Suppose that Stephen is the quality control supervisor for a food distribution company. A shipment containing many thousands of apples has just arrived. Unknown to Stephen, 15% of the apples are damaged due to bruising, worms, or other defects. If Stephen samples 10 apples from the shipment, use the binomial distribution to estimate the probability that his sample will contain at least one damaged apple. Give your answer as a decimal precise to at least four decimal places.

The subject of this question is Physics and the grade level is High School. The article describes a factorial experiment designed to determine factors in a high-energy electron beam process that affect hardness in metals. The experiment focuses on two factors: travel speed and beam current. The aim of the experiment is to determine how these factors affect hardness in metals.

In the given article, the subject of the question is physics, specifically the experimentation involving a high-energy electron beam process that affects hardness in metals. The experiment focuses on two factors: travel speed and beam current. The travel speed is measured in mm/s and has three levels: 10, 20, and 30. The beam current is a continuous variable that can be adjusted and is a factor of two. The experiment aims to determine how these factors affect hardness in metals.

Answer: 90% confidence interval would be (6.022,6.986).

Step-by-step explanation:

Since we have given that

Number of students = 361

Sample mean = 6.504

Sample standard deviation = 5.584

Standard error of the mean = 0.294

At 90% confidence level,

So, α = 0.01

So, z = 1.64

Margin of error is given by

[tex]z\times \text{Standard error}\\\\=1.64\times 0.294\\\\=0.48216[/tex]

So, Lower limit would be

[tex]\bar{x}-0.482\\\\=6.504-0.482\\\\=6.022[/tex]

Upper limit would be

[tex]\bar{x}+0.482\\\\=6.504+0.482\\\\=6.986[/tex]

So, 90% confidence interval would be (6.022,6.986).

Answer:

95.4

Step-by-step explanation:

The overall average score would be the total score divided by the total number of students. The total score is the sum of the product of the average of each student body and their average score

[tex]E(x) = \frac{E_1n_1 + E_2n_2+E_3n_3}{n_1+n_2+n_3}[/tex]

[tex]E(x) = \frac{100*60+140*67+110*77}{250} = \frac{23850}{250} = 95.4[/tex]

I can't really read the formula so I'll give a lecture.

With two vectors we can make a parallelogram.  If the vectors are u and v, and one vertex is the origin O, the others are O+u, O+v, and O+v+u.   If we draw any diagonal of the parallelogram, that divides it into two congruent triangles.  So each triangle is half the area of parallelogram.

The signed area of the parallelogram is given by the two D cross product of the vectors, aka the determinant.   So the area of a triangle with vertices (0,0), (a,b) and (c,d) is

[tex]A = \frac 1 2 |ad - bc| = \left| \frac 1 2 \begin{vmatrix} a & b \\ c & d\end{vmatrix} \right|[/tex]

Applying that to vertices

[tex](0,0), (x_1, 71), (x_2, 92)[/tex]

we get area

[tex]A = \frac 1 2 |\ 92x_1 - 71x_2|[/tex]

That formula is accurate for any values of x₁ and x₂

b. (1,2),(3,-1)

[tex]A=\frac 1 2|1(-1)-2(3)| = \frac 1 2 |-7| = \frac 7 2[/tex]

Answer: 7/2

Answer: Yes, this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood.

Step-by-step explanation:

Since we have given n = 157

x = 86

So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{86}{157}=0.55[/tex]

and we have p = 0.4

So, hypothesis would be

[tex]H_0:p=\hat{p}\\\\H_a:p\neq \hat{p}[/tex]

Since there is 1% level of significance.

So, test statistic value would be

[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.55-0.40}{\sqrt{\dfrac{0.4\times 0.6}{157}}}\\\\z=\dfrac{0.15}{0.039}\\\\z=3.846[/tex]

and the critical value at 1% level of significance , z = 2.58

Since 2.58<3.846.

So, we reject the null hypothesis.

Hence, Yes, this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood.

We conducted a hypothesis test and found that the actual percentage of type A blood donations significantly differs from 40%, leading to the rejection of the null hypothesis at a 0.01 significance level.

Hypothesis Testing for Blood Type Proportion

We will perform a hypothesis test to determine whether the percentage of type A blood donations differs from 40%. The appropriate hypotheses for this test are:

Next, we calculate the test statistic for the sample proportion:

The critical value for a two-tailed test at a significance level of 0.01 is approximately ±2.576.

Since 3.79 > 2.576, we reject the null hypothesis (H0). Therefore, the data suggests that the actual percentage of type A donations differs significantly from 40%.

Answer: a) 0.079589 b) 0.079656

Step-by-step explanation:

Since we have given that

Number of times a coin is flipped = 100 times

Number of times he get exactly head = 50

Probability of getting head = [tex]\dfrac{1}{2}[/tex]

We will use "Binomial distribution":

Probability would be

[tex]^{100}C_{50}(\dfrac{1}{2})^{50}(\dfrac{1}{2})^50\\\\=0.079589[/tex]

Using "Normal approximation":

n = 100

p = 0.5

So, mean = [tex]np=100\times 0.5=50[/tex]

Standard deviation is given by

[tex]\sqrt{np(1-p)}\\\\=\sqrt{50(1.05)}\\\\=\sqrt{50\times 0.5}\\\\=\sqrt{25}\\\\=5[/tex]

So,

[tex]P(X<x)=P(Z<\dfrac{\bar{x}-\mu}{\sigma})\\\\So, P(X=50)=P(49.5<X<50.5)\\\\=P(\dfrac{49.5-50}{5}<Z<\dfrac{50.5-50}{5})\\\\=P(-0.1<Z<0.1)\\\\=P(Z<0.1)-P(Z<-0.1)\\\\=0.539828-0.460172\\\\=0.079656[/tex]

Hence, a) 0.079589 b) 0.079656

Answer:

\mu = 14.5\\

\sigma = 5.071\\

k = 1.084

Step-by-step explanation:

given that a  statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20.

i.e. his findings with respect to probability are

[tex]P(9<x<20) \geq 0.15\\P(|x-14.5|<5.5) \geq 0.15[/tex]

Recall Chebyshev's inequality that

[tex]P(|X-\mu |\geq k\sigma )\leq {\frac {1}{k^{2}}}\\P(|X-\mu |\leq k\sigma )\geq 1-{\frac {1}{k^{2}}}\\[/tex]

Comparing with the Ii equation which is appropriate here we find that

[tex]\mu =14.5[/tex]

Next what we find is

[tex]k\sigma = 5.5\\1-\frac{1}{k^2} =0.15\\\frac{1}{k^2}=0.85\\k=1.084\\1.084 (\sigma) = 5.5\\\sigma = 5.071[/tex]

Thus from the given information we find that

[tex]\mu = 14.5\\\sigma = 5.071\\k = 1.084[/tex]

The Correlation coefficient is r = -0.992324879

Step-by-step explanation:

The figure shown is the calculation of all data

Given data :

X y

1 42.2

2 42.14

3 39.38

4 38.22

5 37.46

6 35.2

7 30.24

8 29.38

9 25.92

10 24.86

11 24.8

12 21.04

13 20.98

14 15.42

15 15.46

16 13.4

Step1: Find X mean, Y mean

Mean  = [tex]\frac{\sum x }{N}[/tex]

X mean = [tex]\frac{1+2+3+4+5+...+16}{16}[/tex]

X mean = 8.5

Similarly, we get

Y mean = [tex]\frac{42.2+42.14+39.38+...+15.42+15.46+13.4}{16}[/tex]

Y mean = 28.50625

Step2: Find Standard deviation

Standard deviation = [tex]\sqrt {\frac{ \sum (Xi-X mean)^{2} }{N-1}}[/tex]

Sx = [tex]\sqrt {\frac{ (-7.5)^{2} +  (-6.5)^{2} +(-5.5)^{2} ....+  (7.5)^{2}  }{16-1}}[/tex]

Sx= 4.760952286

Similarly, we get

Sy= 9.765590527

Step3: Find Correlation coefficient

The formula for the Correlation coefficient is given as  

[tex]r=\frac{1}{N-1} \sum(\frac{Xi-X mean }{Sx} ) (\frac{Yi- Y mean }{Sy} )[/tex]

[tex]r=\frac{1}{16-1} \sum(\frac{Xi-8.5 }{4.760952286} ) (\frac{Yi- 28.50625 }{9.765590527} )[/tex]

[tex]r= -0.992324879[/tex]

Thus, The Correlation coefficient is r = -0.992324879

Answer:

The p value obtained was a low value and using 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 of driver fatalities related to alcohol is less from 0.77 or 77%.  

Step-by-step explanation:

1) Data given and notation n  

n=53 represent the random sample taken

X=35 represent the automobile driver fatalities in a certain county involved with an intoxicated driver

[tex]\hat p=\frac{35}{53}=0.660[/tex] estimated proportion of automobile driver fatalities in a certain county involved with an intoxicated driver

[tex]p_o=0.77[/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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the population proportion of driver fatalities related to alcohol is less than 77% or 0.77 in Kit Carson:  

Null hypothesis:[tex]p\geq 0.77[/tex]  

Alternative hypothesis:[tex]p < 0.77[/tex]  

When we conduct a proportion test we need to use the z statisitc, 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].

3) Calculate the statistic  

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

[tex]z=\frac{0.660 -0.77}{\sqrt{\frac{0.77(1-0.77)}{53}}}=-1.903[/tex]

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This methos 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 an unilateral lower test the p value would be:  

[tex]p_v =P(z<-1.903)=0.0285[/tex]  

So the p value obtained was a low value and using 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 of driver fatalities related to alcohol is less from 0.77 or 77%.  

Answer:

-1/3

Step-by-step explanation:

rise / run = slope

4-5/ 1-(-2)

= -1/3

Final answer:

The experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.

Explanation:

To test if the experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period, we can conduct a hypothesis test.

H0 (null hypothesis): The proportion of batteries that fail during the advertised time period is equal to or greater than 0.2 percent.

H1 (alternative hypothesis): The proportion of batteries that fail during the advertised time period is less than 0.2 percent.

We can use a one-sample proportion test to compare the proportion of batteries that fail in the sample to the claimed proportion. The test statistic for this hypothesis test is the z-test statistic.

The critical value(s) for a one-sided test with α = 0.01 is z = -2.33.

The critical (rejection) region is z < -2.33.

The p-value is calculated by finding the probability of observing 15 or fewer failures out of 5000 batteries assuming that the null hypothesis is true. Using a normal approximation, we can calculate the p-value as the probability of observing x ≤ 15, where x is the number of failures. We find the p-value is less than 0.01.

Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. We have sufficient evidence to support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.

Answer: The 95% confidence interval estimate of the population mean is (1760, 1956) .

Step-by-step explanation:

Formula for confidence interval for population mean([tex](\mu)[/tex]) :

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

, where n= Sample size

[tex]\overline{x}[/tex] = sample mean.

[tex]z^*[/tex] = Two-tailed critical z-value

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

By considering the given information, we have

n= 81

[tex]\sigma=450 [/tex] kilowatt-hours.

[tex]\overline{x}=1858[/tex] kilowatt-hours.

By using the z-value table ,

The critical values for 95% confidence interval : [tex]z^*=\pm1.960[/tex]

Now , the 95% confidence interval estimate of the population mean will be :

[tex]1858\pm (1.960)\dfrac{450}{\sqrt{81}}\\\\=1858\pm(1.960)\dfrac{450}{9}=1858\pm98\\\\=(1858-98,\ 1858+98)\\\\=(1760,\ 1956)[/tex]

Hence, the 95% confidence interval estimate of the population mean is (1760, 1956) .

Answer:

Step-by-step explanation:

Convert 7.2 to grams then divide the grams evenlly to the 24 people then you get 300

a. [tex]X[/tex], [tex]N_1[/tex], and [tex]N_2[/tex] each have mean 0, and by linearity of expectation we have

[tex]E[R_1]=E[X+N_1]=E[X]+E[N_1]=0[/tex]

[tex]E[R_2]=E[X+N_2]=E[X]+E[N_2]=0[/tex]

b. By definition of correlation, we have

[tex]\mathrm{Corr}[R_1,R_2]=\dfrac{\mathrm{Cov}[R_1,R_2]}{{\sigma_{R_1}}{\sigma_{R_2}}}[/tex]

where [tex]\mathrm{Cov}[/tex] denotes the covariance,

[tex]\mathrm{Cov}[R_1,R_2]=E[(R_1-E[R_1])(R_2-E[R_2])][/tex]

[tex]=E[R_1R_2]-E[R_1]E[R_2][/tex]

[tex]=E[R_1R_2][/tex]

[tex]=E[(X+N_1)(X+N_2)][/tex]

[tex]=E[X^2]+E[N_1X]+E[XN_2]+E[N_1N_2][/tex]

Because [tex]X,N_1,N_2[/tex] are mutually independent, the expectation of their products distributes over the factors:

[tex]\mathrm{Cov}[R_1,R_2]=E[X^2]+E[N_1]E[X]+E[X]E[N_2]+E[N_1]E[N_2][/tex]

[tex]=E[X^2][/tex]

and recall that variance is given by

[tex]\mathrm{Var}[X]=E[(X-E[X])^2][/tex]

[tex]=E[X^2]-E[X]^2[/tex]

so that in this case, the second moment [tex]E[X^2][/tex] is exactly the variance of [tex]X[/tex],

[tex]\mathrm{Cov}[R_1,R_2]=E[X^2]={\sigma_X}^2[/tex]

We also have

[tex]{\sigma_{R_1}}^2=\mathrm{Var}[R_1]=\mathrm{Var}[X+N_1]=\mathrm{Var}[X]+\mathrm{Var}[N_1]={\sigma_X}^2+{\sigma_{N_1}}^2[/tex]

and similarly,

[tex]{\sigma_{R_2}}^2={\sigma_X}^2+{\sigma_{N_2}}^2[/tex]

So, the correlation is

[tex]\mathrm{Corr}[R_1,R_2]=\dfrac{{\sigma_X}^2}{\sqrt{\left({\sigma_X}^2+{\sigma_{N_1}}^2\right)\left({\sigma_X}^2+{\sigma_{N_2}}^2\right)}}[/tex]

c. The variance of [tex]R_1+R_2[/tex] is

[tex]{\sigma_{R_1+R_2}}^2=\mathrm{Var}[R_1+R_2][/tex]

[tex]=\mathrm{Var}[2X+N_1+N_2][/tex]

[tex]=4\mathrm{Var}[X]+\mathrm{Var}[N_1]+\mathrm{Var}[N_2][/tex]

[tex]=4{\sigma_X}^2+{\sigma_{N_1}}^2+{\sigma_{N_2}}^2[/tex]

Answer:

P = 1/6 or 0.1667

Step-by-step explanation:

Let A,B and C be the three events and 1,2 and 3 be the three dates. The sample space for this problem is:

{A1,B2,C3} {A1,B3,C2} {A2,B1,C3} {A2,B3,C1} {A3,B1,C2} {A3,B2,C1}

There are six possible outcomes and since there is only one correct answer, the possibility of guessing all 3 questions right is:

P = 1/6

P = 0.1667

Answer:

its 114

Step-by-step explanation:

180-66=114 so that's it

Answer:

0.1587

Step-by-step explanation:

Let X be the random variable that represents a return on long-term government bond. We know that X has a mean of 6.0 and a standard deviation of 9.9, in order to compute the approximate probability that your return on these bonds will be less than -3.9 percent in a given year, we should compute the z-score related to -3.9, i.e., (-3.9-6.0)/9.9 =  -1. Therefore, we are looking for P(Z < -1) = 0.1587

The question is about calculating the mean and standard deviation of a discrete random variable representing the size of the freezer model a customer buys, and then the mean and standard deviation of the price paid, which is a function of the freezer size.

The random variable X in this scenario represents the size of the freezer model (in cubic feet) that the next customer buys from the appliance dealer. This is a discrete random variable, with possible values being the sizes of the three available freezer models: 14.5, 16.9, and 19.1 cubic feet.

To calculate the mean and standard deviation of X, we can use the formula for the mean of a discrete random variable and the formula for the standard deviation respectively, which are given as follows:
Mean (expected value), E(X) = Σ [x * p(x)]
Standard Deviation, σ(X) = sqrt{Σ [(x - μ)² * p(x)]}

The mean value of the second variable (Price) can be calculated by plugging the expected value of X into the given Price = 25X - 8.5 equation, and the standard deviation by applying the transformation rule for standard deviations (σ_Y = |b| * σ_X, where b is the coefficient of X in the original expression for Y, which in this case is 25).

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The question involves calculating the probabilities of different combinations of females and males being selected for positions using the hypergeometric distribution, a concept in mathematics related to probability without replacement from two distinct groups.

The question is related to the concept of probability in mathematics, specifically the use of the hypergeometric distribution, which is appropriate when sampling without replacement from a finite population divided into two groups. To calculate the probability of various combinations of female and male applicants selected for the positions, we need to use the hypergeometric probability formula:

These calculations involve combinations (denoted as C(n, k)) and the formula for hypergeometric probability, which is:

Hypergeometric Probability = [C(group1 size, group1 subset size) * C(group2 size, group2 subset size)] / C(total population size, total subset size)

Answer:

Since p value <0.1 accept the claim that oven I repair costs are more

Step-by-step explanation:

The data given for two types of ovens are summarised below:

Group   Group One     Group Two  

Mean 85.7900 78.6700

SD 15.1300 17.8400

SEM 1.9533 2.3840

N 60       56      

Alpha = 10%

[tex]H_0: \mu_1 - \mu_2 =0\\H_a: \mu_1 - \mu_2> 0[/tex]

(Right tailed test)

The mean of Group One minus Group Two equals 7.1200

df = 114

 standard error of difference = 3.065

 t = 2.3234

p value = 0.0219

If p value <0.10 reject null hypothesis

4) Since p value <0.1 accept the claim that oven I repair costs are more

Answer:

398.411

Step-by-step explanation:

Explanation has been given in the following attachments.

Final answer:

(a) Using the differential equation[tex]\( dy/dt = ky(1 - y/K) \)[/tex]  with[tex]\( y(0) = 2 \times 10^7 \)[/tex] kg, the biomass after one year is approximately [tex]\( 3.11 \times 10^7 \) kg.[/tex]

(b) Solving the equation for [tex]\( y(t) = 4 \times 10^7 \) kg yields \( t \approx 3.41 \) years.[/tex]

Explanation:

(a) To find the biomass a year later, we can use the given initial condition and the provided differential equation. Substituting[tex]\( y(0) = 2 \times 10^7 \) kg, \( k = 0.75 \), and \( K = 9 \times 10^7 \)[/tex] kg into the equation, we get:

[tex]\[ \frac{{dy}}{{dt}} = 0.75y \left(1 - \frac{{y}}{{9 \times 10^7}}\right) \][/tex]

Now, we can solve this first-order ordinary differential equation. One method is the separation of variables, then integration. Integrating both sides gives:

[tex]\[ \int \frac{{dy}}{{y(1 - \frac{{y}}{{9 \times 10^7}})}} = \int 0.75 dt \][/tex]

After solving the integrals and applying the initial condition, we find [tex]\( y(1) \approx 3.11 \times 10^7 \) kg.[/tex]

(b) To determine how long it takes for the biomass to reach [tex]\( 4 \times 10^7 \) kg,[/tex] we can use the same differential equation and solve for [tex]\( t \)[/tex] when [tex]\( y(t) = 4 \times 10^7 \) kg.[/tex]This involves solving a separable differential equation and then finding the time [tex]\( t \)[/tex] that satisfies this condition. By solving this equation, we find[tex]\( t \approx 3.41 \) years.[/tex]

Answer:

Part (A): The total number of ways are [tex]4^{100}[/tex]

Part (B): The total number of ways are [tex]1.6122075076\times10^{57}[/tex]

Step-by-step explanation:

Consider the provided information.

Part(A) a) How many ways are there for her to plan her schedule if there are no restrictions on the number of days she studies each of the four subjects?

She plans on studying four different subjects.

That means she has 4 choices for each day also there is no restriction on the number of days.

Hence, the total number of ways are:

[tex]4\times 4\times4\times4\times......4=4^{100}[/tex]

Part (B) How many ways are there for her to plan her schedule if she decides that the number of days she studies each subject will be the same?

She wants that the number of day she studies each subject will be the same that means she studies for 25 days each subject.

Therefore, the number of ways are:

[tex]^{100}C_{25}\times ^{75}C_{25}\times ^{50}C_{25}\times ^{25}C_{25}=1.6122075076\times10^{57}[/tex]

In the case of no restrictions, there are 4^100 possible study schedules. If each subject is studied the same number of days, the number of possible schedules is found using a permutation with repetitions of multiset formula, resulting in 100! / (25! * 25! * 25! * 25!).

This question focuses on combinatorics which is a branch of mathematics that deals with combinations of objects belonging to a finite set. It features the tasks of counting different objects as per certain restrictions, arrangement of objects with regard to considered conditions, and selection of objects satisfying specific criteria.

For part (a), if there are no restrictions, then she could pick any of the four subjects to study each day. For each day, there are 4 choices, so for 100 days, that results in 4^100 possible schedules.

For part (b), if she wants to study each subject the same number of days, then each subject would be studied for 100/4=25 days. This situation is a permutation with repetitions of multiset, which can be found through the formula n!/(r1! * r2! * ... * rk!) where n is total number of items, and r1, r2, ..., rk are numbers of same elements of multiset. Using this equation, the number of ways would be 100! / (25! * 25! * 25! * 25!).

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Answer:

a) The 90% confidence interval would be given by (33.594;39.206)  

b) Since the 90% confidence interval not contains the value 40 we can say that this value at this confidence level is not the true population mean, because it's outside of the limits for the interval calculated.

Step-by-step explanation:

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".  

Part a

[tex]\bar X=36.4[/tex] represent the sample mean  

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

[tex]\sigma=12.1[/tex] represent the population standard deviation  

n=50 represent the sample size  

90% confidence interval  

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

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

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]  

Now we have everything in order to replace into formula (1):  

[tex]36.4-1.64\frac{12.1}{\sqrt{50}}=33.594[/tex]  

[tex]36.4+1.64\frac{12.1}{\sqrt{50}}=39.206[/tex]  

So on this case the 90% confidence interval would be given by (33.594;39.206)  

Part b

Since the 90% confidence interval not contains the value 40 we can say that this value at this confidence level is not the true population mean, because it's outside of the limits for the interval calculated.

Answer:38 min

Step-by-step explanation:

Given

Cutting time [tex]t_1=14 min[/tex]

Assembly Line time [tex]t_2=24 min[/tex]

[tex]t_3=18[/tex] to move a batch of rugs from cutting to assembly

Value-added refers to a process or phase in a system that turns raw materials or work in progress into much more desirable goods and services for downstream consumers.

thus value added time is [tex]t_1+t_2=14+24 =38 min[/tex]                

Answer: 98% confidence interval is (0.06,0.11).

Step-by-step explanation:

Since we have given that

Number of men = 550

Number of women = 2400

Number of men have color blindness = 48

Number of women have color blindness = 5

So, [tex]p_M=\dfrac{48}{550}=0.087\\\\p_F=\dfrac{5}{2400}=0.0021[/tex]

So, at 98% confidence interval, z = 2.326

so, interval would be

[tex](p_M-P_F)\pm z\sqrt{\dfrac{p_M(1-p_M)}{n_M}+\dfrac{p_F(1-p_F)}{n_F}}\\\\=(0.087-0.0021)\pm 2.326\sqrt{\dfrac{0.087\times 0.912}{550}+\dfrac{0.0021\times 0.9979}{2400}}\\\\=0.0849\pm 2.326\times 0.012\\\\=(0.0849-0.027912, 0.0849+0.027912)\\\\=(0.06,0.11)[/tex]

Hence, 98% confidence interval is (0.06,0.11).

No, there does not appears to be a significant difference.

Answer:

Work = e+24

F is not conservative.

Step-by-step explanation:

To find the work required to move an object in the force field  

[tex]\large F(x,y,z)=(e^{x+y},e^{x+y},ze^{x+y})[/tex]

along the straight line from A(0,0,0) to B(-1,2,-5), we have to parameterize this segment.

Given two points P, Q in any euclidean space, you can always parameterize the segment of line that goes from P to Q with

r(t) = tQ + (1-t)P with 0 ≤ t ≤ 1

so  

r(t) = t(-1,2,-5) + (1-t)(0,0,0) = (-t, 2t, -5t)  with 0≤ t ≤ 1

is a parameterization of the segment.

the work W required to move an object in the force field F along the straight line from A to B is the line integral

[tex]\large W=\int_{C}Fdr [/tex]

where C is the segment that goes from A to B.

[tex]\large \int_{C}Fdr =\int_{0}^{1}F(r(t))\circ r'(t)dt=\int_{0}^{1}F(-t,2t,-5t)\circ (-1,2,-5)dt=\\\\=\int_{0}^{1}(e^t,e^t,-5te^t)\circ (-1,2,-5)dt=\int_{0}^{1}(-e^t+2e^t+25te^t)dt=\\\\\int_{0}^{1}e^tdt-25\int_{0}^{1}te^tdt=(e-1)+25\int_{0}^{1}te^tdt[/tex]

Integrating by parts the last integral:

[tex]\large \int_{0}^{1}te^tdt=e-\int_{0}^{1}e^tdt=e-(e-1)=1[/tex]

and  

[tex]\large \boxed{W=\int_{C}Fdr=e+24}[/tex]

To show that F is not conservative, we could find another path D from A to B such that the work to move the particle from A to B along D is different to e+24

Now, let D be the path consisting on the segment that goes from A to (1,0,0) and then the segment from (1,0,0) to B.

The segment that goes from A to (1,0,0) can be parameterized as  

r(t) = (t,0,0) with 0≤ t ≤ 1

so the work required to move the particle from A to (1,0,0) is

[tex]\large \int_{0}^{1}(e^t,e^t,0)\circ (1,0,0)dt =\int_{0}^{1}e^tdt=e-1 [/tex]

The segment that goes from (1,0,0) to B can be parameterized as  

r(t) = (1-2t,2t,-5t) with 0≤ t ≤ 1

so the work required to move the particle from (1,0,0) to B is

[tex]\large \int_{0}^{1}(e,e,-5et)\circ (-2,2,-5)dt =25e\int_{0}^{1}tdt=\frac{25e}{2}[/tex]

Hence, the work required to move the particle from A to B along D is

e - 1 + (25e)/2 = (27e)/2 -1

since this result differs from e+24, the force field F is not conservative.

Answer:

A recursive rule for the sequence is f(1) = -8; f(n) = -4 (n – 1) for all n ≥ 2 is "FALSE"

An explicit rule for the sequence is f(n) = -8 + 4 (n – 1) is "TRUE"

The tenth term is 28 is "TRUE"

Step-by-step explanation:

Statement (1)

While the first part [f(1) = –8] is TRUE, the second part [f(n) = –4 (n – 1) for all n ≥ 2] would only be true if the sequence ends at the second term.

Check: Since the fifth term of the sequence is 8, then f(5) = 8

From the statement,

f(5) = –4 (5 – 1)

f(5) = –4 × 4 = –16

:. f(5) ≠ 8

Statement (2)

f(n) = –8 + 4 (n – 1) is TRUE

Check: The fifth term of the sequence is 8 [f(5) = 8]

From the statement,

f(5) = –8 + 4 (5 – 1)

f(5) = –8 + 4 (4)

f(5) = –8 + 16 = 8

:. f(5) = 8

Statement (3)

f(10) = 28 is TRUE

Since the explicit rule is TRUE, use to confirm if f(10) = 28:

f(10) = –8 + 4 (10 – 1)

f(10) = –8 + 4 (9)

f(10) = –8 + 36

f(10) = 28

:. f(10) = 28

Answer:

Reject null hypothesis

Step-by-step explanation:

a) [tex]H_0: p =0.05\\H_a: p >0.05[/tex]

(Right tailed test at 5% level)

p = risk proportion

Sample proportion = [tex]\frac{46}{384} \\\\=0.1198[/tex]

Std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.05(0.95)}{384} } \\=0.0111[/tex]

p difference = [tex]0.1198-0.05=0.0698[/tex]

b) Sampling proportion is Normal with mean = 0.05 and std error = 0.0698

c) Z = test statistic = p diff/std error

= 6.29

p value <0.05

d) Since p < alpha we reject null hypothesis.

e) The probability that null hypothesis is rejected when true is negligible =p value

Answer:

Step-by-step explanation:

Reject null hypothesis

Answer:

C i think i already answerd this

Step-by-step explanation:

The music professor attacking the test is validity. College courses in music-related subjects, such as voice, instruments, music appreciation, theory, and performance, are taught by music professors.

College courses in music-related subjects, such as voice, instruments, music appreciation, theory, and performance, are taught by music professors. They might instruct sophisticated, highly specialized courses or wide, basic ones. They design curricula, teach classes, and evaluate students' work.

Other names for music teachers include: music educator, band teacher, choir teacher, piano teacher, violin teacher, guitar teacher, elementary school music teacher, high school music teacher, and vocal teacher.

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To learn more about music professor refer to:

https://brainly.com/question/16462757

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Answer:

We conclude that  the mean number of residents in the retirement community household is less than 3.36 persons.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 3.36

Sample mean, [tex]\bar{x}[/tex] = 2.71

Sample size, n =  25

Alpha, α = 0.10

Sample standard deviation, s = 1.10

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 3.36\text{ residents per household}\\H_A: \mu < 3.36\text{ residents per household}[/tex]

We use One-tailed(left) t test to perform this hypothesis.

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{2.71 - 3.36}{\frac{1.10}{\sqrt{25}} } = -2.95[/tex]

Now, [tex]t_{critical} \text{ at 0.10 level of significance, 24 degree of freedom } =-1.31[/tex]

Since,                  

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

We reject the null hypothesis and fail to accept it.

Thus, we conclude that  the mean number of residents in the retirement community household is less than 3.36 persons.

The null hypothesis for testing if the mean number of residents in Arizona retirement community households is less than the national average is that the mean is equal to or greater than 3.36 residents. The alternate hypothesis is that it is less than 3.36 residents. A one-sample t-test would typically be used to test these hypotheses.

Conducting a hypothesis test to determine if the mean number of residents in retirement community households in Arizona is significantly less than the national average reported by the Census Bureau, which is 3.36 residents per household. To address this,

The null hypothesis (H0) is that the mean number of residents per household in the Arizona retirement communities is equal to or greater than the national average, = 3.36 residents.

The alternate hypothesis (Ha) is that the mean number of residents per household in the Arizona retirement communities is less than the national average, < 3.36 residents.

To test this hypothesis at a 0.10 significance level, we would typically use a one-sample t-test to compare the sample mean (2.71 residents) against the national average (3.36 residents) considering the sample standard deviation (1.10 residents) and sample size (25 households).

The hypothesis test would determine if the observed difference is statistically significant, implying that the mean number of residents in retirement community households is indeed less than 3.36 persons.