A roulette wheel has the numbers 1 through 36, 0, and 00. A bet on five numbers pays 6 to 1 (that is, if one of the five numbers you bet comes up, you get back your $1 plus another $6). How much do you expect to win with a $1 bet on five numbers? HINT [See Example 4.] (Round your answer to the nearest cent.)

Answer:

Step-by-step explanation:

Answer: Our required probability is 0.0153.

Step-by-step explanation:

Since we have given that

probability of sample shipped in small packages P(S) = 53% = 0.53

Probability of sample shipped in large packages P(L) = 47% = 0.47

Probability of sample in small break during transit P(S|B)= 2%=0.02

Probability of sample in large break during transit P(L|B) = 1% = 0.01

so, According to bayes theorem, we get that

Proportion of samples break during shipment is given by

[tex]P(S).P(S|B)+P(L).P(L|B)\\\\=0.53\times 0.02+0.47\times 0.01\\\\=0.0153[/tex]

Hence, our required probability is 0.0153.

Final answer:

To find the proportion of samples that break during shipment, multiply the percentage of samples in each packaging type by their breakage rates and add the results. Doing so, we find that 1.47% of the samples break during shipment.

Explanation:

To calculate the proportion of samples that break during shipment, we need to use the percentages of each type of packaging and their respective breakage rates. For small packages, which account for 47% of shipments, a 2% break rate applies. For large packages that make up 53% of the shipments, we have a 1% break rate. The calculation is as follows:

Now we add these two probabilities together to find the total proportion of samples that break:

Total broken samples proportion = (0.47 * 0.02) + (0.53 * 0.01) = 0.0094 + 0.0053 = 0.0147

Thus, the proportion of broken samples during shipment is 0.0147, or 1.47% when rounded to four decimal places.

Answer:

  $29.87

Step-by-step explanation:

Josh's fee is $12.99 plus $3.75 for each of 4.5 hours, so is ...

  $12.99 + 3.75×4.5 = $29.865 ≈ $29.87

Josh spent $29.87.

Answer:

Probability that we have an ordered pair (x,y) representing two points that satisfy the conditions is 328/400.

Step-by-step explanation:

This is a geometric probability.

set of all possible x in [0,20) and y in (20,40] we obtain a square in the x-y plane (in the first quadrant).

Area of this square is 20x20 = 400.

Set of all (x,y), that satisfy the distance being greater than 12

y > x, this means that (y - x) > 12 which is the same as the inequality y > x + 12.

For inequality, we obtain the region in the plane above the line y = x + 12.  The area of this region is

(1/2)bh = (1/2)(12)(12) = 72.

Thus the area of the set of all points in our square that satisfy the condition (y - x) > 12 is 400 - 72 = 328.

 Probability that we have an ordered pair (x,y) representing two points that satisfy the conditions is 328/400.

Answer:

3) x=10

Step-by-step explanation:

9x-16 +3x+11 +7x -5 =180[ sum of

interior angle of

triangle ]

19x -10 =180

19x =190

x=10

4)

2x-12 +4x+43=9x -26

or, 6x + 31= 9x -26

or, 31 +26 = 9x -6x

or, 57 = 3x

or, 19 =x

reason of this question is given up

Answer:

The particles collide when t = 7 at the point (49, 49, 49).

Step-by-step explanation:

We know the trajectories of the two particles,

[tex]r_1(t)=\langle t^2,16t-63,t^2\rangle\\r_2(t)=\langle 9t-14,t^2,13t-42\rangle[/tex]

To find if the tow particles collide you must:

[tex]t^2=9t-14\\t^2-9t+14=0\\\left(t^2-2t\right)+\left(-7t+14\right)=0\\t\left(t-2\right)-7\left(t-2\right)=0\\\left(t-2\right)\left(t-7\right)=0[/tex]

The solutions to the quadratic equation are:

[tex]t=2,\:t=7[/tex]

[tex]16t-63=t^2\\^2-16t+63=0\\\left(t^2-7t\right)+\left(-9t+63\right)=0\\t\left(t-7\right)-9\left(t-7\right)=0\\\left(t-7\right)\left(t-9\right)=0[/tex]

The solutions to the quadratic equation are:

[tex]t=7,\:t=9[/tex]

[tex]t^2=13t-42\\t^2-13t+42=0\\\left(t^2-6t\right)+\left(-7t+42\right)=0\\t\left(t-6\right)-7\left(t-6\right)=0\\\left(t-6\right)\left(t-7\right)=0[/tex]

The solutions to the quadratic equation are:

[tex]t=6,\:t=7[/tex]

Evaluate the position vectors at the common time. The common solution is when t = 7.

[tex]r_1(7)=\langle 7^2,16(7)-63,7^2\rangle=\langle 49,49,49\rangle\\\\r_2(7)=\langle 9(7)-14,7^2,13(7)-42\rangle=\langle 49,49,49\rangle[/tex]

For two particles to collide, they must be at exactly the same coordinates at exactly the same time.

The particles collide when t = 7 at the point (49, 49, 49).

Answer:

Which is the output of the formula =AND(12>6;6>3;3>9)?

A.

TRUE

B.

FALSE

C.

12

D.

9

Step-by-step explanation:

Answer:  2.  H0: μ = 40 vs. H1: μ < 40

Step-by-step explanation:

Let [tex]\mu[/tex] be the average number of tissues a person uses during the course of a cold.

Given : Researchers claim that 40 tissues is the average number of tissues a person uses during the course of a cold.

i.e. [tex]\mu=40[/tex]

The company who makes Puffs brand tissues thinks that fewer of their tissues are needed.

i.e. [tex]\mu<40[/tex]

Thus , the set of hypothesis would be :-

 [tex]H_0: \mu=40[/tex]

 [tex]H_a: \mu<40[/tex]

Thus , the correct answer is option 2. H0: μ = 40 vs. H1: μ < 40

The null hypothesis (H0) is that an average person uses 40 tissues during a cold (H0: μ = 40). The alternative hypothesis (H1), which Puffs brand wants to prove, is that less than 40 of their tissues are needed during a cold (H1: μ < 40).

The question is related to forming a null hypothesis and an alternative hypothesis in the context of statistics. Here, we are talking about the utilization of tissues during a cold. The researchers claim that on average a person uses 40 tissues (let's call this number 'μ') throughout a cold's duration. However, the Puffs brand makers, as per their assumption, think that less tissues of their brand are needed.

The null hypothesis, denoted by H0, is usually a claim of no effect or no difference. In this case, it is H0: μ = 40, denoting that the average number of tissues used is 40. This is also the claim that the Puffs brand intends to challenge or nullify.

On the other hand, the alternative hypothesis (H1) is the claim that the party interested in the outcome (here, the Puffs brand) wants to validate. Here, it would be H1: μ < 40, which signifies that fewer than 40 Puffs tissues are needed during the course of a cold.

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Answer: 28 * 1

Step-by-step explanation:

Answer:

B)1,1,1,4

Step-by-step explanation:

A.1,2,3,4

Mean=[tex]\bar x=\frac{1+2+3+4}{4}=2.5[/tex]

x    [tex]x-\bar x[/tex]   [tex](x-\bar x)^2[/tex]

1       -1.5                                2.25

2       -0.5                              0.25

3         0.5                              0.25  

4          1.5                               2.25

[tex]\sum(x-\bar x)^2=2.25+0.25+0.25+2.25=5[/tex]

B.

Mean=[tex]\bar x=\frac{1+1+1+4}{4}=1.75[/tex]

[tex](x-\bar x)^2[/tex]

0.5625

0.5625

0.5625

5.0625

[tex]\sum(x-\bar x)^2=0.5626+0.5625+0.5625+5.0625=6.75[/tex]

C.

Mean=[tex]\bar x=\frac{1+2+2+4}{4}=2.25[/tex]

[tex](x-\bar x)^2[/tex]

1.5625

0.0625

0.0625

3.0625

[tex]\sum (x-\bar x)=1.5625+0.0625+0.0625+3.0625=4.75[/tex]

D.[tex]Mean=\bar x=\frac{4+4+4+4}{4}=4[/tex]

[tex](x-\bar x)^2[/tex]

0

0

0

0

[tex]\sum (x-\bar x)^2=0+0+0+0=0[/tex]

We know that

S.D is directly proportional to  [tex]\sum (x-\bar x)^2[/tex].

When [tex]\sum (x-\bar x)^2[/tex] is highest  then the S.D is also highest.

We can see that  the value of  [tex]\sum (x-\bar x)^2[/tex] is highest  in option B.

Therefore, S.D of the date set of  option B is highest.

Final answer:

The data set with the numbers 1 through 4 (option A) likely has the highest standard deviation due to having a greater spread compared to others.

Explanation:

The student has asked which data set has the highest standard deviation without doing calculations. In looking at the provided options, we can determine that the higher the variability in the data set, the higher the standard deviation will be. Data set A has numbers 1 through 4, which are more spread out compared to the other sets. Therefore, option A) 1,2,3,4 is likely to have the highest standard deviation. Option D) 4,4,4,4 will have a standard deviation of zero since all the data points are the same. Standard deviation is a measure of how spread out numbers are and a zero standard deviation occurs when all values in a data set are identical.

Answer:

Step-by-step explanation:

See the attached for a graph of 1/sin(x) = csc(x). Just as sin(x) is an odd function, so is 1/sin(x).

Any odd function is symmetrical about the origin, so the left side of the graph is a copy of the right side, rotated 180° about the origin ("in the  other direction"). That is, csc(x) = -csc(-x). This is true because the function is odd, not because it is periodic.

__

Csc(x) is also a periodic function with a period of 2π. That means ...

  csc(x+2π) ≡ csc(x)

This sort of replication of the function is true for all periodic functions (where the added value, the horizontal translation, is the period of the function).

__

You will note that the relation applicable to an odd function (-f(-x) = f(x)) is different from the relation applicable to a periodic function (f(x) = f(x+period)). An odd periodic function, such as csc(x), will be described by both relations.

Answer:

[tex]t=\frac{\pm\sqrt{dk}}{k}[/tex]

Step-by-step explanation:

Given relation:

[tex]d=k\ t^2[/tex]

We need to solve for [tex]t[/tex]

The given relation can be rearranged by isolating [tex]t[/tex] on one side.

Swapping the sides of the equation we have,

[tex]k\ t^2=d[/tex]

Dividing both sides by [tex]k[/tex] on order to cancel out [tex]k[/tex] on left side.

[tex]\frac{k\ t^2}{k}=\frac{d}{k}[/tex]

[tex]t^2=\frac{d}{k}[/tex]

We have got an expression for [tex]t^2[/tex] but we need to solve for [tex]t[/tex]

So, we take square root both sides to change [tex]t^2[/tex] to [tex]t[/tex]

[tex]\sqrt{t^2}=\sqrt{\frac{d}{k}}[/tex]

[tex]t=\pm\sqrt{\frac{d}{k}}[/tex]

So, we have successfully isolated [tex]t[/tex] on left side.

But the expression we got is a fraction with a square root  in the denominator. Thus we need to rationalize it to make it in simplest form.

The expression can be written by taking square root separately for numerator and denominator :

[tex]t=\pm{\frac{\sqrt{d}}{\sqrt{k}}}[/tex]

Multiplying the numerator and denominator by [tex]\sqrt{k}[/tex]

[tex]t=\pm\frac{\sqrt{d}}{\sqrt{k}}\times\frac{\sqrt{k}}{\sqrt{k}} [/tex]

[tex]t=\pm\frac{\sqrt{dk}}{(\sqrt{k})^2}[/tex]

Square of a square root will remove the square root. Thus we have,

∴ [tex]t=\frac{\pm\sqrt{dk}}{k}[/tex]

Thus we have successfully got the expression in the simplest form.

(a) The displacement of the particle is [tex]\(-\frac{21}{2}\)[/tex] meters.

(b) The distance travelled by the particle during the given time interval is [tex]\(\frac{169}{6}\)[/tex] meters.

(a) To find the displacement of the particle over the time interval [0, 3], we need to calculate the definite integral of the velocity function v(t) over this interval. The displacement is given by the integral of velocity with respect to time.

The velocity function is v(t) = 3t - 8.

The displacement S over the interval [a, b] is given by:

[tex]\[ S = \int_{a}^{b} v(t) \, dt \][/tex]

For our interval [0, 3], we have:

[tex]\[ S = \int_{0}^{3} (3t - 8) \, dt \][/tex]

To compute this integral, we find the antiderivative of the integrand:

[tex]\[ \int (3t - 8) \, dt = \frac{3}{2}t^2 - 8t + C \][/tex]

Now, we evaluate this antiderivative at the endpoints of the interval and subtract:

[tex]\[ S = \left(\frac{3}{2}(3)^2 - 8(3)\right) - \left(\frac{3}{2}(0)^2 - 8(0)\right) \][/tex]

[tex]\[ S = \left(\frac{3}{2} \cdot 9 - 24\right) - (0) \][/tex]

[tex]\[ S = \frac{27}{2} - 24 \][/tex]

[tex]\[ S = \frac{27}{2} - \frac{48}{2} \][/tex]

[tex]\[ S = -\frac{21}{2} \][/tex]

So the displacement of the particle is [tex]\(-\frac{21}{2}\)[/tex] meters.

(b) To find the distance travelled by the particle, we need to consider the intervals where the velocity is positive and where it is negative. The distance travelled is the sum of the distances travelled during each interval, without regard to direction.

First, we find the times when the velocity is zero by setting v(t) equal to zero:

[tex]\[ 3t - 8 = 0 \][/tex]

[tex]\[ 3t = 8 \][/tex]

[tex]\[ t = \frac{8}{3} \][/tex]

Since [tex]\(0 \leq t \leq 3\),[/tex] the velocity changes sign at [tex]\(t = \frac{8}{3}\)[/tex]. This divides the interval [0, 3] into two subintervals: [0, 8/3] and [8/3, 3].

We calculate the distance travelled in each subinterval and sum them:

For [0, 8/3]:

[tex]\[ d_1 = \int_{0}^{\frac{8}{3}} |3t - 8| \, dt \][/tex]

Since the velocity is negative on this interval, we have:

[tex]\[ d_1 = \int_{0}^{\frac{8}{3}} (8 - 3t) \, dt \][/tex]

[tex]\[ d_1 = \left(8t - \frac{3}{2}t^2\right) \Bigg|_{0}^{\frac{8}{3}} \][/tex]

[tex]\[ d_1 = \left(8\left(\frac{8}{3}\right) - \frac{3}{2}\left(\frac{8}{3}\right)^2\right) - (0) \][/tex]

[tex]\[ d_1 = \frac{64}{3} - \frac{3}{2}\left(\frac{64}{9}\right) \][/tex]

[tex]\[ d_1 = \frac{64}{3} - \frac{32}{3} \][/tex]

[tex]\[ d_1 = \frac{32}{3} \][/tex]

For [8/3, 3]:

[tex]\[ d_2 = \int_{\frac{8}{3}}^{3} |3t - 8| \, dt \][/tex]

Since the velocity is positive on this interval, we have:

[tex]\[ d_2 = \int_{\frac{8}{3}}^{3} (3t - 8) \, dt \][/tex]

[tex]\[ d_2 = \left(\frac{3}{2}t^2 - 8t\right) \Bigg|_{\frac{8}{3}}^{3} \][/tex]

[tex]\[ d_2 = \left(\frac{3}{2}(3)^2 - 8(3)\right) - \left(\frac{3}{2}\left(\frac{8}{3}\right)^2 - 8\left(\frac{8}{3}\right)\right) \][/tex]

[tex]\[ d_2 = \left(\frac{27}{2} - 24\right) - \left(\frac{32}{3} - \frac{64}{3}\right) \][/tex]

[tex]\[ d_2 = \frac{27}{2} - 24 + \frac{32}{3} \][/tex]

[tex]\[ d_2 = \frac{81}{6} - \frac{144}{6} + \frac{64}{6} \][/tex]

[tex]\[ d_2 = -\frac{105}{6} \][/tex]

[tex]\[ d_2 = -\frac{35}{2} \][/tex]

Since we cannot have a negative distance, we take the absolute value:

[tex]\[ d_2 = \frac{35}{2} \][/tex]

The total distance travelled is the sum of the distances in each interval:

[tex]\[ D = d_1 + d_2 \][/tex]

[tex]\[ D = \frac{32}{3} + \frac{35}{2} \][/tex]

[tex]\[ D = \frac{64}{6} + \frac{105}{6} \][/tex]

[tex]\[ D = \frac{169}{6} \][/tex]

So the distance travelled by the particle during the given time interval is [tex]\(\frac{169}{6}\)[/tex] meters.

Answer:

with

what maam

Step-by-step explanation:

Answer: the correct option is D

Step-by-step explanation:

The zoologist is recording the loss of wolves in her state and she recorded that the number of wolves, w, in the state on January 1.

One year later, there were 84 wolves in the state which is 23 fewer than the number of wolves in the state a year earlier. This means that a year later, the number of wolves was 23 fewer than w

The number of wolves in the state on January 1 would be the sum of the number of wolves in the state a year later and the number by which it became fewer than the number on January 1. This becomes

w = 84 + 23

w - 23 = 84

Answer:

We find the length of each subinterval dividing the distance between the endpoints of the interval by the quantity of subintervals that we want.

Then

Δx= [tex]\frac{0-(-2)}{4}=\frac{2}{4}=\frac{1}{2}[/tex]

Now, each [tex]x_i[/tex] is found by adding Δx iteratively from the left end of the interval.

So

[tex]x_0=-2\\x_1=-2+\frac{1}{2}=\frac{-3}{2}\\x_2=\frac{-3}{2}+\frac{1}{2}=-1\\x_3=-1+\frac{1}{2}=-\frac{1}{2}\\x_4=\frac{-1}{2}+\frac{1}{2}=0[/tex]

Each subinterval is

[tex]s_1=[-2,-3/2]\\s_2=[-3/2,-1]\\s_3=[-1,-1/2]\\s_4=[-1/2,0][/tex]

The midpoints of the subintervals are

[tex]m_1=\frac{-2-3/2}{2}=\frac{-7/2}{2}=\frac{-7}{4}\\m_2=\frac{-1-3/2}{2}=\frac{-5/2}{2}=\frac{-5}{4}\\m_3=\frac{-1/2-1}{2}=\frac{-3/2}{2}=\frac{-3}{4}\\m_4=\frac{0-1/2}{2}=\frac{-1}{4}[/tex]

The points used for the

1. left Riemann sums are the left endpoints of the subintervals, that is

[tex]x_0=-2, x_1=\frac{-3}{2}, x_2=-1, x_3= \frac{-1}{2}[/tex]

2. right Riemann sums are the right endpoints of the subinterval,

[tex]x_1=-\frac{3}{2}, x_2=-1, x_3=-\frac{1}{2}, x_4=0[/tex]

3. midpoint Riemann sums are the midpoints of each subinterval

[tex]m_1,m_2,m_3,m_4.[/tex]

The subinterval length in the given range is 0.5. The grid points are -2, -1.5, -1, -0.5, and 0. The points used for the left, right, and midpoint Riemann sums vary based on the position of the subinterval.

The interval specified is ​[negative 2​,0​] and is partitioned into n=4 subintervals. The length of each subinterval, Upper Delta x, is determined by subtracting the lower boundary from the upper boundary and dividing by the number of subintervals, in this case (0 - (-2)) / 4 = 0.5. Therefore, the subinterval length is 0.5.

The grid points are calculated by adding the subinterval length to each previous point, starting from the lower boundary: x0​ is -2, for x1, add the subinterval length 0.5 to x0, resulting in -1.5. Repeat this process to find x2, x3 and x4 which are -1, -0.5 and 0 respectively.

Now, for the left Riemann sum, you use the left-hand side points of each subinterval, which are x0​, x1​, x2​, x3. The right Riemann sum uses the right-hand side points, x1​, x2​, x3​, x4. Lastly, the midpoint Riemann sum uses points in the middle of each subinterval. As the subintervals here each have a length of 0.5, their midpoints fall on x0.5, x1.5, x2.5 and x3.5 respectively.

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(a) Hypotheses:

[tex]\[ H_0: \mu = 3 \]\\H_a: \mu \neq 3 \][/tex]

(b) **Test Statistic and P-Value:

[tex]\[ t \approx 1.878 \][/tex]

(c) [tex]\[ \text{P-Value} \approx 0.0678 \][/tex]

Let's perform a hypothesis test to determine whether the number of weeks until a house is sold in Greene County differs from the national average. The average number of weeks to sell a home nationally is 3 weeks.

Hypotheses:

- Null hypothesis [tex](\(H_0\))[/tex]: The average number of weeks to sell a home in Greene County is equal to the national average [tex](\(μ = 3\) weeks)[/tex].

[tex]\[ H_0: \mu = 3 \][/tex]

- Alternative hypothesis [tex](\(H_a\))[/tex]: The average number of weeks to sell a home in Greene County differs from the national average.

[tex]\[ H_a: \mu \neq 3 \][/tex]

Given Information:

- Sample mean [tex](\(\bar{x}\))[/tex] = 3.6 weeks

- Sample standard deviation s = 2 weeks

- Sample size n = 39

- Level of significance [tex](\(\alpha\))[/tex] = 0.05

Test Statistic:

The test statistic for a one-sample t-test is given by:

[tex]\[ t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \][/tex]

Calculation:

[tex]\[ t = \frac{3.6 - 3}{\frac{2}{\sqrt{39}}} \][/tex]

Let's calculate the value of \(t\) and the p-value.

Calculation:

[tex]t = \frac{3.6 - 3}{\frac{2}{\sqrt{39}}} \]\\t \approx \frac{0.6}{\frac{2}{\sqrt{39}}} \]\\t \approx \frac{0.6}{0.319439} \]\\t \approx 1.878 \][/tex]

Degrees of Freedom:

The degrees of freedom for a one-sample t-test is \(n-1\), where \(n\) is the sample size.

[tex]\[ \text{Degrees of Freedom} = 39 - 1 = 38 \][/tex]

P-Value:

Now, we need to find the p-value associated with this t-statistic and degrees of freedom. Since it's a two-tailed test, we're interested in the probability that the t-statistic is greater than 1.878 or less than -1.878.

You can use a t-table or statistical software to find this p-value. For [tex]\(df = 38\)[/tex] and [tex]\(t \approx 1.878\)[/tex], the p-value is approximately 0.0678 (two-tailed).

Conclusion:

Since the p-value (0.0678) is greater than the significance level (\(\alpha = 0.05\)), we do not reject the null hypothesis.

(a) Hypotheses:

[tex]\[ H_0: \mu = 3 \]\\H_a: \mu \neq 3 \][/tex]

(b) Test Statistic and P-Value:

[tex]\[ t \approx 1.878 \][/tex]

[tex]\[ \text{P-Value} \approx 0.0678 \][/tex]

(c) Conclusion:

At a 5% level of significance, there is not enough evidence to conclude that the average number of weeks until a house is sold in Greene County differs from the national average in 2017.

(a)State the null and alternative hypothesis. (Enter != for ≠ as needed.)H0:H0: μ = 3 (weeks) Ha: μ ≠ 3 (weeks)(b) the value of the test statistic. (Round your answer to three decimal places.) the p-value. (Round your answer to four decimal places.)p-value =1.702

To conduct the hypothesis test, calculate the test statistic and the p-value. The test statistic for a hypothesis test comparing a sample mean to a population mean when the population standard deviation is known is calculated using the formula:

[tex]\[ Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} \][/tex]

Where:

[tex]\(\bar{X}\)[/tex]n (3.6 weeks)

\(\mu\) = population mean (3 weeks)

[tex]\(\sigma\)[/tex]n standard deviation (2 weeks)

[tex]\(n\)[/tex]mple size (39)

Plugging in the values:

[tex]\[ Z = \frac{3.6 - 3}{\frac{2}{\sqrt{39}}} \][/tex]

[tex]\[ Z ≈ 1.702 \][/tex]

The p-value associated with this test statistic can be found using a standard normal distribution table or a statistical calculator. With a Z-score of 1.702, the corresponding two-tailed p-value is approximately 0.0883.

The calculated test statistic, \( Z ≈ 1.702 \), falls within the critical region for a two-tailed test at a significance level of α = 0.05. This means that the p-value (0.0883) is greater than the chosen level of significance (0.05). Consequently, there is insufficient evidence to reject the null hypothesis.

Thus, based on the sample data from Greene County, Ohio, in 2017, we do not have enough evidence to conclude that the average number of weeks until a house sold differs significantly from the national average in 2017. This suggests that, statistically, the time it takes for a house to sell in Greene County during that year might not be significantly different from the national average.

Overall, with a p-value of 0.0883 and at a significance level of 0.05, the analysis does not provide enough evidence to reject the null hypothesis, suggesting that the average number of weeks until a house sold in Greene County, Ohio, in 2017 may not differ significantly from the national average in the same year.

Answer: (94.08, 101.92)

Step-by-step explanation:

The confidence interval for unknown population mean[tex](\mu)[/tex] is given by :-

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

, where [tex]\overline{x}[/tex] = Sample mean

[tex]\sigma[/tex] = Population standard deviation

z* = Critical z-value.

Given : A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.

[tex]\sigma= 12[/tex]

[tex]\overline{x}=98[/tex]

n= 36

Confidence interval = 95%

We know that the critical value for 95% Confidence interval : z*=1.96

Then, the 95% confidence interval for the mean of x  will be :-

[tex]98\pm (1.96)\dfrac{12}{\sqrt{36}}[/tex]

[tex]=98\pm (1.96)\dfrac{12}{6}[/tex]

[tex]=98\pm (1.96)(2)[/tex]

[tex]=98\pm 3.92=(98-3.92,\ 98+3.92)\\\\=( 94.08,\ 101.92)[/tex]

Hence, the 95% confidence interval for the mean of x is (94.08, 101.92) .

Answer:  95% confidence interval would be (94.08,101.92).

Step-by-step explanation:

Since we have given that

n = 36

standard deviation = 12

sample mean = 98

At 95% confidence, z = 1.96

So, interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=98\pm 1.96\dfrac{12}{\sqrt{36}}\\\\=98\pm 3.92\\\\=(98-3.92,98+3.92)\\\\=(94.08,101.92)[/tex]

Hence, 95% confidence interval would be (94.08,101.92).

The volume of a box is the amount of space in the box.

(a) The dimension of the box

Let:

The width of the cardboard be x.

So, the length of the cardboard is: 2.4x

When 3 inches is removed, the dimension of the box is:

[tex]\mathbf{L = 2.4x- 6}[/tex] --- length

[tex]\mathbf{W = x - 6}[/tex] ---- width

[tex]\mathbf{H = 3}[/tex] --- height

(b) The restriction on x

When 3 inches is removed, it means that a total of 6 inches will be removed from either sides.

Hence, the value of x must be greater than 6.

So, the restriction on x is:

[tex]\mathbf{x > 6}[/tex]

(c) Function that represents volume.

The volume (V) of a box is:

[tex]\mathbf{V = L \times W \times H}[/tex]

So, we have:

[tex]\mathbf{V = (2.4x - 6) \times (x - 6) \times 3}[/tex]

Simplify

[tex]\mathbf{V = (7.2x - 18) \times (x - 6)}[/tex]

Expand

[tex]\mathbf{V = 7.2x^2 - 43.2x - 18x +108}[/tex]

[tex]\mathbf{V = 7.2x^2 -61.2x +108}[/tex]

(d) The dimension, when the volume is 520

This means that, V = 520

So, we have:

[tex]\mathbf{ 7.2x^2 -61.2x +108 = 520}[/tex]

Collect like terms

[tex]\mathbf{ 7.2x^2 -61.2x +108 - 520 = 0}[/tex]

[tex]\mathbf{ 7.2x^2 -61.2x -412 = 0}[/tex]

Using a calculator, we have:

[tex]\mathbf{x = 12.93\ or\ x = -4.43}[/tex]

Recall that: [tex]\mathbf{x > 6}[/tex]

So, the value of x is:

[tex]\mathbf{x = 12.93}[/tex]

Substitute 12.93 for x in [tex]\mathbf{L = 2.4x- 6}[/tex]  and [tex]\mathbf{W = x - 6}[/tex]

So, we have:

[tex]\mathbf{L = 2.4 \times 12.93 - 6 = 25.03}[/tex]

[tex]\mathbf{W = 12.93 - 6 = 6.93}[/tex]

So, the dimension of the box is: 25.03 by 6.93 by 3

(e) The value of x, when the volume is between 600 and 800

This means that, V = 600 and V = 800

When V = 600, we have:

[tex]\mathbf{ 7.2x^2 -61.2x +108 = 600}[/tex]

Collect like terms

[tex]\mathbf{ 7.2x^2 -61.2x +108 - 600 = 0}[/tex]

[tex]\mathbf{ 7.2x^2 -61.2x -492 = 0}[/tex]

Using a calculator, we have:

[tex]\mathbf{x = 13.54\ or\ x = -5.04}[/tex]

Recall that: [tex]\mathbf{x > 6}[/tex]

So, the value of x is:

[tex]\mathbf{x = 13.54}[/tex]

When V = 800, we have:

[tex]\mathbf{ 7.2x^2 -61.2x +108 = 800}[/tex]

Collect like terms

[tex]\mathbf{ 7.2x^2 -61.2x +108 - 800 = 0}[/tex]

[tex]\mathbf{ 7.2x^2 -61.2x -692 = 0}[/tex]

Using a calculator, we have:

[tex]\mathbf{x = 14.93\ or\ x = -6.45}[/tex]

Recall that: [tex]\mathbf{x > 6}[/tex]

So, the value of x is:

[tex]\mathbf{x = 14.93}[/tex]

If such a box is to have a volume between 600 and 800, the value of x would be between 13.54 and 14.93

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The response provides step-by-step explanations and formulas for finding the length, restrictions, volume, and dimensions of a box made from a cardboard piece. It also gives the dimensions that correspond to a specific volume and the range of values that produce volumes within a given range.

a) The length of the original piece of cardboard can be represented as 2.4 times the width, so it is 2.4x inches long.

b) The restrictions on x are x>6. The dimensions of the bottom rectangular base of the box are (2.4x-6) inches for the length and (x-6) inches for the width.

c) The function V that represents the volume of the box in terms of x is V = (2.4x-6)(x-6)(x-6).

d) To find the dimensions of the bottom of the box when the volume is 520 in^3, we can solve the equation V = 520. The length will be approximately 25.0 inches and the width will be approximately 6.9 inches.

e) To find the values of x for a volume between 600 and 800 in^3, we can solve the inequalities 600 <= V <= 800. The range of values for x is approximately 11.9 to 16.9 inches.

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Answer:

[tex]z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}[/tex]   (1)

[tex]z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057[/tex]  

[tex]p_v =P(Z>0.0057)=0.4977[/tex]  

The p value is a very high value and using the significance level [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.  

Step-by-step explanation:

1) Data given and notation  

[tex]X_{M}=23[/tex] represent the number of men that said they enjoyed the activity of Saturday afternoon shopping

[tex]X_{W}=8[/tex] represent the number of women that said they enjoyed the activity of Saturday afternoon shopping

[tex]n_{M}=66[/tex] sample of male selected

[tex]n_{W}=23[/tex] sample of demale selected

[tex]p_{M}=\frac{23}{66}=0.34848[/tex] represent the proportion of men that said they enjoyed the activity of Saturday afternoon shopping

[tex]p_{W}=\frac{8}{23}=0.34782[/tex] represent the proportion of women with red/green color blindness  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the value for the test (variable of interest)

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment , the system of hypothesis would be:  

Null hypothesis:[tex]p_{M} \leq p_{W}[/tex]  

Alternative hypothesis:[tex]p_{M} > p_{W}[/tex]  

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}[/tex]   (1)

Where [tex]\hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{23+8}{66+23}=0.34831[/tex]

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057[/tex]  

4) Statistical decision

Using the significance level provided [tex]\alpha=0.05[/tex], the next step would be calculate the p value for this test.  

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

[tex]p_v =P(Z>0.0057)=0.4977[/tex]  

So the p value is a very high value and using the significance level [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.  

A two-proportion z-test shows there is insufficient evidence at the 5% significance level to conclude that a higher proportion of men enjoy shopping for electronic equipment compared to women. The p-value is greater than the alpha value, leading to not rejecting the null hypothesis.

A hypothesis test can be used to determine if the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy the same activity. We will perform a two-proportion z-test for this purpose. The hypotheses are as follows:

Null Hypothesis (H0): pm = pw

Alternative Hypothesis (Ha): pm > pw

Here, pm is the proportion of men who enjoy shopping for electronic equipment, and pw is the proportion of women who enjoy shopping for electronic equipment.

Given:

The test statistic for the difference in proportions is calculated as:

z = (pm - pw) / [tex]\sqrt{(p*(1 - p*)(1/nm + 1/nw))}[/tex]

Where p* = (xm + xw) / (nm + nw)

Substituting the values:

The calculated z-value is approximately 0.0130.

The corresponding p-value for a z-value of 0.0130 is approximately 0.4948. Since p-value > alpha (0.4948 > 0.05), we do not reject the null hypothesis.

Conclusion:

At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is significantly higher than that of women.

Answer:

a) 25%

b) 50%

c) 87.5%

Step-by-step explanation:

There are [tex]2^4 = 16[/tex] number of outcome in total.

a)First and last flips come up heads. There are 4 outcomes here. Probability of the event is 4/16 = 1/4, or 25%.

- HHHH

- HHTH

- HTHH

- HTTH

b) at least 2 Consecutive flips that is heads. There are 8 outcomes. Probability of event is 8/16 = 1/2, or 50%

- HHHH

- HHHT

- HHTH

- HHTT

- THHH

- THHT

- HTHH

- TTHH

c) At least 2 consecutive flips that are the same. There are 14 outcomes. Probability of 14/16 = 7/8, or 87.5%

- HHHH

- HHHT

- HHTH

- HHTT

- THHH

- THHT

- HTHH

- TTHH

- TTTT

- TTTH

- TTHT

- HTTT

- HTTH

- THTT

Final answer:

In a) The probability is 1/4 and the event can be expressed as {HH**}, b) The probability is 7/16 and the event can be expressed as {HH**, *HH*, **HH}, c) The probability is 9/16 and the event can be expressed as {HH**, TT**, *HH*, *TT*}

Explanation:

To express the events as sets in roster notation, we need to list all the outcomes that satisfy each event.

(a) The first and last flips come up heads:

This event can be expressed as {HH**} (where ** represents any outcome for the two middle flips). The probability of this event is 1/2 * 1 * 1/2 = 1/4, since the first flip must be heads (1/2 probability), the second and third flips can be either heads or tails (1/2 probability each), and the last flip must be heads (1/2 probability).

(b) There are at least two consecutive flips that come up heads:

This event can be expressed as {HH**, *HH*, **HH}, where * represents any outcome for a single flip. The probability of this event is 1/2 * 1 * 1/2 * 1/2 + 1/2 * 1/2 * 1 * 1/2 + 1/2 * 1/2 * 1/2 * 1 = 7/16, since there are three possible patterns that satisfy this event.

(c) There are at least two consecutive flips that are the same:

This event can be expressed as {HH**, TT**, *HH*, *TT*}, where * represents any outcome for a single flip. The probability of this event is 1/2 * 1 * 1 + 1/2 * 1 * 1 + 1/2 * 1/2 * 1 + 1/2 * 1/2 * 1 = 9/16, since there are four possible patterns that satisfy this event.

Answer:

40.72 gallons

Step-by-step explanation:

Since [tex]x = 18cos(\theta)[/tex] and [tex]y = 18sin(\theta)[/tex]. We can say this base has a shape of a circle of radius r = 18.

We can also calculate h in term of angle θ:

[tex]h(\theta) = 12 + \frac{36cos(\theta) - 18sin(\theta)}{6}[/tex]

[tex]h(\theta) = 12 + 6cos(\theta) - 3sin(\theta)[/tex]

We can calculate the area of the fence if we integrate this h function over θr, which is the chord length. θ ranges from 0 to 2π

A = ∫h(θ)rdθ

A = ∫(12 + 6cos(θ) - 3sin(θ))18d

A = (12θ + 6sin(θ) + 3cos(θ))18

As θ ranges from 0 to 2π as a full circle

A = (12*0 + 6sin(0) + 3cos(0))18 - (12*2π + 6sin(2π) + 3cos(2π))18

A = 3*18 - 18(24π + 3) = 1357.17 square yard

As 1 yard = 3 feet then 1 square yard = 9 square feet

1 gallon of paint can cover 300 square feet or 300 / 9 = 33.33 square yard

So to cover square yard Bob would need

1357.17 / 33.33 = 40.72 gallons of paint.

Answer:

a) 0.8132 or 81.32%

b) 0.1868 or 18.68%

c) 0.6116 or 61.16%

d) 0.7667 minutes

e) 0.8667 minutes

f) 0.1 minutes

Step-by-step explanation:

a)

If f(x) = 0.85-0.35x (0<x<2) is the PDF and X is the random variable that measures the time it takes for a trainee to complete the task, the probability a trainee will complete the task in less than 1.31 minutes is P(X<1.31)

[tex]\large P(X<1.31)=\int_{0}^{1.31}f(x)dx=\int_{0}^{1.31}(0.85-0.35x)dx=\\\\=0.85\int_{0}^{1.31}dx-0.35\int_{0}^{1.31}xdx=\\\\=0.85*1.31-0.35*\frac{(1.31)^2}{2}=0.8132[/tex]

b)

The probability that a trainee will complete the task in more than 1.31 minutes is  

P(X>1.31) = 1 - P(X<1.31) = 1 - 0.8132 = 0.1868

c)

The probability it will take a trainee between 0.25 minutes and 1.31 minutes to complete the task is P(0.25<X<1.31)  

[tex]\large P(0.25<X<1.31)=\int_{0.25}^{1.31}f(x)dx=\int_{0}^{1.31}f(x)dx-\int_{0}^{0.25}f(x)dx=\\\\=0.8132-\int_{0}^{0.25}(0.85-0.35x)dx=0.8132-0.85\int_{0}^{0.25}dx+0.35\int_{0}^{0.25}xdx=\\\\=0.8132-0.85*0.25+0.35*\frac{(0.25)^2}{2}=0.6116[/tex]

d)

the expected time it will take a trainee to complete the task

is E(X)

[tex]\large E(X)=\int_{0}^{2}xf(x)dx=\int_{0}^{2}x(0.85-0.35x)dx=0.85\int_{0}^{2}xdx-0.35\int_{0}^{2}x^2dx=\\\\=0.85*\frac{2^2}{2}-0.35*\frac{2^3}{3}=0.7667\;minutes[/tex]

e)

[tex]\large E(X^2)=\int_{0}^{2}x^2f(x)dx=\int_{0}^{2}x^2(0.85-0.35x)dx=0.85\int_{0}^{2}x^2dx-0.35\int_{0}^{2}x^3dx=\\\\=0.85*\frac{2^3}{3}-0.35*\frac{2^4}{4}=0.8667[/tex]

f)

[tex]\large Var(X)=E(X^2)-(E(X))^2=0.8667-0.7667=0.1000[/tex]

Answer:

The instantaneous velocity for [tex]s(t) = 8t + 19[/tex] when t = 4 is [tex]v(t)=8 \:\frac{ft}{s}[/tex].

Step-by-step explanation:

The average rate of change of function f over the interval is given by this expression:

[tex]\frac{f(b)-f(a)}{b-a}[/tex]

The average velocity is the average rate of change of distance with respect to time

[tex]average \:velocity=\frac{distance \:traveled}{time \:elapsed} =\frac{\Delta s}{\Delta t}[/tex]

The instantaneous rate of change is defined to be the result of computing the average rate of change over smaller and smaller intervals.

The derivative of f with respect to x, is the instantaneous rate of change of f with respect to x and is thus given by the formula

[tex]f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]

For any equation of motion s(t), we define what we call the instantaneous velocity at time t to be the limit of the average velocity, between t and t + Δt, as Δt approaches 0.

[tex]v(t)=\lim_{\Delta t \to 0} \frac{s(t+\Delta t)-s(t)}{\Delta t}[/tex]

We know the equation of motion [tex]s(t) = 8t + 19[/tex] and we want to find the instantaneous velocity for the given value of t = 4.

Applying the definition of instantaneous velocity, we get

[tex]v(t)=\lim_{\Delta t \to 0} \frac{s(t+\Delta t)-s(t)}{\Delta t}\\\\v(4)=\lim_{\Delta t \to 0} \frac{8(4+\Delta t)+19-(8(4)+19)}{\Delta t}\\\\v(t)=\lim_{\Delta t \to 0} \frac{32+8\Delta t+19-32-19}{\Delta t}\\\\v(t)=\lim_{\Delta t \to 0} \frac{8\Delta t}{\Delta t}\\\\v(t)=8 \:\frac{ft}{s}[/tex]

The instantaneous velocity for [tex]s(t) = 8t + 19[/tex] when t = 4 is [tex]v(t)=8 \:\frac{ft}{s}[/tex].

Final answer:

By conducting a One-Sample t-test, with the data provided, it was found that there is a significant decrease in test scores for students who studied with an e-book compared to the overall class average. The calculated t-score exceeded the critical value in a one-tailed test at alpha level of .05, leading to the rejection of the null hypothesis.

Explanation:

To determine if there is a significant decrease in test scores for students who studied using an e-book compared to the overall class average, we need to conduct a hypothesis test. First, let's define our null hypothesis (H0) as there being no difference in means, implying that the e-book does not affect scores. The alternative hypothesis (H1) suggests that the e-book usage leads to lower scores. Given that the overall class average (μ) is 85 and the mean (M) for the 10 students who used the e-book is 77 with a standard deviation (s) of 6, we can use the One-Sample t-test to analyze the data.

To calculate the t-score, we use the formula t = (M - μ) / (s / √n), where n is the sample size, which is 10 in this case. Plugging in our values gives us t = (77 - 85) / (6 / √10), resulting in a t-score of approximately -4.216.

Consulting a t-table or using statistical software with a one-tailed test at α = .05 and df = 9 (n-1), we find that our critical t-value is around -1.833. Since our calculated t-score of -4.216 is more extreme than -1.833, we reject the null hypothesis. This indicates a significant decrease in test scores for students who studied with the e-book, as compared to the class average.

Answer:

Yes it is below 10% of the population

Step-by-step explanation:

10% of 12 million is 1,200,00

One of the conditions is as 2,500

The population is at least 10 times as large as the sample.

here 10*n = 10* 2500 = 25000 < 12000000

So the correct option is "Yes, 10(2,500) = 25,000, which is less than the total population."

5. The Lakeland Post polled 1,231 adults in the city to determine whether they wash their hair before their body while in the shower. Of the respondents, 63% said they washed their hair first. Suppose 51% of all adults actually wash their hair first. What are the mean and standard deviation of the sampling distribution?

Sampling distribution of sample proportion ( \hat p ) is approximately normal for large n with mean and standard deviations are as

mean = p = 0.51

for finding the standard deviation we need to use p = 0.51

б = √p·(1 - p )/ n = √0.51 . 0.49/ 1234 = 0.014248

So the correct choice is "The mean is 0.51 and the standard deviation is 0.0142"

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Final answer:

Researchers are testing the hypothesis that second-hand smoke increases the risk of low birthweight, with a null hypothesis of H0: p = 0.078, and an alternative hypothesis of Ha: p > 0.078.

Explanation:

The researchers are questioning whether second-hand smoke increases the risk of low birthweight in babies. They have a sample where 10.4% of babies born to mothers with extensive exposure to second-hand smoke are categorized as having low birth weight. This suggests the proportion might be higher than the national average of 7.8%. The null hypothesis (H0), which is the hypothesis to be tested, states that the proportion of low birth weight for the population is equal to the national average, thus H0: p = 0.078. The alternative hypothesis (Ha), representing the researchers' suspicion, is that the proportion is greater than the national average, hence Ha: p > 0.078. The last hypothesis option provided, H0: p = 0.104, is not relevant for the null hypothesis in this context since this value represents the sample proportion, not the population proportion they are testing against.

Final answer:

The distance the pitcher must travel to get to the ball is approximately 36 ft.

Explanation:

The distance the pitcher must travel to get to the ball can be found using trigonometry. We can break down the motion of the ball into horizontal and vertical components. The horizontal component of the motion is the distance the ball rolls down the first base line, which is given as 33 ft. The vertical component of the motion can be found using the angle of 25 degrees. We can use the sine function to find the vertical distance:

Vertical distance = sin(25 degrees) * 33 ft = 14.04 ft

The distance the pitcher must travel is the hypotenuse of a right triangle formed by the horizontal and vertical distances. Using the Pythagorean theorem, we can find the distance:

Distance = sqrt((33 ft)^2 + (14.04 ft)^2) = 35.88 ft

Therefore, the pitcher must run approximately 36 ft to get to the ball.

Answer: Margin of error = 6.58

Confidence interval =  (26.91, 40.08)

Step-by-step explanation:

Since we have given that

Sample size n = 8

Sample mean = 33.5 minutes

Population standard deviation = 9.5 minutes

At 95% confidence interval,

α = 0.05

t = 1.96

So, Margin of error is given by

[tex]t\times \dfrac{\sigma}{\sqrt{n}}\\\\=1.96\times \dfrac{9.5}{\sqrt{8}}\\\\=6.58[/tex]

Confidence interval would be

Lower limit:

[tex]\bar{x}-6.58\\\\=33.5-6.58\\\\=26.91[/tex]

Upper limit:

[tex]\bar{x}+6.58\\\\=33.5+6.58\\\\=40.08[/tex]

Hence, the interval would be (26.91, 40.08)

But at standard deviation 7.2 minutes, the confidence interval was (27.5, 39.5)

Confidence interval using the standard normal distribution is wider than the confidence interval using t distribution.

The information regarding the statistics shows that the margin of error is 6.58.

From the information given, the following can be deduced:

Therefore, the margin of error will be:

= 1.96 × 9.5/✓8

= 6.58

Also, the lower limit will be:

= 33.5 - 6.58

= 26.91

The upper limit will be:

= 33.5 + 6.58

= 40.08

Therefore, the interval will be (26.91, 40.08)

When the standard deviation is 7.2 minutes, the confidence interval is (27.5, 39.5).

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