Here are summary statistics for randomly selected weights of newborn​ girls: nequals=174174​, x overbarxequals=30.930.9 ​hg, sequals=7.57.5 hg. Construct a confidence interval estimate of the mean. Use a 9595​% confidence level. Are these results very different from the confidence interval 29.629.6 hgless than

Answer:

Confidence interval:  (1760,1956)

Step-by-step explanation:

We are given the following information in the question:

Sample size, n = 81

Sample mean =

[tex]\bar{x} = 1858 \text{ kWh}[/tex]

Population standard deviation =

[tex]\sigma = 450 \text{ kilowatt-hours}[/tex]

Confidence Level = 95%

Significance level = 5% = 0.05

Confidence interval:

[tex]\bar{x} \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96[/tex]

[tex]1858 \pm 1.96(\displaystyle\frac{450}{\sqrt{81}} ) = 1858 \pm 98 = (1760,1956)[/tex]

Answer:

[tex]x_0=-5\\x_1=-2\\x_2=1\\x_3=4\\x_4=7[/tex]

Δn=3

Step-by-step explanation:

Remember, if we need to divide the interval (a,b) in n equal subinterval, then we need divide the distance (d) between the endpoints of the interval and divide it by n. Then the width Δn of each subinterval is d/n.

We have the interval [-5,7]. The distance between the endpoints of the interval is

[tex]d=7-(-5)=12[/tex].

Now, we divide d by 4 and obtain [tex]\frac{d}{4}=\frac{12}{4}=3[/tex]

Then, Δn=3.

Now, to find the endpoints of each sub-interval, we add 3 from the left end of the interval.

[tex]-5=x_0\\x_0+3=-5+3=-2=x_1\\x_1+3=1=x_2\\x_2+3=4=x_3\\x_3+3=7=x_4[/tex]

So,

[tex]x_0=-5\\x_1=-2\\x_2=1\\x_3=4\\x_4=7[/tex]

The width of each subinterval is 3. The endpoints in ascending order are -5, -2, 1, 4, 7.

To find the width of each subinterval, we need to divide the length of the interval by the number of subintervals. In this case, the interval is [-5,7] and we need to divide it into 4 equal subintervals. The length of the interval is 7 - (-5) = 12. So the width of each subinterval is 12/4 = 3.

Next, we can find the endpoints of the subintervals. Since we have 4 subintervals, we need 5 endpoints. The first endpoint is the left endpoint of the interval, which is -5. Then we add the width of each subinterval to find the next endpoints: -5 + 3 = -2, -2 + 3 = 1, 1 + 3 = 4, and finally 4 + 3 = 7, which is the right endpoint of the interval. Therefore, the endpoints in ascending order are -5, -2, 1, 4, 7.

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Final answer:

By setting up a system of equations based on the given information, we find that 200 Basic plans and 80 Standard plans were purchased by the new subscribers.

Explanation:

A cable network offers two types of plans: a Basic plan for $7.31 per month and a Standard plan which costs $3.00 more per month, totaling $10.31. To find out how many Basic and Standard plans were purchased by 280 new subscribers, who altogether paid $2286.80, we can set up a system of equations. Let x represent the number of Basic plans and y represent the number of Standard plans.

The total amount of plans purchased: x + y = 280

The total amount paid: 7.31x + 10.31y = 2286.80

Solving this system of equations:
Step 1: Multiply the first equation by 7.31 to make the x coefficients equal: 7.31x + 7.31y = 2046.80

Step 2: Subtract this equation from the second equation to eliminate x and solve for y: 3y = 240, thus y = 80. Therefore, 80 Standard plans were purchased.

Step 3: Substitute y back into one of the original equations to solve for x: x + 80 = 280, thus x = 200. Therefore, 200 Basic plans were purchased.

So, there were 200 Basic plans and 80 Standard plans purchased by the new subscribers.

Answer:

The original number is 10 times the new number.

Step-by-step explanation:

Final Answer:

A. approximately-normal; μp = 0.35, σp = 0.030


1. Identify conditions for a normal sampling distribution:

Random sampling: The problem states it's a simple random sample.

Large sample size: n = 250 is greater than 10% of the population (assumed to be large), satisfying the condition.

Success-failure condition: np = 250 * 0.35 = 87.5 and n(1-p) = 250 * 0.65 = 162.5 are both greater than 10, meeting the condition.

2. Calculate the mean and standard deviation of the sampling distribution:

Mean (μp): μp = p = 0.35 (equal to the population proportion)

Standard deviation (σp): σp = sqrt(p(1-p)/n) = sqrt(0.35*0.65/250) ≈ 0.030

Final Answer:

A. approximately-normal; μp = 0.35, σp = 0.030

Explanation:

Answer:

175$

Step-by-step explanation:

Given that Pam  is an employee at a jewelry kiosk in a mall. If Pam works hard, there is a 75% probability that jewelry profits will equal $400 a day and a 25% probability that jewelry profits will equal $200 a day. If Pam shirks, there is a 25% probability that jewelry profits will equal $400 a day and a 75% probability that jewelry profits will equal $200 a day.

Since Pam works hard

expected value of profit for the kiosk = [tex]400(0.75)+200(0.25)\\=350[/tex]

From this value, 50% would be given to Pam

Hence expected gain for Pam for working hard

= [tex]0.5(350)\\=175[/tex]

175 $

Answer:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

(–2975, 175)

Do not reject H₀

Step-by-step explanation:

Hello!

The objective of this experiment is to test if the salaries of elementary school teachers are equal in two districts. You can test this trough the population means of the salaries of the teachers if they are either equal or different or directly test if the difference between the salaries of the two districts is cero or not, symbolically: μ₁ - μ₂ = 0

Remember, in the null hypothesis is usually stated the known information, is the "no change" premise and always carries the = symbol.

The hypothesis is:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

You have two normally distributed variables and you are studying the difference of the means. You can use a pooled Z to make the interval, the formula is:

X[bar]₁-X[bar]₂ ± [tex]Z_{1-\alpha /2}[/tex]*(√(σ₁²/n₁+σ₂²/n₂)

Since the test is two-tailed and at a signification level of 5% I've made the interval at the complementary confidence level of 95% so that I can use it to decide over the hypothesis.

X[bar]₁-X[bar]₂ ± [tex]Z_{1-\alpha /2}[/tex]*(√(σ₁²/n₁+σ₂²/n₂)

[tex]Z_{1-\alpha /2}[/tex] = [tex]Z_{0,975}[/tex] = 1,96

[28900-30300 ± 1.96*(√((2300)²/15+(2100)²/15)]

[-1400 ± 1576,15]

[-2976.15;176,15]

Since I've approximated to two decimal units in the intermediate calculations, the values ​​differ slightly, but the interval is:

(–2975, 175)Now, since the 0 is contained in the Confidence interval, the decision is to not reject the null hypothesis. In other words, the difference in average salaries between the two districts is cero.

I hope it helps!

The appropriate set of hypotheses (H0, H1) is B. µ1 – µ2 ≠ 0, µ1 – µ2 = 0.

The null hypothesis in a test predicts that there's no relationship between the variables. From the information, the null and alternative hypothesis are reflected by µ1 – µ2 ≠ 0, µ1 – µ2 = 0.

The confidence interval will be:

= (28900 - 30300) + 2.13 × ✓(2 × 4850000/15)

= = (28900 - 30300) + 2.13 × 804.15

= 325

Also, (28900 - 30300) - 2.13 × ✓(2 × 4850000/15)

= = (28900 - 30300) - 2.13 × 804.15

= -3125

The confidence interval will be (–3125, 325).

Lastly, the salaries of the teachers from the two districts isn't different as it's equal since the confidence interval has negative and positive values.

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The population of the country decreased by approximately 2.4% between 1990 and 2010, calculated using the percent decrease formula $(167 million - 163 million) / 167 million * 100$.

To find the percent decrease in the country's population, we first need to determine the actual decrease in population. In this case, the population in 1990 was about 167 million and it decreased to 163 million in 2010, so the numeric decrease is 167 million - 163 million = 4 million.

Next, we compute the percent decrease based on the initial number from 1990. The formula for percent decrease is: (Decrease in Value / Original Value) * 100. Therefore, the percent decrease is (4 million / 167 million) * 100 = 2.4% (rounded to one decimal place).

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The decrease in the country's population from 1990 to 2010 is approximately 2.4%.

To find the percentage decrease in the country's population, we first calculate the amount of decrease. This is done by subtracting the population in 2010 from the population in 1990, in other words, 167 million - 163 million, which equals 4 million. Next, we divide this number by the original population(the population in 1990) and times the result by 100%. Therefore, the percentage decrease in population is (4/167) * 100%, approximately equals 2.4% when rounded to one decimal place.

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Answer:

I would assume that the shadow length is proportional to the height of the flagpole. Hence as the building is 5 times higher than the flagpole, the length of the shadow would be 5 times higher, 25x5= 125 ft.

Step-by-step explanation:

The solution is 125 feet

The length of the building's shadow is 125 feet

What is Proportion?

The proportion formula is used to depict if two ratios or fractions are equal. The proportion formula can be given as a: b::c : d = a/b = c/d where a and d are the extreme terms and b and c are the mean terms.

Given data ,

Let the length of the building's shadow be = A

Now , the equation will be

Let the height of the flagpole be = 40 feet

Let the length of the shadow of flagpole be = 25 feet

Let the height of the building be = 200 feet

So , the proportion between the height and length is given by

Height of the flagpole / length of the shadow of flagpole = height of the building / length of the building's shadow

Substituting the values in the equation , we get

40 / 25 = 200 / A

On simplifying the equation , we get

1.6 = 200 / A

Multiply by A on both sides of the equation , we get

1.6A = 200

Divide by 1.6 on both sides of the equation , we get

A = 200/1.6

The value of A = 125 feet

Therefore , the value of A is 125 feet

Hence , the length of the shadow is 125 feet

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Answer:

[tex]P(3\leq X \leq 5)=0.09882[/tex]

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=6, p=0.2)[/tex]

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

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

And we want to find this probability:

[tex]P(3 \leq x \leq 5)=P(X=3)+P(X=4)+P(X=5)[/tex]

[tex]P(X=3)=(6C3)(0.2)^3 (1-0.2)^{6-3}=0.08192[/tex]

[tex]P(X=4)=(6C4)(0.2)^4 (1-0.2)^{6-4}=0.01536[/tex]

[tex]P(X=5)=(6C5)(0.2)^5 (1-0.2)^{6-5}=0.001536[/tex]

[tex]P(3 \leq x \leq 5)=0.08192+0.01536+0.001536=0.09882[/tex]

Answer:

a) The 95% confidence interval would be given by (0.341;0.499)

b) The 99% confidence interval would be given by (0.316;0.524)

c) n=335

d)n=649

Step-by-step explanation:

1) Notation and definitions

[tex]X_{IS}=63[/tex] number of high speed internet users that had been interrupted one or more times in the past month.

[tex]n=150[/tex] random sample taken

[tex]\hat p_{IS}=\frac{63}{150}=0.42[/tex] estimated proportion of high speed internet users that had been interrupted one or more times in the past month.

[tex]p_{IS}[/tex] true population proportion of high speed internet users that had been interrupted one or more times in the past month.

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{\hat p(1-\hat p)}{n}})[/tex]

1) Part a

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]t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96[/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]0.42 - 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.341[/tex]

[tex]0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.499[/tex]

The 95% confidence interval would be given by (0.341;0.499)

2) Part b

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 99% of confidence, our significance level would be given by [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:

[tex]t_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58[/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]0.42 - 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.316[/tex]

[tex]0.42 + 2.58\sqrt{\frac{0.42(1-0.42)}{150}}=0.524[/tex]

The 99% confidence interval would be given by (0.316;0.524)

3) Part c

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] 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{0.42(1-0.42)}{(\frac{0.05}{1.96})^2}=374.32[/tex]

And rounded up we have that n=335

4) Part d

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] 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{0.42(1-0.42)}{(\frac{0.05}{2.58})^2}=648.599[/tex]

And rounded up we have that n=649

Answer:

16

Step-by-step explanation:

Let,

P = movie rated​ PG-13,

P' = movie does not rated PG-13,

E = movie earned over​ $100 million,

E' = movie earned less than $100 million,

According to the question,

[tex]n(P\cap E)=16[/tex]

[tex]n(P\cap E')=29[/tex]

[tex]n(P'\cap E')=39[/tex]

Since,

[tex]n(P\cap E)+n(P\cap E')+n(P'\cap E')+n(P'\cap E)[/tex] = total movies

[tex]16 + 29 + 39 + n(P'\cap E) = 100[/tex]

[tex]n(P'\cap E) = 100 - 84 = 16[/tex]

Hence,

The number of movies that earned over​ $100 million and that were not rated​ PG-13 would be 16.

To find the number of movies that earned over $100 million but were not rated PG-13, subtract the number of PG-13 rated movies that earned over $100 million from the total number of movies that earned over $100 million.

To find the number of movies that earned over $100 million but were not rated PG-13, we need to subtract the number of PG-13 rated movies that earned over $100 million from the total number of movies that earned over $100 million.

Substituting the given values, we have:

Total movies earning over $100 million - 16 = Movies not rated PG-13 earning over $100 million.

Therefore, the number of movies that earned over $100 million and were not rated PG-13 is 100 - 16 = 84.

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Answer:

[tex]z=\frac{0.686 -0.68}{\sqrt{\frac{0.68(1-0.68)}{86}}}=0.119[/tex]  

[tex]p_v =P(z<0.119)=0.547[/tex]  

The p value obtained was a very high value and using the significance level given for example [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion interest is not significantly lower than 0.68.  

Step-by-step explanation:

1) Data given and notation

n=86 represent the random sample taken

X=59 represent the adults that match the condition you are testing

[tex]\hat p=\frac{59}{86}=0.686[/tex] estimated proportion of adults that match the condition you are testing

[tex]p_o=0.68[/tex] is the value that we want to test

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

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 is less than 0.68.:  

Null hypothesis:[tex]p=0.68[/tex]  

Alternative hypothesis:[tex]p < 0.68[/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].

3) Calculate the statistic  

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

[tex]z=\frac{0.686 -0.68}{\sqrt{\frac{0.68(1-0.68)}{86}}}=0.119[/tex]  

4) 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 is not given [tex]\alpha[/tex]. The next step would be calculate the p value for this test.  

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

[tex]p_v =P(z<0.119)=0.547[/tex]  

So the p value obtained was a very high value and using the significance level given for example [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion interest is not significantly lower than 0.68.  

Answer:

a) The distribution for the random variable X is given by:

X       |           -5         |    95        |      745      |        1995     |

P(X)   |  6992/7000  |  5/7000  |     2/7000 |       1/7000  |

b) E(X)=-4.43. That means if we buy an individual ticket by $5 on this lottery the expected value of loss if $4.43.

c) [tex]Sd(X)=\sqrt{Var(X)}=\sqrt{738.947}=27.184[/tex]

Step-by-step explanation:

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.

Part a

The info given is:

N=7000 represent the number of tickets sold

$5 is the price for any ticket

Number of tickets with a prize of $2000 =1

Number of tickets with a prize of $750=2

Number of tickets with a prize of $100=5

Let X represent the random variable net gain when we buy an individual ticket. The possible values that X can assume are:

___________________________

Ticket price    Prize     Net gain (X)

___________________________

5                     2000       1995

5                     750          745

5                     100           95

5                      0              -5

___________________________

Now we can find the probability for each value of X

P(X=1995)=1/7000, since we ave just one prize of $2000

P(X=745)=2/7000, since we have two prizes of $750

P(X=95)=5/7000, since we have 5 prizes of $100

P(X=-5)=6992/7000. since we have 6992 prizes of $0.

So then the random variable is given by this table

X       |           -5         |    95        |      745      |        1995     |

P(X)   |  6992/7000  |  5/7000  |     2/7000 |       1/7000  |

Part b

In order to calculate the expected value we can use the following formula:

[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]

And if we use the values obtained we got:

[tex]E(X)=(-5)*(\frac{6992}{7000})+(95)(\frac{5}{7000})+(745)(\frac{2}{7000})+(1995)(\frac{1}{7000})=\frac{-31000}{7000}=-4.43[/tex]

That means if we buy an individual ticket by $5 on this lottery the expected value of loss if $4.43.

Part c

In order to find the standard deviation we need to find first the second moment, given by :

[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]

And using the formula we got:

[tex]E(X^2)=(25)*(\frac{6992}{7000})+(9025)(\frac{5}{7000})+(555025)(\frac{2}{7000})+(3980025)(\frac{1}{7000})=\frac{5310000}{7000}=758.571[/tex]

Then we can find the variance with the following formula:

[tex]Var(X)=E(X^2)-[E(X)]^2 =758.571-(-4.43)^2 =738.947[/tex]

And then the standard deviation would be given by:

[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{738.947}=27.184[/tex]

Answer:

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean lead concentration of sea water samples

Number of samples. n = 6

Mean, u = 0.903 cc/cubic meter

Standard deviation, s = 0.0566

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

0.903 +/- 1.96 × 0.0566/√6

= 0.903 +/- 1.96 × 0.0566/2.44948974278

= 0.903 +/- 0.045

The lower end of the confidence interval is 0.903 - 0.045 =0.858

The upper end of the confidence interval is 0.903 + 0.045 =0.948

Therefore, with 95% confidence interval, the mean lead concentration of the sea water is between 0.858 cc/cubic meter and 0.948 cc/cubic meter

Answer:

C) A person's height, recorded in inches

Step-by-step explanation:

Quantitative Variable:

A) The color of an automobile

The color of car is not a quantitative variable as its outcome cannot be measured and expressed in value. It is a categorical variable.

B) A person's zip code

Some variables like zip codes take numerical values. But they are not considered quantitative. They are considered as a categorical variable because average of zip codes have no significance.

C) A person's height, recorded in inches

Height is a qualitative variable because it can be measured and its value is expressed in numbers.

C) A person's height, recorded in inches

A variable can be defined as the characteristics of an object.

A quantitative variable is a variable that usually changes it values as a result of either counting or measuring. Examples are time, weight, height.

A qualitative variable is a variable which do not usually change as a result of counting or measurement. Such as the color of an object, zip code, phone number, license number.

Height is a quantitative variable.

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Answer:

Step-by-step explanation:

Perimeter of a rectangle is expressed as (2 length + 2 width) = 2(L + W)

A) The length of rectangle A is y + 1

The width of rectangle A is x

Perimeter of rectangle A = 2(y + 1 + x) = 2y + 2 + 2x

= 2x + 2y + 2

The length of rectangle B is 2x - 2y

The width of rectangle B is x + 1

Perimeter of rectangle B = 2(2x - 2y+ x + 1) = 4x - 4y + 2x + 2) = 4x + 2x - 4y + 2

= 6x - 4y + 2

The length of rectangle C is 3x + 3y

The width of rectangle C is 2x - 3

Perimeter of rectangle C = 2(3x + 3y + 2x - 3) = 6x + 6y + 4x - 6) =

(6x + 6y + 4x - 6)

= 10x + 6y - 6

B) The combined perimeters will be the sum if perimeter of rectangle A, perimeter of rectangle B and perimeter of rectangle C. It becomes

2x + 2y + 2 + 6x - 4y + 2 + 10x + 6y - 6

Collecting like terms

2x + 6x + 10x + 2y + 6y - 4y + 2 + 2 - 6

The combined perimeter = 18x + 4y - 2

Answer:

In bold below.

Step-by-step explanation:

A.

2x + 2(y + 1)

= 2x + 2y + 2.

2(2x - 2y) + 2(x + 1)

= 4x - 4y + 2x + 2

= 6x - 4y + 2.

2(3x + 3y ) + 2(2x - 3)

= 6x + 6y + 4x - 6

= 10x + 6y - 6.

B.

2x + 2y + 2 + 10x + 6y - 6 + 6x - 4y + 2

=  18x + 4y - 2.

Answer:

[tex]\lambda \geq 6.63835[/tex]

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that [tex]X \sim Poisson(\lambda)[/tex]

The probability mass function for the random variable is given by:

[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]

And f(x)=0 for other case.

For this distribution the expected value is the same parameter [tex]\lambda[/tex]

[tex]E(X)=\mu =\lambda[/tex]

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

[tex]P(X\geq 2)=1-P(X<2)=1-P(X\leq 1)=1-[P(X=0)+P(X=1)][/tex]

Using the pmf we can find the individual probabilities like this:

[tex]P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}[/tex]

[tex]P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}[/tex]

And replacing we have this:

[tex]P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[][/tex]

[tex]P(X\geq 2)=1-e^{-\lambda}(1+\lambda)[/tex]

And we want this probability that at least of 99%, so we can set upt the following inequality:

[tex]P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99[/tex]

And now we can solve for [tex]\lambda[/tex]

[tex]0.01 \geq e^{-\lambda}(1+\lambda)[/tex]

Applying natural log on both sides we have:

[tex]ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)[/tex]

[tex]ln(0.01) \geq -\lambda+ln(1+\lambda)[/tex]

[tex]\lambda-ln(1+\lambda)+ln(0.01) \geq 0[/tex]

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

[tex]x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}[/tex]

Where :

[tex]f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)[/tex]

[tex]f'(x_n)=1-\frac{1}{1+\lambda}[/tex]

Iterating as shown on the figure attached we find a final solution given by:

[tex]\lambda \geq 6.63835[/tex]

The problem pertains to Poisson Distribution in probability theory, focusing on finding the smallest mean (λ) such that the probability of having at least two chocolate chips in a cookie is more than 0.99. This involves solving an inequality using the formula for Poisson Distribution.

This problem pertains to the Poisson Distribution, often used in probability theory. In particular, we're looking at the number of events (in this case, the number of chocolate chips) that occur within a fixed interval. Here, the interval under study is a single cookie. The question requires us to find the smallest value of λ (the mean value of the distribution) such that the probability of getting at least two chocolate chips in a cookie is more than 0.99.

Using the formula for Poisson Distribution, the probability of finding k copies of an event is given by:

P(X=k) = λ^k * exp(-λ) / k!

The condition here is that the probability of finding at least 2 copies is more than 0.99. Therefore, you formally need to solve the inequality:

P(X>=2) = 1 - P(X=0) - P(X=1) > 0.99

Substituting the values of P(X=0) and P(X=1) from our standard formula, you will need to calculate and find the smallest value of λ that satisfies this inequality.

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Answer:

a) Null hypothesis:[tex]\mu_{s}-\mu_{n}\geq 10[/tex]

Alternative hypothesis:[tex]\mu_{s} - \mu_{n}<10[/tex]

[tex]p_v =P(Z<-1.904)=0.0284[/tex]

Comparing the p value with the significance level [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly lower than 10.

b) The 90% confidence interval would be given by [tex]-21.035 \leq \mu_1 -\mu_2 \leq 7.685[/tex]

c) If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.

Step-by-step explanation:

Data given and notation

[tex]\bar X_{s}=750.2[/tex] represent the mean for the sample standard  process

[tex]\bar X_{n}=756.875[/tex] represent the mean for the sample with the new process

[tex]s_{s}=19.128[/tex] represent the sample standard deviation for the sample standard process

[tex]s_{n}=21.283[/tex] represent the sample standard deviation for the new process

[tex]\sigma_s=\sigma_n=\sigma=20[/tex] represent the population standard deviation for both samples.

[tex]n_{s}=15[/tex] sample size for the group Cincinnati

[tex]n_{n}=8[/tex] sample size for the group Pittsburgh

z would represent the statistic (variable of interest)

Concepts and formulas to use

(a) Formulate and test an appropriate hypothesis using a 0.10. What are your conclusions?

We need to conduct a hypothesis in order to check if the difference in mean batch viscosity is 10 or less system of hypothesis would be:

Null hypothesis:[tex]\mu_{s}-\mu_{n}\geq 10[/tex]

Alternative hypothesis:[tex]\mu_{s} - \mu_{n}<10[/tex]

We have the population standard deviation, so for this case is better apply a z test to compare means, and the statistic is given by:

[tex]z=\frac{(\bar X_{s}-\bar X_{n})-\Delta}{\sqrt{\frac{\sigma^2_{s}}{n_{s}}+\frac{\sigma^2_{n}}{n_{n}}}}[/tex] (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

With the info given we can replace in formula (1) like this:

[tex]z=\frac{(750.2-756.875)-10}{\sqrt{\frac{20^2}{15}+\frac{20^2}{8}}}}=-1.904[/tex]  

Statistical decision

Since is a one tail left test the p value would be:

[tex]p_v =P(Z<-1.904)=0.0284[/tex]

Comparing the p value with the significance level [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly lower than 10.

(b) Find a 90% confidence interval on the difference in mean batch viscosity resulting from the process change.

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_s -\bar X_n) \pm z_{\alpha/2}\sqrt{\sigma^2(\frac{1}{n_s}+\frac{1}{n_s})}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:

[tex]\bar X_s -\bar X_n =750.2-756.875=-6.675[/tex]

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 table 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]  

The standard error is given by the following formula:

[tex]SE=\sqrt{\sigma^2(\frac{1}{n_s}+\frac{1}{n_n})}[/tex]

And replacing we have:

[tex]SE=\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=8.756[/tex]

Confidence interval

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

[tex]-6.675-1.64\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=-21.035[/tex]  

[tex]-6.675+1.64\sqrt{20^2(\frac{1}{15}+\frac{1}{8})}=7.685[/tex]  

So on this case the 90% confidence interval would be given by [tex]-21.035 \leq \mu_1 -\mu_2 \leq 7.685[/tex]

(c) Compare the results of parts (a) and (b) and discuss your findings.

If we analyze the interval obtained we see that the interval contains the value of -10 so then we agree with the result obtained from the hypothesis, that he alternative hypothesis is true.

Final answer:

The viscosity analysis involves a hypothesis test to check for significant changes in mean batch viscosity and constructing a confidence interval to estimate the range of this difference following a process change in the manufacturing of a polymer.

Explanation:

The question involves conducting a hypothesis test and constructing a confidence interval to analyze the change in mean batch viscosity of a polymer before and after a process change. A hypothesis test will be used to determine if the mean batch viscosity has changed significantly, and the confidence interval will provide a range in which the true difference in means likely falls.

Hypothesis Test:

Calculate the mean of both sets of viscosity measurements.

Set up the null hypothesis (H0) and the alternative hypothesis (H1): H0: μ1 - μ2 <= 10 (no significant change), H1: μ1 - μ2 > 10 (significant change).

Determine the test statistic using a t-test for two independent samples.

Calculate the p-value associated with the test statistic.

Compare the p-value with the significance level (α = 0.10). If p <= α, reject H0; otherwise, fail to reject H0.

Calculate the standard deviation of both sets of measurements.

Determine the standard error of the difference in means.

Find the appropriate t-distribution critical value based on the given confidence level (90%) and degrees of freedom.

Calculate the 90% confidence interval using the difference in means ± the margin of error.

Compare the results of the hypothesis test and the confidence interval. If the hypothesis test leads to rejection of the null hypothesis while the confidence interval for the difference in means includes values greater than 10, there is evidence suggesting a significant change in mean batch viscosity. However, if the hypothesis test does not lead to rejection of H0 and the confidence interval includes 10, it suggests that the process change did not have a significant effect on mean batch viscosity.

To determine the angles of diffraction maxima for a helium-neon laser light incident on a diffraction grating, use the formula d sin(θ) = m λ with d the grating spacing and λ the wavelength. Calculate for each order by substituting m=1, 2, etc., and solve for θ using inverse sine function.

To find the angles at which one would observe the first order maximum, second-order maximum, and so forth for monochromatic light incident on a diffraction grating, we use the diffraction grating equation: d × sin(θ) = m × λ, where d is the grating spacing, λ is the wavelength of light, m is the order of maximum, and θ is the diffraction angle.

For a diffraction grating with 6000 lines/cm, the spacing d is 1/6000 cm, or 1.67 x 10^-4 cm (since 1 cm = 10^-2 m, d = 1.67 x 10^-6 m). Given the wavelength λ = 632.8 nm or 6.328 x 10^-7 m, we can substitute these values into the equation to solve for θ for any order m.

To find the actual angles, use the inverse sine function (arcsin), keeping in mind the units.

Answer:

t=5.86

We can conclude that the population correlation between their assets and pretax profit is higher than 0 at the significance level provided.

Step-by-step explanation:

n= 20 random sample taken

r=0.81 correlation coeffcient obtained

[tex]\alpha=0.025[/tex] significance level obtained

1) System of hypothesis

The system of hypothesis given are:

Null hypothesis :[tex]\rho \leq 0[/tex]

Alternative hypothesis: [tex] \rho >0[/tex]

2) Calculate the statistic

The statistic in order to test an hypothesis for the correlation coefficient is given by:

[tex]t =\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}[/tex]

This statistic follows a t distribution with n-2 degrees of freedom

If we replace the values given we got:

[tex]t =\frac{0.81\sqrt{20-2}}{\sqrt{1-(0.81^2)}}=5.86[/tex]

3) P value

For this case w eneed to calculate first the degrees of freedom

[tex]df=n-2=20-2=18[/tex]

And then analyzing the alternative hypothesis we can calculate the p value on this way:

[tex]p_v =P(t_{18} >5.86) =1-P(t_{18} <5.86)=1-0.99999=7.51x10^{-6}[/tex]

Since the P-value is smaller than the significance level, we have enough evidence to reject the null hypothesis in favor of the alternative. We conclude "there is sufficient evidence at the significance level to conclude that there is a linear relationship in the population between the two variables analyzed."

The decision rule for a 0.025 significance level intends to reject the null hypothesis if the p-value from the test statistic is less than 0.025. Using a standard Z test to compute the test statistic, we substitute the given values to find the result. As the p-value is 0.026, which is greater than the 0.025 significance level, we do not reject the null hypothesis, indicating insufficient evidence to conclude that the population correlation is greater than zero.

To answer your question, let's first state the decision rule for a 0.025 significance level. Considering your null hypothesis (H0: rho ≤ 0) and the alternative hypothesis (H1: rho > 0), we want to reject the null hypothesis if the p-value from the test statistic is less than 0.025.

Since we aren't given a specific formula or context about how the test statistic should be computed, I'll assume that it's via a standard Z test using the formula:  

where r is the sample correlation (0.81 in this case), and n is the number of data points (20 in this case). Let's substitute these values into the formula to find the test statistic.

Based upon the p-value from your computing software as 0.026 and the significance level being 0.025, we do not reject the null hypothesis because the p-value is greater than the significance level. In other words, there is insufficient evidence at the 0.025 significance level to conclude that the population correlation is greater than zero.

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Answer with Step-by-step explanation:

We are given that a set {a,b,c,d}

S={(a,b),(a,c),(c,d),(c,a)}

R={(b,c),(c,b),(a,d),(d,b)]

Composition of relation:Let R and S are two relations on the given set

If ordered pair (a,b) belongs to relation R and (b,c) belongs to S .

Then, SoR={(a,c)}

By using this rule

SoR={(b,d),(b,a)}[/tex]

Because [tex](b,c)\in R[/tex] and [tex](c,d)\in S[/tex].Thus, [tex](b,d)\in SoR[/tex]

[tex](b,c))\in R[/tex] and [tex](c,a)\in S[/tex].Thus, [tex](b,a)\in SoR[/tex]

b.RoS={(a,c),(a,b),(c,b),(c,d)}

Because

[tex](a,b)\in S,(b,c)\in R[/tex] .Therefore, the ordered pair [tex](a,c)\in[/tex] RoS

[tex](a,c)\in S,(c,b)\in R[/tex] .Thus, [tex](a,b)\in RoS[/tex]

[tex](c,d)\in S,(d,b)\in R[/tex].Thus, [tex](c,b)\in RoS[/tex]

[tex](c,a)\in S,(a,d)\in R[/tex].Thus,[tex](c,d)\in RoS[/tex]

c.SoS={(a,d),(a,a),(c,c),(c,b)}

Because

[tex](a,c)\;and\; (c,d)\in S[/tex].Thus, [tex](a,d)\in SoS[/tex]

[tex](c,a),(a,b)\in S[/tex].Thus,[tex](c,b)\in SoS[/tex]

[tex](a,c)\in S[/tex] and [tex](c,a)\in S[/tex].Thus,[tex](a,a)\in SoS[/tex]

[tex](c,a)\in S[/tex] and [tex](a,c)\in S[/tex].Thus ,[tex](c,c)\in SoS[/tex]

Final answer:

Null Hypothesis (H0): P >= 0.80, Alternative Hypothesis (Ha): P < 0.80, Test Statistic (z): -1.1112, P-value: 0.1335, Conclusion: Fail to reject H0.

Explanation:

Null Hypothesis (H0): P ≥ 0.80

Alternative Hypothesis (Ha): P < 0.80

Test Statistic (z): -1.1112

P-value: 0.1335

Conclusion: Fail to reject H0. There is not sufficient evidence to support the claim that polygraph test results are correct less than 80% of the time.

Answer:

16 goes in the blank

Step-by-step explanation:

V(c) = 2*π*r*h

Differentiating boh sides

DV(c)/Dt   =  2πr Dh/Dt    now radius is 8 m

DV(c)/Dt   =  8π Dh/Dt

That expression gives the relation of changes in V and h

DV(c)/Dt  is the speed of growing of the volume

Dh/Dt  is the speed of increase in height

so if the cylinder is filling at a rate of  2 m³/min   the height will increase at a rate of 16π m/min  

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

From the diagram

b is the length of side adjacent to the angle (x) in the question

a is the length of side opposite to the angle

c is the length of hypotenuse

θ represents the measure of the angle in either degree or radians

From the diagram

Cosecant (csc θ) = c/a

Secant (sec θ) = c/b

Cotangent (cot θ) = b/a

The trigonometric functions in a right triangle are defined as the ratios of the sides: sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent.

In a right triangle with sides of lengths a, b, and c (where c is the hypotenuse) and an angle θ, the trigonometric functions are defined as follows:

These functions can easily be remembered using the mnemonic "SOHCAHTOA" - Sine equals Opposite over Hypotenuse, Cosine equals Adjacent over Hypotenuse, and Tangent equals Opposite over Adjacent.

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Answer:

The required probability is 0.988.

Step-by-step explanation:

Consider the provided information.

Based on a​ poll, 67​% of Internet users are more careful about personal information when using a public​ Wi-Fi hotspot.

That means the probability of more careful is 0.67

The probability of not careful is: 1-0.67 = 0.33

We have selected four random Internet​ users. we need to find the probability that at least one is more careful about personal information.

P(At least one careful) = 1 - P(None of them careful)

P(At least one careful) = 1 - (0.33×0.33×0.33×0.33)

P(At least one careful) = 1 - 0.012

P(At least one careful) = 0.988

Hence, the required probability is 0.988.

The result may be higher because of the convenience bias in retrieving the sample. Because the survey subjects volunteered to​ respond not random.

The probability that at least one out of four randomly selected Internet users is more careful about personal information when using a public Wi-Fi hotspot is approximately 98.81%, assuming that each event is independent and the probability of an individual being careful is 67%. However, voluntary responses to the survey might introduce a non-response bias, affecting the accuracy of this probability.

The question asks for the probability that among four randomly selected Internet​ users, at least one is more careful about personal information when using a public​ Wi-Fi hotspot, given that 67% of Internet users are like this. To find the probability of 'at least one', it is easier to calculate the complement—that is, the probability that none of the four users are careful—and subtract it from 1 (the total probability of any outcome).

The probability that a randomly selected Internet user is not more careful is 1 - 0.67 = 0.33. Since we are considering four independent events, we raise the single-event probability to the fourth power:

(0.33)^4 = 0.33 * 0.33 * 0.33 * 0.33

The calculation results in approximately 0.0119. Now, subtract this from 1 to get the probability of at least one person being more careful:

1 - 0.0119 = 0.9881

Therefore, the probability that at least one out of four randomly selected Internet users is more careful about personal information when using a public Wi-Fi hotspot is about 98.81%.

However, the accuracy of the result could be influenced by the fact that the survey subjects volunteered to respond, which can result in a non-response bias. Volunteers might have different behaviors or opinions compared to the general Internet user population, potentially skewing the results of the survey and, consequently, the estimated probability.

Answer:

solution is in the image below

Step-by-step explanation:

Final answer:

To construct confidence intervals for the mean compressive strength of concrete with known variance, formulas involving the z-score are employed. The 95% confidence interval is narrower than the 99% interval, illustrating that higher confidence requires a wider interval to encapsulate the true mean with more assurance.

Explanation:

To construct a confidence interval for the mean compressive strength of concrete, we utilize the formula for the confidence interval of the mean when the population variance (σ2) is known. Given that the variance is 1000 psi2 and the mean (μ) is 3250 psi for a sample of n=12 specimens, the z-score corresponding to a 95% confidence level is approximately 1.96, and for a 99% confidence level, the z-score is approximately 2.576.

95% Confidence Interval:

CI = μ ± z(σ/√n)
= 3250 ± 1.96(/1000/12)
= 3250 ± (1.96)(28.8675)
= 3250 ± 56.61
= (3193.39, 3306.61) psi

99% Confidence Interval:

CI = μ ± z(σ/√n)
= 3250 ± 2.576(/1000/12)
= 3250 ± (2.576)(28.8675)
= 3250 ± 74.36
= (3175.64, 3324.36) psi

Comparing the widths, the 99% confidence interval is wider than the 95% confidence interval, which is a reflection of increased certainty (or confidence level) requiring a wider interval.

To determine the number of cards each class parent helper will receive, divide the total number of cards (24) by the number of helpers (3), resulting in each helper receiving 8 cards.

Ms. Taylor's students are giving out cards to each of the 3 class parent helpers equally, and there are a total of 24 cards. To find out how many cards each helper gets, we can divide the total number of cards by the number of helpers.

Step 1: Determine the total number of cards: 24.

Step 2: Determine the number of class parent helpers: 3.

Step 3: Divide the total number of cards by the number of helpers: 24 ÷ 3.

Step 4: Calculate: 24 ÷ 3 = 8.

So, each class parent helper will receive 8 cards.