Shane receives an hourly wage of $30.40 an hour as an emergency room nurse. How much does he make if he works 30 hours during the day at the normal work rate?​

Answer:

0.25

Step-by-step explanation:

Given that a club can select one member to attend a conference. All of the club officers want to attend. There are a total of four ​officers, and their designated positions within the club are President​ (P), Vice dash President (Upper V )comma Secretary (Upper S )comma nbspand Treasurer (Upper T ).

Sample space would be

a){ {P}, {V}, {S} {T}} is the sample space with notations standing for as given in the question

b) Each sample is equally likely.  Hence we have equal chances for selecting any one out of the four.

If probability of selecting a particular sample of size I is p, the by total probability axiom we have

[tex]4p =1\\p =0.25[/tex]

There are four possible samples, one for each club officer (P, V, S, T). The chance of drawing a particular sample is 1/4 or 25%, considering the selection is random and each officer has an equal chance of being chosen.

For a club with four officers designated as President (P), Vice President (V), Secretary (S), and Treasurer (T) that can send only one member to attend a conference, we first identify all possible samples of size 1. The possible samples are simply each officer as a single-member delegation, so we have:

Since there is an equal chance of each officer being selected, and there are four officers, the chance or probability of any one of them being selected for the sample is 1 divided by the total number of officers.

Probability = 1 / 4 = 0.25 or 25%

Thus, there is a 25% chance or probability that a particular sample of size 1 (one officer) will be drawn for the conference.

Answer:

0.4

Step-by-step explanation:

To calculate the probability that a maintenance operation will take more than 2 hours and 15 minutes. We can first calculate the probability that ALL maintenance operation on 70 of the units will take less than 2 hours and 15 minutes, then subtract it from 1.

So the probability of a maintenance operation that would take less than 2 hours and 15 minutes, or 135 minutes is:

[tex]P(X \leq 135, \mu = 120, \sigma = 60) = 0.6 [/tex]

So the probability that a maintenance operation will take more than 2 hours and 15 minutes is:

[tex] 1 - 0.6 = 0.4[/tex]

Answer:

1 year

Step-by-step explanation:

The population as a function of time is:

[tex]P(t)  =177*4^{\frac{t}{2}}[/tex]

First, find the initial population at t=0:

[tex]P(0)  =177*4^{\frac{0}{2}}\\P(0) = 177[/tex]

Then, double it:

[tex]2P(0) = 2*177 = 354[/tex]

Finally, find the value of 't' for which the population is 354:

[tex]354  =177*4^{\frac{t}{2}}\\4^{\frac{t}{2}}=2\\log(4^{\frac{t}{2}})=log(2)\\\frac{t}{2} *log(4) = log(2)\\t=2*\frac{log(2)}{log(4)}\\t=1[/tex]

The alligator population will double afer 1 year.

To find the doubling time for the population of alligators, we set the population growth function equal to twice the original population and solve for t.

To find the doubling time for the population of alligators, we need to determine the time it takes for the population to double. In this case, the population growth function is given by P(t) = 177 * 4^(t/2), where t represents the number of years from the time of introduction. To find the doubling time, we set P(t) equal to twice the original population: 2 * P(0) = P(t).

Substituting the given population growth function, we have: 2 * 177 = 177 * 4^(t/2). To solve for t, we divide both sides of the equation by 177: 2 =

Next, we take the logarithm (base 4) of both sides of the equation to solve for t: log4(2) = t/2. Multiplying both sides by 2 gives us the doubling time: t = 2 * log4(2). Use a calculator to approximate this value.

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Stokes' theorem equates the line integral of [tex]\vec F[/tex] along the curve to the surface integral of the curl of [tex]\vec F[/tex] over any surface with the given curve as its boundary. The simplest such surface is the triangle with vertices (1,0,1), (0,1,0), and (0,0,1).

Parameterize this triangle (call it [tex]T[/tex]) by

[tex]\vec s(u,v)=(1-v)((1-u)(1,0,1)+u(0,1,0))+v(0,0,1)[/tex]

[tex]\vec s(u,v)=((1-u)(1-v),u(1-v),1-u+uv)[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Take the normal vector to [tex]T[/tex] to be

[tex]\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,1-v,1-v)[/tex]

Divide this vector by its norm to get the unit normal vector. Note that this assumes a "positive" orientation, so that the boundary of [tex]T[/tex] is traversed in the counterclockwise direction when viewed from above.

Compute the curl of [tex]\vec F[/tex]:

[tex]\vec F=(2x,2y,2x+2z)\implies\mathrm{curl}\vec F=(0,-2,0)[/tex]

Then by Stokes' theorem,

[tex]\displaystyle\int_{\partial T}\vec F\cdot\mathrm d\vec r=\iint_T\mathrm{curl}\vec F\cdot\mathrm d\vec S[/tex]

where

[tex]\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\,\mathrm dS[/tex]

[tex]\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv[/tex]

[tex]\mathrm d\vec S=\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv[/tex]

The integral thus reduces to

[tex]\displaystyle\int_0^1\int_0^1(0,-2,0)\cdot(0,1-v,1-v)\,\mathrm du\,\mathrm dv=\int_0^12(v-1)\,\mathrm dv=\boxed{-1}[/tex]

The question requires the use of Stokes' theorem to compute the integral of a given vector field around a defined curve. This involves calculating the curl of the vector field and a surface integral. However, specific computations were not provided.

The question is asking for the use of the Stokes' theorem to compute the integral of the vector field F=(2x,2y,2x+2z)F=(2x,2y,2x+2z) around the curve consisting of the straight lines joining the points (1,0,1), (0,1,0) and (0,0,1). In Stokes theorem, we're interested in the curl of the vector field and the surface integral of that curl over some surface S, with orientation determined by a unit normal vector. The problem also requires the computation of the curl of F which is a vector field whose components are the partial derivatives of the components of F. However, with this given information, we're unable to proceed with computations as detailed integral computations weren't provided. Therefore, a complete answer can't be provided at this time.

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Answer:

We conclude that the residents of Wilmington, Delaware, have more income than the national average.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ =  $50,000

Sample mean, [tex]\bar{x}[/tex] = $60,000

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = $10,000

a) First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 50000\text{ dollars}\\H_A: \mu > 50000\text{ dollars}[/tex]

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

c) Formula:

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

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{60000 - 50000}{\frac{10000}{\sqrt{10}} } = 3.162[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833[/tex]

b) Rejection Rule:

If the calculated t-statistic is greater than the the critical value, we rect the null hypothesis.

Since,                  

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

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

d) There is enough evidence to conclude that the residents of Wilmington, Delaware, have more income than the national average.

Answer:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.0245}{1.96})^2}=1600[/tex]  

c. 1600

Step-by-step explanation:

Previous concepts

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".  

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".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Solution to the problem

In order to solve this problem we need to assume a confidence level. Let's assume that is 95%

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

The margin of error for the proportion interval is given by this formula:  

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]    (a)  

First we need to find the margin of error from the original sample given by:

[tex] ME=1.96\sqrt{\frac{0.5 (1-0.5)}{400}}=0.049[/tex]

And on this case we have that [tex]ME =\pm 0.049/2=0.0245[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  

And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.0245}{1.96})^2}=1600[/tex]  

c. 1600

The minimum sample size necessary to achieve a margin of error half as great as the original is Option c. 1,600.

To determine the minimum sample size necessary to achieve a margin of error that is half as great as the one produced by the initial sample, we need to understand the relationship between sample size and margin of error.

The margin of error for a confidence interval for a proportion is inversely proportional to the square root of the sample size, n. Specifically, if the margin of error for a sample size of 400 is E, then to achieve half that margin of error (E/2), we need a sample size n' such that:

E/2 = (E / √(n'))

Simplifying gives us √(n') = 2 × √(n). Squaring both sides, we get:

n' = 4 × n

Given n = 400, we find:

n' = 4 × 400 = 1600

Therefore, the minimum sample size necessary is Option c. 1,600.

Answer:

[tex]P(0.26 \leq p \leq 0.43)=0.7204-0.0032=0.7172[/tex]

Step-by-step explanation:

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The population proportion have the following distribution

[tex]p \sim N(p=0.4,\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.4(1-0.4)}{91}}=0.0514)[/tex]

And we can solve the problem using the z score on this case given by:

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

We are interested on this probability:

[tex]P(0.26 \leq p \leq 0.43)[/tex]

And we can use the z score formula, and we got this:

[tex]P(\frac{0.26 -0.4}{\sqrt{\frac{0.4(1-0.4)}{91}}} \leq Z \leq \frac{0.43 -0.4}{\sqrt{\frac{0.4(1-0.4)}{91}}})[/tex]

[tex]P(-2.726 \leq Z \leq 0.584)[/tex]

And we can find this probability like this:

[tex]P(-2.726 \leq Z \leq 0.584)=P(Z<0.584)-P(Z<-2.726)=0.7204-0.0032=0.7172[/tex]

Answer:

[tex]C(T(x))=1.15x-5[/tex] and [tex]T(C(x))=1.15x-5.75[/tex]

C(T(x)) represents the conditions of the coupon.

Step-by-step explanation:

The price after the coupon reduction is represented by the function

[tex]C(x)=x-5[/tex]

The price after the tip is applied is represented by the function

[tex]T(x)=1.15x[/tex]

We need to find the composite functions C(T(x)) and T(C(x)).

[tex]C(T(x))=C(1.15x)[/tex]            [tex][\because T(x)=1.15x][/tex]

[tex]C(T(x))=1.15x-5[/tex]            [tex][\because C(x)=x-5][/tex]

This function represents that 15% tip will be added first after that $5 is subtracted.

Similarly,

[tex]T(C(x))=T(x-5)[/tex]           [tex][\because C(x)=x-5][/tex]

[tex]T(C(x))=1.15(x-5)[/tex]            [tex][\because T(x)=1.15x][/tex]

[tex]T(C(x))=1.15x-5.75[/tex]

This function represents that $5 is subtracted first after that 15% tip will be added.

It is given that a coupon for $5 off any lunch price states that a 15% tip will be added to the price before the $5 is subtracted.

It means 15% tip will be added first after that $5 is subtracted. So, C(T(x)) represents the conditions of the coupon.

The functions C(T(x)) and T(C(x)) represent the application of a tip and a coupon to a lunch price, respectively, in different order. The function that correctly represents the specific conditions given by the coupon in the problem statement is T(C(x)).

The composite function C(T(x)) is found by substituting T(x) into the function C(x). So, C(T(x)) = T(x) - 5 = 1.15x - 5.

The composite function T(C(x)) is calculated by substituting C(x) into the function T(x). So, T(C(x)) = 1.15 * (x - 5) = 1.15x - 5.75.

In the context of the coupon conditions, the right composite function is T(C(x)). This composite function first applies the $5 coupon reduction (C(x)) and then the 15% tip (T(x)), which exactly follows the procedure described by the coupon.

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Answer:

Step-by-step explanation:

We will use the product rule from combinatorics.

Answer: the cost of one daylilies is $4

the cost of one ivy is $9

Step-by-step explanation:

Let x represent the cost of one daylilies.

Let y represent the cost of one ivy.

Nicole and Kim bought their supplies from the same store. Nicole spent $99 on 9 daylilies and 7 pots of ivy. This means that

9x + 7y = 99 - - - - - - - -1

Kim spent $144 on 9 daylilies and 12 pots of ivy. This means that

9x + 12y = 144 - - - - - - - - -2

We will eliminate x by subtracting equation 2 from equation 1, it becomes

- 5y = - 45

y = - 45/-5 = 9

Substituting y = 9 into equation 2, it becomes

9x + 12 × 9 = 144

9x + 108 = 144

9x = 144 - 108 = 36

x = 36/9 = 4

To calculate the work required to pump the contents of a tank to a higher level, one needs to set up an integral using the weight of the fluid and the height difference.

A tank in the shape of a surface generated by revolving the parabolic segment y = x^2 for 0 ≤ x ≤ 3 about the y-axis will have a volume that can be determined using calculus and rotational solids concept. To calculate the work required to pump the contents of the tank to a level 5 feet above the top of the tank, we need to set up an integral using the weight of the fluid and the height difference.

Integral setup:

Answer:

B. 0.0918

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 18, \sigma = 1.5[/tex]

What proportion of batteries would be expected to last less than 16 hours?

This is the pvalue of Z when X = 16. So:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{16 - 18}{1.5}[/tex]

[tex]Z = -1.33[/tex]

[tex]Z = -1.33[/tex] has a pvalue of 0.0918.

So the correct answer is:

B. 0.0918

Answer: the correct option is B

Step-by-step explanation:

the durations of the batteries are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = durations of the batteries in hours

u = mean time

s = standard deviation

From the information given,

u = 18 hours

s = 1.5 hours

We want to find the proportion or probability of batteries would be expected to last less than 16 hours. It is expressed as

P(x lesser than 16)

For x = 16,

z = (16 - 18)/1.5 = - 1.33

Looking at the normal distribution table, the probability corresponding to the z score is 0.09176

Approximately 0.0918

Answer:

[tex]\chi^2 =27.356[/tex]

[tex]p_v = P(\chi^2_{9} >27.356)=0.00122[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(27.356,9,TRUE)"

Since the p value is lower than the significance level assumed 0.05 we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

                           Amusement Parks     Mexico    Home    Other    Total

Freshman                     23                         21            43          21         108

Sophomore                  34                         23            14          26        97

Junior                           25                         30            23         26        104

Senior                           27                         33             17         19         96

Total                            109                        107            97         92        405

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the two random variables

H1: There is dependence between the two variables

The level os significance assumed for this case is [tex]\alpha=0.05[/tex]

The statistic to check the hypothesis is given by:

[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]

The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]

And the calculations are given by:

[tex]E_{1} =\frac{109*108}{405}=29.07[/tex]

[tex]E_{2} =\frac{107*108}{405}=28.53[/tex]

[tex]E_{3} =\frac{97*108}{405}=25.87[/tex]

[tex]E_{4} =\frac{92*108}{405}=24.53[/tex]

[tex]E_{5} =\frac{109*97}{405}=26.11[/tex]

[tex]E_{6} =\frac{107*97}{405}=25.63[/tex]

[tex]E_{7} =\frac{97*97}{405}=23.23[/tex]

[tex]E_{8} =\frac{92*97}{405}=22.03[/tex]

[tex]E_{9} =\frac{109*104}{405}=27.99[/tex]

[tex]E_{10} =\frac{107*104}{405}=27.48[/tex]

[tex]E_{11} =\frac{97*104}{405}=24.91[/tex]

[tex]E_{12} =\frac{92*104}{405}=23.62[/tex]

[tex]E_{13} =\frac{109*96}{405}=25.84[/tex]

[tex]E_{14} =\frac{107*96}{405}=25.36[/tex]

[tex]E_{15} =\frac{97*96}{405}=22.99[/tex]

[tex]E_{16} =\frac{92*96}{405}=21.81[/tex]

And the expected values are given by:

                        Amusement Parks     Mexico    Home    Other    Total

Freshman                   29.07                28.53     25.87     24.53      108

Sophomore                26.11                  25.63     23.23     22.03      97

Junior                         27.99                 27.48     24.91      23.62     104

Senior                        25.84                 25.36     22.99      21.81       96

Total                            109                      107         97           92         405

And now we can calculate the statistic:

[tex]\chi^2 =27.356[/tex]

Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=(rows-1)(cols-1)=(4-1)(4-1)=9[/tex]

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{9} >27.356)=0.00122[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(27.356,9,TRUE)"

Since the p value is lower than the significance level assumed 0.05 we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

We want to find a solution to the initial value problem:

[tex]y'' + 18x = 0 \qquad,\qquad y(0) = 5 \qquad,\qquad y'(0)=1.[/tex]

We can start by integrating the equation once:

[tex]\dfrac{\textrm{d}^2 y}{\textrm{d}x^2} + 18 x = 0 \iff \dfrac{\textrm{d}^2 y}{\textrm{d}x^2} = -18 x \iff\\\\\iff \dfrac{\textrm{d}y}{\textrm{d}x} = -18\displaystyle\int x\textrm{ d}x \iff \dfrac{\textrm{d}y}{\textrm{d}x}=-18\dfrac{x^2}{2} + C \iff\\\\\iff \dfrac{\textrm{d}y}{\textrm{d}x} = -9x^2 + C.[/tex]

Using the initial condition [tex]y'(0) = 1[/tex], we can determine the integration constant [tex]C[/tex]:

[tex]\dfrac{\textrm{d}y}{\textrm{d}x}\Big\vert_{x= 0} = 1 \iff -9 \times 0^2 + C = 1 \iff C = 1.[/tex]

Therefore, we have:

[tex]\dfrac{\textrm{d}y}{\textrm{d}x} = -9x^2 + 1[/tex]

We can now integrate again:

[tex]y(x) = \displaystyle\int\dfrac{\textrm{d}y}{\textrm{d}x}\textrm{ d}x = \int\left(-9x^2+1\right)\textrm{d}x = -9\int x^2\textrm{ d}x + \int\textrm{d}x =\\\\= -9\dfrac{x^3}{3} + x + K = -3x^3 + x + K.[/tex]

The integration constant [tex]K[/tex] is determined by using [tex]y(0) = 5[/tex]:

[tex]y(0) = 5 \iff -3 \times 0^3 + 0 + K = 5 \iff K = 5.[/tex]

Finally, the solution is:

[tex]\boxed{y(x) = -3x^3 + x + 5}.[/tex]

By separation of variables, the solution is given by:

[tex]y(x) = -3x^3 + x + 5[/tex]

The differential equation is:

[tex]y^{\prime\prime}(x) + 18x = 0[/tex]

[tex]y^{\prime\prime}(x) = -18x[/tex]

Applying separation of variables:

[tex]\int y^{\prime\prime}(x) = -\int 18x dx[/tex]

[tex]y^{\prime}(x) = -9x^2 + K[/tex]

Since [tex]y^{\prime}(0) = 1, K = 1[/tex]

Thus:

[tex]y^{\prime}(x) = -9x^2 + 1[/tex]

To find y, another separation of variables is appled:

[tex]\int y^{\prime}(x) = \int(-9x^2 + 1)dx[/tex]

[tex]y(x) = -3x^3 + x + K[/tex]

Since y(0) = 5, K = 5, thus, the solution is:

[tex]y(x) = -3x^3 + x + 5[/tex]

A similar problem is given at https://brainly.com/question/13244107

Answer:

n=237

Step-by-step explanation:

Previous concepts

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".

Assuming the X follows a normal distribution

[tex]X \sim N(\mu, \sigma=112.2)[/tex]

We know that the margin of error for a confidence interval is given by:

[tex]Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.96=0.04[/tex] and [tex]\alpha/2=0.02[/tex]

Using the normal standard table, excel or a calculator we see that:

[tex]z_{\alpha/2}=2.054[/tex]

If we solve for n from formula (1) we got:

[tex]\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}[/tex]

[tex]n=(\frac{z_{\alpha/2} \sigma}{Me})^2[/tex]

And we have everything to replace into the formula:

[tex]n=(\frac{2.054(112.2)}{15})^2 =236.05[/tex]

And if we round up the answer we see that the value of n to ensure the margin of error required [tex]\pm=15 min[/tex] is n=237.

Answer:

13/9

y =-22x+32

Step-by-step explanation:

Given that the functions f and g are differentiable for all real numbers x. The table below gives values for the functions and their first derivatives at selected values of x.

x f(x) f'(x) g(x) g'(x)

1 4 -3 5 2

2 -3 -1 4 6

3 π 8 -1 4

4 -5 unknown 0 3

a) [tex]h(x) = \frac{f(x)}{g(x)} \\h'(x) = \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}[/tex]

(using quotient rule)

Substitute 1 for x

[tex]h'(1) = \frac{g(1)f'(1)-f(1)g'(1)}{(g(1))^2}\\=\frac{9-(-4)}{9} \\=\frac{13}{9}[/tex]

b) [tex]r(x) = f(x) g(x)\\r'(x) = f(x) g'(x)+g(x)f'(x)[/tex]

when [tex]x =2, r(x) = r(2) = f(2) g(2) = -12[/tex]

point of contact is (2,-12)

Slope of tangent =[tex]r'(2) = f(2) g'(2)+g(2)f'(2)\\=-18+(-4) \\=-22[/tex]

Using point slope form, tangent is

[tex]y+12 = -22(x-2)\\y = -22x +32[/tex]

a. h'(1) = -23/25.

b. The equation of the tangent line to r(x) = f(x)g(x) at x = 2 is y = -22x + 32.

a. To find h'(1), you need to apply the quotient rule.

The quotient rule states that if you have a function h(x) = f(x) / g(x), then the derivative h'(x) is given by:

h'(x) = (f'(x) * g(x) - f(x) * g'(x)) / (g(x))^2

In your case, h(x) = f(x) / g(x), and you want to find h'(1), so:x = 1

f(1) = 4

f'(1) = -3

g(1) = 5

g'(1) = 2

Now, plug these values into the quotient rule formula:

h'(1) = (f'(1) * g(1) - f(1) * g'(1)) / (g(1))^2

h'(1) = (-3 * 5 - 4 * 2) / (5^2)

h'(1) = (-15 - 8) / 25

h'(1) = -23 / 25

So, h'(1) = -23/25.

b. To find the equation of the tangent line to r(x) = f(x)g(x) at x = 2, you need to follow these steps:

Find r(2):

r(2) = f(2) * g(2)

r(2) = (-3) * 4

r(2) = -12

Find r'(x) using the product rule. The product rule states that if you have a function r(x) = f(x) * g(x), then the derivative r'(x) is given by:

r'(x) = f(x) * g'(x) + f'(x) * g(x)

Plugging in the given values:

r'(2) = (-3) * 6 + (-1) * 4

r'(2) = -18 - 4

r'(2) = -22

Now, you have r(2) and r'(2). You can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is:

y - y1 = m(x - x1)

Where (x1, y1) is the point on the curve (in this case, (2, -12)), and m is the slope (in this case, m = r'(2)).

Plug in the values:

y - (-12) = -22(x - 2)

Simplify and solve for y:

y + 12 = -22(x - 2)

y + 12 = -22x + 44

y = -22x + 44 - 12

y = -22x + 32

So, the equation of the tangent line to r(x) at x = 2 is y = -22x + 32.

For similar question on tangent line.

https://brainly.com/question/7252502

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Final answer:

The equation of the slant asymptote is y = 5x + 4. The equation of the slant asymptote for the given function is found by performing polynomial long division and taking the quotient without the remainder.

Explanation:

To find the equation of the slant asymptote for the function y = (5x4 + x2 + x) / (x3 - x2 + 5), we need to divide the numerator by the denominator using polynomial long division or synthetic division. The quotient, without the remainder, will give us the equation of the slant asymptote because as x approaches infinity, the remainder becomes insignificant compared to the terms in the quotient.

Step 1: Perform the division. (This will be specific to the given function.)

Step 2: The quotient (without the remainder) is the equation of the slant asymptote.

To find the equation of the slant asymptote, we need to divide the numerator (5x^4 + x^2 + x) by the denominator (x^3 - x^2 + 5) using long division.

The quotient is 5x + 4, so the equation of the slant asymptote is y = 5x + 4. It is important to note that the slant asymptote represents the behavior of the function as x approaches positive or negative infinity.

Answer:

n=543

Step-by-step explanation:

1) Notation and definitions

[tex]n[/tex] random sample (variable of interest)

[tex]\hat p[/tex] estimated proportion of interest

[tex]p[/tex] true population proportion of interest

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".  

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".  

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

2) Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by [tex]\alpha=1-0.98=0.02[/tex] and [tex]\alpha/2 =0.01[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-2.33, t_{1-\alpha/2}=2.33[/tex]

The margin of error for the proportion interval is given by this formula:  

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]    (a)  

And on this case we have that [tex]ME =\pm 0.05[/tex] (5% points means 0.05 on fraction) and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  

Since we don't have a prior estimate for [tex]\hat p[/tex] we can use 0.5 as the prior estimate, and replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.05}{2.33})^2}=542.89[/tex]  

And rounded up we have that n=543

Answer:

To test wether this is a significant decrease we have to perform a hypothesis test on the proportions.

The null hypothesis represents the past condition (the proportion of 0.23 accidents/day is equal or bigger) and the alternative hypothesis is what we claim that is happening now (the proportion have lowered).

We want to perform the test in order to know if there is enough evidence that it has changed. The result can be:

- The null hypothesis is rejected: there is enough evidence with this sample that the rate of accidents has lowered from 0.23.

- The null hypothesis failed to be rejected: there is not enough evidence to say that the rate of accidents has lowered, although the sample proportion is lower.

Step-by-step explanation:

Numerical solution:

We have to perform a hypothesis test on proportions. We want to know if there is enough evidence to claim that the number of accidents per day has lowered from 0.23.

The null and alternative hypothesis are:

[tex]H_0: \pi\geq0.23\\\\H_1:\pi<0.23[/tex]

The significance level assumed is 0.05.

The proportion of the sample is:

[tex]p=\frac{3}{30}=0.1[/tex]

The standard deviation is calculated from the population proportion

[tex]\sigma=\sqrt{\pi(1-\pi)/N} =\sqrt{0.23*(1-0.23)/30} =0.077[/tex]

The z-value now can be calculated as

[tex]z=\frac{p-\pi+0.5/N}{\sigma}=\frac{0.10-0.23+0.5/30}{0.077}  =-1.475[/tex]

The P-value for z=-1.475 is P=0.07011. The P-value is greater than the significant level, so the effect is not significant and it failed to reject the null hypothesis.

Answer: a) 984   b) 1068

Step-by-step explanation:

When the prior estimate of the population proportion(p) is available .

Then the formula to find the sample size  :-

[tex]n=p(1-p)(\dfrac{z^*}{E})^2[/tex]

, where E = margin of error

and z* = Critical z-value .

a) p= 0.36

E= 0.03

Critical value for 95% confidence level = z*= 1.96

Required sample size=[tex]n= 0.36(1-0.36)(\dfrac{1.960}{0.03})^2[/tex]

[tex]n= 0.36(0.64)(65.3333333333)^2[/tex]

[tex]n=(0.2304)(4268.44444444)=983.4496\approx984[/tex]

Hence, the required sample size is 984.

b) When the prior estimate of the population proportion is unavailable .

Then we use formula to find the sample size  :-

[tex]n= 0.25(\dfrac{z^*}{E})^2[/tex]

, where E = margin of error

and z* = Critical z-value

Put E= 0.03 and z*= 1.960

Required sample size =[tex]n= 0.25(\dfrac{1.960}{0.03})^2[/tex]

[tex]n= 0.25(65.3333333333)^2[/tex]

[tex]n= 0.25(4268.44444444)=1067.11111111\approx1068[/tex]  

Hence, the required sample size is 1068.

Answer:

1.14 s

Step-by-step explanation:

Time, [tex]t=\frac {d}{s}[/tex]

Here, d is the distance and s is the speed/velocity

Since we're given the velocity, s as 28 ft/s and the distance between the position of the rocket and ground as 32 ft then

[tex]t=\frac {32}{28}=1.142857143\approx 1.14 s[/tex]

Therefore, it needs 1.14 seconds

Note: As you missed to mention the given equation for t seconds and height h, so I am taking a sample equation h(t) =-16t² + 28t  + 40. So, I am explaining your question based on this equation, which would anyways clear your query.

Answer:

It will take 2 seconds for the rocket to return to the ground when is the rocket 32 feet above ground.

Note: Sample equation h(t) =-16t² + 28t  + 40 was used to solve this problem, as you had not mentioned the equation.

Step-by-step explanation:

To determine:

How long will it take for the rocket to return to the ground when is the rocket 32 feet above ground?

Information Fetching and solution steps:

So,

Let us consider the equation h(t) = -16t² + 28t + 40

32 = -16t² + 28t + 40

To find out how long will it take for the rocket to return to the ground when is the rocket 32 feet above ground, plug in h(t) = 32ft, rearrange into quadratic form, and solve:

32 = -16t² + 28t + 40

0 = -16t² + 28t + 8

Step 1: Factor right side of equation

0 = −4(4t + 1)(t − 2)

−4(4t + 1)(t − 2) = 0

Step 2: Set factors equal to 0

4t + 1 = 0 or t − 2 = 0

t = -1/4 or t = 2

As t can not be negative, so t = 2 seconds.

Hence, it will take 2 seconds for the rocket to return to the ground when is the rocket 32 feet above ground.

Keywords: time, height, velocity

Learn more time and height measure from brainly.com/question/1580756

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Answer:

Arc length [tex]=\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx[/tex]

Arc length [tex]=9.75053[/tex]

Step-by-step explanation:

The arc length of the curve is given by [tex]\int_a^b \sqrt{1+[f'(x)]^2}\ dx[/tex]

Here, [tex]f(x)=\int_0^{4.5x}sin(t) \ dt[/tex] interval [tex][0, \pi][/tex]

Now, [tex]f'(x)=\frac{\mathrm{d} }{\mathrm{d} x} \int_0^{4.5x}sin(t) \ dt[/tex]

[tex]f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( [-cos(t)]_0^{4.5x} \right )[/tex]

[tex]f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( -cos(4.5x)+1 \right )[/tex]

[tex]f'(x)=4.5sin(4.5x)[/tex]

Now, the arc length is [tex]\int_0^{\pi} \sqrt{1+[f'(x)]^2}\ dx[/tex]

[tex]\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx[/tex]

After solving, Arc length [tex]=9.75053[/tex]

Answer:

1) A) II and III

2) A) Critical values: r = plusminus 0.396, significant linear correlation

3) Yi= 0.41 + 0.37Xi

Step-by-step explanation:

Hello!

The objective of the linear correlation analysis is to test if there is an association between two study variables (X₁ and X₂).

Pearson's Coefficient of correlation

For Variables with a bivariate normal distribution (X₁, X₂)~N₂(μ₁; μ₂; σ₁²; σ₂²; ρ)

To do so, the study parameter is the population coefficient of correlation (ρ) - Rho- (If you were to make a graphic of the correlation line, Rho represents the slope)

Sample coefficient of correlation: r

It takes values between -1 and 1

This coefficient gives an idea of ​​the degree of correlation between the variables.

If ρ = 0 then there is no linear correlation between X₁ and X₂ Graphically, the slope is cero

If ρ < 0 then there is a negative association between X₁ and X₂ (i.e. when one variable increases the other one decreases) In a graphic, the slope of the line is negative.

If ρ > 0 then there is a positive association between X₁ and X₂ (i.e. Both variables increase and decrease together)

The closer to 1 or -1 the coefficient is, the stronger the association between variables. Using the absolute value of the correlation coefficients you can compare them, the greater the value, the stronger is the association between variables. For example, if you were to have two coefficients r₁= -0.24 and r₂= 0.67 then the absolute values are Ir₁I= 0.24 and Ir₂I= 0.67 you can see that the coefficient of the second sample is bigger than the first sample, that means that there is a stronger correlation in the second sample than the first one.

The non-parametric coefficient of correlation has the same characteristics.

1) Statements:

I: If the linear correlation coefficient for the two variables is zero, then there is no relationship between the variables. FALSE, when r=0 then there is no linear association between the two variables, this doesn't mean that there isn't any other type of association between them.

II: If the slope of the regression line is negative, then the linear correlation coefficient is negative. TRUE

The regression and correlation analyses are closely linked because for a regression equation to be reasonable, the sample points must be linked to the equation and the correlation coefficient between both variables must be large when the degree of association is high and small when The degree of association is low in addition to being independent of the units.

The regression analysis tests whether or not there is an association between both variables and the correlation analysis indicates the degree of that association.

If the slope of the regression is negative, then the correlation coefficient is negative.

III: The value of the linear correlation coefficient always lies between -1 and 1. TRUE, it is one of the characteristics of the correlation coefficient.

0.62 suggests a stronger linear relationship than a linear correlation coefficient of -0.82. FALSE, to check wich correlation coefficient shows a stronger correlation look at their absolute values, the one that is closer to 1 is the stronger, Ir₁I= 0.62 < Ir₂I= 0.82

Correct answer:

A) II and III

2) Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05. r = 0.523, n = 25

For this, you have to use a Table of cumulative probabilities for the linear correlation coefficient. (I've used Pearson)

For a two-tailed test (H₀: ρ=0)

[tex]r_{n-2; \alpha/2}= r_{23; 0.025}=[/tex] ± 0.396

Against r = 0.523, the decision is to reject the null hypothesis. There is a linear correlation between the two study variables.

Correct answer:

A) Critical values: r = plus-minus 0.396, significant linear correlation

3) Construct a scatterplot for the given data. Check 1st attachment for Data and Scatterplot.

Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary.

Equation of regression:

Yi= a + bXi

a= [tex](\frac{sum Yi}{n})[/tex]+b[tex](\frac{sum Xi}{n})[/tex]

b= [tex]\frac{sum XiYi*\frac{(sum Xi)(sum Yi)}{n} }{/(sumXi^2)-\frac{(sumXi)^2}{n} }[/tex]

Using the given Data:

∑Xi= -11

∑Xi²= 201

∑Yi= 0

∑Yi²= 176

Mean X= -1.10

Mean Y= 0

a= 0.41

b= 0.37

Yi= 0.41 + 0.37Xi

4) Managers rate empoyees acording to job performance and attitude. The results fro several randomly selected empoyees are given below.

Performance: 59; 63; 65; 69; 58; 77; 76; 69; 70; 64

Attitude: 72; 67; 78; 82; 73; 87; 92; 83; 87; 78

No question found?

I hope it helps!

Final answer:

The linear correlation coefficient ranges from -1 to 1, indicating strength and direction of a linear relationship. Statements II and III are true concerning the correlation coefficient. A calculated coefficient of 0.523 with a sample of 25 is significant given critical values of ±0.396 at a 0.05 significance level.

Explanation:

1. The linear correlation coefficient, known as the Pearson correlation coefficient and symbolized by r, provides a measure of the strength and direction of the linear relationship between two variables. Statement II is true: If the slope of the regression line is negative, the linear correlation coefficient is also negative. Statement III is accurate as well because the value of the linear correlation coefficient lies between -1 and 1, inclusive.

For Statement I, although a correlation coefficient of zero indicates no linear relationship, there could be a non-linear relationship present. Regarding Statement IV, the strength of the relationship is determined by the absolute value of the correlation coefficient, so -0.82 indicates a stronger relationship than 0.62.

Therefore, the correct answer is A) II and III.

2. Using a Table of Critical Values for the Pearson correlation coefficient or technology like a calculator's LinRegTTest function, we compare the computed value of r to the critical values at a given significance level to determine whether the correlation is significant. For r = 0.523 and n = 25, the degrees of freedom are 23 (n - 2), and by referencing a table or using software, we find the critical values at a significance level of 0.05.

The correct answer is A) Critical values: r = ±0.396, significant linear correlation, because the reported r value of 0.523 exceeds the critical value.

Answer:

a) -882.6296

b) The regression equation overestimated the net income.

c) -547.0104

d) The regression equation overestimated the net income.

Step-by-step explanation:

We are given the following information in the question:

The regression equation for net income is given by:

[tex]\text{Net Income} = y = 2159 + .0312(\text{Revenue})[/tex]

a)  residual for the (x, y) pair ($45,533, $2,697)

Calculated net income =

[tex]2159 + .0312(45533) = 3579.6296[/tex]

Residual = Observed net income - Calculated net income

[tex]\text{Residual} = 2697-3579.6296 = -882.6296[/tex]

b)  The regression equation overestimated the net income.

c)  residual for the (x, y) pair ($60,417, $3,497)

Calculated net income =

[tex]2159 + .0312(60417) = 4044.0104[/tex]

Residual = Observed net income - Calculated net income

[tex]\text{Residual} = 3497-4044.0104 = -547.0104[/tex]

d)  The regression equation overestimated the net income.

The answers are : (a-1) The residual for the [tex](x,y)[/tex] pair [tex](\$45,533, $2,697)[/tex] is [tex]\[\text{Residual} = -881.6136\][/tex]. (a-2) The regression equation overestimated the net income. (b-1) The residual for the x, y pair ([tex]\$60,417, $3,497[/tex]) is [tex]\[\text{Residual} = -546.9704\][/tex]. (b-2) The regression equation overestimated the net income.

1. Calculate the predicted value [tex](\(\hat{y}\))[/tex] using the regression equation:

[tex]\[ \hat{y} = 2{,}159 + 0.0312 \times x \][/tex]

2. Compute the residual by subtracting the predicted value from the actual value:

[tex]\[ \text{Residual} = y - \hat{y} \][/tex]

Part (a-1)

For the [tex]\( x, y \) pair \((45,533, 2,697)\)[/tex]

1. Calculate the predicted net income

[tex]\[ \hat{y} = 2{,}159 + 0.0312 \times 45{,}533 \][/tex]

[tex]\[ \hat{y} = 2{,}159 + 1{,}419.6136 \][/tex]

[tex]\[ \hat{y} = 3{,}578.6136 \][/tex]

2. Compute the residual

[tex]\[ \text{Residual} = y - \hat{y} \][/tex]

[tex]\[ \text{Residual} = 2{,}697 - 3{,}578.6136 \][/tex]

[tex]\[ \text{Residual} = -881.6136 \][/tex]

Part (a-2)

The residual is negative, which means the actual net income is less than the predicted net income. Therefore, the regression equation overestimated the net income.

Part (b-1)

For the [tex]\( x, y \) pair \((60,417, 3,497)\)[/tex]

1. Calculate the predicted net income

[tex]\[ \hat{y} = 2{,}159 + 0.0312 \times 60{,}417 \][/tex]

[tex]\[ \hat{y} = 2{,}159 + 1{,}884.9704 \][/tex]

[tex]\[ \hat{y} = 4{,}043.9704 \][/tex]

2. Compute the residual

[tex]\[ \text{Residual} = y - \hat{y} \][/tex]

[tex]\[ \text{Residual} = 3{,}497 - 4{,}043.9704 \][/tex]

[tex]\[ \text{Residual} = -546.9704 \][/tex]

Part (b-2)

The residual is negative, which means the actual net income is less than the predicted net income. Therefore, the regression equation overestimated the net income.

Answer:

0.1024

Step-by-step explanation:

Given that in a country there is only one of every 5 births is a girl.

i.e probability of a child born being a girl = 0.20

EAch birth is independent of the other and there are only two outcomes

Hence X no of girls will be binomial.

Required probability

= Probability that the first girl they have is the fourth baby

= Probability for first three children to be boys and fourth be a girl

Since each birth is independent of other, we have

Required probability

=[tex]0.8^3*0.2\\=0.1024[/tex]

Answer:

a) [tex]z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097[/tex]

[tex]p_v =P(Z>2.097)=0.018[/tex]  

If we compare the p value obtained and the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

b) The 90% confidence interval would be given by (0.512;0.596)

Step-by-step explanation:

Part a

Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

[tex]\hat p=\frac{209}{377}=0.554[/tex] estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:  

Null hypothesis:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(Z>2.097)=0.018[/tex]  

If we compare the p value obtained and the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Part b

The population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:

[tex]t_{\alpha/2}=-1.64, t_{1-\alpha/2}=1.64[/tex]

The confidence interval for the mean is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

If we replace the values obtained we got:

[tex]0.554 - 1.64\sqrt{\frac{0.554(1-0.554)}{377}}=0.512[/tex]

[tex]0.554 + 1.64\sqrt{\frac{0.554(1-0.554)}{377}}=0.596[/tex]

The 90% confidence interval would be given by (0.512;0.596)

Answer: (30.49 years, 42.31 years)

Step-by-step explanation:

The formula to find the confidence interval is given by :-

[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}.[/tex]

, where [tex]\overline{x}[/tex] = Sample mean

z* = Critical value.

[tex]\sigma[/tex] = Population standard deviation.

n= Sample size.

As per given , we have

[tex]\overline{x}=36.4[/tex]

[tex]\sigma=14.5[/tex]

n= 40

We know that the critical value for 99% confidence interval : z* = 2.576 (By z-table)

A 99 percent confidence interval for µ, the true mean age of guests will be :

[tex]36.4\pm (2.576)\dfrac{14.5}{\sqrt{40}}\\\\ 36.4\pm (2.576)2.29265130362\\\\=36.4\pm5.90586975813\\\\\approx36.4\pm5.91\\\\=(36.4-5.91,\ 36.4+5.91)\\\\=(30.49,\ 42.31) [/tex]

∴ a 99 percent confidence interval for µ, the true mean age of guests  = (30.49 years, 42.31 years)

The 99% confidence interval for the true mean age of guests at the ski resort, with a sample mean of 36.4 years and standard deviation of 14.5 years from 40 guests, is approximately (30.5, 42.3) years.

The question is about calculating a 99 percent confidence interval for the true mean age of guests at a ski mountain resort, where the ages have a right-skewed distribution, and the population standard deviation is 14.5 years. Given a sample mean of 36.4 years from a sample of 40 guests, we can use the following formula to calculate the confidence interval:

CI = µ ± (z * (σ / √n))

Where CI is the confidence interval, µ is the sample mean, z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and n is the sample size.

For a 99% confidence interval with a sample size of 40, we need to find the z-score that corresponds to the middle 99% of the normal distribution. The z-score for a 99% confidence level is typically 2.576. Then, the margin of error (ME) can be calculated as follows:

ME = 2.576 * (14.5 / √40) ≈ 2.576 * 2.29 ≈ 5.90

Thus, the confidence interval is:

CI = 36.4 ± 5.90

So, the 99% confidence interval for the true mean would be approximately (30.5, 42.3) years.

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Answer:

56 NOT -56

Step-by-step explanation:

That's just the answer

Answer:

56

Step-by-step explanation:

14 - (-21) + (-31) - (-25) - (-27)

= 14 +21 -31 + 25 +27

= 56

Answer:

0.33

Step-by-step explanation:

To solve this example we use this rule :

Δx/Δy

x is amount that changed. so Δx=0,99.

How we get 0.99.

Our storm is to 3 to 6 so we find difference for x : for 3rd and 6th member of table... 3.88-2.89=0.99

Now we have :

0.99/Δy

Because we need to find from 3 to 6 , Δy=3 ,

When we find both, we can find rate of change with :

0.99/3=0.33

Answer:

0.33 lol

Step-by-step explanation:

Answer:

The capacity is 18 transistors, 43 resistors and 4 computer chips.

Step-by-step explanation:

This problem can be solved by a system of equations.

I am going to say that:

x is the number of transistors that are made.

y is the number of resistors that are made.

z is the number of computer chips that are made.

Building the system:

A transistor requires 1 unit of copper. A resistor requires 2 units of copper. There are 104 units of copper. So [tex]x + 2y = 104[/tex]

A transistor requires 2 units of zinc. A resistor requires 3 units of zinc. A chip requires 1 unit of zing. There are 169 units of zinc. So [tex]2x + 3y + z = 169[/tex].

A transistor requires 3 units of glass. A chip requires 5 units of glass. There are 74 units of glass. So [tex]3x + 5z = 74[/tex]

We have the following system:

[tex]x + 2y = 104[/tex]

[tex]2x + 3y + z = 169[/tex]

[tex]3x + 5z = 74[/tex]

The solution is:

[tex]x = 18, y = 43, z = 4[/tex]

The capacity is 18 transistors, 43 resistors and 4 computer chips.