1 inch = 2.54 centimeters

800 centimeters= _______ inches

please round to nearest tenth please

I have been getting wrong answers

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.

There is a loss of $4,400 on retirement. So option (b) is correct.

To calculate the gain or loss on retirement of the bonds, we need to compare the carrying value of the bonds with the price at which they are being called.

Given:

Par value of the bonds = [tex]$121,000$[/tex]

Carrying value of the bonds = [tex]$109,900$[/tex]

Call price = [tex]$105,500$[/tex]

The gain or loss on retirement is calculated as the difference between the carrying value and the call price.

Loss on retirement = Carrying value - Call price

Substitute the given values:

Loss on retirement = $109,900 - $105,500

Loss on retirement = $4,400

However, since the call price is lower than the carrying value, the loss is incurred by the company. Thus, the correct answer is a [tex]$\$4,400$[/tex] loss.

So, the correct option is: (b) $4,400 loss.

Clabber Company has bonds outstanding with a par value of [tex]$\$ 121,000$[/tex] and a carrying value of [tex]$\$ 109,900$[/tex]. If the company calls these bonds at a price of [tex]$\$ 105,500$[/tex], the gain or loss on retirement is:

Multiple Choice

(a) [tex]$\$ 15,500$[/tex] loss.

(b) [tex]$\$ 4,400$[/tex] loss.

(c) [tex]$\$ 11,100$[/tex] loss.

(d) [tex]$\$ 4,400$[/tex] gain.

The gain on retirement is $15,500.

To calculate the gain or loss on retirement of the bonds, we first need to understand the definitions involved:

1. Par value: This is the face value of the bonds, which is $121,000 in this case.

2. Carrying value: This is the book value of the bonds on the company's balance sheet, which is $109,900.

3. Call price: This is the price at which the company is redeeming (calling) the bonds, which is $105,500.

Now, let's calculate the gain or loss:

- Gain/Loss = Par Value - Call Price

If the call price is less than the carrying value, it results in a gain. If the call price is greater than the carrying value, it results in a loss.

Given:

- Par value = $121,000

- Carrying value = $109,900

- Call price = $105,500

[tex]\[ \text{Gain/Loss} = \text{Par Value} - \text{Call Price} \]\[ \text{Gain/Loss} = \$121,000 - \$105,500 \]\[ \text{Gain/Loss} = \$15,500 \][/tex]

Since the call price is less than the carrying value, it results in a gain.

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:

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:

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

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.

3(-3 + 5x) -1 (4 - 4x)

Simplify each term by first using the distributive property with each set of parenthesis:

3*-3 + 3*5x - 1*4 -1* -4x

Now do the multiplications:

-9 + 15x -4 + 4x

Combine like terms

15x +4x - 9 -4

19x - 13

Answer: The description are as follows:

Step-by-step explanation:

Correlation coefficients is a statistical measure that measures the relationship between the two variables.

(a) r = 1, it means that there is a Perfect positive relationship between the two variables. If there is positive increase in one variable then other variable also increases with a fixed proportion.

(b) r = -1, it means that there is a perfect negative relationship between the two variables. If there is positive increase in one variable then other variable decreases with a fixed proportion.

(c) r = 0, this is a situation which shows that there is no relationship between the two variables.

(d) r = 0.86, this is a situation  which shows that there is a fairly strong positive relationship between the two variables.

(e) r = 0.06, it is nearly zero which represents that either there is a very minor positive relationship between the two variables or there is no relationship between them.

(f) r = -0.89,  this is a situation  which shows that there is a fairly strong negative relationship between the two variables.

Final answer:

The correlation coefficient, r, measures the strength and direction of the linear relationship between x and y values in a sample.

Explanation:

The correlation coefficient, r, measures the strength and direction of the linear relationship between x and y values in a sample. Let's analyze each correlation coefficient:

a. r=1: This indicates a perfect positive correlation between x and y values, meaning that as x increases, y also increases at a constant rate. All the data points fall on a straight line with a positive slope.

b. r=-1: This indicates a perfect negative correlation between x and y values, meaning that as x increases, y decreases at a constant rate. All the data points fall on a straight line with a negative slope.

c. r=0: This indicates no linear relationship between x and y values. The data points are scattered randomly, and there is no consistent pattern or trend between the variables.

d. r=0.86: This indicates a strong positive correlation between x and y values. As x increases, y also increases, but not at a perfect constant rate. The data points approximately fall on a line with a positive slope.

e. r=0.06: This indicates a weak positive correlation between x and y values. As x increases, y also increases, but the relationship is not very strong. The data points have a scattered pattern around a slightly positive sloped line.

f. r=-0.89: This indicates a strong negative correlation between x and y values. As x increases, y decreases, but not at a perfect constant rate. The data points approximately fall on a line with a negative slope.

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]

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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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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Parameterize this surface (call it [tex]S[/tex]) by

[tex]\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(9-u^2)\,\vec k[/tex]

with [tex]0\le u\le3[/tex] and [tex]0\le v\le2\pi[/tex]. Take the normal vector to [tex]S[/tex] to be

[tex]\vec r_u\times\vec r_v=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k[/tex]

Then the area of [tex]S[/tex] is

[tex]\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle\int_0^{2\pi}\int_0^3u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv[/tex]

[tex]=\displaystyle2\pi\int_0^3u\sqrt{1+4u^2}\,\mathrm du=\boxed{\frac{37\sqrt{37}-1}6\pi}[/tex]

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:  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 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:

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:

  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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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:

  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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Answer:

The total result of the investment after 4 years is $395.98

Step-by-step explanation:

Great Question, since we are talking about compounded interest we can use the Exponential Growth Formula to calculate the total value of the investment after 4 years. The Formula is the following,

[tex]y = a*(1+\frac{x}{n})^{nt}[/tex]

Where:

Now we can plug in the values given to us in the question and solve for the total amount (y).

[tex]y = 300*(1+\frac{0.07}{4})^{4*4}[/tex]

[tex]y = 300*(1.0175)^{16}[/tex]

[tex]y = 300*1.3199[/tex]

[tex]y = 395.98 [/tex] ... rounded to the nearest hundredth

Now we can see that the total result of the investment after 4 years is $395.98

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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]

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:

[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

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:94

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:

The correct option is A.

Step-by-step explanation:

The formula of probability is

[tex]P=\frac{a}{b}[/tex]

Where, a≤b, a is total favorable outcomes and b is total possible outcomes.

If an event is certain to occur, then a=b and the probability of an event that is certain to occur is

[tex]P=\frac{a}{a}=1[/tex]

If an event is impossible, then a=0 and the probability of an impossible event is

[tex]P=\frac{0}{a}=0[/tex]

Since total favorable outcomes a and total possible outcomes b can not be negative, a is always less than of equal to b. So,

[tex]0\leq \frac{a}{b}\leq 1[/tex]

[tex]0\leq P\leq 1[/tex]

Therefore the probability of any event is between 0 and 1 inclusive.

All events are not equally likely in any probability procedure. So, the statement "All events are equally likely in any probability procedure" is not true.

Therefore the correct option is A.

The statement 'All events are equally likely in any probability procedure' is NOT a fundamental principle of probability. Events in a probability procedure can have different probabilities based on the situation.

The question is asking to identify which of the given options is NOT a principle of probability. Here, the principles of probability suggest that the probability of an event that is certain to occur is 1 (option B), the probability of an impossible event is 0 (option C), and that the probability of any event is between 0 and 1 inclusive (option D). These are well-established principles of probability and hold true in most situations.

However, option A, 'All events are equally likely in any probability procedure', is NOT a fundamental principle of probability. This is not always true as the likelihood of events can vary greatly depending on the scenario. For instance, if you roll a fair six-sided dice, the probability of landing a 1 is 1/6, but in sampling with or without replacement, probabilities of different outcomes can differ. Thus, it is not a rule that all outcomes are always equally likely in any probabilistic process.

A simple real-life application of this can be seen in card games. If you draw one card from a standard deck of 52, the probability of drawing a heart is 1/4, not equal to the probability of drawing a specific number card like the 7 of clubs, which is 1/52. Therefore, the statement that 'all events are equally likely in any probability procedure' is not always true.

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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.