The weight (in pounds) of ``medium-size'' watermelons is normally distributed with mean 15 and variance 4. A packing container for several melons has a nominal capacity of 140 pounds. What is the maximum number of melons that should be placed in a single packing container if the nominal weight limit is to be exceeded only 5\% of the time? Use the results of question 8 to help you answer this one.

Final answer:

The expected number of girls in a family with five children, given a 51% probability of having a girl, is 2.6 when rounded to the nearest tenth.

Explanation:

The question involves calculating the expected number of girls in a family with five children, given that the probability of having a girl is 51%. The expected value in probability theory is calculated by multiplying each possible outcome by its probability and then summing these values. In this case, since there are five children and each child has a 51% chance of being a girl, the expected number of girls can be found using the formula for the expectation:

Expected number of girls = Total number of children × Probability of having a girl

Therefore, the expected number of girls = 5 × 0.51 = 2.55. Rounding this to the nearest tenth, we get 2.6 girls.

To construct a confidence interval for the difference between two proportions, you use an unbiased estimate of the difference between the two proportions.

When constructing a confidence interval for the difference between two proportions, an unbiased estimate of the difference between the two proportions is obtained using the formula:

(p1 - p2) ± ME

Where:

The margin of error is calculated as:

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

Where:

Using this formula, you can calculate the confidence interval for the difference between two proportions.

Answer:

a. 20.23

b. 14.1

c. 8.3, 9.2, 9.8, 12.1, 13.5, 14.7, 26.5, 34.7, 36.5, 37

d. 22.65

Step-by-step explanation:

Sum = 202.3

Count = 10

Mean = 202.3 / 10

Mean = 20.23

Since there is an even number of data values the median is the mean of the two data values in the middle, calculated as follows;

8.3, 9.2, 9.8, 12.1, 13.5, 14.7, 26.5, 34.7, 36.5, 37

Median = (13.5 + 14.7) / 2

Median = 14.1

Mode is the value or values in the data set that occur most frequently. Since all are occurred once so all the values will be considered as Mode;

Mode = 8.3, 9.2, 9.8, 12.1, 13.5, 14.7, 26.5, 34.7, 36.5, 37

Midrange = (Highest Value + Lowest Value) / 2

Midrange = (37 + 8.3) / 2

Midrange = 22.65

Answer:

Option D) The prisoner is actually guilty and the jury sets him free.

Step-by-step explanation:

We are given the following in the question:

Null Hypothesis:

The null hypothesis states that the prisoner is innocent

Alternate Hypothesis:

The alternate hypothesis states that the prisoner is guilty and not innocent.

Type II error:

It is the type error made when we fail to reject the ll hypothesis when it is actually false.

That is we accept a false null hypothesis.

Thus a type II error in the above scenario will be to accept that the prisoner is innocent (accepting the null hypothesis) when actually he is guilty( the alternate hypothesis)

Thus, type II error would be setting free a guilty prisoner.

Option D) The prisoner is actually guilty and the jury sets him free.

"The prisoner is actually guilty and the jury sets him free" represents the risk involved in making a Type II error. The correct option is d) The prisoner is actually guilty and the jury sets him free.

In statistical hypothesis testing, particularly in the context of a trial, we have two competing hypotheses:

Null Hypothesis (H₀): The prisoner is innocent (presumed innocent until proven guilty).

Alternative Hypothesis (H₁): The prisoner is guilty.

A Type II error, in this scenario, occurs when the jury fails to reject the null hypothesis (fails to convict the prisoner) when the alternative hypothesis is true (the prisoner is actually guilty).

Let's analyze the options in the context of Type II error:

(a) The prisoner is actually guilty and the jury sends him to jail. This is not a Type II error. This describes a correct decision where the jury convicts the guilty prisoner.

(b) The prisoner is actually innocent and the jury sends him to jail. This is a Type I error, where the jury convicts an innocent person (rejects the null hypothesis when it is true).

(c) The prisoner is actually innocent and the jury sets him free. This is the correct outcome when the jury correctly fails to reject the null hypothesis (the prisoner is innocent).

(d) The prisoner is actually guilty and the jury sets him free. This is a Type II error. Here, the jury fails to reject the null hypothesis (prisoner is innocent) when it is false (prisoner is guilty).

(e) The prisoner is sent to jail. This could happen in both correct decisions (prisoner is guilty) and Type I errors (prisoner is innocent but convicted).

Therefore, the answer that represents the risk involved in making a Type II error is d. The prisoner is actually guilty and the jury sets him free.

Answer:

A solution curve pass through the point (0,4) when [tex]c_{1} = -\frac{4}{3}[/tex].

There is not a solution curve passing through the point(0,1).

Step-by-step explanation:

We have the following solution:

[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]

Does any solution curve pass through the point (0, 4)?

We have to see if P = 4 when t = 0.

[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]

[tex]4 = \frac{c_{1}}{1 + c_{1}}[/tex]

[tex]4 + 4c_{1} = c_{1}[/tex]

[tex]c_{1} = -\frac{4}{3}[/tex]

A solution curve pass through the point (0,4) when [tex]c_{1} = -\frac{4}{3}[/tex].

Through the point (0, 1)?

Same thing as above

[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]

[tex]1 = \frac{c_{1}}{1 + c_{1}}[/tex]

[tex]1 + c_{1} = c_{1}[/tex]

[tex]0c_{1} = 1[/tex]

No solution.

So there is not a solution curve passing through the point(0,1).

Answer:

(D) 21

Explanation:

3(2x−1)−10=8+5x

Step 1: Simplify both sides of the equation.

3(2x−1)−10=8+5x

(3)(2x)+(3)(−1)+−10=8+5x(Distribute)

6x+−3+−10=8+5x

(6x)+(−3+−10)=5x+8(Combine Like Terms)

6x+−13=5x+8

6x−13=5x+8

Step 2: Subtract 5x from both sides.

6x−13−5x=5x+8−5x

x−13=8

Step 3: Add 13 to both sides.

x−13+13=8+13

x=21

Answer:

D. 21

Step-by-step explanation:

Given equation: [tex]\[3(2x – 1) – 10 = 8 + 5x\] [/tex]

Simplifying:

[tex]\[6x – 3 – 10 = 8 + 5x\] [/tex]

=> [tex]\[6x – 13 = 8 + 5x\] [/tex]

=> [tex]\[6x – 5x = 8 + 13\] [/tex]

=> [tex]\[(6 – 5)x = 21\] [/tex]

=> [tex]\[x = 21\] [/tex]

Verifying by substituting the value of x in the equation:

Left Hand Side: [tex]\[3(2x – 1) – 10\][/tex]

[tex]\[3(2*21 – 1) – 10\][/tex]

= [tex]\[3(42 – 1) – 10\][/tex]

= [tex]\[3(41) – 10\][/tex]

= [tex]\[123 – 10\][/tex]

= [tex]\[113\][/tex]

Right Hand Side: [tex]\[8 + 5x\][/tex]

= [tex]\[8 + 5*21\][/tex]

= [tex]\[8 + 105\][/tex]

= [tex]\[113\][/tex]

So, Left Hand Side = Right Hand Side when x=21.

Answer:

21.75 square units

Step-by-step explanation:

Draw a radial line from O to the point where AB intersects the circle.  We'll call this point P.

Draw another radial line from O to the point where arc BC intersects the circle.  We'll call this point Q.

OQ is equal to the radius of the circle, 2.  And AQ is equal to the radius of the sector, 8.  Therefore, the length of AO is 6.

Next, OP is equal to the radius of the circle, 2.  Since AB is tangent to the circle, it is perpendicular to OP.

So we have a right triangle with a hypotenuse of 6 and a short leg of 2.  Finding the angle ∠PAO:

sin ∠PAO = 2/6

∠PAO = asin(1/3)

That means the angle ∠BAC is double that:

∠BAC = 2 asin(1/3)

∠BAC ≈ 38.94°

Therefore, the area of the sector is:

A = (θ/360) πr²

A = (38.94/360) π(8)²

A ≈ 21.75

The area of the sector is approximately 21.75 square units.

Answer:

a) [tex]y = 3.8 x +100[/tex]

b) [tex]Abs. change= |168.4-167.5|=0.9[/tex]

So the calculated value is 0.9 points above the actual value.

[tex]Relative. Change =\frac{|168.4 -167.5|}{167.5}x100 =0.537[/tex]%

And the calculated value it's 0.537% higher than the actual value.

c) For this case we can use the slope obtained from the linear model to answer this question, and we can conclude that the CPI is increasing at approximate 3.8 units per year.

Step-by-step explanation:

Data given

1982 , CPI=100

1986, CPI = 191.2

Notation

Let CPI the dependent variable y. And the time th independent variable x.

For this case we want to adjust a linear model givn by the following expression:

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

Solution to the problem

Part a

For this case we can find the slope with the following formula:

[tex] m =\frac{CPI_{2006}-CPI_{1982}}{2006-1982}[/tex]

And if we replace we got:

[tex] m =\frac{191.2-100}{2006-1982}=3.8[/tex]

Let X represent the number of years after. Then for 1982 t = 0, and if we replace we can find b:

[tex] 100 = 3.8(0)+b[/tex]

And then [tex]b=100[/tex]

So then our linear model is given by:

[tex]y = 3.8 x +100[/tex]

Part b

For this case we need to find the years since 1982 and we got x = 2000-1982=18, and if we rpelace this into our linear model we got:

[tex]y = 3.8(18) +100=168.4[/tex]

And the actual value is 167.5 we can compare the result using absolute change or relative change like this:

[tex]Abs. change= |168.4-167.5|=0.9[/tex]

So the calculated value is 0.9 points above the actual value.

And we can find also the relative change like this:

[tex]Relative. Change =\frac{|Calculated -Real|}{Real}x100[/tex]

And if we replace we got:

[tex]Relative. Change =\frac{|168.4 -167.5|}{167.5}x100 =0.537[/tex]%

And the calculated value it's 0.537% higher than the actual value.

Part c

For this case we can use the slope obtained from the linear model to answer this question, and we can conclude that the CPI is increasing at approximate 3.8 units per year.

Answer:

yes,def=(2399/3713)x100=64.61%

yes,probably= (774/3713)x100=20.85%

yes,probably no= (322/3713)x100=8.67%

yes,def.no= (218/3713)x100=5.87%

Step-by-step explanation:

Data given

We assume the following frequency table:

Type                       Frequency

Yes, definitely          2399

Yes, probably            774

No, probably not       322

No, probably not        218

Total                           3713

Solution to the problem

And for this case in order to find the percentages we can use the definition of percentage frequency with the following formula:

[tex] \% = \frac{Number of people with characteristic}{Total} x100[/tex]

And using this formula we can find the percentages like this:

yes,def=(2399/3713)x100=64.61%

yes,probably= (774/3713)x100=20.85%

yes,probably no= (322/3713)x100=8.67%

yes,def.no= (218/3713)x100=5.87%

Answer:

Option a is right

Step-by-step explanation:

Given that a random sample of 250 men yielded 175 who said they'd ridden a motorcycle at some time in their lives, while a similar sample of 215 women yielded only 43 that had done so.

For proportions since binomial and sample size large we can use  z critical values.

Sample             I            II

N                    250        215     465

X                     175           43    218

p                       0.7          0.2    0.4688

p difference = 0.5

Std error of difference = [tex]\sqrt{p(1-p)(\frac{1}{n_1}+  \frac{1}{n_2} }\\=\sqrt{0.4688*0.5312)(\frac{1}{250} + \frac{1}{215} )}\\=0.0409[/tex]

Margin of error for 99% = 2.58*std error =  0.105

Confidence interval 99% = (0.5±0.105)            

Option a is right.

Answer:

Step-by-step explanation:

Let X(t) denote the number of machines breakdown at time t.

The givenn problem follows birth-death process with finite space

S={0, 1, 2, 3} with

[tex] \lambda_0=\frac{3}{10}, \mu_1=\frac{1}{8}\\\\ \lambda_1=\frac{2}{10}, \mu_2=\frac{2}{8}\\\\ \lambda_2=\frac{1}{10}, \mu_3=\frac{2}{8}[/tex]

The birth-death process having balance equations [tex]\lambda_sP_i=\mu_{s+1}P_{i+1},i=0,1,2[/tex]

since, state  rate at which leave = rate at which enter

            0      [tex]\lambda_0P_0=\mu_1P_1[/tex]

             1     [tex](\lambda_1+\mu_1)P_1= \mu_2P_2 + \lambda_0P_0[/tex]

             2   [tex](\lambda_2+\mu_2)P_2= \mu_3P_3 + \lambda_1P_1[/tex]

[tex]P_1=\frac{12}{5}P_0=P_0=\frac{5}{12}P_1\\\\P_2=\frac{48}{25}P_0=P_0=\frac{25}{48}P_2\\\\P_3=\frac{192}{250}P_0=P_0=\frac{250}{192}P_3[/tex]

Since [tex]\sum\limits^3_{i=0} {P_i=1}\\\\p_0=[1+\frac{5}{12}+\frac{48}{25}+\frac{192}{250}]^{-1}=\frac{250}{1522}[/tex]

a)

Average number not in use equals the mean of the stationary distribution [tex]P_1+2P_2+3P_3=\frac{2136}{751}[/tex]

b)

Proportion of time both repairmen are busy [tex]P_2+P_3=\frac{672}{1522}=\frac{336}{761}[/tex]

The average number of machines not in use is 0.5, and the repairmen are both busy 64% of the time. This has been found under the assumption of exponential distribution for both longevity of machines and repair time. The scenario represents a M/M/2 queue in operations research.

In this scenario, the life of the machines and the repair time are governed by exponential distributions. Exponential distribution is often used to model the amount of time until an event occurs, such as machine failure in this case.

(a) To find the average number of machines not in use, we need to consider the rate of machine breakdown and repair. A machine works an average of 10 hours before failure, which translates to a failure rate of 1/10. A single repairman can fix a machine in an average of 8 hours, meaning a rate of 1/8 per repairman, or 1/4 for two repairmen combined. As there are three machines, the average number of machines in use is the ratio of the arrival rate to the service rate: (1/10) / (1/4) = 2.5 machines. This implies that on average, 0.5 machines are not in use.

(b) Both repairmen will be busy when there are at least two machines that require fixing. The proportion of time in which this is the case is obtained by calculating the probability that the number of machines failed is two or more. This is a problem of queueing theory, in particular an M/M/2 queue. The formula for this probability is P(X >= k) = (1 - rho) * rho^(k) / (1 - rho^(c+1)), where rho = arrival rate / service rate, k = 2, and c = 2 (service channels). Substituting rho = 2.5, we obtain P(X >= 2) = 0.64, meaning that the repairmen are both busy 64% of the time.

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Answer:

No. He would need $1,080 to perform a test with this error.

Step-by-step explanation:

To answer this question, we have to calculate the sample size. This is the sample size that allow to estimate the average amount of sugar per can with 95% confidence (95% CI).

The difference between the upper and lower limit of the CI have to be equal or less to e=2 mg. The z value for a 95% CI is z=1.96.

[tex]e\leq z\sigma/\sqrt{n}[/tex]

[tex]e=z\sigma/\sqrt{n}\\\\\sqrt{n}=z\sigma/e=\\\\n=(z\sigma/e)^2=(1.96*15/2)^2=14.7^2=216\\\\n=216[/tex]

The minimum sample size needed for this error is 216. At a cost of $5/test, this sample size would cost [tex]n*p=216*5=\$ 1,080[/tex].

This is over the budget for this experiment ($1000).

Answer:

Paired  t test

Step-by-step explanation:

Given that you collect a random sample of 28 adult golfers and record two scores for each: one taken before the subject receives professional coaching and one taken after.

Here subject of interest is to study whether the professional coaching really improves the scores.

For this two groups are taken from golfers and they were given chances to play and scores were recorded before and after coaching.  The same subject with two different scores recorded and the difference calculated.  Hence here the appropriate test is paired t test

Statistic should be t because population std devition is not known.

Hence paired t test for comparison of mean differences before and after should be done.

Answer:

There is enough evidence to conclude that the sample is from a population of songs with a mean greater than 210 sec.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 210 seconds

Sample mean, [tex]\bar{x}[/tex] = 234.1 sec

Sample size, n = 40

Sample standard deviation, s = 54.52 sec

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu \leq 210\text{ seconds}\\H_A: \mu > 210\text{ seconds}[/tex]

We use one-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{234.1 - 210}{\frac{54.52}{\sqrt{40}} } = 2.795[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.684[/tex]

We calculate the p-value with the help of standard normal table.

P-value = 0.004005

Since,                    

[tex]t_{stat} > t_{critical}[/tex]

We fail to accept the null hypothesis and accept the alternate hypothesis.

Thus, there is enough evidence to conclude that the sample is from a population of songs with a mean greater than 210 sec.

The given data shows that  the sample mean length of 234.1 sec results in a low p-value, it reject the claim that songs must be no longer than 210 seconds.

Explanation of hypothesis testing for mean song length compared to a specific value with given data.

Null hypothesis (H0): The mean length of songs is 210 seconds or less (μ ≤ 210 sec).

Alternative hypothesis (Ha): The mean length of songs is greater than 210 seconds (μ > 210 sec).

Test statistic: Calculate the z-score using the sample mean, population mean, standard deviation, and sample size.

P-value: Using the z-score, find the p-value from the standard normal distribution table.

Conclusion: If the p-value is less than the significance level (0.05), reject the null hypothesis. In this case, if the sample mean length of 234.1 sec results in a low p-value, we reject the claim that songs must be no longer than 210 seconds.

Answer:

93

Step-by-step explanation:

Using long division, this is the correct answer. Here are the steps:

remove decimal points: 744 ÷ 8

divide using long division: 93

Answer:

93

Step-by-step explanation:

Given number = 74.4

Divisor = 0.8

[tex]\[\frac{74.4}{0.8} = \frac{74.4*10}{0.8*10}\][/tex]

[tex]\[= \frac{744}{8}\][/tex]

Simplifying,

[tex]\[= \frac{744\div 4}{8\div 4}\][/tex]

[tex]\[= \frac{186}{2}\][/tex]

[tex]\[= 93\][/tex]

Validating by multiplying the quotient and divisor,

[tex]\[93 * 0.8\][/tex]

[tex]\[= 74.4\][/tex]

This is equal to the dividend.

To find the closest point on a line to another point, convert the problem into finding a minimum of a distance function, then use calculus (derivatives) to find the minimum.

To solve this problem, you need to identify a goal and an approach using calculus. The goal is to find a point (x, y) on the given line 5x + y = 7 that is closest to the point (−3, 2). This translates into minimizing the distance between (x, y) and (-3, 2) with respect to (x, y).

The distance formula between two points (x1, y1) and (x2, y2) is: √[(x2-x1)²+(y2-y1)²]. For our purposes x1 = -3, y1 = 2 and x2 = x, y2 = y. However, since y is from our given line (5x + y = 7), we can substitute for y = 7 - 5x. Thus, our new distance will become a function in terms of x: d(x) = √[(x - -3)²+(7 - 5x - 2)²].

To minimize it, we should take the derivative of this function, and set it equal to zero, which will give an extreme point (either minimum or maximum). Using the chain-rule results in the minimizing x value, and substituting it back to the line gives us the y value. This resultant point (x, y) would be our answer to the problem.

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Answer:

3.573 to 4.127

Step-by-step explanation:

Given

Sample size = 15

Mean = Sum of ratings/ sample size

Mean = 57.7/15

Mean = 3.85

Degree of freedom = sample size - 1

Degree of freedom = 15 - 1 = 14

df = 14

Then we calculate the standard deviation

(x - mean)² ||

(3.6 - 3.85)² || 0.0625

(2.9 - 3.85)² || 0.9025

(3.8 - 3.85)² || 0.0025

(4.5 - 3.85)² || 0.4225

(3.2 - 3.85)² || 0.4225

( 3.9 - 3.85)² || 0.0025

( 3.3 - 3.85)² || 0.3025

( 4.6 - 3.85)² || 0.5625

(4.1 - 3.85)² || 0.0625

(4.3 - 3.85)² || 0.2025

4.4 - 3.85)² || 0.3025

( 3.9 - 3.85)² || 0.0025

(3.2 - 3.85)² || 0.4225

( 4.2 - 3.85)² || 0.1225

( 3.8 - 3.85)² || 0.0025

Total || 3.7975

Variance = 3.7975/15 = 0.253167

Standard Deviation = √0.253167 = 0.50315703314174194

Standard Deviation = 0.5 ------- Approximated

The next step is to subtract the confidence level from 1, then divide by two.

i.e (1 - 0.95)/2 = 0.025

α = 0.025

Then we look up this answer to step in the t-distribution table.

For 14 degrees of freedom (df) and α = 0.025, my result is 2.145

The next step is to divide the sample standard deviation by the square root of the sample size.

0.5 / √15 = 0.129

Next is to multiply this result by step 2.145 (from the t table)

0.129 × 2.45 = 0.277

For the lower end of the range, subtract 0.277 from the sample mean.

3.85 – 0.277 = 3.573

Step 7: For the upper end of the range, add step 0.277 to the sample mean.

3.85 + 0.277 = 4.127

Answer:

f(x) = 7x⁶ + 3x³ + 12

Step-by-step explanation:

The number of roots a function has corresponds to the highest exponent value.

In the function f(x) = 7x⁶ + 3x³ + 12 ,

the number 7x⁶ has the highest exponent value of 6, so it'll have 6 roots.

Green's theorem says the circulation of [tex]\vec F[/tex] along the rectangle's border [tex]C[/tex] is equal to the integral of the curl of [tex]\vec F[/tex] over the rectangle's interior [tex]D[/tex].

Given [tex]\vec F(x,y)=2xy\,\vec\imath[/tex], its curl is the determinant

[tex]\det\begin{bmatrix}\frac\partial{\partial x}&\frac\partial{\partial y}\\2xy&0\end{bmatrix}=\dfrac{\partial(0)}{\partial x}-\dfrac{\partial(2xy)}{\partial y}=-2x[/tex]

So we have

[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_D-2x\,\mathrm dx\,\mathrm dy=-2\int_0^3\int_0^8x\,\mathrm dx\,\mathrm dy=\boxed{-192}[/tex]

The circulation of the vector field 2xyi around the given rectangle, as computed via Green's Theorem, is 0 due to the curl of F being 0.

To use Green's Theorem to calculate the circulation around a rectangle, first we should realize that Green's Theorem states that the line integral around a simple closed curve C of F.dr is equal to the double integral over the region D enclosed by C of the curl of F. Here, F is the vector field defined as 2xyi. The given rectangle is oriented counterclockwise and the values of x and y are given as 0≤x≤8 and 0≤y≤3 respectively. The line integral denotes the circulation of the field.

The circulation is thus the double integral over the rectangle of ∇ x F. But in this case, since F = 2xyi, we get ∇ x F = 0. Hence, the circulation of F around the given rectangle is 0.

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Answer:

All of them are polynomial functions

Step-by-step explanation:

Remember that a polynomial function of x is a function whose value f(x) is always equal to [tex]f(x)=a_0+a_1x+a_2x^2+\cdots a_nx^n[/tex] for a fixed n≥0 (the degree of f) and fixed coefficients [tex]a_i\in\mathbb{R}[/tex]

For example, [tex]f(x)=x^2+3x[/tex] is a polynomial function, but [tex]g(x)=2^x+x[/tex] is not because [tex]2^n[/tex] is not a nonnegative power of x. Another example of a non-polynomial function is [tex]g(x)=x^{-1}=\frac{1}{x}[/tex].

f(x)=4⋅11x is polynomial with degree 1 and [tex]a_0=0,a_1=4\cdot 11[/tex]. For the same reasons, f(x)=3⋅18x and f(x)=10⋅17x are polynomial functions.

f(x)=−4x³−4x²+5x+1 is a polynomial function of degree 3 with [tex]a_0=1,a_1=5, a_2=a_3=-4[/tex]. and f(x)=−2x−1 is a polynomial function of degree 1 and coefficients [tex]a_0=-1,a_1=-2[/tex].

In mathematics, a polynomial function involves operations of addition, subtraction, multiplication, and non-negative, whole-number exponents of variables. Here, the polynomial functions from the given list are: f(x)=−4x3−4x2+5x+1 and f(x)=−2x−1.

In the realm of mathematics, a polynomial function is an expression consisting of variables and coefficients, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents of variables. When asked to determine which of the given functions are polynomial functions, we can apply this definition. Your options were: f(x)=4⋅11x, f(x)=3⋅18x, f(x)=10⋅17x, f(x)=−4x3−4x2+5x+1, and f(x)=−2x−1.

By our definition, the polynomial functions from the list are:  f(x)=−4x3−4x2+5x+1 and f(x)=−2x−1. The other three functions are not polynomial functions as they do not comply with the characteristics of polynomial functions.

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Answer:

The dimensions of the newly formed rectangle is length is 41 cm and width is 9 cm.

Step-by-step explanation:

Given,

A 1 meter steel rod is bent into a rectangle.

That means the perimeter of the rectangle is 1 meter.

So, Perimeter = [tex]1\ m=100\ cm[/tex]

Let the width of the rectangle be 'x'.

Now According to question, length exceeds quadruple its width by 5 cm.

Hence framing the above sentence in equation form, we get;

[tex]Length=4x+5[/tex]

Now we use the formula of  perimeter of rectangle,

[tex]Perimeter=2(Length+width)[/tex]

On substituting the values, we get;

[tex]2(4x+5+x)=100\\\\(5x+5)=\frac{100}{2}\\\\5x+5=50\\\\5x=50-5\\\\5x=45\\\\x=\frac{45}{5}=9[/tex]

Now, we substitute the value of x to get the value of length;

[tex]Length=4x+5=4\times9+5=36+5=41\ cm[/tex]

Thus, Width=9 cm    Length=41 cm

Hence the dimensions of the newly formed rectangle is length is 41 cm and width is 9 cm.

Answer:

a) Null hypothesis:[tex]\mu_{1} \leq \mu_{2}[/tex]

Alternative hypothesis:[tex]\mu_{1} > \mu_{2}[/tex]

b) [tex]z_{crit}=2.33[/tex]

And since the calculated value is lower than the critical value we have enough evidence at 0.01 of significance to FAIL to reject the null hypothesis.

Step-by-step explanation:

Part a

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

We want to test this:

Null hypothesis:[tex]\mu_{1} \leq \mu_{2}[/tex]

Alternative hypothesis:[tex]\mu_{1} > \mu_{2}[/tex]

Part b

Golf course designer Roberto Langabeer is evaluating two sites, Palmetto Dunes and Ocean Greens, for his next golf course. He wants to prove that Palmetto Dunes residents (population 1) play golf more often than Ocean Greens residents (population 2). Roberto commissions a market survey to test this hypothesis. The market researcher used a random sample of 64 individuals from each suburb, and reported the following:  X 1 = 15  times per month and  X 2 = 14  times per month. Assume that  σ 1 = 2  and  σ 2 = 3 . With  α = .01 , the critical z value is _____.

Data given and notation

[tex]\bar X_{1}=15[/tex] represent the mean for the sample 1

[tex]\bar X_{2}=14[/tex] represent the mean for the sample 2

[tex]\sigma_{1}=2[/tex] represent the population deviation for 1

[tex]\sigma_{2}=3[/tex] represent the population deviation for 2

[tex]n_{1}=64[/tex] sample size selected for 1

[tex]n_{2}=64[/tex] sample size selected for 2

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

The statistic is given by:

[tex]z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex]     (1)

[tex]z=\frac{15-14}{\sqrt{\frac{2^2}{64}+\frac{3^2}{64}}}}=2.22[/tex]  

In order to find the critical value we need a value that accumulates 0.01 of the area on the right tail, since we are conducting a right tailed test. And the critical value is:

[tex]z_{crit}=2.33[/tex]

And since the calculated value is lower than the critical value we have enough evidence at 0.01 of significance to FAIL to reject the null hypothesis.

Answer:

a) The 90% confidence interval would be given by (71.689;78.311)  

The 95% confidence interval would be given by (71.012;78.988)  

b) [tex]p_v =2*P(t_{24}<-2.583)=0.016[/tex]  

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.  So we have a significant effect.

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

[tex]\bar X=75[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)  

[tex]s=9.68[/tex] represent the sample standard deviation  

n=25 represent the sample size  

Assuming the X follows a normal distribution  

[tex]X \sim N(\mu, \sigma=0.15[/tex]

The sample mean [tex]\bar X[/tex] is distributed on this way:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]  

Part a

The confidence interval on this case is given by:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

We can find the degrees of freedom like this:

[tex]df=n-1=25-1=24[/tex]

confidence 90%

The next step would be find the value of [tex]\t_{\alpha/2}[/tex], [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex]  

Using the t distribution with 24 df, excel or a calculator we see that:  

[tex]t_{\alpha/2}=1.71[/tex]

Since we have all the values we can replace:

[tex]75 - 1.71\frac{9.68}{\sqrt{25}}=71.689[/tex]  

[tex]75 + 1.71\frac{9.68}{\sqrt{25}}=78.311[/tex]  

So on this case the 90% confidence interval would be given by (71.689;78.311)  

Confidence 95%

The next step would be find the value of [tex]\t_{\alpha/2}[/tex], [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex]  

Using the t distribution with 24 df, excel or a calculator we see that:  

[tex]t_{\alpha/2}=2.06[/tex]

Since we have all the values we can replace:

[tex]75 - 2.06\frac{9.68}{\sqrt{25}}=71.012[/tex]  

[tex]75 + 2.06\frac{9.68}{\sqrt{25}}=78.988[/tex]  

So on this case the 95% confidence interval would be given by (71.012;78.988)  

Part b

Null hypothesis:[tex]\mu = 80[/tex]  

Alternative hypothesis:[tex]\mu \neq 80[/tex]  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]t=\frac{75-80}{\frac{9.68}{\sqrt{25}}}=-2.583[/tex]  

P-value  

The degrees of freedom are 25-1=24

Since is a two tailed test the p value would given by:  

[tex]p_v =2*P(t_{24}<-2.583)=0.016[/tex]  

Conclusion  

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.  So we have a significant effect.

Answer:

B. Rational functions

Step-by-step explanation:

The partial fraction decomposition is used for functions there are described by fractions, and for which the substitution method is not possible. These are rational functions, in which both the numerator and the denominator are polynomials.

So the correct answer is:

B. Rational functions

Partial fraction decomposition is used to simplify complex rational functions to make them easier to integrate. Rational function is the correct answer.

The correct answer is B. Rational functions. Partial fraction decomposition is a mathematical technique used essentially to simplify complex rational functions. A rational function is a function that can be defined as the ratio of two polynomials. By breaking down a complex rational function into simpler fractions (which is what partial fraction decomposition does), integration becomes more manageable. For example, it's easier to integrate simple fractions like 1/x or 2/x^2, which would be the result of a partial fraction decomposition, than complex, intertwined expressions.

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Answer:[tex]\pm2.131[/tex]

Step-by-step explanation:

The critical value in the t-distribution for a two-tailed hypothesis test is the t-value in the students's t-distribution table corresponding to the [tex]\alpha/2[/tex] and df , where [tex]\alpha=[/tex] significance level and df = Degree of freedom.

We are given , [tex]\alpha=0.05[/tex]

Then,  [tex]\alpha/2=0.025[/tex]

df = 15

Now , from the students's t-distribution table

The  critical value in the t-distribution for a two-tailed hypothesis test is [tex]t=\pm2.131449\approx\pm2.131[/tex]

Hence, the required t-value = [tex]\pm2.131[/tex]

Answer:

1.86 < t < 2.31

Step-by-step explanation:

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean height actually is higher\lower than an specified value, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq \mu_o[/tex]  

Alternative hypothesis:[tex]\mu > \mu_o[/tex]

Or like this:

Null hypothesis:[tex]\mu \geq \mu_o[/tex]  

Alternative hypothesis:[tex]\mu < \mu_o[/tex]

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We don't know on this case the calculated value and with the p value we can find it.

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=9-1=8[/tex]  

Since is a one side test the p value would be:  

[tex]p_v =P(t_{(8)}>t_o)=0.035[/tex]

And we can find the critical value with the following excel code:

" =T.INV(1-0.035,8)" and we got [tex]t_p =2.09[/tex]

And cannot be a lower tail test since all the options have positive values for the statistic. So on this case the best option is:

1.86 < t < 2.31

Answer:

Null hypothesis:[tex]\mu \leq 8000[/tex]  

Alternative hypothesis:[tex]\mu > 8000[/tex]  

[tex]z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2[/tex]  

[tex]p_v =P(Z>2)=0.0228[/tex]  

Step-by-step explanation:

1) Data given and notation  

[tex]\bar X=8300[/tex] represent the sample mean  

[tex]\sigma=1200[/tex] represent the population standard deviation  

[tex]n=64[/tex] sample size  

[tex]\mu_o =800[/tex] represent the value that we want to test  

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 8000, the system of hypothesis are :  

Null hypothesis:[tex]\mu \leq 8000[/tex]  

Alternative hypothesis:[tex]\mu > 8000[/tex]  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2[/tex]  

4) P-value  

Since is a one-side upper test the p value would given by:  

[tex]p_v =P(Z>2)=0.0228[/tex]  

5) Conclusion  

If we compare the p value and the significance level assumed, for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than $8000.  

Answer:

a) [tex]x=a\cdot cos(b), y=-a\cdot sin(b), 0 \leq b \leq 2\pi[/tex]

b) [tex]x=a\cdot cos(b), y=a\cdot sin(b), 0 \leq b \leq 2\pi[/tex]

c) [tex]x=a\cdot cos(b), y=-a\cdot sin(b), 0 \leq b \leq 6\pi[/tex]

d) [tex]x=a\cdot cos(b), y=a\cdot sin(b), 0 \leq b \leq 6\pi[/tex]

Step-by-step explanation:

The parametric equation for a circle is:

[tex]x=a\cdot cos(b)[/tex]

[tex]y=a\cdot sin(b)[/tex]

Where a is the radius and b is the angular displacement.

a) If a is negative in y and 0 ≤ b ≤ 2π, we have clockwise moves.

[tex]x=a\cdot cos(b), y=-a\cdot sin(b), 0 \leq b \leq 2\pi[/tex]

b)  If a is positive in y and 0 ≤ b ≤ 2π, we have counterclockwise moves.

[tex]x=a\cdot cos(b), y=a\cdot sin(b), 0 \leq b \leq 2\pi[/tex]

c) If a is negative in y and 0 ≤ b ≤ 6π, we have three times clockwise moves.

[tex]x=a\cdot cos(b), y=-a\cdot sin(b), 0 \leq b \leq 6\pi[/tex]

d) If a is positive in y and 0 ≤ b ≤ 6π, we have three times counterclockwise moves.

[tex]x=a\cdot cos(b), y=a\cdot sin(b), 0 \leq b \leq 6\pi[/tex]

Have a nice day!

The question requires finding parametric equations for a particle's motion along a circle of radius a, both clockwise and counterclockwise for one and three complete cycles. The equations are based on the standard circle parametric equations with adjustments to the angle parameter \(\theta\) and the interval according to the direction and number of cycles.

The student is asking for the parametric equations and parameter intervals for the motion of a particle tracing a circle, both clockwise and counterclockwise, once and three times. A circle with radius a and center at the origin has standard parametric equations x = a cos(\theta) and y = a sin(\theta), where \(\theta\) is the parameter usually representing an angle in radians.

Therefore, the appropriate parametric equations for each situation are:

Answer:

[tex]p = \frac{1564}{5020} = 0.3116[/tex]

Step-by-step explanation:

The point estimate for the population proportion is the number of sucesses divided by the size of the sample.

In this problem, we have that:

A success is being a tradition hogan in the reservation(the population). In a sample of 5020, 1564 are traditional hogans.

So [tex]p = \frac{1564}{5020} = 0.3116[/tex]

In this exercise we have to use the knowledge of variance to calculate the value of n, so we have that:

the sample is n=306

Organizing the information given in the statement we have that:

So given by the equation we have:

[tex]Z' = t(0.075)= 1.44\\n = (Z'*S/E)^2\\n = ( 1.44 * 0.85/0.07)^2\\n = (17.4857)^2\\n = 305.75\\n = 306[/tex]

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The minimum number of males over age 25 they must include in their sample is 512.

Given, Desired confidence level: 85%

Maximum error (E): 0.07 liters

Variance ([tex]\sigma^{2}[/tex]): 1.21 liters

Standard deviation ([tex]\(\sigma\)[/tex]): [tex]\(\sigma\)[/tex] = [tex]\sqrt{1.21}[/tex] = 1.1

Z-value for 85% confidence level (lookup Z-value for 0.425 in the standard normal distribution): [tex]\[ Z \approx 1.44 \][/tex]

n= [tex]\left(\frac{Z \sigma}{E}\right)^2[/tex]

[tex]\[ n = \left(\frac{1.44 \times 1.1}{0.07}\right)^2 \][/tex]

n= (1.584/0.07)² = 511.986

n = 512