qual das inequações seguintes e equivalente à inequação -2x <4?
a) x<-2
b) x>-2
c) x<2
d) x>2​

The answer is (-5, -1), option B is correct.

Given that vector u has initial point at (3,9) and terminal point at (-7,5) and vector v has initial point at (1, -4) and terminal point at (6, -1). We need to find u + v in component form.The component form of the vector is obtained by subtracting the initial point from the terminal point. The result is the vector in component form. The components of vector u are:u = (-7 - 3, 5 - 9) = (-10, -4)The components of vector v are:v = (6 - 1, -1 - (-4)) = (5, 3)Now, we can add the vectors in component form. u + v = (-10, -4) + (5, 3) = (-10 + 5, -4 + 3) = (-5, -1)Hence, the answer is (-5, -1).Therefore, option B is correct.

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We can model the time between telephone orders using an exponential distribution with a rate parameter of λ = 5.1 orders per hour.

The probability of at least 35 minutes (0.5833 hours) elapsing before the next order is the same as the probability that the time until the next order is greater than 0.5833 hours.

Let X be the time until the next order, then X is exponentially distributed with parameter λ = 5.1. The probability we want to find is:

P(X > 0.5833) = e^(-λ * 0.5833)

Substituting λ = 5.1, we get:

P(X > 0.5833) = e^(-5.1 * 0.5833) = 0.3239

Therefore, the probability that at least 35 minutes will elapse before the next telephone order is 0.3239, rounded to 4 decimal places.

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Option B. The function that represents the profit, P(m), where m is the number of minutes the platform uses on advertisements is: P(m) = -1.6(x - 150)² + 10000.

The capability that addresses the benefit, P(m), where m is the quantity of minutes the stage utilizes on promotions is:

P(m) = - 1.6(x - 150)² + 10000

This is on the grounds that we know that the greatest benefit of $10,000 happens when the stage utilizes 150 minutes daily on notices, and the benefit capability ought to have a most extreme as of now. The capability is in the vertex structure, which is P(m) = a(x - h)² + k, where (h,k) is the vertex of the parabola and a decides if the parabola opens upwards or downwards.

The negative worth of an in the capability shows that the parabola opens downwards and has a most extreme worth at the vertex (h,k). The vertex is at (150,10000), and that implies that the most extreme benefit of $10,000 happens when the stage utilizes 150 minutes daily on ads.

In this way, the capability that addresses the benefit, P(m), where m is the quantity of minutes the stage utilizes on ads is P(m) = - 1.6(x - 150)² + 10000. The other given capabilities don't match the given circumstances for the most extreme benefit, and in this way, they are not fitting to address the benefit capability of JP Creations.

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Given that $Y\sim\text{Exp}(A)$, the probability density function of $Y$ is $f_Y(y)=Ae^{-Ay}$ for $y\geq 0$.

Let $X\sim\text{Poisson}(Y)$. Then, the conditional probability

mass function of $X$ given $Y=y$ is

P(X=k∣Y=y)=e−yykk!,k=0,1,2,…

To find the mean and variance of $X$, we first find $E[XY]$ and $E[X^2Y]$.

\begin{align*}

E[XY] &= \int_{0}^{\infty} E[XY|Y=y]f_Y(y)dy \

&= \int_{0}^{\infty} E[Xy]Ae^{-Ay}dy \

&= \int_{0}^{\infty} ye^{-y}\sum_{k=0}^{\infty}k\frac{y^k}{k!}Ae^{-Ay}dy \

&= \int_{0}^{\infty} ye^{-y}\sum_{k=1}^{\infty}\frac{y^{k-1}}{(k-1)!}Ae^{-Ay}dy \

&= A\int_{0}^{\infty} y\sum_{k=1}^{\infty}\frac{(Ay)^{k-1}}{(k-1)!}e^{-Ay}e^{-y}dy \ &= A\int_{0}^{\infty} y\sum_{k=0}^{\infty}\frac{(Ay)^{k}}{k!}e^{-Ay}e^{-y}dy \

&= A\int_{0}^{\infty} ye^{-(A+1)y}\sum_{k=0}^{\infty}\frac{(Ay)^{k}}{k!}dy \

&= A\int_{0}^{\infty} ye^{-(A+1)y}e^{Ay}dy \

&= \frac{A}{(A+1)^2} \end{align*}

Similarly, we can find $E[X^2Y]$ as:

\begin{align*}

E[X^2Y] &= \int_{0}^{\infty} E[X^2Y|Y=y]f_Y(y)dy \

&= \int_{0}^{\infty} E[X^2y]Ae^{-Ay}dy \

&= \int_{0}^{\infty} y^2e^{-y}\sum_{k=0}^{\infty}k^2\frac{y^k}{k!}Ae^{-Ay}dy \

&= \int_{0}^{\infty} y^2e^{-y}\sum_{k=2}^{\infty}\frac{k(k-1)y^{k-2}}{(k-2)!}Ae^{-Ay}dy \

&= A\int_{0}^{\infty} y^2\sum_{k=0}^{\infty}\frac{(Ay)^{k}}{k!}e^{-Ay}e^{-y}dy \ &= A\int_{0}^{\infty} y^2e^{-(A+1)y}\sum_{k=0}^{\infty}\frac{(Ay)^{k}}{k!}dy \

&= A\int_{0}^{\

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The sum of the expressions is 2(3x + 4)

To determine the sum of the expressions, we need to know that algebraic expressions are described as those expressions that are made up of terms, variables, constants, coefficients and factors.

Algebraic expressions are also those expressions that are known to consist of different arithmetic operations.

These arithmetic operations are enumerated thus;

From the information given, we have that;

3x + 4 + 4 + 3x

collect the like terms, we have;

3x + 3x + 4 + 4

Add the like terms, we get;

6x + 8

Factorize

2(3x + 4)

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Complete question:

What is the sum when 3x+4 is added to 4+3x?

The area of the composite shape is 51 sq. in.

First, let's calculate the area of the rectangle:

Area of a rectangle = length × width

Given that the length of the rectangle is 6 inches and the width is 7 inches, the area of the rectangle is:

Area of rectangle = 6 × 7 = 42 sq. in.

Next, let's calculate the area of the triangle:

Area of a triangle = 1/2(base × height)

The base of the triangle is 3 inches, and we need to determine the height. Unfortunately, the height is not given, so we cannot calculate the area of the triangle accurately. Let's consider it as an incomplete shape for now.

Now, let's find the total area of the composite shape by adding the area of the rectangle and the area of the triangle:

Area of composite shape = area of rectangle + area of triangle

Substituting the known values:

Area of composite shape = 42 sq. in. + 1/2(3 × 6)

Simplifying the expression:

Area of composite shape = 42 sq. in. + 1/2(18)

Area of composite shape = 42 sq. in. + 9 sq. in.

Area of composite shape = 51 sq. in.

Therefore, the area of the composite shape is 51 sq. in.

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The Airy differential equation is a second-order linear ordinary differential equation given by Fourier Transform:

-d^2u/dx^2 = x*u

To find a bounded solution to this equation, we can use the Fourier transform. The Fourier transform of a function f(x) is given by:

F(ω) = ∫ f(x) e^(-iωx) dx

Using the Fourier transform, we can convert the differential equation into an algebraic equation in terms of the Fourier transform F(ω):

-ω^2 F(ω) = ∫ x*u(x) e^(-iωx) dx

We can rewrite the integral on the right-hand side using integration by parts:

∫ x*u(x) e^(-iωx) dx = -∫ u(x) d/dx(e^(-iωx) dx)

= -iω∫ u(x) e^(-iωx) dx + [u(x) e^(-iωx)]^∞_0

Since we are looking for a bounded solution, the term [u(x) e^(-iωx)]^∞_0 must be equal to zero. Therefore, we have:

ω^2 F(ω) = iω∫ u(x) e^(-iωx) dx

We can then solve for the Fourier transform F(ω):

F(ω) = i/ω ∫ u(x) e^(-iωx) dx

Finally, we can take the inverse Fourier transform to find the solution u(x):

u(x) = (1/2π) ∫ F(ω) e^(iωx) dω

Substituting the expression for F(ω), we have:

u(x) = i/(2πω) ∫ ∫ u(y) e^(-iω(y-x)) dy dω

This gives us an integral formula for a bounded solution to the Airy differential equation in terms of the Fourier transform.

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The word bank to fill in the blanks in the given sentences is messenger, transcribed, nucleus, RNA, codons, ribosome, transfer, amino acid, and protein. The amino acids are added to the growing protein chain as specified by the mRNA sequence, which is read by the ribosome.

This occurs in the nucleus. Second, the RNA gets translated into a sequence of amino acids. The ribosome reads the mRNA every three bases known as codons. The transfer RNA carries amino acids to the ribosome to build the protein. In the first step of the central dogma, transcription takes place. Transcription is the process of DNA being copied into RNA. The DNA is present in the nucleus of a cell.

The RNA is formed through the transcription process. mRNA is produced from the DNA molecule during transcription. mRNA stands for messenger RNA. The transcription process is divided into three stages: initiation, elongation, and termination. In the second step of the central dogma, translation takes place. In the process of translation, mRNA is translated into a protein. Amino acids are linked together to form a protein chain, which is determined by the sequence of codons in the mRNA molecule. Ribosomes are the sites of translation. Transfer RNA (tRNA) molecules carry amino acids to the ribosome. Each amino acid is attached to a specific tRNA molecule.

The amino acids are added to the growing protein chain as specified by the mRNA sequence, which is read by the ribosome.

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The encrypted message for "snowfall" using Vigenere cipher with key "blue" is "TYPAGKL".

To use the Vigenere cipher with key "blue" to encrypt the message "snowfall," we follow these steps:

Write the key repeatedly below the plaintext message:

Key:   blueblu

Plain: snowfal

Convert each letter in the plaintext message to a number using a simple substitution, such as A=0, B=1, C=2, etc.:

Key:   blueblu

Plain: snowfal

Nums:  18 13 14 22 5 0 11

Convert each letter in the key to a number using the same substitution:

Key:   blueblu

Nums:  1 11 20 4 1 11 20

Add the corresponding numbers in the plaintext and key, modulo 26 (i.e. wrap around to 0 after 25):

Key:   blueblu

Plain: snowfal

Nums:  18 13 14 22 5 0 11

Key:   1 11 20 4 1 11 20

Enc:   19 24 8 0 6 11 5

Convert the resulting numbers back to letters using the same substitution:

Encrypted message: TYPAGKL

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The smallest value of n that works is 11.

We can expand the given equation using the binomial theorem:

(x^2 + 1.2)^n = ∑(k=0 to n) [n choose k] x^(2(n-k)) (1.2)^k

The coefficient of x^6 in the expansion will be given by the term where k = n - 3, i.e.,

[n choose n-3] x^(2(3)) (1.2)^(n-3) = (n(n-1)(n-2)/(3!)) x^6 (1.2)^(n-3)

We want this coefficient to be larger than 200000, so we have:

(n(n-1)(n-2)/(3!)) (1.2)^(n-3) > 200000/16

Simplifying and taking the logarithm of both sides:

(n-1)log(1.2) + log(n(n-1)(n-2)) - log(3!) > log(12500)

Using the fact that log(n) < n for all n > 0, we can approximate log(n(n-1)(n-2)) by n log(n) and simplify further:

(n-1)log(1.2) + 3log(n) - log(3) > log(12500)

Now, we can use trial and error to find the smallest value of n that satisfies this inequality. We can start with n = 10 and increase it until we get a value that works:

For n = 10: (9)log(1.2) + 3log(10) - log(3) ≈ 2.413, which is not greater than log(12500) ≈ 4.819.

For n = 11: (10)log(1.2) + 3log(11) - log(3) ≈ 4.299, which is greater than log(12500).

Therefore, the smallest value of n that works is 11.

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Joe had the correct answer, and Robert and Myra made mistakes in their addition of the fractions.

a) To determine who had the correct answer, we compare the given answers of Robert, Myra, and Joe.

Robert's answer: 5 1/3

Myra's answer: 2 1/12

Joe's answer: 4 5/6

To compare these mixed numbers, it's helpful to convert them to improper fractions:

Robert's answer: 5 1/3 = (5 * 3 + 1) / 3 = 16/3

Myra's answer: 2 1/12 = (2 * 12 + 1) / 12 = 25/12

Joe's answer: 4 5/6 = (4 * 6 + 5) / 6 = 29/6

Comparing the improper fractions, we can see that Joe's answer of 29/6 is the largest.

Therefore, Joe had the correct answer.

b) Let's analyze how Robert and Myra obtained their answers and where they went wrong:

Robert's answer of 5 1/3 = 16/3:

It seems that Robert incorrectly added the whole number and the fraction separately without considering the common denominator. . The correct sum would be (5 * 3 + 1) / 3 = 16/3, which is Joe's answer.

Myra's answer of 2 1/12 = 25/12:

Myra's mistake appears to be similar to Robert's mistake. She may have added 2 and 1 to get 3 and then added 1/12 to get 1 1/12.

However, the correct addition should be done by finding a common denominator, which in this case is 12, and adding the fractions. The correct sum would be (2 * 12 + 1) / 12 = 25/12, which is not the correct answer.

In conclusion, Joe had the correct answer, and Robert and Myra made mistakes in their addition of the fractions.

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Assume x and y are functions of t, the value of dy/dt is 1.

To evaluate dy/dt for the given equation y^3 = 2x^2 + 2, with dx/dt = 3, x = 1, and y = 2, we first need to apply the Chain Rule for differentiation with respect to t.
Step 1: Differentiate both sides of the equation with respect to t.
d(y^3)/dt = d(2x^2 + 2)/dt
Step 2: Apply the Chain Rule.
3y^2(dy/dt) = 4x(dx/dt)
Step 3: Plug in the given values for x, y, and dx/dt.
3(2^2)(dy/dt) = 4(1)(3)
Step 4: Simplify the equation.
12(dy/dt) = 12
Step 5: Solve for dy/dt.
(dy/dt) = 12/12
(dy/dt) = 1
So, the value of dy/dt is 1.

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a. to make f(x) continuous at x = 11, we need to define f(11) = 1/22. b. there is a solution for the equation f(x) = 0 on the interval [1, 2].

a) f(11)=5

To make the function f(x) continuous at x = 11, we need to remove the infinite discontinuity at x = 11. We can do this by factoring out (x-11) from the numerator and simplifying the expression. After factoring out (x-11), we get (sqrt(x) + 5 + 4)/(x - 11) = (sqrt(x) + 9)/(x - 11). We can see that this expression is undefined at x = 11, so we need to determine the limit of the expression as x approaches 11. We can use L'Hopital's rule to find that the limit is 1/22. Therefore, to make f(x) continuous at x = 11, we need to define f(11) = 1/22.

b) No. because f(1)<0 and f(2)<0.

The intermediate value theorem states that if f(x) is continuous on the closed interval [a, b] and if k is any number between f(a) and f(b), then there exists at least one number c in the open interval (a, b) such that f(c) = k. In this case, f(x) = 2x^3 - sqrt(6x + 2) is continuous on the closed interval [1, 2]. We can see that f(1) is negative and f(2) is also negative. Therefore, by the intermediate value theorem, we cannot conclude that there is a solution for the equation f(x) = 0 on the interval [1, 2].

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It does appears that elite soccer players are more likely to develop arthritis of the hip or knee than comparable recreational soccer players.

To determine whether elite soccer players are more likely to develop arthritis of the hip or knee than recreational soccer players, we can conduct a hypothesis test using a two-proportion z-test.

Let p1 be the proportion of former elite soccer players who developed arthritis of the hip or knee and p2 be the proportion of recreational soccer players who developed arthritis of the hip or knee. The null hypothesis is that the two proportions are equal (p1 = p2) and the alternative hypothesis is that the proportion for the elite soccer players is higher than that for the recreational soccer players (p1 > p2).

The test statistic for the two-proportion z-test is:

z = (p1 - p2) / sqrt((p_hat * (1 - p_hat) / n1) + (p_hat * (1 - p_hat) / n2))

where p_hat is the pooled proportion, n1 and n2 are the sample sizes, and the standard error of the difference in proportions is:

SE = sqrt((p_hat * (1 - p_hat) / n1) + (p_hat * (1 - p_hat) / n2))

Using the given information, we have:

n1 = n2 = 500

x1 = 0.38 * 500 = 190

x2 = 0.32 * 500 = 160

p_hat = (x1 + x2) / (n1 + n2) = 0.35

Therefore, the test statistic is:

z = (0.38 - 0.32) / sqrt((0.35 * (1 - 0.35) / 500) + (0.35 * (1 - 0.35) / 500)) = 1.737

Using a standard normal distribution table, we find the critical value for a one-tailed test with a = 0.05 to be 1.645. Since the test statistic (1.737) is greater than the critical value (1.645), we reject the null hypothesis and conclude that there is evidence to suggest that the proportion of former elite soccer players who develop arthritis of the hip or knee is higher than that for recreational soccer players.

Therefore, it appears that elite soccer players are more likely to develop arthritis of the hip or knee than comparable recreational soccer players.

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The given differential equation dy/dx = sin(x)/y is a first-order separable differential equation.

A separable differential equation is one that can be expressed in the form g(y)dy = f(x)dx, where g(y) and f(x) are functions of y and x, respectively. In this case, we have dy/dx = sin(x)/y, which can be rewritten as ydy = sin(x)dx.

To solve this separable differential equation, we can integrate both sides:

∫ydy = ∫sin(x)dx

Integrating the left side with respect to y gives (1/2)y^2, and integrating the right side with respect to x gives -cos(x) + C, where C is the constant of integration.

Therefore, we have (1/2)y^2 = -cos(x) + C.

The separable differential equation dy/dx = sin(x)/y can be solved by integrating both sides. The solution is given by (1/2)y^2 = -cos(x) + C, where C is the constant of integration.

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Answer:

C

Step-by-step explanation:

first , calculate the measure of ∠ A

the sum of the 3 angles in a triangle = 180° , that is

∠ A + 80° + 60° = 180°

∠ A + 140° = 180° ( subtract 140° from both sides )

∠ A = 40°

the shortest side in the triangle is the side opposite the smallest angle in the triangle.

the smallest angle is ∠ A = 40° , then

the shortest side is the side opposite ∠ A , that is BC

a. The series has form an where an = 1/n(n+1)

b. The first five terms in the sequence {an} are 1/2, 1/6, 1/12, 1/20, 1/30

c. The first five terms in the sequence of partial sums for this series are 1/2, 2/6, 3/12, 4/20, 5/30

d. The general formula for the partial sum Sn is Sn = 1 - 1/(n+1)

e. The sum of a series is defined as the limit of the sequence of partial sums, which means lim as n approaches infinity of Sn = 1.

f. True.

g. False.

a. Each term in the series is given by an = 1/n(n+1), which simplifies to an = 1/n - 1/(n+1). This form is a telescoping series.

b. The first five terms in the sequence {an} are obtained by plugging in n = 1, 2, 3, 4, 5 into the formula for an, respectively. Thus, we have a sequence of {an} = {1/2, 1/6, 1/12, 1/20, 1/30}.

c. The sequence of partial sums is obtained by summing the first n terms of the series. Thus, we have S1 = 1/2, S2 = 2/6, S3 = 3/12, S4 = 4/20, S5 = 5/30.

d. To find a general formula for the nth partial sum Sn, we can use the telescoping property of the series. We have:

Sn = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... - 1/(n+1) + 1/(n+1)

Simplifying, we obtain:

Sn = 1 - 1/(n+1)

e. The sum of the series is defined as the limit of the sequence of partial sums as n approaches infinity. Thus, we have:

lim as n approaches infinity of Sn = lim as n approaches infinity of (1 - 1/(n+1))

= 1 - 0 = 1

f. True, the sequence {an} converges to 0 since each term approaches 0 as n approaches infinity.

g. False, the sequence {an} converges to 0, not to 1.

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The sum of the infinite geometric series is 64/7 or approximately 9.143.

To find the sum of an infinite geometric series, we need to determine if the series is convergent or divergent. A geometric series is convergent if the common ratio, denoted by "r", lies between -1 and 1.

In the given series, the common ratio can be calculated by dividing any term by its preceding term. Let's calculate the common ratio:

r = [tex](-12) / 16 = -3/4[/tex]

Since the absolute value of the common ratio, |r| = 3/4, is less than 1, the series is convergent.

The sum of an infinite geometric series can be calculated using the formula: S = a / (1 - r), where "a" is the first term of the series.

Using the given series, a = 16 and r = -3/4, we can calculate the sum:

S = [tex]16 / (1 - (-3/4)) = 16 / (1 + 3/4) = 16 / (7/4) = 16 * (4/7) = 64/7[/tex]

Therefore, the sum of the infinite geometric series is 64/7 or approximately 9.143.

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Based on the measured Auv value of 2.3, the new estimate for the Eddington ratio of the AGN would be lower than previously thought. The Eddington ratio represents the balance between the accretion rate onto the supermassive black hole at the center of the AGN and the radiation pressure that is generated. A higher Eddington ratio indicates that the black hole is accreting material at a rate that is approaching or exceeding the maximum limit set by radiation pressure. The presence of dust in the galaxy's center has attenuated the light from the AGN, which has impacted our interpretation of its behavior by obscuring the true level of accretion onto the black hole.

To provide a new estimate of the Eddington ratio for this AGN, considering the value of Auv 2.3 toward the nucleus of the galaxy, you should follow these steps:

1. Determine the intrinsic luminosity of the AGN by correcting the observed luminosity for dust extinction. Use the given Auv value (2.3) to find the extinction factor and calculate the intrinsic luminosity (L_intrinsic = L_observed * extinction factor).
2. Calculate the Eddington luminosity (L_Eddington) for the AGN, which is the maximum luminosity it can achieve while still being stable. You will need to know the mass of the black hole at the center of the galaxy for this calculation.
3. Divide the intrinsic luminosity by the Eddington luminosity to get the Eddington ratio: Eddington ratio = L_intrinsic / L_Eddington.

The Eddington ratio provides insight into the accretion rate and radiative efficiency of the AGN. A higher Eddington ratio indicates that the AGN is accreting material at a faster rate, leading to more intense radiation. The presence of dust has impacted your interpretation of the AGN's behavior by attenuating the light from the AGN, causing you to underestimate its true luminosity and, consequently, the Eddington ratio. Correcting for this dust extinction provides a more accurate estimate of the AGN's accretion rate and radiative efficiency.

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The correct answer is F. None of the above. Data analysts prefer to deal with random sampling error rather than statistical bias for non of the reasons.

Data analysts prefer to deal with random sampling error rather than statistical bias because random sampling error is a type of error that occurs by chance and can be reduced through larger sample sizes or better sampling methods.

Thus, Data analysts prefer to deal with random sampling error rather than statistical bias for non of the reasons.

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The water usage of New York State per person per day can be calculated using the given information.

The population of New York State is given to be 1.9 × 10⁷, while more than 8.9 × 10⁹ gallons of water are withdrawn each day from the water sources.

The daily water usage per person in New York State is as follows:

Number of gallons of water withdrawn each day from all sources of water = More than 8.9 × 10⁹

Number of persons living in New York State = 1.9 × 10⁷

Now, we can calculate the daily water usage per person in New York State as follows:

Daily water usage per person =

Number of gallons of water withdrawn each day / Number of persons living in New York State

= (8.9 × 10⁹) / (1.9 × 10⁷)

≈ 468 gallons (rounded to the nearest whole number)

Therefore, the daily water usage per person in New York State is approximately 468 gallons.

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The best point estimate for the mean amount of d-glucose in cockroach hindguts under these conditions is approximately 44.24 micrograms.

To find the best point estimate for the mean, we calculate the average (or the arithmetic mean) of the given data points. Adding up the amounts of d-glucose in the hindguts of the 5 cockroaches and dividing by the total number of cockroaches (which is 5 in this case), we get:

(55.95 + 68.24 + 52.73 + 21.50 + 23.78) / 5 ≈ 44.24

Therefore, the best point estimate for the mean amount of d-glucose in cockroach hindguts, based on the given sample, is approximately 44.24 micrograms.

The best point estimate for the mean is obtained by calculating the average of the observed values in the sample. This provides a single value that represents the central tendency of the data. In this case, we add up the amounts of d-glucose in the hindguts of the 5 cockroaches and divide by the total number of cockroaches to find the mean. Rounding the result to the nearest hundredth, we obtain 44.24 micrograms as the best point estimate for the mean amount of d-glucose in cockroach hindguts under the given conditions.

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a) This is a nonlinear differential equation of the form dy/dx = f(x,y). We can rewrite it as:

dy/(y^2 - 4) = (x^2 - 4)/(y^2 - 4) dx

Integrating both sides, we get:

-1/2 arctan(y/2) = (1/3) x^3 - 4x + C

where C is the constant of integration.

b) This is a linear first-order differential equation of the form dy/dx + P(x)y = Q(x). We can rewrite it as:

dy/dx + (1-x)/(x-1) y = e^(4x)/(x-1)

This is a homogeneous equation with integrating factor mu(x) = e^(-ln(x-1)) = 1/(x-1). Multiplying both sides by mu(x), we get:

(1/(x-1)) dy/dx + y/(x-1) = e^(4x)/((x-1)^2)

Using the product rule for differentiation, we can rewrite the left-hand side as:

d/dx (y/(x-1)) = e^(4x)/((x-1)^2)

Integrating both sides, we get:

y/(x-1) = -(1/4)e^(4x) + C

where C is the constant of integration.

c) This is a homogeneous first-order differential equation of the form M(x,y) dx + N(x,y) dy = 0, where M(x,y) = 7x - 3y and N(x,y) = 6y - 3x. We can check if it is exact by computing the partial derivatives:

dM/dy = -3

dN/dx = -3

Since dM/dy is not equal to dN/dx, the equation is not exact. We can find an integrating factor mu(x,y) by dividing one partial derivative by the other:

mu(x,y) = e^(int ((dN/dx - dM/dy)/M) dx) = e^(-3x/2 + 2ln|y|)

Multiplying both sides of the equation by mu(x,y), we get:

(7xy - 3y^2)e^(-3x/2 + 2ln|y|) dx + (6y^2 - 3xy) e^(-3x/2 + 2ln|y|) dy = 0

This equation is exact, so we can find the solution by integrating M(x,y) with respect to x and N(x,y) with respect to y:

(7/2)x^2y - 3y^3 ln|y| + f(y) = C

where f(y) is the constant of integration.

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The equation that gives total savings y as a function of the number of months x is y = $300x

Given that Mrs. White started saving $300 a month. After 3 months, she had $1200. Now, we need to write an equation that gives total savings y as a function of the number of months x
Let us consider that the total savings Mrs. White saved after x months = y
From the given data, we can see that the amount of saving she does each month = $300
So, at the end of 3 months, she had saved an amount of= $300 × 3 = $900
Total savings after 3 months, y = $1200
Thus, we can say that; the total amount she saves, increases every month by $300$300$300 ×x= $y (total savings)
We can write this equation as the function of total savings y as a function of the number of months
x:y = $300x

Thus, the equation that gives total savings y as a function of the number of months x is y = $300x.

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To find the area under the standard normal curve between z = -0.62 and z = 1.47, we need to use a standard normal distribution table or a calculator with a standard normal distribution function.

Using a standard normal distribution table, we can find the area to the left of z = -0.62 and z = 1.47, and then subtract the smaller area from the larger area to find the area between the two z-scores.

From the table, we find:

The area to the left of z = -0.62 is 0.2676

The area to the left of z = 1.47 is 0.9292

Therefore, the area between z = -0.62 and z = 1.47 is:

0.9292 - 0.2676 = 0.6616

Rounding this answer to four decimal places, we get:

Area between z = -0.62 and z = 1.47 ≈ 0.6616

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(a) Proved If [tex]a = a^-1[/tex], then det(a) = ±1.

(b) The determinant must be +1 or -1

When a square matrix a is equal to its inverse [tex]a^-1[/tex], the determinant of a is either +1 or -1. This can be explained as follows:

(a) In the case where [tex]a = a^-1[/tex], we can multiply both sides of the equation by a to obtain [tex]a^2 = I[/tex], where I is the identity matrix.

Taking the determinant of both sides, we have[tex]det(a^2) = det(I)[/tex], and since [tex]det(I) = 1,[/tex] we get [tex](det(a))^2 = 1.[/tex]

This implies that det(a) is either +1 or -1.

(b) The determinant of a matrix represents the scaling factor of the transformation it represents.

If [tex]a = a^-1[/tex], it means that applying the transformation twice results in the identity transformation, which preserves the shape and orientation of vectors.

Therefore, the determinant must be +1 or -1 to maintain this property.

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The common factor for 13%, 0.125, 1/5, and 10% is that they can be expressed as fractions with denominators of 100.

All of the numbers can be converted to fractions with a denominator of 100.

To convert 13% to a fraction with a denominator of 100, we need to divide 13 by 100, which gives us 0.13.

To convert 0.125 to a fraction with a denominator of 100, we multiply both the numerator and denominator by 100 to get 12.5/100.

To convert 1/5 to a fraction with a denominator of 100, we multiply the numerator and denominator by 20, which gives us 20/100.

To convert 10% to a fraction with a denominator of 100, we divide 10 by 100, which gives us 0.1.

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We have W = (X1 + ... + X500) - (Y1 + ... + Y500), where X1, Y1, X2, Y2, ... are independent and uniformly distributed in the unit interval [0, 1].

The mean of each X and Y variable is 1/2, and the variance of each variable is 1/12. By linearity of expectation, we have E[W] = E[X1 + ... + X500] - E[Y1 + ... + Y500] = 0, and by independence, Var(W) = Var(X1 + ... + X500) + Var(Y1 + ... + Y500) = 500/12 + 500/12 = 250/6.

By the Central Limit Theorem, we know that the distribution of W is approximately normal with mean 0 and variance 250/6. Therefore, we can standardize W as follows:

Z = (W - E[W]) / sqrt(Var(W)) = W / (sqrt(250/6)).

Then, we can approximate P(W - E[W] < 0.01) as:

P(W < 0.01) = P(Z < 0.01 / sqrt(250/6)).

Using a standard normal table or calculator, we find that P(Z < 0.01 / sqrt(250/6)) is approximately 0.122. Therefore, a good approximation to the probability P(W - E[W] < 0.01) is 0.122.

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Points A and D are 6 units away from the origin. Therefore, the coordinates of point M are (3, 0) and (0, 6).

Given that point M is 6 units away from the origin. We are to find out which pair of the given possible coordinates corresponds to point M. Let the coordinates of point M be (x, y).The distance formula to find the distance between two points, say A(x1, y1) and B(x2, y2) is given by AB=√((x2−x1)²+(y2−y1)²)If point M is 6 units away from the origin, we can write the following equation.6=√((x−0)²+(y−0)²)6²=(x−0)²+(y−0)²36=x²+y²From the given coordinates, we can check each one by substituting their respective values for x and y and see if the resulting equation is true or false.

A (3.0): 36=3²+0² ⟹ 36=9+0 ⟹ 36=9+0 ➡ TrueB. (4,2): 36=4²+2² ⟹ 36=16+4 ⟹ 36=20 ➡ FalseC. (5,5): 36=5²+5² ⟹ 36=25+25 ⟹ 36=50 ➡ FalseD. (0,6): 36=0²+6² ⟹ 36=0+36 ⟹ 36=36 ➡ TrueE. (4,4): 36=4²+4² ⟹ 36=16+16 ⟹ 36=32 ➡ FalseF. (1,5): 36=1²+5² ⟹ 36=1+25 ⟹ 36=26 ➡ FalseTherefore, points A and D are 6 units away from the origin. Therefore, the coordinates of point M are (3, 0) and (0, 6).

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