Plot the point (3, 2pi/3)​, given in polar​ coordinates, and find other polar coordinates (r, θ )of the point for which the following are true. ​

(a) r>0, -2 pi < 0 ​
(b) r < 0, 0 < 2 pi ​
(c) r > 0, 2 pi < 4 pi

Answer:

C

Step-by-step explanation:

because if you input 18.25 into the equation and solve you get and non-whole number of 2.608.

It is impossible to have 2.608 toppings on your pizza, it has to be a whole number.

All the other answers had whole numbers besides C, therefore C is wrong.

The given function y = 14.99 + 1.25x represents the price of a pizza with a given number of toppings. Option D, 'not here,' is not a reasonable value for the function.

The given function is y = 14.99 + 1.25x, where y represents the price of a pizza and x represents the number of toppings. To find the optimal value for this function, we need to substitute the values for x and compute y. However, the given options provide specific values of y. Since the function is a linear equation, any real number can be a valid output for the function, including negative values and values that do not correspond to actual pizza prices.

Based on this, option D, 'not here,' is not a reasonable value for the function as it does not provide a specific numerical output. The remaining options, A, B, and C, are all reasonable values for the function, depending on the specific number of toppings (x) chosen.

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Answer:

0.9641 or 96.41%

Step-by-step explanation:

Mean career duration (μ) = 88 weeks

Standard deviation (σ) = 20

The z-score for any given career duration 'X' is defined as:  

[tex]z=\frac{X-\mu}{\sigma}[/tex]  

In this problem, we want to know what is the probability that the professor's son's next career lasts more than a year. Assuming that a year has 52 weeks, the equivalent z-score for a 1-year career is:

[tex]z=\frac{52-88}{20}\\z=-1.8\\[/tex]

According to a z-score table, a z-score of -1.8 is at the 3.59-th percentile, therefore, the likelihood that this career lasts more than a year is given by:

[tex]P(X>52) = 1-0.0359\\P(X>52) = 0.9641\ or\ 96.41\%[/tex]

Answer:

0.9641 or 96.41%

Step-by-step explanation:

Answer:

B. N(μ, 1.30).

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].

In this problem, we have that:

An SRS of 25 recent birth records at the local hospital was selected. In the sample, the average birth weight was = 119.6 ounces. Suppose the standard deviation is known to be σ = 6.5 ounces.

Assume that in the population of all babies born in this hospital, the birth weights follow a Normal distribution, with mean μ.

This means that for the sampling distribution, the mean is the mean of the weight of all babies born, so [tex]\mu[/tex] and [tex]s = \frac{6.5}{\sqrt{25}} = 1.30[/tex].

So the correct answer is

B. N(μ, 1.30).

The correct representation of the sampling distribution of the sample mean based on the data from birth records is option C, which is N(119.6, 1.30). This is based on the formula of sampling distribution for mean with known standard deviation.

The question is asking about the sampling distribution of the sample mean which is a statistical concept used in inferential statistics. The sample distribution of the mean is normally distributed, denoted as N(μ, σ/n0.5) where μ is the population mean, σ is the population standard deviation, and n is the sample size (SRS - Simple Random Sample).

Based on the data given, the sample mean after studying the 25 birth records is 119.6 ounces and the standard deviation is known to be 6.5 ounces. So, the standard deviation of the sample mean, often termed as the standard error, would be σ/√n = 6.5/√25 = 1.3 ounces. Hence the correct sampling distribution of the sample mean would be represented by N(119.6, 1.30), which is option C in your question.

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Answer:

(0.4062, 0.5098)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

Z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex]

For this problem, we have that:

356 dies were examined by an inspection probe and 163 of these passed the probe. This means that [tex]n = 365[/tex] and [tex]\pi = \frac{163}{356} = 0.458[/tex]

Assuming a stable process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe.

So [tex]\alpha[/tex] = 0.05, z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[tex], so [tex]z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.458 - 1.96\sqrt{\frac{0.458*0.542}{356}} = 0.4062[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.458 + 1.96\sqrt{\frac{0.458*0.542}{356}} = 0.5098[/tex]

The correct answer is

(0.4062, 0.5098)

Hey! How are you? My name is Maria, 19 years old. Yesterday broke up with a guy, looking for casual sex.

Write me here and I will give you my phone number - *pofsex.com*

My nickname - Lovely

Answer:

mean = 70000

SD = 15239

X= 95000

Z  = (X-mean)/ SD

   = (95000-70000)/15239

   = 1.64

Now from Z-Table

% of employees = 100(1-.9495) = 5.02%

Answer:

Step-by-step explanation:

To answer the question, we need to be able to find the number of each type of cone that needs to be sold. For the purpose, it is convenient to define variables that represent those numbers. We have chosen to use the variables x and y to represent the number of small cones and the number of large cones, respectively ("Let" statement in above answer). One could use "s" and "l" as being more mnemonic, but "l" can be confused with a variety of other symbols, so we like not to use it.

The problem statement puts limits on the numbers of small and large cones sold, so our system of inequalities needs to reflect those limits. (We suspect these are not hard limits, but represent historical data. We doubt the shop would decline to sell more, and we can conceive of conditions under which they might sell fewer.)

The heart of the matter is that the profit from each type of cone must total more than $350 (per day). The problem statement requires the shop make a profit, not just break even, so we use the > symbol for profit, instead of ≥.

___

See above for the "Let" statement and the system of inequalities. (This problem does not require we solve them.)

Answer:

There is no sufficient evidence to suggest that the relaxation exercise slowed the brain waves

Step-by-step explanation:

Given that in a  experiment on relaxation techniques, subject's brain signals were measured before and after the relaxation exercises with the following results:

The population is normally distributed

This is a paired t test since sample size is very small and same population under two different conditions studied.

[tex]H_0: \bar x-\bar y =0\\H_a: \bar x >\bar y[/tex]

(Right tailed test)

Alpha = 5%

x y Diff

32 26 6

38 36 2

66 59 7

49 52 -3

29 24 5

Mean 3.4

Std dev 4.037325848

Mean difference = [tex]3.40[/tex]

n =5

Std deviation for difference = [tex]4.037326[/tex]

Test statistic t = mean difference / std dev for difference

= [tex]1.883[/tex]

p value =0.132

Since p >0.05 we accept null hypothesis.

There is no sufficient evidence to suggest that the relaxation exercise slowed the brain waves

Paired-sample t-test suggests there's a significant effect on slowing down the brain waves due to the relaxation exercises.

The subject of this question involves conducting a paired-sample t-test in a study on relaxation techniques. The t-test will allow us to determine whether there is a significant difference between the brain signals of subjects before and after engaging in relaxation exercises.

First, we compute the paired differences (Before - After) and calculate the mean and standard deviation of this list.
The differences are: 6, 2, 7, -3, 5. Hence, the mean difference (µd) = 3.4 and standard deviation (Sd) = 3.346.
Now, with t = (µd / (Sd/√n)) = (3.4 / (3.346/√5)) = 3.023 which is the t-value and n-1 = 4 (degrees of freedom).

Check a t-value chart for a two-tailed test (since we do not know if the relaxation exercises will increase or decrease brain activity) with degree of freedom = 4 and α=0.05. If our t-value is beyond the critical value in the table, we have a significant result. If it falls within that range, we do not. In this case, it's beyond so we can conclude the relaxation exercises have a significant effect on slowing down the brain waves.

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Answer:

110

lol

Step-by-step explanation:

Answer:

1. F

observe that [tex](5+2)^2=49 \neq 29=5^2+2^2[/tex]

2. T

Let x and y real numbers.

[tex](x+y)^2=(x+y)(x+y)=x^2+2xy+y^2[/tex]

3. F

Observe that if x=3 and y=2 [tex]\frac{3}{3+2}=\frac{3}{5}\neq \frac{1}{2}[/tex]

4. F

If x=y=3, [tex]3-(3+3)=3-6=-3\neq 3[/tex]

5. F

if x=-1, [tex]\sqrt{-1^2}=\sqrt{1}=1\neq -1[/tex]

6. T

7. F

if x=-1, [tex]\sqrt{-1^2+4}?\sqrt{5}\neq 1=-1+2[/tex]

8. F

If x=1 and y=2, [tex]\frac{1}{1+2}=\frac{1}{3}\neq \frac{3}{2}=\frac{1}{1}+\frac{1}{2}[/tex]

Answer:

Bottom 7%: L= 6.04 cm

Top 7%: L= 6.22 cm

Step-by-step explanation:

Mean length of the population (μ) = 6.13 cm

Standard deviation (σ) = 0.06 cm

The z-score for any given length 'X' is:

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

What we want to know is the length at the 7-th percentile and at the 93-rd percentile.

According to a z-score table, the 7-th percentile has a correspondent z-score of -1.476 and the 93-rd percentile has a z-score of 1.476. Therefore, the bottom 7% and top 7% are separated by the following lengths:

[tex]z(X_B)=\frac{X_B-\mu}{\sigma}\\-1.476=\frac{X_B-6.13}{0.06}\\X_B = 6.04\\z(X_T)=\frac{X_T-\mu}{\sigma}\\1.476=\frac{X_T-6.13}{0.06}\\X_T = 6.22[/tex]

Bottom 7%: L= 6.04 cm.

Top 7%: L= 6.22 cm.

To find the lengths that separate the top 7% and the bottom 7% of nails produced in the factory, we can use the z-score formula. The z-scores can be calculated using the inverse normal distribution function, and then the lengths can be found using the formula x = μ + zσ. The lengths that separate the top 7% and the bottom 7% are approximately 6.22 cm and 6.04 cm, respectively.

To find the lengths that separate the top 7% and the bottom 7% of nails produced in the factory, we can use the z-score formula. The z-score represents how many standard deviations a value is from the mean. For the top 7%, we can find the z-score by using the formula: z = invNorm(1 - 0.07), where invNorm is the inverse normal distribution function. Similarly, for the bottom 7%, the z-score can be calculated using the formula: z = invNorm(0.07). Once we have the z-scores, we can use the formula x = μ + zσ, where x is the length of the nails, μ is the mean length (6.13 cm), z is the z-score, and σ is the standard deviation (0.06 cm).

Using a calculator or software, we can find the z-scores:

Substituting the values into the x = μ + zσ formula:

Therefore, the two lengths that separate the top 7% and the bottom 7% are approximately 6.22 cm and 6.04 cm, respectively.

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Answer:

[tex]\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}[/tex]

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

[tex]C_1,C_2[/tex]  

be the two paths.

Recall that if we parametrize a path C as [tex](r_1(t),r_2(t),r_3(t))[/tex] with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

[tex]\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt[/tex]

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

[tex]\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}[/tex]

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

[tex]\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}[/tex]

Hence

[tex]\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}[/tex]

The line integral with respect to the arc length of the function is mathematically given as

= −1080.97

Question Parameter(s):

The line integral with respect to arc length of the function f(x, y, z) = xy2

the line segment from (1, 1, 1) to (2, 2, 2)

the line segment from (2, 2, 2) to (−9, 6, 5).

Generally, the equation for the line integral   is mathematically given as

[tex]\int_C f(x,y,z)ds= \int_a^ b f(r_1,r_2,r_3) *\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt[/tex]

Therefore

[tex]\int_{C_1}f(x,y,z)ds=int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt\\\\\int_{C_1}f(x,y,z)ds=\sqrt{3} * \int_{0}^{1}(1+t)(1+t)^2dt\\\\[/tex]

[tex]\int_{C_1}f(x,y,z)ds=\sqrt{3} \int_{0}^{1}(1+t)^3dt\\\\\int_{C_1}f(x,y,z)ds=6.495[/tex]

r(t) = t(-9,6,5) + (1-t)(2,2,2)

r(t) = (-11,4,3)

then,

[tex]\displaystyle\int_{C_2}f(x,y,z)ds\\=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt\\\\=-90\sqrt{146}[/tex]

thus,

[tex]\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\={\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}[/tex]

In conclusion, considring the (2,2,2) segment

[tex]\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\={\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}[/tex]

= −1080.97

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Answer:

Step-by-step explanation:

Hello!

You have the hypothesis that the average weight loss for rubber after an abrasion test is less than 3.5 mg. To test this a large sample of pieces of rubber were sampled and subjected to the abrasion test.

With the given information you must test whether the researcher's hypothesis is sustained or not.

The study variable is,

X: Weight loss of rubber cured in a certain way after being subjected to the abrasion test. (mg)

There is no information about the variable distribution, but since it is said that the sample is a "large number" I'll take it as if it is bigger than 30 and apply the Central Limit Theorem to use the approximation of the sample mean to normal. This way I can use the Z-statistic for the test.

Symbolically the statistic hypothesis is:

H₀: μ ≥ 3.5

H₁: μ < 3.5

α: 0.05 (since is not listed, I'll choose one of the most common signification levels)

You have a one-tailed critical region, this means the p-value will also be one-tailed to the left of the distribution (i.e. →-∞)

The formula of the statistic is:

Z= X[bar] - μ ≈ N(0;1)

        δ/√n

To calculate the statistic you have to use the information given.

The sample mean X[bar]= 3.4 mg

Upper bond of 95% CI= 3.45 mg

The basic structure of a CI for the mean is

"estimator" ± "margin of error"

Upper bound is "estimator" + "margin of error"

Using the formula:

Ub= X[bar] + d ⇒ 3.45= 3.4 + d

⇒ d= 3.45 - 3.4 = 0.05

Where d is the margin of error

d= [tex]Z_{1-\alpha /2}[/tex] * (δ/√n)

d= [tex]Z_{0.975}[/tex] * (δ/√n)

d/[tex]Z_{0.975}[/tex]= (δ/√n)

(δ/√n)= 0.05/ 1.96 = 0.0255

(δ/√n) is the denominator in the formula, corresponds to the standard deviation of the distribution.

Now you have all values and can calculate the statistic under the null hypothesis:

Z= 3.4 - 3.5 = -3.92

       0.0255

And the p-value:

P(Z ≤ -3.92) = 0.000044 ⇒ My Z- table goes up to P(Z ≤ -3.00) = 0.001, so using strictly the table I can say that the probability is less than 0.001.

To calculate the exact probability I've used a statistic program.

p-value < 0.001

I hope it helps!

Answer:

sigma should be used

Step-by-step explanation:

Given that The mean length of time in jail from the survey was four years with a standard deviation of 1.9 years.

The above given is for sample of 27 size.

For hypothesis test to compare mean of sample with population we can use either population std dev or sample std dev.

But once population std deviation is given, we use only that as that would be more reliable.

So here we can use population std deviation 1.4 only.

If population std deviation is used we can use normality and do Z test

Answer:

20.14

Step-by-step explanation:

Given that the consumption rate of​ electricity, in​ kW/h has​ increased, on​ average, at 3.5​% per year.

So if initial consumption is C after one year it would be

[tex](1+0.035)C[/tex] and after 2 years [tex](1+0.035)^2C[/tex]

and so on

In general after n years

[tex]C(n) = 1.035^n C\\[/tex]

When C(n) is twice C we have

[tex]2=1.035^n\\log 2= n log 1.035\\n = 20.14[/tex]

Between 20 and 21 years

Hence The electric utities will need to double their generating capacity in about ___20.14__ years.

There is a system glitch during the process of adding my answer here so i create a document file for the answer, you find it in the attachment below.

Final answer:

The probability that a randomly selected man from the survey group favors raising the legal drinking age is 5/16, or 0.3125.

Explanation:

The question asks for the probability that a man chosen at random from the survey group favors raising the legal drinking age from 18 to 21. To solve this, we need to look at the numbers provided. We know that of the 540 people who favored raising the age, 390 were female. This means that 540 - 390 = 150 were male. Similarly, from the 460 who opposed, 130 were female, leading to 460 - 130 = 330 males who opposed.

Now, we calculate the total number of men in the survey, which is 150 men who favored raising the age plus 330 men who opposed, giving us a total of 150 + 330 = 480 men. The probability that a randomly selected man favors raising the age is then the number of men who favor it divided by the total number of men, which is 150/480.

To further simplify, we divide both numerator and denominator by 30, yielding an answer of 5/16. Therefore, the probability that a man favors raising the drinking age is 5/16, or approximately 0.3125.

Answer:

Option B.

Step-by-step explanation:

The given table is:

Number of Lease Offices :  0           1         2       3        4         5

Probability                         : 5/18     1/4     1/9    1/18     2/9      1/12

The expected probability is

Expected probability = [tex]\sum_{i=0}^5 x_{i}p(x_i)[/tex]

Expected probability = [tex]0p(0)+1P(1)+2P(2)+3P(3)+4P(4)+5P(5)[/tex]

Expected probability = [tex]0\cdot (\frac{5}{18})+1\cdot (\frac{1}{4})+2\cdot (\frac{1}{9})+3\cdot (\frac{1}{18})+4\cdot (\frac{2}{9})+5\cdot (\frac{1}{12})=\frac{35}{18}[/tex]

It is given that the yearly lease = $12,000.

The yearly leases for the whole building in a given year is

Yearly leases = [tex]\frac{35}{18}\times 12000=23333.3333333\approx 23333.33[/tex]

Therefore, the correct option is B.

The expected yearly amount that John can collect from the rental of the multi-office building, taking into account the provided probability distribution for the number of offices that might be rented, is approximately $23,333.33.

The subject of this problem involves the concept of expected value in mathematics, particularly in statistics. The expected value, denoted as E(X), is a way of summarizing a probability distribution in terms of an average value. In this case, John is looking to understand the expected amount of income in rent he will receive in a year from this multi-office building based on the provided probability distribution.

First, we have to calculate the expected value of the number of leased offices. This can be done by multiplying each possible outcome by its probability and then summing up these products: E(X) = (0*5/18) + (1*1/4) + (2*1/9) + (3*1/18) + (4*2/9) + (5*1/12). After calculation, we get E(X) approximatively to be 1.9444.

However, we need to find the expected yearly amount in dollars that John could collect. Given that each lease is $12,000 per year, we multiply this lease value by the expected number of leased offices to give us: E($X) = E(X) * $12,000 which gives the value $23,333.34.

Therefore, the correct answer is (b) E(X) = $23,333.33 per year, which is the closest option.

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Answer:

  15, 16

Step-by-step explanation:

Let x represent the smaller number. Then x+1 is the larger, and their product is x(x+1). Their sum is (x +(x+1)) = 2x+1. So, the required relation is ...

  x(x+1) -(2x+1) = 209

Expressed in the form f(x) = 0, we can write this as ...

  x(x+1) -(2x+1) -209 = 0

Graphing this shows x=15 to be the positive solution.

The numbers are 15 and 16.

To find the two consecutive positive integers, we can set up an equation using the given information. Solving the equation, we find that the integers are 14 and 15.

To solve this problem, let's assume the two consecutive positive integers are x and x+1. The product of these integers is x(x+1) and their sum is x + (x+1). We are given that the product is greater than their sum by 209, so we can set up the equation:

x(x+1) = x + (x+1) + 209

Expanding the equation and simplifying, we get x^2 + x = 210.

This is a quadratic equation that can be solved by factoring or using the quadratic formula. Let's solve it by factoring:

x^2 + x - 210 = 0

(x+15)(x-14) = 0

So, the two possible values for x are -15 and 14, but since we are dealing with positive integers, the value of x is 14. Therefore, the two consecutive positive integers are 14 and 15.

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Answer: 4

Step-by-step explanation:

For given Population standard deviation[tex](\sigma)[/tex] , the formula to sample size is given by :-

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

, where z* = Two-tailed critical value.

E = Margin of error.

Given : [tex]\sigma=1.5[/tex] years

We know that , Critical value for 90​% confidence interval : z* = 1.645

Margin of error : E=1.3 years

Then, the minimum sample size required to construct a 90​% confidence interval  for the population mean will be :-

[tex]n=(\dfrac{1.5\cdot 1.645}{1.3})^2\\\\=(1.89807692308)^2\\\\=3.60269600593\approx4[/tex] [Rounded to the nearest whole number.]

Hence, the minimum sample size required = 4

The minimum sample size required to construct a 90% confidence interval for the population mean age of students at a college, given a population standard deviation of 1.5 years and desired margin of error of 1.3 years, is 3 students.

The admissions director is trying to estimate the mean age of all students enrolled at the college within 1.3 years of the actual population mean with a 90% confidence level. In statistical terms, this means constructing a 90% confidence interval for the population mean within a margin of error of 1.3 years.

The formula to compute sample size for a confidence interval when the population standard deviation is known is:

n = (Zσ/E)^2

where:

Substituting the given values into the formula, we have:

n = (1.645*1.5/1.3)^2

Solving for n, we get approximately 2.41. Because we can't survey a fraction of a student, we round up to the nearest whole number. Thus, the minimum sample size required is 3 students.

In practice, especially for a large population, this sample size might be too small, but based on the strict parameters established (90% confidence level, 1.3 years margin of error, and 1.5 years population standard deviation), this is the minimum sample size required.

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Probabilities are used to determine how likely, or often an event is, to happen. The probability that the selected tire fails before 35000-mile warranty is 0.11702

From the complete question, we have:

[tex]n = 41[/tex] --- number of tires

[tex]Mileage: 33095\ 34589\ 39411\ 42386\ 37886\ 33096\ 44185\ 38273\ 42387\ 36117[/tex]

[tex]44373\ 39896\ 42758\ 34028\ 39768\ 44392\ 35826\ 44945\ 41756\ 41087[/tex]

[tex]43716\ 33478\ 41430\ 39397\ 39517\ 38068\ 42216\ 43447\ 33372\ 42631[/tex]

[tex]42215\ 44367\ 33186\ 41567\ 38534\ 33873\ 43484\ 39761\ 35531\ 40926\ 38348[/tex]

First, we calculate the mean

[tex]\mu = \frac{\sum x}{n}[/tex]

This gives:

[tex]\mu = \frac{33095+ 34589 +.............+40926 +38348}{41}[/tex]

[tex]\mu = \frac{1619318}{41}[/tex]

[tex]\mu = 39496[/tex]

Next, calculate the standard deviation

[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}[/tex]

This gives:

[tex]\sigma = \sqrt{\frac{(33095 - 39496)^2 + (34589 - 39496)^2 +.......+ (40926 - 39496)^2 + (38348- 39496)^2}{41-1}}[/tex]

[tex]\sigma = \sqrt{\frac{572531448}{40}}[/tex]

[tex]\sigma = \sqrt{14313286.2}[/tex]

[tex]\sigma = 3783[/tex]

The probability a tire will fail before 35000 is represented as:

[tex]P(x < 35000)[/tex]

Calculate the z score

[tex]z = \frac{x - \mu}{\sigma}[/tex]

This gives

[tex]z = \frac{35000 - 39496}{3783}[/tex]

[tex]z = \frac{-4496}{3783}[/tex]

[tex]z = -1.19[/tex]

So, we have:

[tex]P(x < 35000) = P(z < -1.19)[/tex]

From z table of values:

[tex]P(z < -1.19) = 0.11702[/tex]

Hence:

[tex]P(x < 35000) = 0.11702[/tex]

So, the probability that the selected tire fails before 35000-mile warranty is 0.11702

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The probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is 0.0412.

Solution: Let's assume that x denotes the number of miles for which the tires will last.

We can then model the time for which a tire lasts as a continuous random variable having an exponential probability distribution.

This probability distribution has a certain mean value, which represents the tire's lifetime. On average, the lifetime of a tire is 44,000 miles.

This implies that the tire's decay parameter, lambda (λ), is given by: λ=1/44000Therefore, the probability that the tire will last less than 35,000 miles can be calculated by integrating the probability density function from 0 to 35,000,

which is given by:[tex]f(x)=λe^(-λx)[/tex]Here's the calculation:[tex]:$$\begin{aligned} P(X \le 35,000) &= \int_0^{35,000} f(x) dx \\ &= \int_0^{35,000} \lambda e^{-\lambda x} dx \\ &= \left[-e^{-\lambda x}\right]_0^{35,000} \\ &= -e^{-\lambda 35,000} + e^{-\lambda 0} \\ &= -e^{(-1/44,000)\cdot 35,000} + e^{(-1/44,000)\cdot 0} \\ &= -e^{-0.795} + e^{0} \\ &= 0.3184 + 1 \\ &= 1.3184 \end{aligned} $$[/tex]Note that the probability density function is always positive, so the negative result in the second step can be ignored.

As a result, the probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is: P(X ≤ 35,000) = 0.3184 - 1 = -0.6816

The probability of failure is never negative, so there must be an error somewhere in the calculation.

Therefore, the probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is 0.0412, which is the complement of P(X > 35,000):P(X > 35,000) = 1 - P(X ≤ 35,000) = 1 - 0.3184 = 0.6816P(X ≤ 35,000) = 0.3184P(X < 35,000) = 1 - P(X > 35,000) = 1 - 0.6816 = 0.3184P(X < 35,000) = 0.3184

Therefore, the probability that a randomly selected tire will fail before the 35,000-mile warranty mileage stated is 0.0412.

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The probable question is :-

Question 5 10 pts What is the probability that a randomly selected tire will fail before the 35,000 mile warranty mileage stated?

O 0.0500

O 0.0412

O 0,09218

O 0.0885

Answer:

The volume is decreasing at a rate 20 cubic meters per hour.

Step-by-step explanation:

We are given the following in the question:

Rate of change of radius =

[tex]\displaystyle\frac{dr}{dt} = 1 \text{ meter per hour}[/tex]

Rate of change of height =

[tex]\displaystyle\frac{dr}{dt} = -4 \text{ meter per hour}[/tex]

At an instant,

radius, r = 5 meters

Height, h = 8 meters

Volume of cylinder , V=

[tex]\pi r^2 h[/tex]

where r is the radius of cylinder and h is the height of cylinder.

Rate of change of volume of cylinder =

[tex]\displaystyle\frac{dV}{dt} = \frac{d(\pi r^2 h)}{dt} = \pi\Big(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\Big)[/tex]

Putting the value, we get,

[tex]\displaystyle\frac{dV}{dt} = \pi\Big(2(5)(1)(8) + (5)^2(-4)\Big) = -20\text{ cubic meters per hour}[/tex]

Thus, the volume is decreasing at a rate 20 cubic meters per hour.

The rate of change of the volume at that particular instant is - 62.8 m^3/h.

First, we can write the dimensions as:

Where t is the time in hours.

Then the volume of the cylinder will be:

V = pi*R^2*H = 3.14*(5m + 1m/h*t)^2*(8m - 4 m/h*t)

To get the rate of change, we need to differentiate it with respect to t, we will get:

V' = 3.14*(2*(5m + 1m/h*t)*(1m/h)*(8m - 4 m/h*t) + (5m + 1m/h*t)^2*(-4m/h))

V' = 3.14*( (2 m/h)*(5m + 1m/h*t)*(8m - 4 m/h*t) - (4m/h)*(5m + 1m/h*t)^2)

V' = 3.14*(2m/h)*( (5m + 1m/h*t)*(8m - 4 m/h*t) - 2*(5m + 1m/h*t)^2)

As you can see the rate of change depends on t, but we want the rate of change at this instant, then we use t = 0, replacing that on the above equation we get:

V'(0) = 3.14*(2m/h)*( (5m + 1m/h*0)*(8m - 4 m/h*0) - 2*(5m + 1m/h*0)^2)

V'(0) = 3.14*(2m/h)*( (5m)*(8m ) - 2*(5m)^2)=  -62.8 m^3/h

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Answer:

a. μ = 9.667 hours

b. σ = 1.972 hours

c. SE = 0.805 hours

Step-by-step explanation:

Sample size (n) = 6

Sample data (xi) = 10, 8, 9, 7, 11, 13

a. Mean time spent in a week for this course by students:

Sample mean is given by:

[tex]\mu = \frac{\sum x_i}{n} \\\mu = \frac{10+9+7+11+13}{6}\\\mu=9.667[/tex]

Mean time spent in a week per student is 9.667 hours

b. Standard deviation of the time spent in a week for this course by students:

Standard deviation is given by:

[tex]\sigma = \sqrt{\frac{\sum(x_i - \mu)^2}{n}}\\\sigma = \sqrt{\frac{(10- 9.667)^2+(8- 9.667)^2+(9- 9.667)^2+(7- 9.667)^2+(11- 9.667)^2+(13- 9.667)^2}{6}}\\\sigma =1.972[/tex]

c. Standard error of the estimated mean time spent in a week for this course by students:

Standard error is given by:

[tex]SE = \frac{\sigma}{\sqrt n}\\SE = \frac{1.972}{\sqrt 6}\\SE=0.805[/tex]

Answer:

93.03%

Step-by-step explanation:

Population mean (μ) = 66.1 inches

Standard deviation (σ) = 2.7 inches

The z-score for a given 'X' value is:

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

For X = 62 inches

[tex]z = \frac{62- 66.1}{2.7}\\z=-1.5185[/tex]

A z-score of -1.5185 corresponds to the 6.44-th percentile of a normal distribution.

For X = 62 inches

[tex]z = \frac{73- 66.1}{2.7}\\z=2.5555[/tex]

A z-score of 2.5555 corresponds to the 99.47-th percentile of a normal distribution.

The total percentage of men eligible for the military is the percentage within those two values, therefore:

[tex]E = 99.47-6.44\\E=93.03 \%[/tex]

93.03%  this​ country's men are eligible for the military based on height.

Final answer:

To determine the percentage of men eligible for the military based on height, we need to calculate the proportion of men whose heights fall between 62 inches and 73 inches. First, convert the height values to z-scores using the formula z = (x - mean) / standard deviation. Then, look up the corresponding z-scores in the standard normal distribution table to find the proportion of men within the desired height range.

Explanation:

To determine the percentage of men eligible for the military based on height, we need to calculate the proportion of men whose heights fall between 62 inches and 73 inches.

First, we convert the height values to z-scores using the formula z = (x - mean) / standard deviation.

Then, we look up the corresponding z-scores in the standard normal distribution table to find the proportion of men within the desired height range. We can subtract this proportion from 1 to find the percentage of men who are eligible for the military based on height.

The test statistic is determined for the random sample conducted by rhe psychologist and the values are:

a) t ≈ 2.12

b) P-value ≈ P(T > 2.12)

c) Do not reject H0 because the P-value is less than the α = 0.05 level of significance.

d) It is not possible to make a definitive conclusion about the IQs of children born to mothers who listen to Mozart.

Given data:

To test whether the children in the study have higher IQs,conduct a one-sample t-test.

a) To conduct the t-test, set up the null and alternative hypotheses:

Null hypothesis (H0):

The population mean IQ of the children is equal to 100.

Alternative hypothesis (Ha):

The population mean IQ of the children is greater than 100.

b)

Calculate the P-value:

Determine the P-value using the t-distribution with the sample mean, sample standard deviation, and sample size.

P-value = P(T > t), where T follows a t-distribution with (n-1) degrees of freedom.

Using the given information, calculate the t-value:

t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))

t = (104.1 - 100) / (15 / sqrt(20))

t ≈ 2.12

Next, we need to find the P-value associated with this t-value using the t-distribution.

For a one-tailed test, the P-value is the probability of observing a t-value greater than 2.12.

P-value ≈ P(T > 2.12)

c)

State the conclusion for the test:

To make a conclusion, we compare the P-value to the significance level (α = 0.05).

Since the P-value is not provided, we cannot make a definitive conclusion. However, based on the options given, the correct answer would be:

c. Do not reject H0 because the P-value is less than the α = 0.05 level of significance.

d)

State the conclusion in the context of the problem:

Based on the incomplete information provided, we cannot make a definitive conclusion about the IQs of children born to mothers who listen to Mozart.

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To determine if there is evidence that the children have higher IQs, a p-value is calculated from the sample using a t-test. If the p-value is less than the significance level of 0.05, the null hypothesis is rejected, indicating evidence of higher IQs; otherwise, there is insufficient evidence.

The question asks whether the children have higher IQs than the normal mean of 100, given a sample mean IQ of 104.1 with a standard deviation of 15 from a sample of 20 children.

To calculate the p-value for a one-sample t-test:

The exact calculation is not provided here, but once you have the p-value:

We would either conclude that there is or is not sufficient evidence at the α=0.05 level to suggest that mothers who listen to Mozart have children with higher IQs, depending on whether the p-value is less than or greater than 0.05.

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To find the probability, use the Central Limit Theorem to calculate the z-scores for the lower and upper limits of the sample mean. Look up these z-scores in the standard normal distribution table to find the probabilities. Subtract the probability for the lower z-score from the probability for the higher z-score to find the probability that the mean of the sample falls between the two values.

To find the probability that the mean of a sample will be between 94.8 inches and 95.8 inches, we can use the Central Limit Theorem. We start by calculating the z-scores for these values using the formula: z = (x - μ) / (σ / √n), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. With the given values, the z-score for 94.8 inches is -3.85 and the z-score for 95.8 inches is 1.92.

Next, we look up these z-scores in the standard normal distribution table to find the corresponding probabilities. The probability for a z-score of -3.85 is approximately 0.00005 and the probability for a z-score of 1.92 is approximately 0.97128. To find the probability that the mean of the sample falls between these two values, we subtract the probability for the lower z-score from the probability for the higher z-score: 0.97128 - 0.00005 = 0.97123.

Therefore, the probability that the mean of the sample will be between 94.8 inches and 95.8 inches is approximately 0.97123, or 97.12%.

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Answer:

[293.21;372.28]

Step-by-step explanation:

Hello!

You are asked to construct a 90% Confidence Interval for the mean duration of the foreclosure process in Boston.

The Study variable is X: duration of foreclosure in Boston. X≈N(μ;σ²)

Sample n=8

To study the population sample you can use either a Z or a t-statistic. Since the sample is less than 10, I'll choose a Student t-statistic, because it is more potent with small samples than the Z, but either one is a good choice.

X[bar]= 332.75

S= 59.01

[tex]X[bar] ± t_{n-1; 1-\alpha /2}*\frac{S}{\sqrt{n} }[/tex]

[tex]332.75 ± 1.895*\frac{59,01}{\sqrt{8} }[/tex]

[293.21;372.28]

With a confidence level of 90% you'd expect the interval [293.21;372.28] to contain the population mean of the duration of the foreclosure process in Boston.

I hope you have a SUPER day!

The question asks for the parametrization of boundary edges of a region in the 1st quadrant and the computation of a line integral over it. The region is bounded by the x-axis, quarter-circle of radius 7, and the y-axis. This is done by parameterizing each edge and applying Green's theorem to calculate the related integral.

This exercise is essentially setting up and evaluating a line integral over closed path, which tells you about the interaction between a vector field and a curve in its domain. The region in question is in the first quadrant, and is bounded by the x-axis (C_1), a quarter circle (arc) of radius 7 (C_2), and the y-axis (C_3). The oriented counterclockwise is a convention means going from the origin along the x-axis, then following the quarter-circle (arc) around, and then starting back along the y-axis towards the origin.

The parametrization of the constant-speed edge C_1 is x_1(t)=7t, y_1(t)=0, for 0 <= t <= 1. For edge C_2 is x_2(t)=7cos(π/2t), y_2(t)=7sin(π/2t), for 0 <= t <= 1 and for C_3 is x_3(t)=0, y_3(t)=7(1-t), for 0 <= t <= 1. As the integral of a scalar function over C (the quarter-circle location) is equal to the sum of integrals over the three parts (C1, C2, C3), we can break it into three simpler parts and integrate separately.

The Vector Field F related to the integral can be represented by F = y^2xi + x^2yj based on the given equation. To find the resulting integral via Green's theorem, you would just take the divergence of the field F, yielding a double integral over the region.

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Answer:

  290.6 backpacks per year

Step-by-step explanation:

The average of 5 numbers is 1/5 of their sum:

  average = (278 +310 +320 +242 +303)/5 = 290.6

The average number of backpacks given out in the last 5 years is 290.6, about 291.

Answer:

There is no evidence that the spirituality category proportions are different for natural and social scientists.

Step-by-step explanation:

To solve this question we must perform a Chi square test calculating the expected values of the observed behavior.

We start by completing the table given by adding the totals:

Observed    Very   Moderate   Slightly   Not at all      Total

Natural Sc    54           161            195            216          626

Social Sc      55           220          239           242         756

Total            109           381           434           458         1382

Now, the chi square value is calculated with the following formula:

[tex]x^{2}[/tex]=∑∑[tex]\frac{O_{ij}-E_{ij} }{E_{ij} }[/tex]

Where:

[tex]O_{ij}[/tex]: Observed value (the ones we have in our table)

[tex]E_{ij}[/tex]: Expected value

The expected value of every observation (ij) is calculated as it follows:

[tex]E_{ij}=\frac{n_{i}c_{j} }{N}[/tex]

Where,

[tex]n_{i}[/tex]: marginal total by rows

[tex]c_{j}[/tex]: marginal total by columns

N: Total of observations

Now, for the expected observations we obtain the following table:

Expected     Very     Moderate   Slightly     Not at all      Total

Natural Sc    49.37      172.58       196.59      207.46        626

Social Sc      59.63      208.42      237.41      250.54        756

Total            109           381           434           458              1382

Having the expected values we now can calculate [tex]x^{2}[/tex] by first calculating [tex]\frac{O_{ij}-E_{ij} }{E_{ij} }[/tex] for each observation:

Chi sq           Very     Moderate   Slightly     Not at all      Total

Natural Sc    0.434      0.777         0.013        0.352           1.575

Social Sc      0.359      0.643        0.011         0.291            1.304

Total             0.793      1.420         0.024       0.643           2.879

[tex]x^{2}=2.879[/tex]

We may consider our null hypothesis as it follows:

[tex]H_{0}:[/tex] The degree of spirituality category proportions are the same for natural and social scientists.

To prove this we have:

[tex]P(x^{2}\geq 2.879)[/tex]

α=0.01

α: significance level  

Calculating this value with a chi square table (or with statistical software like R) we obtain:

P-value=0.4105

Because the p-value is larger than α the null hypothesis is accepted. Which means we cannot say that the spirituality category proportions are different.