In a certain country the heights of adult men are normally distributed with a mean of 66.1 inches and a standard deviation of 2.7 inches. The​ country's military requires that men have heights between 62 inches and 73 inches. Determine what percentage of this​ country's men are eligible for the military based on height.

Answer:

We conclude that the lifetime of tires is less than 30,000 miles.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 30,000 miles

Sample mean, [tex]\bar{x}[/tex] = 29,400 miles

Sample size, n = 54

Alpha, α = 0.05

Population standard deviation, σ =1200 miles

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 30000\text{ miles}\\H_A: \mu < 30000\text{ miles}[/tex]

We use One-tailed z test to perform this hypothesis.

Formula:

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

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{29400 - 30000}{\frac{1200}{\sqrt{54}} } = -3.6742[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = -1.64[/tex]

Since,  

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

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that the lifetime of tires is less than 30,000 miles.

To maximize and minimize the quantity given by p^2q^250 with the constraint p+q=10, express q in terms of p, derive a single-variable function, optimize using differentiation, find critical points, and evaluate at the endpoints of the domain [0, 10]. The difference between the maximum and minimum values would be the answer.

The student asks how to maximize and minimize the quantity p2q250, given the constraint that p + q = 10 and that p and q are nonnegative. To find the maximum and minimum values of this expression, we can use the concept of optimization in algebra. First, express q in terms of p, using the given equation p + q = 10. We get q = 10 - p. Substitute this into the original expression to obtain a single-variable function f(p) = p2(10 - p)250. Now, we can differentiate this function with respect to p and find its critical points.

To find the difference between the maximum and minimum values, evaluate f(p) at the critical points and at the endpoints of the domain (since p and q have to be nonnegative, the domain is [0, 10]). The required difference is the greatest value of f(p) minus the smallest value of f(p), which is the solution to the student's problem. This can be found by tedious algebra and the final answer would be expressed as a fraction.

Answer:

It was originally $91.25

Answer:

92 meters

Step-by-step explanation:

This is a problem that can be solved by the concept of "similar triangles", where we have two right angle triangles that share also another common angle" the angle that the rays of the sun form with them (see attached image).

The smaller triangle is formed by the limiting rays of the sun, the pole, and its shadow: it has a height of 3.3 m (the length of the pole) and a base of 1.69 m (the pole's shadow).

The larger triangle is formed by the rays of the sun, the building and its shadow: it has a base of 47.25 m and an unknown height that we named as "x" (our unknown).

The ratio of the bases of such similar triangles must be in the same proportion as the ratio of their heights, so we can create a simple  equation that equals such ratios, and then solve for the unknown "x":

[tex]\frac{H}{h} =\frac{B}{b}\\\frac{x}{3.3} =\frac{47.25}{1.69}\\x=\frac{47.25\,*\,3.3}{1.69} \\x=92.26331\, m[/tex]

which we can round to the nearest meter as: 92 m

Answer:

(a)C(x) = 2500 + 3x

(b)R(x) = 5.5x

(c)x = 1000

Step-by-step explanation:

(a)His cost function as a function of x

C(x) = 2500 + 3x

(b)His revenue R function as a function of x

R(x) = 5.5x

(c)When revenue R equals to Cost C

C(x) = R(x)

2500 + 3x = 5.5x

2.5x = 2500

x = 1000

(d) Refer to attachment. We can see that the 2 lines intercepts at x = 1000 and y = 2500. That's the point of turning to profit.

Answer:

  |x| < √7

Step-by-step explanation:

The product of the roots of the given quadratic equation is 2/a, so the other root (the one not given) is 2/(7a). It will be negative, since "a" is negative.

The roots of ax^2 +bx +2 = 0 are the values of x^2 that satisfy ax^4 +bx^2 +2 = 0. That is, roots of the latter equation will be the square root of the roots of the former equation.

We know that two of the zeros of the quartic are ±√7, and the other two are complex, as they are the square roots of a negative number. So, the graph of the quartic opens downward (because a < 0), and has real zeros at x=±√7.

The solution to the inequality must be ...

  -√7 < x < √7

_____

The graph shows an example of the quadratic (green) and quartic (black). The ripple in the quartic changes amplitude with different values of "a", but the locations of the zeros do not change.

Answer:

The 98% confidence interval for the mean repair cost for the dryers is (83.4161, 103.3039).

Step-by-step explanation:

We have a small sample size n = 25, [tex]\bar{x} = 93.36[/tex] and s = 19.95. The confidence interval is given by  [tex]\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})[/tex] where [tex]t_{\alpha/2}[/tex] is the [tex]\alpha/2[/tex]th quantile of the t distribution with n - 1 = 25 - 1 = 24 degrees of freedom. As we want the 98% confidence interval, we have that [tex]\alpha = 0.02[/tex] and the confidence interval is [tex]93.36\pm t_{0.01}(\frac{19.95}{\sqrt{25}})[/tex] where [tex]t_{0.01}[/tex] is the 1st quantile of the t distribution with 24 df, i.e., [tex]t_{0.01} = -2.4922[/tex]. Then, we have [tex]93.36\pm (-2.4922)(\frac{19.95}{\sqrt{25}})[/tex] and the 98% confidence interval is given by (83.4161, 103.3039).

Answer:

Step-by-step explanation:

Our aim is to determine a 98% confidence interval for the mean repair cost for the dryers

Number of samples. n = 25

Mean, u = $93.36

Standard deviation, s = $19.95

For a confidence level of 98%, the corresponding z value is 2.33. This is determined from the normal distribution table.

We will apply the formula,

Confidence interval

= mean ± z × standard deviation/√n

It becomes

93.36 ± 2.33 × 19.95/√25

= 93.36 ± 2.33 × 3.99

= 93.36 ± 9.2967

The lower boundary of the confidence interval is 93.36 - 9.2967 =84.0633

The upper boundary of the confidence interval is 93.36 + 9.2967 = 102.6567

Therefore, with 98% confidence interval, the mean repair costs for the dryers is between $84.0633 and $102.6567

Answer:

36

Step-by-step explanation:

multiply the width & the height

Answer:

  b. y-y1 = m(x-x1)

Step-by-step explanation:

It's a matter of definition. There are perhaps a dozen useful forms of equations for a line. Each has its own name (and use). Here are some of them.

Adding y1 to the point-slope form puts it in an alternate form that is useful for getting to slope-intercept form faster: y = m(x -x1) +y1. I use this when asked to write the equation of a line with given slope through a point, with the result in slope-intercept form.

The equation of the line, in point-slope form, is given by:

[tex]y - y_1 = m(x - x_1)[/tex]

Option b.

-----------------------------------------

The equation of a line, in point-slope form, is given by:

[tex]y - y_1 = m(x - x_1)[/tex]

In which

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Answer:

80 ft x 40 ft

Step-by-step explanation:

Let 'L' be the length of the longer side and 'W' be the length of the shorter side (or the width).

The equations that compose the linear system are:

[tex]L+2W=160\\L=2W[/tex]

Solving the system:

[tex]L+L=160\\L=80\\W=\frac{L}{2}\\W=40[/tex]

The garden is a rectangle with dimensions 80 ft x 40 ft.

The dimensions of the garden are approximately 53.33 feet in width and 106.66 feet in length according to the given linear equations and conditions.

Given that the fencing of 160 feet encloses a rectangular garden on three sides and one side of the rectangle is parallel with the barn, we can derive the linear equations based on these parameters. Let's denote the width of the rectangle as x and the length of the rectangle as 2x, as stated the length is twice the width.

The perimeter of this garden will equal the amount of fencing, which is 160 feet. Because only three sides of the rectangle are enclosed by the fence, the equation for the perimeter becomes 2x + x = 160. Simplifying this, we get 3x = 160. Solving for x, we divide both sides by 3, hence, x = 53.33 ft approximately. The width of the garden is thus around 53.33 feet and the length (being twice the width) would be 2x = 2*53.33 = 106.66 ft.

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To find the probability that the average tax paid on the sample forms is greater than $1980, standardize the sample mean using the Central Limit Theorem. To find the probability that more than 60 of the sampled forms have a tax greater than $3200, use a normal distribution approximation.

To solve this problem, we can use the Central Limit Theorem.

a. To find the probability that the average tax paid on the sample forms is greater than $1980, we need to standardize the sample mean. We calculate the z-score by subtracting the population mean from the sample mean and dividing by the population standard deviation divided by the square root of the sample size. Then, we can use a z-table or a calculator to find the probability.

b. To find the probability that more than 60 of the sampled forms have a tax greater than $3200, we can use a normal distribution approximation. We can calculate the z-score for 60 forms by subtracting the population mean from 60 and dividing by the square root of the population mean multiplied by (1 - the population mean) divided by the sample size. Then, we can use a z-table or a calculator to find the probability.

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Answer:

20 inches

Step-by-step explanation:

Take 2/3 of an inch 30 times.

The height of the model is then 30 times 2/3 inches = 10*2 = 20 inches

Answer:

The probability that the mean life a random sample of 25 such blenders falls between 4.6 and 5.1 years is 0.6687.

Step-by-step explanation:

We have given :

The average life a manufacturer's blender is 5 years i.e. [tex]\mu=5[/tex]

The standard deviation is [tex]\sigma=1[/tex]

Number of sample n=25.

To find : The probability that the mean life falls between 4.6 and 5.1 years ?

Solution :

Using z-score formula, [tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

The probability that the mean life falls between 4.6 and 5.1 years is given by, [tex]P(4.6<X<5.1)[/tex]

[tex]P(4.6<X<5.1)=P(\frac{4.6-5}{\frac{1}{\sqrt{25}}}<Z<\frac{5.1-5}{\frac{1}{\sqrt{25}}})[/tex]

[tex]P(4.6<X<5.1)=P(\frac{-0.4}{\frac{1}{5}}<Z<\frac{0.1}{\frac{1}{5}})[/tex]

[tex]P(4.6<X<5.1)=P(-2<Z<0.5)[/tex]

[tex]P(4.6<X<5.1)=P(Z<0.5)-P(Z<-2)[/tex]

Using z-table,

[tex]P(4.6<X<5.1)=0.6915-0.0228[/tex]

[tex]P(4.6<X<5.1)=0.6687[/tex]

Therefore, the probability that the mean life a random sample of 25 such blenders falls between 4.6 and 5.1 years is 0.6687.

The probability that the mean life of a random sample of 25 blenders falls between 4.6 and 5.1 years is approximately 0.3585, or 35.85%.

To find the probability that the mean life of a random sample of 25 blenders falls between 4.6 and 5.1 years, we can use the standard normal distribution. First, we need to standardize the values using the z-score formula. The z-score for 4.6 years is (4.6 - 5) / 1 = -0.4, and the z-score for 5.1 years is (5.1 - 5) / 1 = 0.1. We can then look up the corresponding probabilities in the standard normal distribution table or use a calculator to find the area under the curve between these two z-scores. The probability that the mean life falls in this range is approximately 0.3585, or 35.85%.

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Answer:

Step-by-step explanation:

Hello!

You have X~Bi (n;ρ)

Where:

n=10

ρ= 0.10

For all asked probabilities you need to use a Binomial distribution table. Remember this table has the information of the cummulative probabilities P(X≤x).

a. f(0) ⇒ P(X=0) = 0.3487

b. f(2) ⇒ P(X=2) ⇒ P(X≤2) - P(X≤1) = 0.9298 - 0.7361 = 0.1937

c. P(X≤2) = 0.9298

d. P(X ≥ 1) = 1 - P(X ≤ 1) = 1 - 0.7361 = 0.2639

e. E(X)= nρ = 10*0.10 = 1

f. V(X)= nρ(1-ρ) = 10*0.1*0.9 = 0.9

σ= √V(X) = √0.9 = 0.9487

I hope it helps!

The binomial experiment depicts that f(0) will be 0.3487.

From the information given, it can be noted that:

n = 10

p = 0.10

q = 1 - 0.10 = 0.90

f(0) ⇒ P(X=0) = 0.3487

f(2) = P(X=2) ⇒ P(X≤2) - P(X≤1)

= 0.9298 - 0.7361

= 0.1937

P(X≤2) = 0.9298

P(X ≥ 1) = 1 - P(X ≤ 1)

= 1 - 0.7361 = 0.2639

E(X) = nρ

= 10 × 0.10 = 1

V(X) = nρ(1-ρ)

= 10 × 0.1 × 0.9

= 0.9

σ = √V(X) = √0.9

= 0.9487

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For testing if a coin is fair, a binomial test is used to compare the observed heads to what's expected under a fair coin scenario. Differences in p-values for the same sample proportion across different sample sizes are due to the increased precision of larger samples.

In hypothesis testing, when we want to determine whether a coin (or in this case, a hypothetical lizard) is fair, we employ a binomial test. This test assesses the number of 'successes' (heads in our case), comparing it to what we would expect under the null hypothesis of p=0.5, indicating a fair coin.

For part (a), where we get 56 heads out of 100 tosses, we would calculate the p-value by looking at both the proportion of heads obtained and the standard error for a binomial distribution. Although not calculated here directly, if this p-value is less than the chosen significance level, typically 0.05, we reject the null hypothesis, concluding the coin is not fair.

For part (b), getting 560 heads out of 1000 tosses yields the same sample proportion as part (a). However, the larger sample size increases the power of the test, often resulting in a smaller p-value if the observed proportion is different from the expected 0.5.

Comparing p-values from (a) and (b) reveals that despite having the same sample proportions, the p-values differ due to the sample sizes. Larger sample sizes yield more precise estimates of the population proportion, thus potentially resulting in smaller p-values for the same sample proportion deviation from the null hypothesis.

Answer:

option 0.8%

Step-by-step explanation:

Data provided in the question:

Mean = 5.7 years

Standard deviation, s = 1.8 years

Now,

P(the employee has worked at the store for over 10 years)

= P(X > 10 years)

= [tex]P (Z > \frac{X-Mean}{\sigma})[/tex]

or

= [tex]P (Z > \frac{10-5.7}{1.8})[/tex]

= P (Z > 2.389 )

or

= 0.008447     [from standard  z table]

or

= 0.008447 × 100% = 0.84% ≈ 0.8%

Hence,

the correct answer is option 0.8%

To find the probability that an employee has worked at a large department store for over 10 years, we can use the z-score formula and a standard normal distribution table or calculator.

To find the probability that an employee has worked at the store for over 10 years, we can use the z-score formula:

z = (x - μ) / σ

where x is the value we are interested in (10 years), μ is the mean (5.7 years), and σ is the standard deviation (1.8 years).

Plugging in the values, z = (10 - 5.7) / 1.8 = 2.39

Using a standard normal distribution table or a calculator, we can find that the probability of a z-score greater than 2.39 is approximately 0.008, or 0.8%.

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Answer:

Step-by-step explanation:

(a) The average number of times a customer carries out banking transactions per week is 13 give or take 5 or so.

(b) The 90% confidence interval for the average number of banking operations per week is given by:

(Assuming a normal distribution of the number of banking operations)

CI=\overline{X}\pm 1.645\times \sigma/\sqrt{n}

CI=13\pm 1.645\times 5/\sqrt{350}

CI=13\pm 0.439645

CI=(12.560355, 13.439645)

The 99% confidence interval for the average number of banking operations per week is given by:

(Assuming a normal distribution of the number of banking operations)

CI=\overline{X}\pm 2.576\times \sigma/\sqrt{n}

CI=13\pm 2.576\times 5/\sqrt{350}

CI=13\pm 0.688465

CI=(12.311535, 13.688465)

The difference in the margin of error is 0.688465-0.439645=0.24882

The difference in the margin of error will make the confidence interval wider in the second case(99% confidence interval) as compared to the first case(90% confidence interval).

(c) Since, both our confident interval contains the value 10 hence we have sufficient evidence to conclude that the apparent difference in banking habits between the nation (It is given that the national survey suggest that on an average 10 transactions are performed per week by a single person) and the customer of the bank is not significant or is not real and is just due to the random errors/variations in sampling.

(d) A 95% confidence interval gives a range of values for the (A) Population Average which are plausible according to the observed data.

(The confidence interval always gives the range of possible values for the population values)

(e) The sample standard deviation measures how far (A) number of bank operations is from sample average.

The standard deviation for the sample average measures how far (B) Average number of bank operations is from the population average for typical (D) samples.

Answer: f(-) = - 12

Step-by-step explanation:

The function is expressed as

f(x)= -x^2+10x-3

To determine f(-1), we will substitute

x = - 1 into the given function. It becomes

f(-1) = (- 1)^2 + 10(-1) - 3

f(-1) = 1 - 10 - 3

f(-1) = - 12

Answer:

Less.

Step-by-step explanation:

Most tables are rectangle so hope it helps!

Answer: 7

Step-by-step explanation: Linear equations can be written in the form y = mx + b where the multiplier, m, represents the slope of the line and the base, b, represents the y-intercept of the line.

The linear equation y = -6x + 7 has a y-intercept equal to 7.

Answer:

Amount of mercury in population is in between 0.2841 , 1.1531

Step-by-step explanation:

Step by step explanation is given in the attachments

The P(69 ≤ X ≤ 71) when n = 16 is approximately 0.9312.

Here's the calculation for P(69 ≤ X ≤ 71) when n = 16:

1. Find the standard error (SE):

SE = σ / √n = 2.2 GPa / √16 = 0.55 GPa

2. Standardize the values:

Z1 = (69 - 70) / 0.55 = -1.82

Z2 = (71 - 70) / 0.55 = 1.82

3. Use a standard normal table or calculator to find the probabilities:

P(Z ≤ 1.82) = 0.9656

P(Z ≤ -1.82) = 0.0344

4. Calculate the probability of the interval:

P(69 ≤ X ≤ 71) = P(-1.82 ≤ Z ≤ 1.82) = P(Z ≤ 1.82) - P(Z ≤ -1.82)

= 0.9656 - 0.0344 = 0.9312

Therefore, P(69 ≤ X ≤ 71) when n = 16 is approximately 0.9312.

Answer:

skew lines

Step-by-step explanation:

we are given 2 lines in parametric form as

L1:x=18+6t,y=7+3t,z=13+3t and L2:x=−14+7ty=−12+5tz=−8+6t[tex]L1:x=18+6t,y=7+3t,z=13+3t \\ L2:x=-14+7t,y=-12+5t,z=-8+6t[/tex]

If the lines intersect then the two points must be equal for one value of t.

Let us try equating x,y and z coordinate.

[tex]18+6t = -14+7t\\t=32\\[/tex]

when we equate y coordinate we get

[tex]7+3t =-12+5t\\2t =19\\t =9.5[/tex]

Since we get two different t we find that these two lines cannot intersect.

Comparing direction ratios we have

I line has direction ratios as (6,3,3) and second line (7,5,6)

These two are not proportional and hence not parallel

So these lines are skew lines

After analyzing the direction vectors of lines L1 and L2, it is clear that they are neither parallel nor do they intersect, which means the lines are skew.

To determine whether the lines L1 and L2 intersect, are skew, or are parallel, we need to compare the direction vectors and check if they can be expressed as multiples of each other. If we consider the components of each direction vector given by the 't' terms in the equations of L1 and L2, they are (6, 3, 3) for L1 and (7, 5, 6) for L2 respectively.

To check for parallelism, we find a constant 'k' such that (6, 3, 3) = k*(7, 5, 6). After trying to solve for 'k', we immediately see that no such constant can exist, as 6/7 is not equal to 3/5 or 3/6. Therefore, these lines are not parallel.

To check for intersection, we need to find a common point that satisfies both line equations for some value of the parameter 't'. However, since an attempt to find such values results in inconsistent equations, it suggests that no such point exists and these lines do not intersect.

Given that the lines are neither parallel nor do they intersect, they must be skew lines. Skew lines are lines that do not intersect and are not parallel, lying in different planes.

Answer: [tex]0.12< \sigma<0.42[/tex]

Step-by-step explanation:

Confidence interval for standard deviation is given by :-

[tex]\sqrt{\dfrac{s^2(n-1)}{\chi^2_{\alpha/2}}}< \sigma<\sqrt{\dfrac{s^2(n-1)}{\chi^2_{1-\alpha/2}}}[/tex]

Given : Confidence level : [tex]1-\alpha=0.95[/tex]

⇒[tex]\alpha=0.05[/tex]

Sample size : n= 7

Degree of freedom = 6    (df= n-1)

sample standard deviation : s= 0.19

Critical values by using chi-square distribution table :

[tex]\chi^2_{\alpha/2, df}}=\chi^2_{0.025, 6}}=14.4494\\\\\chi^2_{1-\alpha/2, df}}=\chi^2_{0.975, 6}}=1.2373[/tex]

Confidence interval for standard deviation of the weights of the packages prepared by the machine is given by :-

[tex]\sqrt{\dfrac{ 0.19^2(6)}{14.4494}}< \sigma<\sqrt{\dfrac{ 0.19^2(6)}{1.2373}}[/tex]

[tex]\Rightarrow0.12243< \sigma<0.418400[/tex]

[tex]\approx0.12< \sigma<0.42[/tex]

Hence, the 95% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine.  :

[tex]0.12< \sigma<0.42[/tex]

Answer:

  31.747 GW

Step-by-step explanation:

The total cell capacity (C) manufactured in the time period is the integral of the rate of manufacture. So, we have ...

  [tex]C=\displaystyle\int\limits^3_0 {3.7e^{0.61t}} \, dt=\frac{3.7}{0.61}(e^{0.61\cdot 3}-1)\approx 31.747 \quad\text{GW}[/tex]

About 31.747 GW of generating capacity was manufactured in that time interval.

Answer:

can you provide the list?

Step-by-step explanation:

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].

Step-by-step explanation:

1) Data given and notation

[tex]s^2 =16[/tex] represent the sample variance

s=4 represent the sample standard deviation

[tex]\bar x[/tex] represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

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 mean or variance 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.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

[tex]df=n-1=20-1=19[/tex]

Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=36.191[/tex]

[tex]\chi^2_{1- \alpha/2}=7.633[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}[/tex]

[tex] 8.400 \leq \sigma^2 \leq 39.827[/tex]

So the 98% confidence interval for the variance in the pounds of impurities would be [tex]8.400 \leq \sigma^2 \leq 39.827[/tex].

The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points

Standard deviation = √variance = √16 = 4

The z score of  98% confidence interval is 2.326

The margin of error (E) is:

[tex]E = Z_\frac{\alpha }{2} *\frac{standard\ deviation}{\sqrt{sample\ size} } =2.326*\frac{4}{\sqrt{20} } =2.08[/tex]

The confidence interval = mean ± E = 72 ± 2.08 = (69.92, 74.08)

The 98% confidence interval for the mean change in score is between 69.92 points to 74.08 points

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Answer:

Step-by-step explanation:

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

[tex]T(t)=T_a+(T_0-T_a)e^{-kt}[/tex]

where [tex]T_a[/tex] is the ambient temperature, [tex]T_0[/tex] is the initial temperature, [tex]t[/tex] is the time and [tex]k[/tex] is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say [tex]T_0[/tex]

We know that the ambient temperature is [tex]T_a=65[/tex], so

[tex]T(t)=65+(T_0-65)e^{-kt}[/tex]

We also know that when [tex]t=5 \:min[/tex] the temperature is [tex]T(5)=35[/tex] and when [tex]t=10 \:min[/tex] the temperature is [tex]T(10)=50[/tex] which gives:

[tex]T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}[/tex]

[tex]T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}[/tex]

Rearranging,

[tex]35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}[/tex]

[tex]50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}[/tex]

If we divide these two equations we get

[tex]\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}[/tex]

[tex]\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}[/tex]

Now, that we know the value of [tex]k[/tex] we can use it to find the initial temperature of the beam,

[tex]35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5[/tex]

so the beam started out at 5 °F.

Using Newton's Law of Cooling, the initial temperature of the metal beam is estimated to be 5°F.

Estimating the Initial Temperature of the Beam Using Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its current temperature and the ambient temperature of its surroundings. Mathematically, it can be expressed as:

where:

Given:

Step-by-Step Solution:

Therefore, solving for k: k = ln(1/2) / 5 ≈ -0.1386

Finally, To = 65 - 60 = 5°F

Answer:

t= 32.327

Would be a significant difference in the average amount of saturated fat in solid and liquid fats.

Step-by-step explanation:

1) Data given and notation

Stick:[26.2,25.6,25.5,26.1,26.5,26.7]

Liquid:[16.3,16.4,16.3,16.7,16.6,17.7]

[tex]\bar X_{stick}[/tex] represent the mean for the sample Stick

[tex]\bar X_{Liquid}[/tex] represent the mean for the sample Liquid

[tex]s_{stick}[/tex] represent the sample standard deviation for the sample Stick

[tex]s_{Liquid}[/tex] represent the sample standard deviation for the sample Liquid

[tex]n_{stick}=6[/tex] sample size for the group Stick

[tex]n_{Liquid}=6[/tex] sample size for the group Liquid

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null Hypothesis :[tex]\mu_{stick}=\mu_{Liquid}[/tex]

Alternative Hypothesis :[tex]\mu_{stick} \neq \mu_{Liquid}[/tex]

If we analyze the size for the samples both are < 30 so for this case we can apply a t test to compare means, and the statistic formula is:

[tex]t=\frac{\bar X_{stick}-\bar X_{Liquid}}{\sqrt{\frac{s^2_{stick}}{n_{stick}}+\frac{s^2_{Liquid}}{n_{Liquid}}}}[/tex] (1)

t-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.

In order to calculate the mean and the sample deviation we can use the following formulas:

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

3) Calculate the statistic

First we need to calculate the mean and deviation for each sample, after apply the formulas (2) and (3) we got the following results:

[tex]\bar X_{stick}=26.10[/tex] [tex]s_{stick}=0.477[/tex]

[tex]\bar X_{Liquid}=16.67[/tex] [tex]s_{Liquid}=0.532[/tex]

And with this we can replace in formula (1) like this:

[tex]t=\frac{26.10-16.67}{\sqrt{\frac{0.477^2}{6}+\frac{0.532^2}{6}}}}=32.327[/tex]  

4) Statistical decision

For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

[tex]df=n_{stick}+n_{liquid}-2=6+6-2=10[/tex]

Since is a bilateral test the p value would be:

[tex]p_v =2*P(t_{(10)}>32.327)=1.8x10^{-11}[/tex]

So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and a would be a significant difference in the average amount of saturated fat in solid and liquid fats.