 by 4

5:20

The number of next 20 drinks which may be Americanos is 5.

Probability is simply the possibility of getting an event. Or in other words, we are predicting the chance of getting an event.

The value of probability will be always in the range from 0 to 1.

Given that,

Total drinks sold = 8

Number of drinks that is Americanos = 2

Probability of finding Americano = 2/8 = 1/4

If the total number of drinks next is 20,

Number of Americanos expected = Probability of Americanos × Number of drinks

= 1/4 × 20

= 5

Hence the number of Americanos expected is 5.

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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.

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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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.

For values of n other than 0 or 1 in a Bernoulli differential equation, the substitution [tex]u = y^{(1-n)[/tex] is used to transform it into a linear equation.

A Bernoulli differential equation is given by the form:

dy + P(x)y dx = Q(x)[tex]y^n[/tex] (*)

If we consider the case when n = 0 or n = 1, the Bernoulli equation becomes linear. Let's examine each case:

When n = 0:

Substituting[tex]u = y^{(-n) }= y^{(-0)} = 1[/tex], the differential equation becomes:

[tex]dy + P(x)y dx = Q(x)y^0[/tex]

dy + P(x)y dx = Q(x)

This is a linear differential equation of the first order.

When n = 1:

Substituting [tex]u = y^{(-n) }= y^{(-1)},[/tex] we have:

[tex]u = y^{(-1)[/tex]

Taking the derivative of both sides with respect to x:

[tex]du/dx = -y^{(-2)} \times dy/dx[/tex]

Rearranging the equation:

[tex]dy/dx = -y^2\times du/dx[/tex]

Now substituting the expression for dy/dx in the original Bernoulli equation:

[tex]dy + P(x)y dx = Q(x)y^1\\-y^2 \times du/dx + P(x)y dx = Q(x)y\\-y \times du + P(x)y^3 dx = Q(x)y[/tex]

This equation is also a linear differential equation of the first order, but with the variable u instead of y.

In summary, when n is equal to 0 or 1, the Bernoulli equation becomes linear. For other values of n, a substitution u = y^(-n) is typically used to transform the Bernoulli equation into a linear differential equation, allowing for easier analysis and solution.

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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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(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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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 subspace topology for Z, which is a subset of R with its usual (standard) topology, is the set of open sets in Z.

In other words, the subspace topology on Z is obtained by considering the intersection of Z with open sets in R.

To find the subspace topology for Z, we need to determine which subsets of Z are open. In the usual topology on R, an open set is a set that can be represented as a union of open intervals. Since Z is a subset of R, its open sets will be the intersection of Z with open intervals in R.

For example, let's consider the open interval (a, b) in R. The intersection of (a, b) with Z will be the set of integers between a and b (inclusive) that belong to Z. This intersection is an open set in Z.

By considering all possible open intervals in R and their intersections with Z, we can generate the collection of open sets that form the subspace topology for Z. This collection of open sets will satisfy the axioms of a topology, including the properties of openness, closure under unions, and closure under finite intersections.

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Answer:

(4,2)

Step-by-step explanation:

Answer:

(4, 2)

Step-by-step explanation:

Answer:

1

Step-by-step explanation:

The y-intercept in the equation is 1 because the equation uses the format y=mx+b. The b in y=mx+b represents the y-intercept So, in this equation the y-intercept is 1 because b=1.

Answer:

the full answer is 215.859885inches cubed

Step-by-step explanation:

times the length, width and height together

jesse has never used the sliding board at his daycare because he is afraid. His teacher has encouraged him, but he refuses to slide down. one day his mother stands at the bottom And say's Jesse let's go and slides into his mother's arms Laughing The situation is a example of Social referencing. Option (d) is correct.

Situation refers to a group of conditions or a current state of events.

Social reference is the method through which newborns control their behavior toward surrounding items, people, and circumstances by observing the emotive displays of an adult.

For adaptive social functioning to occur, one must recognize and make use of the emotional communication of others. The ability to negotiate complicated and frequently ambiguous settings is known as social referencing in the developmental literature and social appraisal in adult studies.

Therefore, Option (d) is correct. The situation is a example of Social referencing.

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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°

Answer: 2 1/2

Step-by-step explanation:

the answer is D i took the test here is proof

Answer:

False

Step-by-step explanation:

A P E X

(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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