Consider the system of differential equations dxdt=−4ydydt=−4x. Convert this system to a second order differential equation in y by differentiating the second equation with respect to t and substituting for x from the first equation. Solve the equation you obtained for y as a function of t; hence find x as a function of t. If we also require x(0)=4 and y(0)=5, what are x and y?

Answer:

Step-by-step explanation:94

Answer:  212.88

Step-by-step explanation:

Given : The probability that a daily average over a given month is greater than x = [tex]2.5\%=0.025[/tex]

The probability that corresponds to  0.025 from a Normal distribution table is 1.96.

Mean : [tex]\mu = 109[/tex]

Standard deviation : [tex]\sigma = 53[/tex]

The formula for z-score : -[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

[tex]\Rightarrow\ 1.96=\dfrac{x-109}{53}\\\\\Rightarrow\ x=53\times1.96+109\\\\\Rightarrow\ x=212.88[/tex]

Z scores (converted value in standard normal distribution) can be mapped to probabilities by z tables. The value of x is 212.88 approx.

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

[tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

[tex]P(Z \leq z) = P(Z < z) )[/tex]

Also, know that if we look for Z = z in z tables, the p value we get is

[tex]P(Z \leq z) = \rm p \: value[/tex]

For the given case, let the random variable X tracks the number of dinners at given restaurant. Assuming normal distribution being pertained by X, we get:

[tex]X \sim N(109, 53)[/tex]

The given data shows that:

2.5% of all daily averages records lie bigger than value X = x

or

P(X > x) = 2.5%  0.025

Converting it to standard normal distribution(so that we can use z tables and p values to get the unknown x), we get:

[tex]z = \dfrac{x-\mu}{\sigma} = \dfrac{x - 109}{53}[/tex]

The given probability statement is expressed as:

[tex]P(Z > z) = 2.5\% = 0.025\\P(Z \leq z) = 1 - 0.025 = 0.975[/tex]

Seeing the z tables, we will try to find at what value of z, the p value is obtained near to 0.975

We get z = 1.96.

Thus,

[tex]z = 1.96 = \dfrac{x - 109}{53}\\\\x = 1.96 \times 53} + 109 = 212.88[/tex]

Thus,

The value of x is 212.88 approx.

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Answer:

The test statistic for the above hypothesis test of proportion of fish that are infected is 0.472.

Step-by-step explanation:

[tex]\text{consider the provided information}[/tex]

It is given that the total sample space is 50 fish. Out of which we have found that 6 of the fish sampled are infected. Therefore,

n is 50 and x = 6

[tex]\text{The hypotheses is}[/tex]

[tex]H_0: P=0.10, H_a: P<0.10[/tex]

Now, calculate the sample proportion by dividing the infected sample by sample space as shown:

[tex]\hat{p}=\frac{6}{50}=0.12[/tex]

The standard deviation of proportion can be calculated by using the formula:

[tex]\sigma=\sqrt{\frac{p(1-p)}{n}}[/tex]

[tex]\text{Now substitute the respective values in the above formula}[/tex]

[tex]\sigma=\sqrt{\frac{0.10(1-0.10)}{50}}[/tex]

[tex]\sigma=\sqrt{\frac{0.10(0.9)}{50}}[/tex]

[tex]\sigma=\sqrt{0.0018}[/tex]

[tex]\sigma=0.0424[/tex]

[tex]\text{The test statistic is:}[/tex]

[tex]z=\frac{\hat{p}-p}{\sigma}[/tex]

[tex]\text{Now substitute the respective values in the above formula}[/tex]

[tex]z=\frac{0.12-0.10}{0.0424}[/tex]

[tex]z=\frac{0.02}{0.0424}[/tex]

[tex]z=0.472\ approximately [/tex]

Hence, the test statistic for the above hypothesis test of proportion of fish that are infected is 0.472.

In a hypothesis test for a fishing farm regarding the proportion of fish that are infected, using the given sample size of 50 fish and the number of successes (infected fish) as 6, we calculate a sample proportion of 0.12. In using a normal approximation given the conditions are met, the test statistic (Z-score) is calculated as 0.67 when rounded to two decimal places.

From your question, it seems we are testing a claim about a population proportion in a fishing farm. Here the null hypothesis (H0) would be that p ≥ 0.10 (i.e., 10% or more of the fish are infected) and the alternative hypothesis (H1) would be that p < 0.10 (i.e., less than 10% of the fish are infected).

We're given a sample (n) of 50 fish, of which 6 are infected (successes, x). Since both np and nq are larger than 5 (6 > 5 and 44 > 5), we can use the normal approximation to the binomial distribution. Thus, we calculate the sample proportion (p') as x/n = 6/50 = 0.12.

The test statistic is calculated using the formula Z = (p' - p) / sqrt [ p(1-p) / n ].

Therefore, the test statistic for the above hypothesis test of the proportion of fish that are infected is 0.67.

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Answer:

The slope is: 3

The y-intercept is: [tex]\frac{2}{3}[/tex] or [tex]0.66[/tex]

Step-by-step explanation:

The equation of the line in Slope-Intercept form is:

[tex]y=mx+b[/tex]

Where "m" is the slope of the line and "b" is the y-intercept.

To write the given equation in this form, we need to solve for "y":

[tex]9x - 3y = -2\\\\- 3y = -9x-2\\\\y=3x+\frac{2}{3}[/tex]

Therefore, you can identify that the slope of this line is:

[tex]m=3[/tex]

And the y-intercept is:

[tex]b=\frac{2}{3}=0.66[/tex]

Answer:

The given statement is true.

Step-by-step explanation:

Conferences and conventions are resources that should be explored as part of a job search.

Yes this statement is true.

One can choose the various conferences being held around them through sources like internet, social media, newspapers etc. These are very helpful for job seekers as one can get a lot of job related advice.

Answer:

See below.

Step-by-step explanation:

5/6 is a fraction. The 5 is in the numerator, and the 6 is in the denominator.

The denominator is the number of parts the unit was divided into. In this case, the denominator is 6. That means one unit, 1, was divided into 6 equal parts. Each part is one-sixth.

The numerator is the number of those parts that you use. 5 in the numerator means to use 5 of those parts, each of which is 1/6 of 1.

In other words, 5/6 means divide 1 into 6 equal parts, and take 5 of those parts.

5/6 is the same as 5 divided by 6, so as a decimal it is 0.8333...

As a percent it is 83.333...%

Answer:

x ≥ -2

Step-by-step explanation:

We need to solve 3 x plus 10 greater than or equal to 4

3x + 10 ≥ 4

Solving and finding the value of x

Adding -10 on both sides

3x + 10 -10 ≥ 4 -10

3x ≥ -6

Divide by 3

3x/3 ≥ -6/3

x ≥ -2

So, the solution is x ≥ -2

Answer:

x [tex]\geq[/tex] - 2

Step-by-step explanation:

We need to solve 3 x plus 10 greater than or equal to 4

3x + 10 ≥ 4

Solving and finding the value of x

Adding -10 on both sides

3x + 10 -10 ≥ 4 -10

3x ≥ -6

Divide by 3

3x/3 ≥ -6/3

x ≥ -2

So, the solution is x ≥ -2

Answer:

Given expressions are,

217 x 328 x 11

213 x 345 x 74,

Since, 217 = 7 × 31

328 = 2 × 2 × 2 × 41,

11 = 1 × 11,

So, we can write, 217 x 328 x 11 = 7 × 31 × 2 × 2 × 2 × 41 ×  1 × 11

Now, 213 = 3 × 71

345 = 3 × 5 × 23,

74 = 2 × 37,

So, 213 x 345 x 74 = 3 × 71 × 3 × 5 × 23 × 3 × 5 × 23

Thus, GCF ( greatest common factor ) of the given expressions = 1 ( because there are no common factors )

We know that if two numbers have GCF 1 then their LCM is obtained by multiplying them,

Hence, LCM ( least common multiple ) of the given expressions = 217 x 328 x 11 x 213 x 345 x 74

Answer:

[tex]y=\frac{2x}{9}-2[/tex]

Step-by-step explanation:

The given equation is:

2x-9y=18

To solve for y means we need to isolate y on one side of the equation, carrying all the other variables and terms to the other side so that we get a formula for y. This can be done as shown below:

2x - 9y = 18

Subtracting 2x from both sides, we get:

-9y = 18 - 2x

Dividing both sides by -9, we get:

[tex]\frac{-9y}{-9}=\frac{18}{-9}-\frac{2x}{-9}\\\\ y=-2+\frac{2x}{9}\\\\ y=\frac{2x}{9}-2[/tex]

Answer:

[tex] y = \frac { 2 ( x - 9 ) } { 9 } [/tex]

Step-by-step explanation:

We are given the following expression and we are to solve it for the indicated variable (y):

[tex] 2 x - 9 y = 1 8 [/tex]

Making [tex] y [/tex] the subject of the equation and simplifying it to get:

[tex] 2 x - 1 8 = 9 y \\\\ 9 y = 2 x - 1 8 \\\\ y = \frac { 2 x - 1 8 } { 9 } \\\\ y = \frac { 2 ( x - 9 ) } { 9 } [/tex]

Answer:

Standard deviation is 7.16

Step-by-step explanation:

We have given a set of numbers :

73, 76, 79, 82, 84, 84, 97

To calculate the standard deviation of the given data set, first we have to work out the mean.

Mean = [tex]\frac{(73+76+79+82+84+84+97)}{7}[/tex]

         = [tex]\frac{575}{7}[/tex] = 82.14

Now for each number subtract the mean and square the result

(73 - 82.14)²  = (-9.14)² = 83.54

(76 - 82.14)² =  (-6.14)² = 37.70

(79 - 82.14)² = (-3.14)²  = 9.86

(82 - 82.14)² = (0.14)²  = 0.02

(84 - 82.14)² = (1.86)²  = 3.46

(84 - 82.14)² = (1.86)²  = 3.46

(97 - 82.14)² = (14.86)²= 220.82

Now we calculate the mean from of those squared differences :

Mean = [tex]\frac{83.54+37.70+9.86+0.02+3.46+3.46+220.82}{7}[/tex]

         = [tex]\frac{358.86}{7}[/tex]

         = 51.27

Now square root of this mean = standard deviation = √51.27 = 7.16

Therefore, Standard deviation is 7.16

The ODE is exact because [tex]M_y=N_x[/tex]. Then

[tex]F_x=8x+8y\implies F(x,y)=4x^2+8xy+g(y)[/tex]

[tex]F_y=8x+g'(y)=8x+4y\implies g'(y)=4y\implies g(y)=2y^2+C[/tex]

So we have

[tex]F(x,y)=4x^2+8xy+2y^2=C[/tex]

Final answer:

The given differential equation is exact because the partial derivatives My and Nx are equal, both being 8. Integrating and finding the potential function F(x, y), we get F(x, y) = 4x² + 8xy + 2y².

Explanation:

The equation given is (8x + 8y)dx + (8x + 4y)dy = 0. To determine if this differential equation is exact, we will compare the partial derivatives of M with respect to y (My) and N with respect to x (Nx). In this case, we have M(x, y) = 8x + 8y and N(x, y) = 8x + 4y. Calculating the partial derivatives, we get My = 8 and Nx = 8. Since My equals Nx, the differential equation is exact, implying there is a function F(x, y) such that its differential dF = M dx + N dy.

To find F(x, y), we integrate M with respect to x, yielding F(x, y) = 4x² + 8xy + h(y), where h(y) is an arbitrary function of y. To determine h(y), we differentiate F(x, y) with respect to y and equate it to N: Fy = 8x + h'(y) = 8x + 4y. From this equation, it follows that h'(y) = 4y, so integrating with respect to y gives h(y) = 2y². Therefore, the potential function F(x, y) that makes the differential exact is F(x, y) = 4x² + 8xy + 2y².

Answer:

  17, 1

Step-by-step explanation:

You can find the yards by multiplying by the conversion factor, then determining what the fraction or remainder means. Since there are 3 ft in 1 yd, 1/3 yd is 1 ft.

  52 ft = (52 ft) × (1 yd)/(3 ft) = 52/3 yd = 17 1/3 yd = 17 yd 1 ft

To convert feet to yards, you divide by 3. Therefore, 52 feet is equivalent to 17 yards and 1 foot.

In mathematics, particularly in the subject of measurement conversions, it is useful to know that 1 yard (yd) is equal to 3 feet (ft). So, to convert from feet to yards, you divide the number of feet by 3. Applying this principle to the question, you will take the 52 feet and divide it by 3.

52 feet ÷ 3 = 17.3 repeating

However, since the options provided do not include a decimal, we take the whole number part of the answer, which is 17 yards. There is a remainder when you divide 52 by 3, which indicates there are additional feet not making up a full yard. In this case, it is 1 foot (the decimal part times 3), making the final conversion 17 yards 1 foot.

So, 52 ft equals 17 yards and 1 foot.

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The confidence interval for mean of the "before-after" differences is [tex]\fbox{(-0.4037,1.8037)}[/tex]

Further explanation:

Find the difference between the before pain and the after pain.

Difference = before-after

Kindly refer to the Table for the difference of between the before and after pain.

Sum of difference = [tex]5.6[/tex]

Total number of observation = [tex]8[/tex]

Mean of difference = [tex]0.7[/tex]

Sample standard deviation [tex]s[/tex] = [tex]1.3201[/tex]

Level of significance = [tex]5\%[/tex]

Formula for confidence interval = [tex]\left( \bar{X} \pm t_{n-1, \frac{\alpha}{2}\%} \frac{s}{\sqrt{n}} \right)[/tex]

confidence interval = [tex]\left( 0.7 \pm t_{8-1, \frac{5}{2}\%} \frac{1.3201}{\sqrt{8}} \right)[/tex]

confidence interval = [tex]\left( 0.7 \pm t_{7, \frac{5}{2}\%} \frac{1.3201}{\sqrt{8}} \right)[/tex]

From the t-table.

The value of [tex]t_{7, \frac{5}{2}\%[/tex]=[tex]2.365[/tex]

Confidence interval = [tex]( 0.7 \pm 2.365}\times \frac{1.3201}{\sqrt{8}}) \right)[/tex]

Confidence interval = [tex]\left( 0.7 - 2.365}\times \0.4667,0.7 + 2.365}\times \0.4667) \right[/tex]

Confidence interval = [tex](0.7-1.1037,0.7+1.1037)[/tex]

Confidence interval = [tex]\fbox{(-0.4037,1.8037)}[/tex]

The [tex]95\%[/tex] confidence interval tells us about that [tex]95\%[/tex] chances of the true mean or population mean lies in the interval.

Yes, the hypnotism appear to be effective in reducing pain as confidence interval include includes the positive deviation from the mean.

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Answer Details:

Grade: College Statistics

Subject: Mathematics

Chapter: Confidence Interval

Keywords:

Probability, Statistics, Speed dating, Females rating, Confidence interval, t-test, Level of significance , Normal distribution, Central Limit Theorem, t-table, Population mean, Sample mean, Standard deviation, Symmetric, Variance.

Answer:

The 95% confidence interval for the mean of the “before-after” difference is (-0.4039,1.8039)

No, Hypnotism doesn’t appear to be effective in reducing pain.

Further explanation:

Given: The table of measure the effectiveness of hypnotism in reducing pain.

Before : 6.4    2.6    7.7   10.5    11.7    5.8    4.3    2.8

After    : 6.7    2.4    7.4     8.1     8.6    6.4    3.9    2.7

we make the table of difference between “before-after”

(Before-After) :

6.4-6.7  2.6-2.4  7.7-7.4  10.5-8.1  11.7-8.6  5.8-6.4  4.3-3.9  2.8-2.7

   -0.3        0.2       0.3         2.4         3.1        -0.6       0.4         0.1

Now, we find the sample mean and sample standard deviation of above table.

[tex]\text{Sample Mean, }\bar{x}=\dfrac{\text{Sum of number}}{\text{number of observation}}[/tex]

[tex]=\dfrac{-0.3+0.2+0.3+2.4+3.1-0.6+0.4+0.1}{8}[/tex]

[tex]=\dfrac{5.6}{8}=0.7[/tex]

[tex]\text{Sample Standard deviation, s} = \sqrt{\dfrac{(x-\bar{x})^2}{n-1}}[/tex]

[tex]=\sqrt{\dfrac{(-0.3-0.7)^2+(0.2-0.7)^2+(0.3-0.7)^2 ...+(0.1-0.7)^2}{8-1}}[/tex]

[tex]=\dfrac{12.2}{7}=1.3201[/tex]

For 95% confidence interval [tex]\mu_d[/tex] using t-distribution

[tex]\text{Marginal Error, E}=t_{\frac{\alpha}{2},df}\times \dfrac{s}{\sqrt{n}}[/tex]

Where,

For critical value,

[tex]\Rightarrow t_{\frac{\alpha}{2},df}[/tex]

[tex]\Rightarrow t_{\frac{0.05}{2},7}[/tex]

[tex]\Rightarrow t_{0.025,7}[/tex]

using t-distribution two-tailed table,

[tex]t_{0.025,7}=2.365[/tex]

Substitute the values into formula and calculate E

[tex]E=2.65\times \dfrac{1.301}{\sqrt{8}}[/tex]

Therefore, Marginal error, E=1.1039

95% confidence interval given by:

[tex]=\bar{x}\pm E[/tex]

[tex]=0.7\pm 1.1039[/tex]

Therefore, 95% confidence interval using t-distribution: (-0.4039,1.8039)

This interval contains 0

Therefore, Hypnotism doesn’t appear to be effective in reducing pain.  

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Keywords:  

T-distribution, Sample mean, sample standard deviation, Critical value of t, degree of freedom, t-test, confidence interval, significance level.  

Answer:

Step-by-step explanation:

Given x,y,a&b are greater than zero

also [tex]\frac{x}{y}[/tex][tex]\leq [/tex][tex]\frac{a}{b}[/tex]

since x,y,a&b are greater than zero therefore we can cross multiply them without changing the inequality

therefore

[tex]\frac{x}{a}[/tex][tex]\leq [/tex][tex]\frac{y}{b}[/tex]

adding 1 on both sides we get

[tex]\frac{x}{a}[/tex]+1[tex]\leq [/tex][tex]\frac{y}{b}[/tex]+1

[tex]\frac{x+a}{a}[/tex][tex]\leq [/tex][tex]\frac{y+b}{b}[/tex]

rearranging

[tex]\frac{x+a}{y+b}[/tex][tex]\leq [/tex][tex]\frac{a}{b}[/tex]

Answer:

Step-by-step explanation:

Let's simplify step-by-step.

4x3−6x2−24x+(7)(−23)

=4x3+−6x2+−24x+−161

Answer:

=4x3−6x2−24x−161

Answer:

56.25 feet.

Step-by-step explanation:

h(t) = 60t - 16t^2

Differentiating to find the velocity:

v(t) = 60 -32t

This  equals zero when  the ball reaches its maximum height, so

60-32t = 0

t = 60/32 = 1.875 seconds

So the maximum height is  h(1.875)

= 60* 1.875 - 16(1.875)^2

= 56.25 feet.

Answer: 56.25 feet.

Step-by-step explanation:

For a Quadratic function in the form [tex]f(x)=ax^2+bx+c[/tex], if [tex]a<0[/tex] then the parabola opens downward.

Rewriting the given function as:

[tex]h(t) = - 16t^2+60t[/tex]

You can identify that [tex]a=-16[/tex]

Since [tex]a<0[/tex] then the parabola opens downward.

Therefore, we need to find the vertex.

Find the x-coordinate of the vertex with this formula:

[tex]x=\frac{-b}{2a}[/tex]

Substitute values:

[tex]x=\frac{-60}{2(-16)}=1.875[/tex]

Substitute the value of "t" into the function to find the height in feet that the ball will reach. Then:

 [tex]h(1.875)=- 16(1.875)^2+60(1.875)=56.25ft[/tex]

Answer:

The probability of getting the toy in any given cereal box is [tex]\frac{1}{49}[/tex].

Step-by-step explanation:

Given,

On average, we get a toy in every 49th cereal box,

That is, in every 49 boxes there is a toy,

So, the total outcomes = 49,

Favourable outcomes ( getting a toy ) = 1

Since, we know that,

[tex]\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex]

Hence, the probability of getting the toy in any given cereal box = [tex]\frac{1}{49}[/tex]

Answer:

The probability of getting the toy you want in any given cereal box is of 0.0204 = 2.04%.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected number of trials for r sucesses is:

[tex]E = \frac{r}{p}[/tex]

If, on average, you get the toy you want in every 49th cereal box, what is the probability of getting the toy you want in any given cereal box?

This means that [tex]E = 49, r = 1[/tex]

So

[tex]49 = \frac{1}{p}[/tex]

[tex]49p = 1[/tex]

[tex]p = \frac{1}{49}[/tex]

[tex]p = 0.0204[/tex]

The probability of getting the toy you want in any given cereal box is of 0.0204 = 2.04%.

Answer:

After 5.31 months sales will drop below 1000 and 5 months after the end of the campaign sales will be 1282.09

Step-by-step explanation:

Let's find the solutions for the two questions.

First question: How many months after the end of the campaign will sales drop below 1000.

Because the problem asks for how many months, and since 'x' represents month variable, then the problem is asking for 'x'.

Using the same equation for sales we can observe the following:

[tex]S=70000*e^{-0.8X}[/tex], but we have S which is 1000, so:

[tex]1000=70000*e^{-0.8X}[/tex] which is equal to:

[tex]1000/70000=e^{-0.8X}[/tex] which is equal to:

[tex]1/70=e^{-0.8X}[/tex] by applying ln(x) properties:

[tex]ln(1/70)=ln(e^{-0.8X})[/tex] which is equal to:

[tex]ln(1/70)=-0.8X[/tex] which is equal to:

[tex]ln(1/70)/(-0.8)=X[/tex] so:

[tex]X=5.31 months[/tex]

Second question: what will be the sales 5 months after the end of the campaign.

Because the problem asks for what will be the sales, and since 'S' represents the sales, then the problem is asking for 'S'.

Using the same equation for sales we can observe the following:

[tex]S=70000*e^{-0.8X}[/tex], but we have x which is 5 months, so:

[tex]S=70000*e^{-0.8*5}[/tex] which is equal to

[tex]S=1282.09[/tex]

In conclusion, after 5.31 months sales will drop below 1000 and 5 months after the end of the campaign sales will be 1282.09.

[tex]2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0[/tex]

Divide both sides by [tex]x^2\,\mathrm dx[/tex] to get

[tex]2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}[/tex]

Substitute [tex]v(x)=\dfrac{y(x)}x[/tex], so that [tex]\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}[/tex]. Then

[tex]x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}[/tex]

[tex]x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}[/tex]

[tex]x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}[/tex]

The remaining ODE is separable. Separating the variables gives

[tex]\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x[/tex]

Integrate both sides. On the left, split up the integrand into partial fractions.

[tex]\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}[/tex]

[tex]\implies v^2+2v+1=a(v^2+1)+(bv+c)v[/tex]

[tex]\implies v^2+2v+1=(a+b)v^2+cv+a[/tex]

[tex]\implies a=1,b=0,c=2[/tex]

Then

[tex]\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v[/tex]

On the right, we have

[tex]\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C[/tex]

Solving for [tex]v(x)[/tex] explicitly is unlikely to succeed, so we leave the solution in implicit form,

[tex]\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C[/tex]

and finally solve in terms of [tex]y(x)[/tex] by replacing [tex]v(x)=\dfrac{y(x)}x[/tex]:

[tex]\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C[/tex]

[tex]\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C[/tex]

[tex]\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}[/tex]

Answer:

[tex]\overrightarrow{r}=2\widehat{i}-0.75\widehat{j}[/tex]

Step-by-step explanation:

Stagnation point is point of zero velocity thus each component of the velocity must also be zero

[tex]Given\\u=(1-0.5x)[/tex]

[tex]\therefore[/tex] u=0 at x=2m

Similarly

[tex]\\v=(-1.5-2y)\\\\\therefore v=0 \\\\y=\frac{1.5}{-2}\\\\y=-0.75[/tex]

Thus point of stagnation is (2,-0.75)

Thus it's position vector is given by

[tex]\overrightarrow{r}=2\widehat{i}-0.75\widehat{j}[/tex]

Answer:

  7.86 km

Step-by-step explanation:

Let x represent the distance point P lies east of the refinery. (We assume this direction is downriver from the refinery.)

The cost of laying pipe to P from the refinery (in millions of $) will be ...

  0.5√(1² +x²)

The cost of laying pipe under the river from P to the storage facility will be ...

  1.0√(2² +(9-x)²) = √(85 -18x +x²)

We want to minimize the total cost c. That total cost is ...

  c = 0.5√(x² +1) +√(x² -18x +85)

The minimum value is best found using technology. (Differentiating c with respect to x results in a messy radical equation that has no algebraic solution.) A graphing calculator shows it to be at about x ≈ 7.86 km.

Point P should be located about 7.86 km downriver from the refinery.

The cost minimization problem of constructing a pipeline from a refinery to the storage tanks on the other side of the river can be solved by differential calculus. The distance of point P down the river from the refinery can be determined by differentiating the total cost function and equating it to zero to minimize the cost.

This is a math problem related to cost minimization and deals with the principles of trigonometry. Let's denote the distance downriver from the refinery to point P as x. Since the refinery, point P, and the oil tank form a right triangle, we can apply the Pythagorean theorem. The length of the pipeline from point P to the tanks is the hypotenuse of the triangle, which is sqr((9-x)^2+2^2). Hence, the total cost of the pipeline is C = $500,000*x + $1,000,000*sqr((9-x)^2+2^2).

To find the minimum cost, we need to differentiate the total cost function C with respect to x and equate it to zero. By solving the derivative of C equals to zero, we can find the optimal distance x value. Therefore, solving this derivative can give us the solution in terms of miles.

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Answer:

a.[tex]y(x)=C_1x^{-1}+C_2x^{-8}+\frac{1}{30}x^2[/tex]

b.[tex]y(x)=x^2(C_1cos (3lnx)+C_2sin(3lnx))+\frac{4}{13}+\frac{3}{10}x[/tex]

Step-by-step explanation:

1.[tex]x^2y''+10xy'+8y =x^2[/tex]

It is Cauchy-Euler equation where [tex]x=e^t[/tex]

Then auxillary equation

[tex]D'(D'-1)+10D'+8=0[/tex]

[tex]D'^2+9D'+8=0[/tex]

[tex](D'+1)(D'+8)=0[/tex]

D'=-1 and D'=-8

Hence, C.F=[tex]C_1e^{-t}+C_2e^{-8t}[/tex]

C.F=[tex]C_1\frac{1}{x}+C_2\frac{1}{x^8}[/tex]

P.I=[tex]\frac{e^{2t}}{D'^2+9D'+8}=\frac{e^{2t}}{4+18+8}[/tex]

Where D'=2

[tex]P.I=\frac{1}{30}e^{2t}=\frac{1}{30}x^2[/tex]

[tex]y(x)=C_1x^{-1}+C_2x^{-8}+\frac{1}{30}x^2[/tex]

b.[tex]x^2y''-3xy'+13y=4+3x[/tex]

Same method apply

Auxillary equation

[tex] D'^2-D'-3D'+13=0[/tex]

[tex]D'^2-4D'+13=0[/tex]

[tex]D'=2\pm3i[/tex]

C.F=[tex]e^{2t}(C_1cos 3t+C_2sin 3t)[/tex]

C.F=[tex]x^2(C_1cos (3lnx)+C_2sin(3lnx))[/tex]

[tex]e^t=x[/tex]

P.I=[tex]\frac{4e^{0t}}{D'^2-4D'+13}+3\frac{e^t}{D'^2-4D'+13}[/tex]

Substitute D'=0 where[tex] e^{0t}[/tex] and D'=1 where [tex]e^t[/tex]

P.I=[tex]\frac{4}{13}+\frac{3}{10}e^t[/tex]

P.I=[tex]\frac{4}{13}+\frac{3}{10}x[/tex]

[tex]y(x)=C.F+P.I=x^2(C_1cos (3lnx)+C_2sin(3lnx))+\frac{4}{13}+\frac{3}{10}x[/tex]

Refer the attachment...

Hope it helps you...

Answer:

ABDE is a parallelogram.

CE=2.7 cm  Reason:  The diagonals bisect each other. CE=AC.

DC=3.08 cm Reason: The diagonals bisect each other. This is half of DB. Just like CB is half of DB.

mAngleABE=104 degrees Reason: Opposite angles are congruent in a parallelogram.

mAngleDEB=76 degrees Reason: Consecutive angles in a parallelogram are supplementary.

Step-by-step explanation:

You have a parallelogram because both of your pairs of opposite sides are parallel.

CE=2.7 cm because CE is congruent to AC.  The diagonals of a parallelogram bisect each other.  Bisect means to cut into equal halves.

DC=3.08 cm because DC is congruent to CB which means the measurement of DC is half the length of DB.

mAngleABE=mAngleADE because opposite angles of a parallelogram are congruent. So the mAgnleABE=104 degrees.

AngleDEB is supplementary to AngleADE because they are consecutive angles in a parallelogram. This means mAngleDEB=180-104=76 degrees.

Answer:

The value of [tex]x_2=-1.6923[/tex].

Step-by-step explanation:

Consider the provided information.

The provided formula is [tex]f(x)=x^3+x+6[/tex]

Substitute [tex]x_1=-2[/tex] in above equation.

[tex]f(x_1)=(-2)^3+(-2)+6[/tex]

[tex]f(x_1)=-8-2+6[/tex]

[tex]f(x_1)=-4[/tex]

Differentiate the provided function and calculate the value of [tex]f'(x_1)[/tex]

[tex]f'(x)=3x^2+1[/tex]

[tex]f'(x)=3(-2)^2+1[/tex]

[tex]f'(x)=13[/tex]

The Newton iteration formula: [tex]x_2=x_1-\frac{f(x_1)}{f'(x_1)}[/tex]

Substitute the respective values in the above formula.

[tex]x_2=-2-\frac{(-4)}{13}[/tex]

[tex]x_2=-2+0.3077[/tex]

[tex]x_2=-1.6923[/tex]

Hence, the value of [tex]x_2=-1.6923[/tex].

To use Newton's method with an initial approximation of -2 on the equation x3 + x + 6 = 0, we identify the function and its derivative. We then substitute into Newton's method's formula, x_2 = x_1 - f(x_1) / f'(x_1). This will provide us with an approximate second root, which can then be refined further.

The subject of the question pertains to Newton's method for root finding, which is a method for finding successively better approximations to the roots (or zeroes) of a real-valued function. The function in question is x3 + x + 6 = 0 and the initial approximation provided is -2.

First, in order to use Newton's method, we need to identify the function and its derivative. The function (f(x)) is x3 + x + 6. The derivative (f'(x)) would be 3x2 + 1.

Newton's method follows the formula: x_(n+1) = x_n - f(x_n) / f'(x_n).

With x_1 as -2, we substitute into the Newton method's formula to find x_2. Hence, x_2 = -2 - f(-2) / f'(-2), which results in an approximation of the root that can be further refined. Remember to round your answer to four decimal places after calculations.

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Answer:

A(t) = 300 -260e^(-t/50)

Step-by-step explanation:

The rate of change of A(t) is ...

A'(t) = 6 -6/300·A(t)

Rewriting, we have ...

A'(t) +(1/50)A(t) = 6

This has solution ...

A(t) = p + qe^-(t/50)

We need to find the values of p and q. Using the differential equation, we ahve ...

A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50

0 = 6 -p/50

p = 300

From the initial condition, ...

A(0) = 300 +q = 40

q = -260

So, the complete solution is ...

A(t) = 300 -260e^(-t/50)

___

The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.

The number of grams of salt in the tank at any time t is 40 grams. The inflow and outflow of brine do not affect the amount of salt in the tank because the solution is well-mixed, and the salt concentration remains constant.

To solve this problem, we need to set up a differential equation that describes the rate of change of salt in the tank over time. Let A(t) represent the number of grams of salt in the tank at time t.

Let's break down the components affecting the rate of change of salt in the tank:

Salt inflow rate: The brine is being pumped into the tank at a constant rate of 6 liters per minute, and it contains 1 gram of salt per liter. So, the rate of salt inflow is 6 grams per minute.

Salt outflow rate: The solution in the tank is being pumped out at the same rate of 6 liters per minute, which means the rate of salt outflow is also 6 grams per minute.

Mixing of the solution: Since the tank is well-mixed, the concentration of salt remains uniform throughout the tank.

Now, let's set up the differential equation for A(t):

dA/dt = Rate of salt inflow - Rate of salt outflow

dA/dt = 6 grams/min - 6 grams/min

dA/dt = 0

The above equation shows that the rate of change of salt in the tank is constant and equal to zero. This means the number of grams of salt in the tank remains constant over time.

Now, let's find the constant value of A(t) using the initial condition where the tank initially contains 40 grams of salt.

When t = 0, A(0) = 40 grams

Since the rate of change is zero, A(t) will be the same as the initial amount of salt in the tank at any time t:

A(t) = 40 grams

So, the number of grams of salt in the tank at any time t is 40 grams. The inflow and outflow of brine do not affect the amount of salt in the tank because the solution is well-mixed, and the salt concentration remains constant.

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Answer:

Below.

Step-by-step explanation:

To prove this for mathematical induction, we will need to prove:

Part 1)  That [tex]7^n-1[/tex] is a multiple of 6 for n=1.

Part 2) That, if by assuming [tex]7^{n}-1[/tex] is a multiple of 6, then showing [tex]7^{n+1}-1[/tex] is a multiple of 6.

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

Part 1) If n=1, we have [tex]7^n-1=7^1-1=7-1=6[/tex] where 6 is a multiple of 6 since 6 times 1 is 6.

Part 2) A multiple of 6 is the product of 6 and k where k is an integer.  So let's assume that there is a value k such that [tex]7^n-1=6k[/tex] for some number natural number [tex]n[/tex].

We want to show that [tex]7^{n+1}-1[/tex] is a multiple of 6.

[tex]7^{n+1}-1[/tex]

[tex]7^n7^1-1[/tex]

[tex](7)7^n-1[/tex]

[tex](7)7^{n}-7+6[/tex]

[tex]7(7^{n}-1)+6[/tex]

[tex]7(6k)+6[/tex] (this is where I applied my assumption)

[tex]6[7k+1][/tex] (factoring with the distributive property)

Since 7k+1 is an integer then 6(7k+1) means that [tex]7^{n+1}-1[/tex] is a multiple of 6.

This proves that [tex]7^n-1[/tex] is a multiple of 6 for all natural n.

The p​-value in a test of the claim that the mean College Algebra final exam score is 0.1336.

Given data:

To find the p-value in a test of the claim that the mean College Algebra final exam score of engineering majors is equal to 88, use the test statistic and the standard normal distribution.

The test statistic z = 1.50 represents how many standard deviations the sample mean is away from the hypothesized population mean.

The p-value is the probability of observing a test statistic as extreme or more extreme than the observed test statistic, assuming the null hypothesis is true.

Since the alternative hypothesis is not specified, assume a two-tailed hypothesis.

To find the p-value, we need to calculate the probability of observing a test statistic as extreme or more extreme than z = 1.50 in a standard normal distribution.

For a two-tailed test, we will find the probability in both tails.

The probability in the right tail is given by:

P(Z > 1.50) = 1 - P(Z < 1.50)

Using a standard normal distribution table, we find that P(Z < 1.50) is approximately 0.9332.

Therefore, P(Z > 1.50) = 1 - 0.9332 = 0.0668.

To find the p-value for the left tail, use the symmetry of the standard normal distribution.

P(Z < -1.50) = P(Z > 1.50) = 0.0668.

Since this is a two-tailed test, sum the probabilities of both tails to find the p-value:

p-value = 2 * 0.0668

p-value = 0.1336.

Hence, the p-value in this test is approximately 0.1336.

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In the given question, the p-value for a test statistic (z) of 1.50 can be computed as 0.0668. This means that if the null hypothesis is true (the mean final exam score being 88), there is a 6.68 percent chance we would observe a test statistic this extreme or more. This p-value is computed by subtracting the left tailed probability for z=1.50 from 1.

The p-value (Probability Value) is a statistical measure used in hypothesis testing to determine the significance of the obtained results. It denotes the probability of obtaining the observed sample data given that the null hypothesis is true. In this case, we are interested in the p-value associated with a test statistic z=1.50 under the claim that the mean College Algebra final exam score for engineering majors is 88.

To find this p-value, we reference a standard normal (z) distribution table or use statistical software. Look up the value corresponding to z=1.50 which will give us the cumulative probability P(Z ≤ 1.50). However, since we want P(Z > 1.50), we subtract the obtained value from 1. This is due to the fact that the total probability under the curve of the standard normal distribution equals 1.

For a z score of 1.50 the standard z table gives a left tailed probability of approximately 0.9332. Therefore, P(Z > 1.50) = 1 - P(Z ≤ 1.50) = 1 - 0.9332 = 0.0668. So the p-value is approximately 0.0668.

The interpretation of this value would be: if the null hypothesis is true (the mean final exam score is 88), then there is a 0.0668 probability, or 6.68 percent, that we would observe a test statistic greater than or equal to 1.50.

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a. Since [tex]i=e^{i\pi/2}[/tex], we have

[tex]i^i=(e^{i\pi/2})^i=e^{i^2\pi/2}=e^{-\pi/2}[/tex]

b. Since [tex]1=e^0[/tex], we have

[tex]-1^{-i}=-(e^0)^{-i}=-(e^0)=-1[/tex]

c. Yes, for the reason illustrated in part b. [tex]1=e^0[/tex], and raising this to any power [tex]z\in\mathbb C[/tex] results in [tex]e^{0z}=e^0=1[/tex].

Answer: The seasonal relatives is calculated are as follows:

Step-by-step explanation:

Given that,

restaurant only open from Wednesday to Saturday,

∴ The remaining 19% of its business he does on Wednesday

Now, suppose that total production of sales in a given week be 'y'

So, average sales in a week = [tex]\frac{y}{4}[/tex]

If we assume that y = 1

hence, average sales in a week = [tex]\frac{1}{4}[/tex]

= 0.25

Now, we have to calculate the seasonal relatives,

that is,

=  [tex]\frac{Sales in a given day}{average sales in a week}[/tex]

Wednesday:

= [tex]\frac{0.19}{0.25}[/tex]

= 0.76

Thursday:

= [tex]\frac{0.21}{0.25}[/tex]

= 0.84

Friday:

= [tex]\frac{0.29}{0.25}[/tex]

= 1.16

Saturday:

= [tex]\frac{0.31}{0.25}[/tex]

= 1.24

- Wednesday:0.76

- Thursday:0.84

- Friday:1.16

- Saturday:1.24

To determine the seasonal relatives for each night, we need to express the business done each night as a percentage of the total business for the week. The given percentages are:

- Friday: 29%

- Saturday: 31%

- Thursday: 21%

First, let's find the total percentage accounted for by Wednesday, Thursday, Friday, and Saturday. Since we're missing Wednesday's percentage, we can sum the given percentages and subtract from 100%.

[tex]\[\text{Total percentage} = 29\% + 31\% + 21\% = 81\%\][/tex]

The remaining percentage for Wednesday is:

[tex]\[\text{Wednesday's percentage} = 100\% - 81\% = 19\%\][/tex]

Now, we'll convert these percentages into seasonal relatives. Seasonal relatives are the ratios of each night's business to the average nightly business across the four nights.

First, compute the average nightly business percentage:

[tex]\[\text{Average nightly business percentage} = \frac{100\%}{4} = 25\%\][/tex]

Next, calculate the seasonal relatives by dividing each night's percentage by the average nightly business percentage:

1. Wednesday:

[tex]\[ \text{Wednesday's seasonal relative} = \frac{19\%}{25\%} = 0.76 \][/tex]

2. Thursday:

[tex]\[ \text{Thursday's seasonal relative} = \frac{21\%}{25\%} = 0.84 \][/tex]

3. Friday:

 [tex]\[ \text{Friday's seasonal relative} = \frac{29\%}{25\%} = 1.16 \][/tex]

4. Saturday:

[tex]\[ \text{Saturday's seasonal relative} = \frac{31\%}{25\%} = 1.24 \][/tex]

Rounded to two decimal places, the seasonal relatives are:

- Wednesday:0.76

- Thursday:0.84

- Friday:1.16

- Saturday:1.24

Answer: Option 'e' is correct.

Step-by-step explanation:

Qualitative data refers to those data which does not include any numerical value.

Average rainfall on 25 days of the month is a quantitative data as it would represented as numerical value.

Number of accidents occurred in the month of September is quantitative data.

Height of 40 students in a class is quantitative data too.

Weight of 35 students in a class is quantitative data too.

But political affiliation of 2250 randomly selected voters is qualitative data as it has not included any numerical value.

Hence, option 'e' is correct.

Option e, 'Political affiliation of 2,250 randomly selected voters,' represents qualitative data because it categorizes voters based on a non-numeric characteristic, their political affiliation.

The question 'Which of the following is an example of qualitative data? e. Political affiliation of 2,250 randomly selected voters' is asking us to identify a type of data among the given options. Qualitative data refers to information that is categorized based on attributes or qualities rather than numerical values. In contrast, quantitative data involve numbers and can either be discrete, which are countable, or continuous, which can take on any value within a range.

To answer the question, option e, 'Political affiliation of 2,250 randomly selected voters,' represents qualitative data because it describes a characteristic (political affiliation) that is non-numeric and is usually expressed in words or categories. Other options such as average rainfall, number of accidents, weight, and height involve numerical measurements and are therefore examples of quantitative data, which can either be discrete or continuous depending on the nature of the measurement.

Option e is unique as the only example of qualitative data in the options provided, because political affiliation does not result from a measurement or count, but rather, it is a category used to describe a voter's preferred political party or stance.