One question in the survey asked how much time per year the children spent in volunteer activities. The sample mean was 14.76 hours and the sample standard deviation was 16.54 hours.Required:

a. Based on the reported sample mean and sample standard deviation, explain why it is not reasonable to think that the distribution of volunteer times for the population of South Korean middle school students is approximately normal.
b. The sample size was not given in the paper, but the sample size was described as large. Suppose that the sample size was 500. Explain why it is reasonable to use a one-sample t confidence interval to estimate the population mean even though the population distribution is not approximately normal.
c. Calculate and interpret a confidence interval for the mean number of hours spent in volunteer activities per year for South Korean middle school children.

Answers

Answer 1

Answer:

Answer:

a. If the distribution was normal, many values would be negative, what is incompatible with the response variable (hours dedicated to volunteer activities).

b. If the sample is big, accordingly to the Central Limit Theorem, the sampling distribution shape tends to be normally-like, so we can apply a one-sample t-test.

c. The 95% confidence interval for the mean is (13.307, 16.213).

Step-by-step explanation:

a. If the distribution was normal, the values with one or more standard deviation below the mean would be negative, what is incoherent for this case. This, in a normal distribution, represents approximately 16% of the values.

If we calculate the probabilty for a normal distribution with the sample parameters, the probability of having "negative hours" is 18.6% (see picture attached).

b. If the sample is big, accordingly to the Central Limit Theorem, the sampling distribution shape tends to be normally-like, so we can apply a one-sample t-test.

The sampling distribution standard deviation is also reduced by a factor of 1/√n.

c. We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=14.76.

The sample size is N=500.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=(s)/(√(N))=(16.54)/(√(500))=(16.54)/(22.3607)=0.7397$

The t-value for a 95% confidence interval is t=1.965.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.965 \cdot 0.7397=1.453$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 14.76-1.453=13.307\n\nUL=M+t \cdot s_M = 14.76+1.453=16.213$

The 95% confidence interval for the mean is (13.307, 16.213).

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Answers

Answer:

it is point B

Step-by-step explanation:

because im smart

Answer:

Point B

Step-by-step explanation:

√11 is 3.3

And A is about 2. Something and inbetween D and E is 4 which means C will be 3.5

So it leaves it at B

Which ordered pair describes the location of point x

Answers

x , y
If I understand the question correctly

Compute the area of the region D bounded by xy=1, xy=16, xy2=1, xy2=36 in the first quadrant of the xy-plane. Using the non-linear change of variables u=xy and v=xy2, find x and y as functions of u and v.x=x(u,v)= ?

y=y(u,v)=?

Find the determinant of the Jacobian for this change of variables.

∣∣∣∂(x,y)/∂(u,v)∣∣∣=det=?

Using the change of variables, set up a double integral for calculating the area of the region D.

∫∫Ddxdy=?

Evaluate the double integral and compute the area of the region D.

Area =

Answers

Answer:

53.7528

Step-by-step explanation:

Notice that when

$xy = 1 ,\,\,\, xy = 16 , \,\,\, xy^2 = 1 \,\,\,, xy^2 = 36 \n\n$

If you set

$u = xy , v = xy^2$

as they suggest, then

$y = (v)/(u)} \,\,\,\, \text{and} \,\,\,\, \n\n{\displaystyle x = (u)/(y) = (u)/(v/u) = (u^2)/(v)$

Then

${\diplaystyle (\partial(x,y))/(\partial(u,v))} =\det \begin{pmatrix} 2u/v && -u^2/v^2 \n -v/u^2 && 1/u \end{pmatrix} = (1)/(v) }$

Therefore

$\iint\limits_(D) dx\,dy = \int\limits_(1)^(36)\int\limits_(1)^(16) (1)/(v) \, du \, dv = 15 \ln(36) = 53.7528$

A Jacobian matrix is formed by the first partial derivatives of a multivariate function that utilizes a training algorithm, and further calculation as follows:

Jacobian:

To evaluate the integral, cover the bounds, the integrand, and the differential area dA.

Transform the four equations in terms of u and v, notice that $u= xy \ \ and \ \ xy = 1, xy = 16$

implies that $1\leq u \leq 16.$

Similarly, $v= xy^2\ \ and\ \ xy^2= 1 , xy^2= 25$ implies that $1 \leq v \leq 25$

so write this integration region as $S= {(u,v) |1 \leq u \leq 18, 1 \leq v \leq 25}.$

Translate the equations from uv - plane to xy- plane. It is obtained by solving,

$u= xy, y= xy^2 \n\n\left.\begin{matrix}u=xy & \n v=xy^2& \end{matrix}\right\} \to \left.\begin{matrix}u^2=x^2y^2 & \n v=xy^2& \end{matrix}\right\} \n\n\to x=(u^2)/(v), y=(v)/(u)$

Convert dA part of the integral , using is $dA= |(\partial (x,y))/(\partial(u,v))| dudv.$

That is, $dA= \begin{vmatrix}(\partial x)/(\partial u) & (\partial x)/(\partial v)\n (\partial y)/(\partial u) & (\partial y)/(\partial v) \end{vmatrix} \ du dv \n\n$

Sampule the partial derivatives to find the Jacobian.

$dA=\begin{vmatrix}(2u)/(v) &-(u^2)/(v^2) \n -(v)/(u^2) &(1)/(u) \end{vmatrix} \ dudv\n\n=[((2u)/(v)) ((1)/(u)) -(- (u^2)/(v^2))(-(v)/(u^2))]\ du dv\n\n=((2)/(v)- (1)/(v)) \ dudv\n\n=(1)/(v)\ du dv\n\n$

The Jacobian the transformation is $dA= (1)/(v)dudv$

The region is $S={(u,v) |1\leq u \leq 16, 1\leq v\leq 25}.$

Rewrite the integral, using the transformation: $S,\ x=(u^2)/(v) =, y=(v)/(u) \ \ and\ \ dA=(1)/(v) dudv\n\n\int\int_R 1dA =\int \int_S (1)/(v)\ dudv= \int^(25)_(1) \int^(16)_(1) \ (1)/(v) \ dudv\n\n$

Evaluate the inner integral with respect to u.

$\to \int\int_R 1dA = \int^(25)_(1) \int^(16)_(1) \ (1)/(v) \ dudv\n\n$

by solving the value we get

$= 30 \ ln (5) \approx 48.28$

Find out more about the Jacobians here:

brainly.com/question/9381576

There are approximately 2.6 million deaths per year in country A. Express this quantity as deaths per minute.

Answers

In one year, there are 365 days
365x24=8760 hours
8760x60=525 600 minutes

Dividing 2.6 million to the last number (525 600), we find death per minute in this country, which gives us more than 4 deaths in a minute.

Take 2.7 million divided by the number of minutes in a year. 1 hour= 60 minutes 1 day= 24 hours So you take 24 hours × 60 minutes =1440 minutes 1 year= 365 per minute.

Is a quadrilateral in which each side is parallel to the side oppositeA Rhombo
B. Rectangle
C. Trapezium
D. Parallelogram

Answers

Answer:

parallelogram

Step-by-step explanation:

parallelograms are parallel.

Please help. Dont put a random answer plz.

Answers

Answer:

Explination:

Work:

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