What are the quotient and remainder of (3x^4+ 2x^2 - 6x + 1) /(x + 1)

Answer:

The coefficient of variation for A is 24.6%.

The coefficient of variation for B is 33.7%.

Step-by-step explanation:

The coefficient of variation (CV) is well defined as the ratio of the standard deviation to the mean. It exhibits the degree of variation in association to the mean of the population.

The formula to compute the coefficient of variation is,

[tex]CV=\frac{SD}{Mean}\times 100\%[/tex]

Consider the data set A.

Compute the mean of the data set A as follows:

[tex]Mean_{A}=\frac{1}{n}\sum X[/tex]

            [tex]=\frac{1}{14}\times [36900+19400+...+26000+38400]\\=30064.2857[/tex]

Compute the standard deviation of the data set A as follows:

[tex]SD_{A}= \sqrt{ \frac{ \sum{\left(x_i - Mean_{A}\right)^2 }}{n-1} }[/tex]

        [tex]= \sqrt{ \frac{ 712852142.8571 }{ 14 - 1} } \\\approx 7405.051[/tex]

Compute the coefficient of variation for A as follows:

[tex]CV=\frac{SD_{A}}{Mean_{A}}\times 100\%[/tex]

      [tex]=\frac{7405.051}{30064.2857}\times 100\%\\=24.6\%[/tex]

The coefficient of variation for A is 24.6%.

Consider the data set B.

Compute the mean of the data set B as follows:

[tex]Mean_{B}=\frac{1}{n}\sum X[/tex]

            [tex]=\frac{1}{11}\times [2.1+5.0+...+4.1+1.7]\\=3.2455[/tex]

Compute the standard deviation of the data set B as follows:

[tex]SD_{B}= \sqrt{ \frac{ \sum{\left(x_i - Mean_{B}\right)^2 }}{n-1} }[/tex]

        [tex]= \sqrt{ \frac{ 11.9873 }{ 11 - 1} } \\\approx 1.0949[/tex]

Compute the coefficient of variation for B as follows:

[tex]CV=\frac{SD_{B}}{Mean_{B}}\times 100\%[/tex]

      [tex]=\frac{1.0949}{3.2455}\times 100\%\\=33.7\%[/tex]

The coefficient of variation for B is 33.7%.

Answer:

5.8 years

Step-by-step explanation:

The sampling distribution of the sample means has a mean that is equal to mean of the population from which the sample has been drawn.

study of elementary school students reports that the mean age at which children begin reading is 5.8 years with a standard deviation of 0.7 years.

This means the population mean is

[tex] \mu = 5.8 \: years[/tex]

and the population standard deviation is

[tex] \sigma = 0.7 \: years[/tex]

If a sampling distribution is created using samples of the ages at which 61 children begin reading, the mean of the sampling distribution of the sample mean will be

[tex]5.8 \: years[/tex]

The mean of the sampling distribution of sample means can be calculated by dividing the mean of the population by the square root of the sample size.

To find the mean of the sampling distribution of sample means, we need to divide the mean of the population by the square root of the sample size. In this case, the mean of the population is 5.8 years and the sample size is 61 children. So, the mean of the sampling distribution of sample means can be calculated as:

Mean of sampling distribution of sample means = 5.8 years / sqrt(61) ≈ 0.745 years

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Answer:3+4-1

Step-by-step explanation:

Answer:

9

Step-by-step explanation:

[tex] \frac{(15 + 39)}{6} \\ \frac{54}{6 } \\ ans = 9 \\ [/tex]

Answer:

The dimensions are 32 ft by 16 ft

Step-by-step explanation:

Area  of the coffee and recreation room=512 square foot

LB=512

L=512/B

Perimeter of the Room = Perimeter of north wall+Perimeter of east wall+ Perimeter of west wall =L+2B (West and East are opposite)

Cost of the Perimeter=16L+4(2B)

[tex]C(B)=16(\frac{512}{B})+8B\\C(B)=\frac{8192+8B^2}{B}[/tex]

To minimize cost, first, we take the derivative of C(B)

Using quotient rule

[tex]C^{'}(B)=\frac{8B^2-8192}{B^2}[/tex]

Setting the derivative equal to zero

[tex]8B^2-8192=0\\8B^2=8192\\B^2=1024\\B=32 ft[/tex]

[tex]L=\frac{512}{B}=\frac{512}{32}=16ft[/tex]

The dimension of the cheapest recreation area will be 32 ft by 16 ft

Answer:

-A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate.

-This confidence interval is not valid since the distribution of spending in the sample data is right skewed.

-The margin of error is $4.4.

-This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436.

-We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.

Step-by-step explanation:

A 90% confidence interval would be narrower than the 95% confidence interval if we don't need to be as sure about our estimate.

TRUE. The 90% confidence is less strict in its probability of having the mean within the interval, so it is narrower than the 95% CI. It relies more in the information given by the sample.

In order to decrease the margin of error of a 95% confidence interval to a third of what is is now, we would need to use a sample 3 times larger.

FALSE. The margin of error is z*σ/(n^0.5). So to reduce it by two thirds, the sample size n needs to be 3^2=9 times larger.

This confidence interval is not valid since the distribution of spending in the sample data is right skewed.

FALSE. There is no information about the skewness in the sample.

The margin of error is $4.4.

TRUE. The margin of error is (89.11-80.31)/2=$4.4.

We are 95% confident that the average spending of these 435 American adults over this holiday season is between $80.31 and $89.11.

FALSE. The CI is related to the populations mean. We are 95% confident that the average spending of the population is between $80.31 and $89.11.

This confidence interval is valid since the sampling distribution of sample mean would be approximately normal with sample size of 436.

TRUE. This happens accordingly to the Central Limit Theorem.

95% of random samples have a sample mean between $80.31 and $89.11.

FALSE. The confidence interval refers to the population mean.

We are 95% confident that the average spending of all American adults over this holiday season is between $80.31 and $89.11.

TRUE. This is the conclusion that is looked for when constructing a confidence interval.

The answer is 17 free throws because I said so and that’s the correct answer

Answer:

my answer was gunna be his but that person answered first so ill just sound like i copied that person

Step-by-step explanation:

Answer:

pi is the ratio if the circumference over the diameter

Step-by-step explanation:

Answer:

The standard error of the sample mean is _17.677_ psi.

Step-by-step explanation:

Explanation:-

A random sample of n = 8 specimens is collected.

Given sample size is n = 8

Given mean of the population 'μ' = 2500 psi

standard deviation 'σ' = 50 psi

Let x⁻ is the mean of the observed sample

Standard error of the sample mean = [tex]\frac{S.D}{\sqrt{n} }[/tex]      ...(i)

Given Population of standard error (S.D) 'σ' = 50 psi

Now substitute all values in (i)

[tex]S.E = \frac{50}{\sqrt{8} } =17.677[/tex]

Conclusion:-

The standard error of the sample mean is _17.677psi.

The standard error of the sample mean for a sample of 8 concrete specimens is approximately 17.68 psi

To calculate the standard error of the sample mean, we use the formula:

[tex]\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} \][/tex]

Where:

- [tex]\sigma[/tex] is the standard deviation of the population.

- n is the sample size.

Given:

- [tex]\sigma[/tex] = 50 psi

- n = 8

Substituting these values into the formula:

[tex]\[ \text{Standard Error} = \frac{50}{\sqrt{8}} \][/tex]
[tex]\[ \text{Standard Error} = \frac{50}{2.828} \][/tex]
[tex]\[ \text{Standard Error} \approx 17.68 \][/tex]

So, the standard error of the sample mean is approximately 17.68 psi.

Answer:

   y = 6/19 = 0.316

Step-by-step explanation:

:)

Answer:

y=6/19

Step-by-step explanation:

15y − 1 = 11/2y + 2

15y-11/2y=3

30/2y-11/2y=3

19/2y=3

19y=6

y=6/19

Complete Question:

The typical lifespan for various mammal species in captivity (L) in years has been related to average adult size (M) in kilograms according to the regression equation seen below.

[tex]ln L = 2.468 + 0.2 (lnM)[/tex]

A typical adult Meerkat weighs about 0.9 kilograms. What is the predicted lifespan of a Meerkat incaptivity, according to this equation? * 2 points 11.6 years 2.4 years 14.1 years 2.6 years 28.8 years

Answer:

Option A) L = 11.6 years

Step-by-step explanation:

From the given equation:

[tex]ln L = 2.468 + 0.2 (lnM)[/tex]..........(1)

Average adult size, M = 0.9 kg

Putting the value of M into the regression equation in (1)

[tex]ln L = 2.468 + 0.2 (ln0.9)\\ln L = 2.468 + (-0.02107)\\ln L = 2.447\\L = e^{2.447} \\L = 11.55 years[/tex]

The question is an illustration of regressions.

The lifespan of a typical adult Meerkat that weighs about 0.9 kilograms is (a) 11.6 years

The regression equation is given as:

[tex]\ln(L) = 2.468 + 0.2\ln(M)[/tex]

From the question, we have:

M = 0.9

Substitute 0.9 for M in [tex]\ln(L) = 2.468 + 0.2\ln(M)[/tex]

[tex]\ln(L) = 2.468 + 0.2\ln(0.9)[/tex]

Take natural logarithm of 0.9

[tex]\ln(L) = 2.468 + 0.2\times -0.1054[/tex]

[tex]\ln(L) = 2.468 -0.02108[/tex]

[tex]\ln(L) = 2.44692[/tex]

Take exponents of both sides

[tex]L = e^{2.44692}[/tex]

[tex]L = 11.553[/tex]

Approximate

[tex]L = 11.6[/tex]

Hence, the lifespan is (a) 11.6 years

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Answer:

-7

Step-by-step explanation:

Given  : sequence  56, 49, 42, 35, ...

To find : Find the common difference.

Solution : We have given that 56, 49, 42, 35, ...

By the common difference of an arithmetic sequence is:

d = a_{2} -a_{1}a2−a1

Where, a_{1}a1 is the first term and  a_{1}a1 is the second term.

Then common difference =  49 -56

                                          = -7 .

Answer:

28 - 4x

Step-by-step explanation:

[tex]18 -2 [x + (x - 5)] \\ = 18 - 2 [x + x - 5] \\ = 18 - 2 [2x - 5] \\ = 18 -4x + 10 \\ = 28 - 4x \\ [/tex]

Answer:

a) to remove bias in the study

b) The number of barbell curls performed by group who was encourged was higher than the group who was not encouraged

c) H0: mean of barbell curls for group encouraged= mean of barbell curls for group not encouraged

Ha: mean of barbell curls for group encouraged≠ mean of barbell curls for group not encouraged

d)  two sample t-test.

Requirement has been met as there are two separate groups whose means have to be compared. Median of box plot is equal to mean. Standard Deviation is equal to range divided by four. Sample size will have to be assumed

e) Encouragement affects the number of barbell curl performed by an individua

Step-by-step explanation:

a) random assignemnt removes bias in any study.

b) box plot is attached

c) null hypthesis and alternate hypothesis would compare means of the two distributions

d) As we are comparing two groups, we will use two sample t-test.

The median of box plot is equal to mean for both groups.

The rnge/4 is equal to standard deviation.

mean of encouraged group= 30

mean of not encoouraged group=19

SD of encouraged group= (75-5)/4= 15

SD of not encouraged group= (60-0)/4= 15

Assume number of inidivudals in encouraged group =15

assume number of individuals in not encouraged group = 14

e) test statistic= (mean1-mean2)/√(SD1²/(N1-1)+ SD2²/(N2-1))

                      = (30-19)/(√(15²/14+ 15²/13)

                       = 1.9734

Since test statistic is greater than p-value, null hypothesis is rejected.

Encouragement affects the number of barbell curl performed by an individual

Answer:

The solution to the equation is

x = -4

Step-by-step explanation:

We want to find the solution to the equation

(1/4)x - 1/8 = 7/8 + (1/2)x

First, add -(1/2)x + 1/8 to both sides of the equation.

(1/4)x - 1/8 - (1/2)x + 1/8 = 7/8 + (1/2)x - (1/2)x + 1/8

[1/4 - 1/2]x = 7/8 + 1/8

x(1 - 2)/4 = (7 + 1)/8

-(1/4)x = 8/8

-(1/4)x = 1

Multiply both sides by -4

x = -4

Answer:

x=-4

Step-by-step explanation:

To solve for an in the equation 3ab = c, divide both sides by 3b, yielding a = c / (3b), with the condition that b is not zero.

To solve the equation 3ab = c for a, we need to isolate an on one side of the equation. Starting with the original equation:

3ab = c

We divide both sides of the equation by 3b to get:

a = c / (3b)

This is the solution for an in terms of b and c. It's important to note that b cannot be zero, as division by zero is undefined.

Answer:

E

Step-by-step explanation:

It's E because you have to add 13 and 14 and as you can see there's 14 boys so it's E.

The probability that the person chosen is a boy is 14/27.

Number of girls = 13

Number of boys = 14

Total students = 27

Probability is a measure of the likelihood of occurrence of an event.

P(event) = Favorable outcomes / total outcomes

Favourable outcomes = 14 because there are 14 boys

Total outcomes =27

So, the probability that the person chosen is a boy will be given by:

P(boy)=14/27

So, the probability that the person chosen is a boy is 14/27.

Hence, the probability that the person chosen is a boy is 14/27.

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Answer:

The correct option is (D).

Step-by-step explanation:

To construct the (1 - α)% confidence interval for population proportion the distribution of proportions must be approximated by the normal distribution.

A Normal approximation to binomial can be applied to approximate the distribution of proportion p, if the following conditions are satisfied:

In this case p is defined as the proportions of students who ride a bike to campus.

A sample of n = 125 students are selected. Of these 125 students X = 6 ride a bike to campus.

Compute the sample proportion as follows:

[tex]\hat p=\frac{X}{n}=\frac{6}{125}=0.048[/tex]

Check whether the conditions of Normal approximation are satisfied:

[tex]n\hat p =125\times 0.048=6<10\\n(1-\hat p) =125\times (1-0.048)=119>10[/tex]

Since [tex]n\hat p <10[/tex], the Normal approximation to Binomial cannot be applied.

Thus, the confidence interval cannot be used to estimate the proportion of all students who ride a bike to campus.

Thus, the correct option is (D).

Answer:

Possible options:

A.$182,343

B.$122,792

C.$109,406

D.$270,629

Answer C

Step-by-step explanation:

Under this method the percentage completion done in the Beginning Inventory is ignored while calculating the equivalent completed units during the current period. Beginning Inventory Units are treated as fresh units introduced for production.

Cost Accounted for:

- The cost of opening work-in-progress and cost of the current period are aggregated and the aggregate cost is divided by output in terms of completed equivalent units.

- The units of Beginning Inventory of WIP and their cost are taken in full under this method.

Answer:

It's either B or A

Step-by-step explanation:

If I had to choose one I would pick B because I did the math; and logically, that answer makes the most sense.

In the first study, moose drool reduces the toxin levels on grass. In the second study, canned soup consumption increases urinary BPA concentrations. In the third study, overeating for four weeks leads to long-term weight gain.

Question 1:

The notation for the quantity being estimated is the mean level of the toxin ergovaline on the grass (in ppm). Scientists estimate the mean level after treatment with moose drool to be 0.183 ppm.

The quantity that gives the best estimate is the mean level of the toxin ergovaline on the grass after treatment with moose drool (0.183 ppm).

A 95% confidence interval for the quantity being estimated is (0.183 - 1.96 * 0.016, 0.183 + 1.96 * 0.016), which is approximately (0.152, 0.214) ppm. This means there is a 95% chance that the true mean level of the toxin ergovaline on the grass after treatment with moose drool is between 0.152 and 0.214 ppm.

Question 2:

This is a matched pairs experiment because each participant is subjected to both treatments (canned soup and fresh soup). It was used to eliminate individual differences and control for variables that might affect the urinary BPA concentrations.

The parameter being estimated is the difference in means (canned minus fresh) of urinary BPA concentrations.

The 95% confidence interval for the difference in means is (19.6, 25.5) μg/L. This means that we are 95% confident that the true difference in mean urinary BPA concentrations between canned soup and fresh soup is between 19.6 and 25.5 μg/L.

If the study had included 500 participants instead of 75, we would expect the confidence interval to be narrower because a larger sample size reduces the standard error, leading to a more precise estimate.

Question 3:

The relevant parameter is the mean weight gain for participants two and a half years after the experiment.

The actual exact value of the parameter cannot be found unless we measure the weight gain for all individuals in the population.

A 95% confidence interval for the parameter is (6.8 - 1.96 * 1.2, 6.8 + 1.96 * 1.2), which is approximately (4.44, 9.16) lbs. This means there is a 95% chance that the true mean weight gain for participants two and a half years after the experiment is between 4.44 and 9.16 lbs.

The margin of error is 1.96 * 1.2 lbs, which is approximately 2.35 lbs. This means that the estimated mean weight gain may vary by up to 2.35 lbs from the true mean weight gain.

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Answer:

e= 0.8

Step-by-step explanation:

Get e by itself, so subtrct 1.2 from 2 to get 0.8

The solution to the equation [tex]\(e + 1.2 = 2\) is \(e = 0.8\).[/tex]

To solve the equation \(e + 1.2 = 2\), we need to isolate the variable \(e\) on one side of the equation. Here's how you can do it step by step:

1. **Subtract 1.2 from both sides:**

\[e + 1.2 - 1.2 = 2 - 1.2\]

\[e = 0.8\]

In the context of real numbers, this means that when you substitute \(e = 0.8\) back into the original equation, it holds true:

\[0.8 + 1.2 = 2\]

This equation verifies that \(e = 0.8\) is indeed the solution. In a broader sense, solving equations like this one is a fundamental concept in algebra and mathematics. It allows us to find unknown values in various real-life situations, making it a crucial skill in fields such as physics, engineering, economics, and many others.

Understanding how to manipulate equations helps us model and solve complex problems, making mathematics an essential tool in problem-solving and critical thinking.

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Answer:

use the distance formula,

by using it, prove the adjacent sides of the quadrilateral DEFG

hence, DEFG would be a rhombus

Answer:

The value for p <0.1  i.e. 0.02088 that means ,we had success in test on potassium content

Step-by-step explanation:

Given:

Means: 138 mg

sample size =40

New mean=136.9 mg

To find : Hypothesis on test.

Solution:

We know that ,

Z=(X-mean)/standard deviation)/{(sqrt(sample size)}

Consider the standard deviation of 3.00 mg

Z=(136.9-138)/(3/sqrt(40))

=-1.1/0.4743

Z=-2.3192

two tailed test

=2{(1-p(<z))}

P value for Z=2.3192 is 0.98956

=2{1-0.98956}

=0.02088

The value for p <0.1

we had success in test on potassium content.

The hypothesis test for the sports drink's potassium content involves a null hypothesis (H0) that the mean is equal to the claimed 138 mg, and an alternative hypothesis (Ha) that it is not equal.

The question involves conducting a hypothesis test for a sports drink's potassium content.

Establishing hypotheses for a two-tailed test involves setting a null hypothesis (H0) that the mean potassium content is equal to the claimed amount, and an alternative hypothesis (Ha) that the mean potassium content is not equal to the claimed amount.

Specifically:

This is a basic procedure in statistical testing where the null hypothesis often represents a statement of 'no effect' or 'no difference', and the alternative hypothesis contradicts this assumption.

Answer: 265

Step-by-step explanation:

Add up all the numbers and divide this value by the total number of terms,in this case 6 terms.

(153+ 176+ 249+ 150+ 620+ 242)/6 =265

Answer:

In the equation of a straight line (when the equation is written as "y = mx + b"), the slope is the number "m" that is multiplied on the x, and "b" is the y-intercept (that is, the point where the line crosses the vertical y-axis). This useful form of the line equation is sensibly named the "slope-intercept form".

Step-by-step explanation:

This is just something to keep in mind

Answer:

-2

Step-by-step explanation:

Well, lt me hope the expression is complete amd correct.

Let's go on with the information you submitted.

y² + 7y + 4

When y= -6.

Let's substitute the value, y= -6 in the expression. We'll have to be careful with the signs.

y² + 7y + 4

-6² + 7(-6) + 4

36 - 42 + 4

40 - 42 =

-2

Answer:

a) P ( X ≤ 26 ) = 0.0134

b) The significance of the cell phone causing brain cancer is 1.34 or 0.0134

Step-by-step explanation:

Solution:-

- From the entire population of cell phone users os size N = 346,145 media reported n = 26 people who developed cancer of brain or nervous system.

- The probability of a person developing cancer of the brain or nervous system, assuming that cell phones have no effect, is p = 0.000113.

- Denote a random variable X: The number people who developed cancer of brain or nervous system are normally distributed.

- The expected or mean number of people who developed cancer of brain or nervous system are u = 40.

- The standard deviation ( s ) for the distribution would be:

                          [tex]s = \sqrt{u*( 1 - p ) }\\\\s = \sqrt{40*( 1 - 0.000113 ) }\\\\s = 6.3242\\[/tex]

- The random variate X follows normal distribution with parameters:

                         X ~ Norm ( 40 , 6.3242^2 )

- The probability of X ≤ 26 cases from the total population of N = 346,145 can be determined by evaluating the Z-score standard value of the test statistics:

                        [tex]P ( X \leq 26 ) = P ( Z \leq [ \frac{X - u }{s} ) \\\\P ( Z \leq [ \frac{ 26 - 40 }{6.3242} ) = P ( Z \leq -2.21371 )[/tex]

- Using standard normal table compute the probability on the left side of Z-score value -2.21371. Hence,

                       P ( X ≤ 26 ) = 0.0134

- The probability of the less than 26 number of cases who developed cancer of the brain or nervous system is 0.0134 as per media reports.

- So 1.34% of population is a case brain cancer due to cell phones according to media reports.

Which is significant less than the expected number of cases ( 40 ) that occur regardless of cell phone effect.

Answer:  The significance of the cell phone causing brain cancer is 1.34 or 0.0134.