Suppose that the average number of cars abandoned weekly on a certain highway is 2.2. Approximate the probability that there will be:

a. No abandoned cars in the next week
b. At least 2 abandoned cars in the next week.

The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .

The Laplace transform exist when s > 0 .

Here, the given function is f(t) = sin²(wt) .

The Laplace transform of the the function f(t),

F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }

F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }

F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }

F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]

Next,

The above Laplace transform exist if s > 0 .

Know more about Laplace transform,

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Answer:

97% Confidence interval = (12.62, 18.98)

Step-by-step explanation:

Complete Question

Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 97% confidence interval for the mean germination time.

Solution

We first compute the sample mean and standard deviation for this sample distribution

Sample mean = (Σx)/N = (158/10) = 15.8

Standard deviation = √{[Σ(x - xbar)²]/(N-1)} = 3.3598941782278 = 3.36

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 15.8

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 10 - 1 = 9

Significance level for 97% confidence interval

(100% - 97%)/2 = 1.5% = 0.015

t (0.015, 9) = 2.9982 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 3.36

n = sample size = 10

σₓ = (3.36/√10) = 1.0625

97% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 15.8 ± (2.9982 × 1.0625)

CI = 15.8 ± 3.1809882246

97% CI = (12.6190117754, 18.9809882246)

97% Confidence interval = (12.62, 18.98)

Hope this Helps!!!

Answer:

1)

A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552

C) P(X> 9)= 0.36317

D) IQR= 0.4422

2)

A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)

B) P( [tex]\frac{}{X}[/tex]<30)= 0.50

C) P₉₅= 32.60

D) P( [tex]\frac{}{X}[/tex]>36)= 0

E) Q₃: 31.0586

Step-by-step explanation:

Hello!

1)

The variable of interest is

X: pollutants found in waterways near a large city. (ppm)

This variable has a normal distribution:

X~N(μ;σ²)

μ= 8.5 ppm

σ= 1.4 ppm

A sample of 18 large cities were studied.

A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

The population mean is the same as the mean of the variable

μ= 8.5 ppm

The standard deviation is

σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108

So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B)

P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)

To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.

Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]

Then using the Z table you'll find the probability of

P(Z≤1.51)= 0.93448

Then

1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552

C)

In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:

P(X> 9)= 1 - P(X ≤ 9)

Z= (X-μ)/δ= (9-8.5)/1.44

Z= 0.347= 0.35

P(Z≤0.35)= 0.63683

Then

P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317

D)

The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:

Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25

Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:

P(Z≤z₁)= 0.25

Using the table you have to identify the value of Z that accumulates 0.25 of probability:

z₁= -0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)

z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)

[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5=  8.2789 ppm

The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

Using the table you have to identify the value of Z that accumulates 0.75 of probability:

z₃= 0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5=  8.7211 ppm

IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422

2)

A)

X ~ N(30,10)

For n=4

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

Population mean μ= 30

Population variance σ²/n= 10/4= 2.5

Population standard deviation σ/√n= √2.5= 1.58

[tex]\frac{}{X}[/tex] ~ N(30;2.5)

B)

P( [tex]\frac{}{X}[/tex]<30)

First you have to standardize the value and then look for the probability:

Z=  ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0

P(Z<0)= 0.50

Then

P( [tex]\frac{}{X}[/tex]<30)= 0.50

Which is no surprise since 30 y the value of the mean of the distribution.

C)

P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95

P( Z≤ z₀)= 0.95

z₀= 1.645

Now you have to reverse the standardization:

z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)

z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)

[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60

P₉₅= 32.60

D)

P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0

E)

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

z₃= 0.67

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586

Q₃: 31.0586

Answer:

a)[tex]A(t)=2000e^{0.085t}[/tex]

b)[tex]A'(t)=170e^{0.085t}[/tex]

c)$3059.1808

d)t=4.77 years

e) The balance growing is $254.99/year

Step-by-step explanation:

We are given that Two thousand dollars is deposited into a savings account at 8.5​% interest compounded continuously.

Principal = $2000

Rate of interest = 8.5%

a) What is the formula for​ A(t), the balance after t​ years? ​

Formula [tex]A(t)=Pe^{rt}[/tex]

So,[tex]A(t)=2000e^{0.085t}[/tex]

B)What differential equation is satisfied by​ A(t), the balance after t​ years?

So, [tex]A'(t)=2000 \times 0.085 e^{0.085t}[/tex]

[tex]A'(t)=170e^{0.085t}[/tex]

c)How much money will be in the account after 5 ​years? ​

Substitute t = 5 in the formula "

[tex]A(t)=2000e^{0.085t}\\A(5)=2000e^{0.085(5)}\\A(5)=3059.1808[/tex]

d)When will the balance reach ​$3000​?

Substitute A(t)=3000

So, [tex]3000=2000e^{0.085t}[/tex]

t=4.77

The balance reach $3000 in 4.77 years

e)How fast is the balance growing when it reaches ​$3000​?

Substitute the value of t = 4.77 in derivative formula :

[tex]A'(t)=170e^{0.085t}\\A'(t)=170e^{0.085 \times 4.77}\\A'(t)=254.99[/tex]

Hence the balance growing is $254.99/year

Answer:

His weekly exercise time goal is 168 minutes.

Step-by-step explanation:

This question can be solved using a rule of three.

42 minutes is 25% = 0.25 of the total

x minutes is 100% = 1 of the total.

Then

42 minutes - 0.25

x minutes - 1

[tex]0.25x = 42[/tex]

[tex]x = \frac{42}{0.25}[/tex]

[tex]x =  168[/tex]

His weekly exercise time goal is 168 minutes.

Answer:

The degrees of freedom are given by:

[tex]df=n-1=317-1=316[/tex]

And replaicing we got:

[tex]29-2.1=26.9[/tex]    

[tex]29+2.1=31.1[/tex]    

The 95% confidence interval would be between 26.9 and 31.1

Step-by-step explanation:

Information given

[tex]\bar X= 29[/tex] represent the sample mean

[tex]\mu[/tex] population mean

s represent the sample standard deviation

[tex] ME= 2.1[/tex] represent the margin of error

n represent the sample size  

Solution

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

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

And this formula is equivalent to:

[tex] \bar X \pm ME[/te]x

The degrees of freedom are given by:

[tex]df=n-1=317-1=316[/tex]

And replaicing we got:

[tex]29-2.1=26.9[/tex]    

[tex]29+2.1=31.1[/tex]    

The 95% confidence interval would be between 26.9 and 31.1

Answer:

[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=40-1=39[/tex]  

Thep value for this case would be given by:

[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]  

Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.

Step-by-step explanation:

Information provided

[tex]\bar X=5.2[/tex] represent the sample mean

[tex]s=0.8[/tex] represent the sample standard deviation

[tex]n=40[/tex] sample size  

[tex]\mu_o =5[/tex] represent the value to verify

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

t would represent the statistic

[tex]p_v[/tex] represent the p value

Hypothesis to test

We want to verify if the true mean is higher than 5, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 5[/tex]  

Alternative hypothesis:[tex]\mu > 5[/tex]  

The statistic is given by:

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

Replacing the info given we got:

[tex]t=\frac{5.2-5}{\frac{0.8}{\sqrt{40}}}=1.58[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=40-1=39[/tex]  

Thep value for this case would be given by:

[tex]p_v =P(t_{(39)}>1.58)=0.061[/tex]  

Since the p value is higher than the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that true mean is not significantly higher than 5.

Answer:

2.86=28 tenths+6 hundredths.

Step-by-step explanation:

2.86=2 ones+8 tenths+6 hundredths.

2.86=28 tenths+6 hundredths.

Answer:

see

Step-by-step explanation:

ones .  tenths  hundredths

2      .     8          6

8 tenths

If we are not using the ones place

we have 28 tenths

Answer:

9.375 in^2

Step-by-step explanation:

Answer:

114

Step-by-step explanation:

Answer:

144Step-by-step explanation:

Answer:

a. 0.0668

b. 0.9545

Step-by-step explanation:

We have the following information:

mean (m) = 10000

standard deviation (sd) = 100

(a)

We must calculate the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength as follows:

P (x> 10150) = P [(x - m) / sd> (10150 - 1000 /) 100]

P (x> 10150) = P (z> 1.5)

P (x> 10150) = 1 - P (z <1.5)

P (x> 10150) = 1 - 0.9332 (attached table)

P (x> 10150) = 0.0668

Therefore the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength is 0.0668

(b)

We must calculate the proportion of all components has tensile strength between 9800 and 10200, as follows:

P (9800 <x <10200) = P [(9800 - 1000 /) 100 <(x - m) / sd <(10200 - 1000 /) 100]

P (9800 <x <10200) = P (-2 <z <2)

P (9800 <x <10200) = P (z <2) - P (z <-2)

P (9800 <x <10200) = 0.9773 - 0.0228 (attached table)

P (9800 <x <10200) = 0.9545

the proportion of pieces that would expect to scrap is 0.9545

Answer:

Option A is correct.

A uniform distribution.

Step-by-step explanation:

Complete Question

T-Mobile sells 6 different models of cell phones and have found that they sell an equal number of each model. The probability distribution that would describe this random variable is called:

A) Uniform Distribution

B) Continuous Distribution

C) Poisson Distribution

D) Relative Frequency Distribution

Solution

A uniform distribution is one in which all the variables have the same probability of occurring.

It is also known as a rectangular distribution, as every portion of the sample space has an equal chance of occurring, with equal length on the probability curve, leading to a rectangular probability curve.

And for this question, 6 different models of phones sell an equal number, hence, the probability of selling each model is equal to one another, hence, this is evidently a uniform distribution.

Hope this Helps!!!

Answer:

x = 3.4

Step-by-step explanation:

Step 1: Isolate x

0.5x = 1.7

Step 2: Divide both sides by 0.5

x = 3.4

And we have our final answer!

Answer: x=3.4

Step-by-step explanation:

[tex]0.5x+4.2=5.9[/tex]

multiply both sides by 10

[tex]5x+42=59[/tex]

subtract 42 on both sides

[tex]5x=17[/tex]

divide 5 on both sides

[tex]x=\frac{17}{5}[/tex] or

Simplify

x= 3.4

Answer: B.

x ≈2.5

Step-by-step explanation:

[tex]-\left(u\right)^{-1}-6=-u+10[/tex]

[tex]u=8-\sqrt{65},\:u=8+\sqrt{65}[/tex]

[tex]x=\frac{\ln \left(8+\sqrt{65}\right)}{\ln \left(3\right)}[/tex]

x=2.52...

Answer:

x=2.5

Step-by-step explanation:

Answer:

The statistical test to be used is the paired t-test.

Step-by-step explanation:

The dependent t-test (also known as the paired t-test or paired-samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference between these two means.

We use the paired t-test if we have two measurements on the same item, person or thing. We should also use this test if we have two items that are being measured with a unique condition.

For instance, an experimenter tests the effect of a medicine on a group of patients before and after giving the doses.

Similarly, in this case a paired t-test would be used to deter whether there was any changes in the cholesterol level within each group as result of the treatment.

Thus, the statistical test to be used is the paired t-test.

Answer:

-30000/100

300 0/1

Step-by-step explanation:

We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.

16 2/3 = 16.67

We do the multiplication:

−18⋅16 2/3 = -300

We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:

-30000/100

How mixed number would it be:

300 0/1

Answer:

slope = -4/5

Step-by-step explanation:

A line passes two points (x1, y1) and (x2, y2).

The slope of this line can be calculate by the formula:

s = (y2 - y1)/(x2 - x1)

=>The line that passes A(-4, 8) and B(-9, 12) has the slope:

s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5

Hope this helps!

Answer:

r = 7.5

Step-by-step explanation:

Circle equation: [tex](x - h)^2 + (y - k)^2 = r^2[/tex]

Since we are already give r², we simply just take the square root of 56.25, and we should get 7.5 as our final answer!

Answer:

2/5

Step-by-step explanation:

Answer:

0.03125 = 3.125% probability that the person flipped 5 heads

Step-by-step explanation:

For each coin, there are only two possible outcomes. Either it was heads, or it was tails. The result of a coin toss is independent of other coin tosses. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

Five coins:

This means that n = 5.

Fair coin:

Equally as likely to be heads or tails, so p = 0.5.

What is the probability that the person flipped 5 heads?

This is P(X = 5).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125[/tex]

0.03125 = 3.125% probability that the person flipped 5 heads

Answer:

I am assuming the underlined value is 25%. It is a parameter

Step-by-step explanation:

The value is is a parameter. This is because the parameter is a value that describes the population.

The survey carried out was a national survey of which there were 25% respondents who reported that someone had offered, sold, or given them an illegal drug on school property. It is not a statistics because a sample was not taken out of the population and a survey made on the sample.

The underlined 25% value is the value that summarizes the entire population of high school students

Answer:

x= 8.00 Interest rate on $14000

y= 7.50 Interest rate on $7000

Step-by-step explanation:

Let interest rate of $14000 be x%

and Interest rate for $7000 be y %

According to the first condition

14000 * x% - 7000 * y% = 595

multiply by 100

14000x-7000y = 59500

/700

20x-10y=85.................(1)

II condition

x%=y%+0.5%

x=y+0.5

x-y=0.5..................................(2)

solve (1) & (2)

20 x -10 y = 85 .............1

Total value

1 x -1 y = 0.50 .............2

Eliminate y

multiply (1)by 1

Multiply (2) by -10

20.00 x -10.00 y = 85.00

-10.00 x + 10.00 y = -5.00

Add the two equations

10.00 x = 80.00

/ 10.00

x = 8.00

plug value of x in (1)

20.00 x -10.00 y = 85.00

160.00 -10.00 y = 85.00

-10.00 y = 85.00 -160.00

-10.00 y = -75.00

y = 7.50

x= 8.00 Interest rate on $14000

y= 7.50 Interest rate on $7000

Answer:

A' - (8,2)

B' - (5,1)

C' - (4,2)

D' - (4,5)

Step-by-step explanation:

See attachment for the missing figure.

We can see that the vertices of the polygon ABCD have coordinates A(4,6), B(5,3), C(4,2) and D(1,2)

Polygon ABCD is rotated 90° clockwise about point C to form polygon A′B′C′D′ (see attached diagram), then

A'(8,2);

B'(5,1);

C' is the same as C, thus, C'(4,2);

D'(4,5).

Answer:

Check below, please

Step-by-step explanation:

Step-by-step explanation:

1.For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.)

When the derivative of a function is equal to zero, then it occurs when we have either a local minimum or a local maximum point. So for our x-coordinates we can say

 [tex]f'(x)=0\: at \:x=2, and\: x=-2[/tex]

2. For which values of x is f '(x) positive?

Whenever we have  

 [tex]f'(x)>0[/tex]

then function is increasing. Since if we could start tracing tangent lines over that graph, those tangent lines would point up.

 [tex]f'(x)>0 \:at [-4,-2) \:and\:(2, \infty)[/tex]

3. For which values of x is f '(x) negative?  

On the other hand, every time the function is decreasing its derivative would be negative. The opposite case of the previous explanation. So

 [tex]f'(x) <0 \: at\: [-2,2][/tex]

4.What do these values mean?

 [tex]f(x) \:is \:increasing\:when\:f'(x) >0\\\\f(x)\:is\:decreasing\:when f'(x)<0[/tex]

5.(b) For which values of x is f ''(x) zero?

In its inflection points, i.e. when the concavity of the curve changes. Since the function was not provided. There's no way to be precise, but roughly

at x=-4 and x=4

Answer:

Step-by-step explanation:

Given that,

[tex]n_1=9,x=520,s^2_x=38\\\\n_2=8,y=540,s^2_y=20[/tex]

a) Under null hypothesis H₀ : there is no difference between the variability of night shift and day shift workers

i.e [tex]H_0:\sigma^2_x=\sigma^2_y=\sigma^2[/tex]

Alternative hypothesis [tex]H_1:\sigma_x^2>\sigma_y^2[/tex]

Level of significance = 5% = 0.05

b) The test statistic

[tex]F=\frac{S^2_x}{S_y^2} \sim F(n_1-1,n_2-1)\\\\=\frac{38}{20}\\\\=1.9[/tex]

Table value of [tex]F_{0.05}(n_1-1,n_2-1)[/tex]

[tex]=F_{0.05}(9-1,8-1))\\\\=F_{0.05}(8,7)\\\\=3.726[/tex]

[tex]\therefore F_{calculated}=1.9<Tab F_{0.05}(8,7)=3.726[/tex]

[tex]H_0[/tex] is accepted at 5% level of significance

Therefore ,there is no difference between the variability of night shift and day shift worker

c) The P-value is 0.206356

The result is not significant at P < 0.05

d) At 95% confidence interval

[tex]\frac{\frac{S^2_x}{S^2_y} }{F_{t-\alpha/2(8,7)} } <\frac{\sigma^2_x}{\sigma^2_y}\frac{\frac{S^2_x}{S^2_y} }{F_{\alpha/2(8,7)} } \\\\\Rightarrow\frac{1.9}{F_{t-(0.05/2)}} <\frac{\sigma^2_x}{\sigma^2_y} <\frac{1.9}{F_{(0.05/2)}} \\\\\Rightarrow\frac{1.9}{F_{0.975}} <\frac{\sigma^2_x}{\sigma^2_y} <\frac{1.9}{F_{(0.025)}} \\\\\Rightarrow\frac{1.9}{F_{3.726}} <\frac{\sigma^2_x}{\sigma^2_y} <\frac{1.9}{F_{(0.286)}}[/tex]

Variance -ratio lies betwee (0.51,6.643)

Conclusion: There is not sufficient evidence to support the claim  that the night shift worker show more variability in their output levels, than the day workers at α =0.05

Answer:

Step-by-step explanation:

Hello!

The claim is that the variability of the output levels of the night-shift is greater than the variability in the output levels of the day workers.

Be

X₁: output level of night shift workers.

n₁= 9

X[bar]₁= 520

S₁= 38

X₂: output level of day shift workers.

n₂= 8

X[bar]₂= 540

S₂= 20

Considering both variables have a normal distribution, the parameters of interest are the population variances.

a)

H₀: σ₁² ≤ σ₂²

H₁: σ₁² > σ₂²

b)

To compare both variances you have to conduct a variance ratio test with statistic:

[tex]F= (\frac{S^2_1}{S^2_2} )*(\frac{Sigma^2_1}{Sigma^2_2} )~~F_{n_1-1;n_2-1}[/tex]

[tex]F_{H_0}= (\frac{1444}{400} )*1=3.61[/tex]

c)

The test is one tailed  to the right, the p-value will have the same direction, i.e. it will be in the right tail of the distribution. The F distribution has degrees of freedom:

n₁ - 1= 9 - 1= 8

n₂ - 1= 8 - 1= 7

P(F₈,₇ ≥ 3.61) = 1 - P(F₈,₇ < 3.61) = 1 - 0.9461= 0.0539

The p-value of this test is 0.0539

d)

The CI for the variance ratio is:

[tex][\frac{S^2_1/S_2^2}{F_{n_1-1;n_2-1;1-\alpha /2}}; \frac{S^2_1/S_2^2}{F_{n_1-1;n_2-1;\alpha /2}}][/tex]

[tex]F_{n_1-1;n_2-1;1-\alpha /2}= F_{8;7;0.975}= 4.90[/tex]

[tex]F_{n_1-1;n_2-1;\alpha /2}= F_{8;7;0.025}= 0.22[/tex]

[tex][\frac{1444/400}{4.90}}; \frac{1444/400}{0.22}}][/tex]

[0.736; 16.409]

Using the level of significance complementary to the confidence level of the interval, you can compare it to the p-value calculated in item c.

p-value: 0.0539

α: 0.05

The p-value is less than the significance level, the decision is to reject the null hypothesis. Using a 5% significance level you can conclude that the variance in the output levels of the night shift workers is greater than the variance in the output levels of the day shift workers.

Answer:

tiStep-by-step explanaon:

Answer: Rs. 40,000

Step-by-step explanation:

Let say Marked price of the Watch  

= M  Rs

Discount = 20 %

Discount = (20/100)M = 0.2M   Rs

Price after Discount = M - 0.2M =

Rs 0.8M

13 % Value added tax

=> VAT  = (13/100) * 0.8M  = 0.104M  Rs

Value of Watch = 0.8M + 0.104M    = 0.904M Rs

0.904M = 36160

=> M = 40000 

Marked Price of Watch = Rs 40,000

Answer:

a) Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.  

Step-by-step explanation:

a) The accountant, as he wants to see if there is evidence to support the claim that the mean guest bill has increased significanty, should write the hypothesis like that:

Null hypothesis (H0): the mean guest bill for a weekend is $600.

Alternative hypothesis (Ha): the mean guest bill for a weekend is significantly bigger than $600.

A sample of bills of the period in study needs to be taken in order to have a representation of the actual population of bills and then perform a t-test, as the sample mean and standard deviation will be used to perform the test.

b) When H0 can not be rejected, the conclusion is that there is no enough evidence to claim that the mean guest bill had increased from $600. If the P-value was low but not enough, they may take another sample to perform the test again or leave it like that.

c) When the H0 is rejected, they have enough evidence to claim that the mean guest bill is significantly bigger than $600.  

Answer:

9.42 in²

Step-by-step explanation:

The area of whole circle S=pi*R²    , where pi is appr. 3.14,  R= 6 in

S= 3.14*6² =113.04 in²

The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²

The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.

To find the area of a sector, you can use the formula:

Area of sector = (θ/360) × π × r²

where θ is the central angle in degrees, r is the radius of the sector.

The central angle is 30 degrees and the radius is 6 inches.

Plugging these values into the formula:

Area of sector = (30/360) × π × 6²

= (1/12) × π × 36

= (1/12) × 3.14159 × 36

= 9.42478 square inches

To learn more on Area of sector click:

https://brainly.com/question/29055300

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