Write the equation of the circle with center(-3,-2) and (4,5) a point on the circle

Answer:

a) solved in attachment

b)  If disk is running properly time between failure will exceed then Upper 90 % would be good indicator.

Step-by-step explanation:

Answer:

a) Null Hypothesis:[tex]\mu \geq 409[/tex]

b) Alternative hypothesis:[tex]\mu <409[/tex]

Step-by-step explanation:

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

On this case we are insterested on the claim that "Agriculture scientists believe that the new fertilizer may decrease the yield". So the system of hypothesis for this case would be:

a) Null Hypothesis:[tex]\mu \geq 409[/tex]

b) Alternative hypothesis:[tex]\mu <409[/tex]

Answer:

0.77,0.60

Step-by-step explanation:

Given that in a population of 1000 subjects, 770 possess a certain characteristic.

A sample of 40 subjects selected from this population has 24 subjects who possess the same characteristic

To find out sample proportion:

Sample size n = 40

Favourable x = 24

Sample proportion p = [tex]\frac{24}{40} =0.60[/tex]

To find out population proportion:

Total population N = 1000

Favourable X = 770

population proportion P = [tex]\frac{770}{1000} =0.77[/tex]

Creating a 95% confidence interval involves capturing 95% of the probability deviation in a normal distribution. The formula for doing so is (x ± (standard deviation/√n)x critical value). By adjusting the sample size or confidence level accordingly, different intervals can be created.

The question inquires about the construction of a

confidence interval

for the true average stray-load loss for a certain type of induction motor. When constructing a 95% confidence interval, we are looking to capture the central 95 percent of the probability on a normal distribution, excluding 2.5 percent on each tail of the distribution.

For point one, with n = 25, x = 58.3, and o = 3.0, use the following formula to construct a 95% Confidence Interval: (x ± (standard deviation/√n)x critical value), resulting in (58.3 ± (3/√25)x 1.96). The same method applies to the rest of the points, only changing the sample size, n, or confidence level.

For the final query, the required sample size, n, can be found by rearranging the formula for confidence intervals and solving for n with the half-width of the interval set at 0.5.

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(1). 95% CI (n=25): Standard deviation of 3.0, sample mean of 58.3, yielding a 95% confidence interval of (57.124, 59.476).

(2). 95% CI (n=100): With the same standard deviation and mean, a larger sample size leads to a narrower 95% confidence interval of (57.712, 58.888).

(3). 99% CI (n=100): Expanding the confidence level to 99% increases the margin of error, resulting in a wider interval of (57.5272, 59.0728).

(4). Sample Size (99% CI, ME=0.5): To achieve a half-width of 0.5 for a 99% confidence interval, approximately 956 samples are required.

Given the problem requirements, we will calculate the confidence intervals based on the provided data:

Using these values, we’ll calculate the 95% and 99% confidence intervals (CIs) for the population mean, μ:

1. Constructing a 95% CI when n = 25

SE = σ/√n

= 3.0/√25

= 3.0/5

= 0.6

= 1.96 * 0.6

= 1.176

2. Constructing a 95% CI when n = 100

SE = σ/√n

= 3.0/√100

= 3.0/10

= 0.3

3. Constructing a 99% CI when n = 100

4. Determining the sample size n for a half-width of 0.5 in the 99% CI

Therefore, n ≈ 956 to achieve a half-width of 0.5 for a 99% confidence interval.

Answer:

Null hypothesis: [tex]\mu_1 \geq \mu_2[/tex]

Alternative hypothesis: [tex]\mu_1 <\mu_2[/tex]

[tex]t=-1.329[/tex]

[tex]p_v =P(t_{30}<-1.329) =0.0969[/tex]

With the p value obtained and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the mena of the group 1 is significantly lower than the mean for the group 2.  

Step-by-step explanation:

When we have two independnet samples from two normal distributions with equal variances we are assuming that

[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]

And the statistic is given by this formula:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}[/tex]

Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:

[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]

This last one is an unbiased estimator of the common variance [tex]\simga^2[/tex]

The system of hypothesis on this case are:

Null hypothesis: [tex]\mu_1 \geq \mu_2[/tex]

Alternative hypothesis: [tex]\mu_1 <\mu_2[/tex]

Or equivalently:

Null hypothesis: [tex]\mu_1 - \mu_2 \geq 0[/tex]

Alternative hypothesis: [tex]\mu_1 -\mu_2<0[/tex]

Our notation on this case :

[tex]n_1 =14[/tex] represent the sample size for group 1

[tex]n_2 =18[/tex] represent the sample size for group 2

[tex]\bar X_1 =565[/tex] represent the sample mean for the group 1

[tex]\bar X_2 =578[/tex] represent the sample mean for the group 2

[tex]s_1=28.9[/tex] represent the sample standard deviation for group 1

[tex]s_2=26.3[/tex] represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

[tex]\S^2_p =\frac{(14-1)(28.9)^2 +(18 -1)(26.3)^2}{14 +18 -2}=753.882[/tex]

And the deviation would be just the square root of the variance:

[tex]S_p=27.457[/tex]

And now we can calculate the statistic:

[tex]t=\frac{(565 -578)-(0)}{27.457\sqrt{\frac{1}{14}}+\frac{1}{18}}=-1.329[/tex]

Now we can calculate the degrees of freedom given by:

[tex]df=14+18-2=30[/tex]

And now we can calculate the p value using the altenative hypothesis:

[tex]p_v =P(t_{30}<-1.329) =0.0969[/tex]

So with the p value obtained and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the mena of the group 1 is significantly lower than the mean for the group 2.  

Answer: It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

Step-by-step explanation:

Since we have given that

n = 84

x = 11

So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{11}{84}=0.13[/tex]

p = 0.20

So, hypothesis:

[tex]H_0:p=\hat{p}\\\\H_a:\hat{p}<p[/tex]

so, test statistic value would be

[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.13-0.20}{\sqrt{\dfrac{0.2\times 0.8}{84}}}\\\\z=-1.604[/tex]

At 1% level of significance, critical value would be

z= 2.58

Since 2.58>-1.604

So, We will accept the null hypothesis.

Hence, It is believed that exactly 20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

20% of Evergreen Valley college students attended the opening night midnight showing of the latest harry potter movie.

Answer:

A. 2.5 standard deviations below the average salary.

Step-by-step explanation:

Hello!

Usually, when you calculate a probability of an X₀ value from a normally distributed variable you need to standardize it to reach the proper value. In this case, you already have the Z-value and need to reverse the standardization to obtain the X₀ value.

Z= X₀ - μ

δ

δ*Z = X₀ - μ

X₀= (δ*Z) + μ

replace Z= -2.50

X₀= (δ*-2.50) + μ

X₀= μ - (δ*2.50)

You can replace the Z value any time. I've just choose to do it at the end because it's more confortable for me.

I hope it helps!

The relative risk of anxiety associated with coffee use is 5.

The relative risk of anxiety associated with coffee use can be calculated by comparing the incidence rates of anxiety among coffee drinkers and non-coffee drinkers. In this study, among 10,000 coffee drinkers, 500 developed anxiety, while among 20,000 non-coffee drinkers, 200 developed anxiety.

To calculate the relative risk, we can use the formula:

Relative Risk = (Incidence Rate of Anxiety among Coffee Drinkers) / (Incidence Rate of Anxiety among Non-Coffee Drinkers)

In this case, the incidence rate of anxiety among coffee drinkers is 500/10,000 = 0.05, and the incidence rate of anxiety among non-coffee drinkers is 200/20,000 = 0.01. Therefore, the relative risk of anxiety associated with coffee use is 0.05/0.01 = 5.

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Final answer:

To calculate the relative risk of anxiety associated with coffee use, compare the incidence rates of anxiety in coffee drinkers (500/10,000) and non-coffee drinkers (200/20,000) which results in a relative risk of 5.0, indicating that coffee drinkers have a fivefold increased risk of anxiety.

Explanation:

The relative risk of anxiety associated with coffee use can be calculated by comparing the incidence of anxiety among coffee drinkers versus non-coffee drinkers. In the provided scenario, among 10,000 coffee drinkers, 500 developed anxiety, giving us an incidence rate of 500/10,000 or 0.05. Among 20,000 non-coffee drinkers, 200 developed anxiety, with an incidence rate of 200/20,000 or 0.01. Therefore, the relative risk (RR) is calculated as the incidence rate among the exposed (coffee drinkers) divided by the incidence rate among the unexposed (non-coffee drinkers), which is 0.05/0.01 or 5.0. This suggests that the coffee drinkers have a five times higher risk of developing anxiety compared to non-coffee drinkers.

Answer:

Step-by-step explanation:

Hello!

The population variance is symbolized σ² and the standard deviation σ (remember that for estimations and statistics test, the parameter of study is always the variance)

The sentence "The standard deviation of pulse rates of adult males is less than 12 bpm." is symbolized σ <12

The sample variance is symbolized S² and the standard deviation is S.

The sample standard deviation is S= 11.5.

I hope it helps!

The symbol σ is used to represent the standard deviation in statistics. So, the standard deviation of pulse rates of adult males being less than 12 bpm in symbolic form is σ < 12. This value tells us that the dispersion of pulse rates among adult males is quite small.

The original claim is that the standard deviation of pulse rates of adult males is less than 12 bpm. This can be expressed in symbolic form as follows:

σ < 12

where σ represents the standard deviation.

In the context of this question, the standard deviation is a measure of the dispersion or spread in the pulse rates of adult males. The sample that was taken had a standard deviation of 11.5 bpm, which is less than 12 bpm, thus supporting the original claim.

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Answer:

(4, 12)

Step-by-step explanation:

Simplify (remember to flip the sign when dividing or multiplying by a negative number).

2 − 4 |8 − x| > -14

-4 |8 − x| > -16

|8 − x| < 4

There are two solutions.  The first:

8 − x < 4

-x < -4

x > 4

The second solution:

8 − x > -4

-x > -12

x < 12

Therefore, the interval is (4, 12).

Approximately 55.39% of females would not meet a college requirement of scoring at least 1025 on the SAT. This is found by calculating the z-score for the minimum required score and looking up the corresponding percentile.

To answer this question, we need to calculate the z-score — a statistic that tells us how many standard deviations away a score is from the mean. This formula is Z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation.

In this case, X is 1025 (the minimum required score), μ is 998 (the average SAT score), and σ is 202 (the standard deviation). So the z-score is Z = (1025 - 998) / 202 = 0.1336.

This z-score is positive, which means the minimum required score for the college is above the average score. Next, we check a z-score table or use a statistical calculator to find the percentage of scores that fall below this z-score, which gives us the percentage of females not meeting the requirement. For Z = 0.1336, the percentage is approximately 55.39% (using a statistical calculator or z-table). Therefore, roughly 55.39% of females would not meet a college requirement of scoring at least 1025 on the SAT.

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To find the percentage of females who do not satisfy the minimum score requirement of 1025 on the SAT-I test, we calculate the z-score for the minimum score and find the area under the normal distribution curve. Approximately 55.22% of females do not satisfy the minimum score requirement.

To find the percentage of females who do not satisfy the minimum score requirement of 1025 on the SAT-I test, we need to calculate the z-score for the minimum score and then find the area under the normal distribution curve to the left of that z-score. The formula for calculating the z-score is z = (x - μ) / σ, where x is the minimum score, μ is the mean, and σ is the standard deviation. So, z = (1025 - 998) / 202 = 0.1337.

Next, we can use a z-table or a calculator to find the area under the normal distribution curve to the left of a z-score of 0.1337. The area represents the percentage of females who do not satisfy the minimum score requirement. Using a z-table, we can find that the area is approximately 0.5522 or 55.22%.

Therefore, approximately 55.22% of females do not satisfy the minimum score requirement of 1025 on the SAT-I test.

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Answer:

Two, one for the 14 responses (number of visits) by the adults who fear going to the dentist and one for the 31 responses (number of visits) by the adults who do not fear going to the dentist.

Step-by-step explanation:

Hello!

1)

You want to test if the average visits to the dentist of people who fear to visit it are greater than the average visits of people that don't fear it.

In this case, the statistic to use is a pooled Student t-test. The reason I've to choose this test is that one of your sample sizes is small (n₁= 14) and the t-test is more accurate for small samples. Even if the second sample is greater than 30, if both variables are normally distributed, the pooled t-test is the one to use.

H₀: μ₁ = μ₂

H₁: μ₁ > μ₂

α: 0.10

t= (X₁[bar]-X₂[bar]) - (μ₁ - μ₂) ~ t[tex]_{n₁+n₂-2}[/tex]

        Sₐ√(1/n₁+1/n₂)

Where

X₁[bar] and X₂[bar] are the sample means of both groups

Sₐ is the pooled standard deviation

This is a one-tailed test, you will reject the null hypothesis to big numbers of t. Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis), and in this case, is also one-tailed.

P(t[tex]_{n₁+n₂-2}[/tex] ≥ t[tex]_{H0}[/tex]) = 1 - P(t[tex]_{n₁+n₂-2}[/tex] < t[tex]_{H0}[/tex])

Where t[tex]_{H0}[/tex] is the value of the calculated statistic.

Since you didn't copy the data of both samples, I cannot calculate it.

2)

Well there was one sample taken and separated in two following the criteria "fears the dentist" and "doesn't fear the dentist" making two different samples, so this is a test for two independent samples. To check if both variables are normally distributed you need to make two QQplots.

I hope it helps!

The p-value represents the probability of the observed data under the null hypothesis. Jessie's study regarding dentist visits requires a p-value sketch based on a test statistic of 1.71 standard errors. Two QQ plots are necessary to check the normality condition of the sample data for both groups.

The p-value is a statistical measure that indicates the probability of the observed data or something more extreme occurring under the assumption that the null hypothesis is true. In Researcher Jessie's study, the null hypothesis (H0) is that the mean number of dentist visits for adults who fear going to the dentist (Group 1) is equal to the mean number of visits for those who do not fear going to the dentist (Group 2).

The alternative hypothesis (Ha) is that the mean for Group 1 is greater than the mean for Group 2. Jessie has found that the sample mean for Group 1 is 1.71 pooled standard errors above the mean for Group 2. To sketch the p-value, we need to find the area to the right of the test statistic (1.71 standard errors above the mean) on a normal distribution curve. This area represents the p-value, which we could find using statistical software or by consulting a T table with appropriate degrees of freedom.

Regarding the condition about the normal model, we need to check whether the data is normally distributed to correctly apply the t-test. We would create two QQ plots: one for the 14 responses from adults who fear going to the dentist, and one for the 31 responses from adults who do not.

The QQ plots will help determine if the data for each group deviates from a normal distribution, which is important given the sample sizes are less than 30 and normality cannot be assumed based on the Central Limit Theorem (CLT).

Answer:

3,359

Step-by-step explanation:

Let A, B, C and D the four groups of people.

Let us denote with |A| the number of elements in a set A.

Then the number of elements of A∪B∪C∪D is the sum of the elements in each group subtracting the elements that have been counted twice.

That is,

|A∪B∪C∪D |=|A|+|B|+|C|+|D| - |A∩B| - |A∩C| - |A∩D| - |B∩C| -  |B∩D|- |C∩D| - |A∩B∩C| - |A∩B∩D| - |A∩C∩D| - |B∩C∩ D| - |A∩ B∩C∩ D| =

1000+1000+1000+1000 - 100 - 100 - 100 - 100 - 100 - 100 - 10- 10- 10- 10 - 1 = 3359

Answer:

The Riemann Sum for [tex]\int\limits^5_4 {x^2+4} \, dx[/tex] with n = 4 using midpoints is about 24.328125.

Step-by-step explanation:

We want to find the Riemann Sum for [tex]\int\limits^5_4 {x^2+4} \, dx[/tex] with n = 4 using midpoints.

The Midpoint Sum uses the midpoints of a sub-interval:

[tex]\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)[/tex]

where [tex]\Delta{x}=\frac{b-a}{n}[/tex]

We know that a = 4, b = 5, n = 4.

Therefore, [tex]\Delta{x}=\frac{5-4}{4}=\frac{1}{4}[/tex]

Divide the interval [4, 5] into n = 4 sub-intervals of length [tex]\Delta{x}=\frac{1}{4}[/tex]

[tex]\left[4, \frac{17}{4}\right], \left[\frac{17}{4}, \frac{9}{2}\right], \left[\frac{9}{2}, \frac{19}{4}\right], \left[\frac{19}{4}, 5\right][/tex]

Now, we just evaluate the function at the midpoints:

[tex]f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(4\right)+\left(\frac{17}{4}\right)}{2}\right)=f\left(\frac{33}{8}\right)=\frac{1345}{64}=21.015625[/tex]

[tex]f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(\frac{17}{4}\right)+\left(\frac{9}{2}\right)}{2}\right)=f\left(\frac{35}{8}\right)=\frac{1481}{64}=23.140625[/tex]

[tex]f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(\frac{9}{2}\right)+\left(\frac{19}{4}\right)}{2}\right)=f\left(\frac{37}{8}\right)=\frac{1625}{64}=25.390625[/tex]

[tex]f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(\frac{19}{4}\right)+\left(5\right)}{2}\right)=f\left(\frac{39}{8}\right)=\frac{1777}{64}=27.765625[/tex]

Finally, use the Midpoint Sum formula

[tex]\frac{1}{4}(21.015625+23.140625+25.390625+27.765625)=24.328125[/tex]

This is the sketch of the function and the approximating rectangles.

Final answer:

To construct the 95% confidence interval for the average time spent in a particular job at General Motors, we calculated the margin of error and then added and subtracted it from the sample mean, resulting in a confidence interval of approximately (6.1331, 6.8669).

Explanation:

To construct a 95% confidence interval for the average time spent in a particular job at General Motors, we utilize the sample mean, standard deviation, and the sample size to compute the margin of error and the confidence interval. Given that the sample mean is 6.5 years and the standard deviation is 1.7 years with a sample size of 85, we can calculate the confidence interval as follows:

This confidence interval suggests that we are 95% confident that the true average time spent in the job before promotion at General Motors lies between approximately 6.1 and 6.9 years.

Answer:

a) 0.0176 or 1.76%

b) 0.2270 or 22.70%

c) 0.6268 or 62.68%

Step-by-step explanation:

Democrats: P(D) = 39%

Republicans: P(R) = 26%

Others: P(O) = 7%

Independent: P(I) = 100-39-26-7 =28%

a) Probability of talking to all​ Republicans in three calls:

[tex]P(N_R=3) =P(R)*P(R)*P(R)\\P(N_R=3) =0.26^3 = 0.0176[/tex]

a) Probability of talking to no democrats in three calls:

[tex]P(N_D=0) =(1-P(D)*(1-P(D)*(1-P(D)\\P(N_D=0) =(1-0.39)^3 = 0.2270[/tex]

c) Probability of talking to at least one independent in three calls:

[tex]P(N_I\geq1)=1-P(N_I=0)\\P(N_I\geq1) = 1-(1-0.28)*(1-0.28)*(1-0.28)\\P(N_I\geq1) = 0.6268[/tex]

a) The probability of talking to all Republicans in the first three calls is 1.7576%. b) The probability of talking to no Democrats in the first three calls is 22.6981%. c) The probability of talking to at least one Independent is 19.5643%.

a) To find the probability of talking to all Republicans in the first three calls, we need to multiply the probability of talking to a Republican on each call. The probability of talking to a Republican on the first call is 26%, on the second call is also 26%, and on the third call is still 26%. Therefore, the probability of talking to all Republicans in the first three calls is 0.26 * 0.26 * 0.26 = 0.017576 or 1.7576%.

b) To find the probability of talking to no Democrats in the first three calls, we need to multiply the probability of not talking to a Democrat on each call. The probability of not talking to a Democrat on the first call is 1 - 39% = 61%, on the second call is also 61%, and on the third call is still 61%. Therefore, the probability of talking to no Democrats in the first three calls is 0.61 * 0.61 * 0.61 = 0.226981 or 22.6981%.

c) To find the probability of talking to at least one Independent, we need to subtract the probability of not talking to any Independents from 1. The probability of not talking to an Independent on the first call is 1 - 7% = 93%, on the second call is also 93%, and on the third call is still 93%. Therefore, the probability of not talking to any Independents in the first three calls is 0.93 * 0.93 * 0.93 = 0.804357 or 80.4357%. Since we want the probability of talking to at least one Independent, we subtract this result from 1: 1 - 0.804357 = 0.195643 or 19.5643%.

Answer:

The correct option is D) [tex]f'(t)-g'(t) > 0[/tex]

Step-by-step explanation:

Consider the provided information.

People are entering a building at a rate modeled by f (t) people per hour and exiting the building at a rate modeled by g (t) people per hour,

The change of number of people in building is:

[tex]h(x)=f(t)-g(t)[/tex]

Where f(t) is people entering in building and g(t) is exiting from the building.

It is given that "The functions f and g are non negative and differentiable for all times t."

We need to find the the rate of change of the number of people in the building.

Differentiate the above function with respect to time:

[tex]h'(x)=\frac{d}{dt}[f(t)-g(t)][/tex]

[tex]h'(x)=f'(t)-g'(t)[/tex]

It is given that the rate of change of the number of people in the building is increasing at time t.

That means [tex]h'(x)>0[/tex]

Therefore, [tex]f'(t)-g'(t)>0[/tex]

Hence, the correct option is D) [tex]f'(t)-g'(t) > 0[/tex]

The rate of change of the number of people in the building is increasing at time t and with the help of this statement the correct option is D).

Given :

The change of the number of people in the building is given by:

[tex]h(x) = f(t) - g(t)[/tex]

To determine the inequality, differentiate the above equation with respect to time.

[tex]h'(x)=f'(t)-g'(t)[/tex]

Now, it is given that the rate of change of the number of people in the building is increasing at time t. That means:

[tex]h'(x)>0[/tex]

[tex]f'(t)-g'(t)>0[/tex]

Therefore, the correct option is D).

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Answer:

The equation for sequential circuit shown in the attachment.

Step-by-step explanation:

Answer:

[tex]z=\frac{0.48 -0.47}{\sqrt{\frac{0.47(1-0.47)}{1004}}}=0.635[/tex]  

[tex]p_v =P(z>0.635)=1-P(z<0.635)=1-0.737=0.263[/tex]  

Step-by-step explanation:

1) Data given and notation n  

n=1004 represent the random sample taken

X=482 represent the Chevrolet owners who said they would buy another Chevrolet.

[tex]\hat p=\frac{482}{1004}=0.480[/tex] estimated proportion of Chevrolet owners who said they would buy another Chevrolet.

[tex]p_o=0.47[/tex] is the value that we want to test

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

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of Chevrolet owners who said they would buy another Chevrolet is higher than 47%:  

Null hypothesis:[tex]p\leq 0.47[/tex]  

Alternative hypothesis:[tex]p > 0.47[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

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

[tex]z=\frac{0.48 -0.47}{\sqrt{\frac{0.47(1-0.47)}{1004}}}=0.635[/tex]  

4) Statistical decision  

P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

If we use the significance level provided, [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

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

[tex]p_v =P(z>0.635)=1-P(z<0.635)=1-0.737=0.263[/tex]  

So based on the p value obtained and using the significance level assumed [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of Chevrolet owners who said they would buy another Chevroletis not significantly higher than 0.47 or 47% .  

Answer:

We accept  H₀

We don´t have enough evidence to reject H₀

Step-by-step explanation:

Nomal Distribution

population mean     μ₀   =  39000

sample size     n  = 200

sample mean      μ   =  40000

sample standard deviation     s = 12000

Test hypothesis

As we are just interested in look if the difference was just chance, we will do a one tail-test (right)

1.- Hypothesis

H₀    null hypothesis                      μ₀  =  39000

Hₐ Alternative hypothesis             μ₀  >  39000

2.-We considered the confidence interval of 95 % then

α  =  0,05     and     z(c)   =   1.64

3.-Compute z(s)

z(s)  =  [ (  μ  -  μ₀ ) ] / 12000/√200     z(s)  =  [ 40000- 39000)* √200] / 12000

z(s)  = 1000*14,14/ 12000     z(s)  =  1.1783

4.-Compare

z(s)  and  z(c)

1.1783  <  1.64  

z(s)  is inside acceptance region  we accep H₀  

Answer: d) one can be 90% confident that the mean drive through service time of this fast food chain is between 178.2 seconds and 181.6 seconds.

Step-by-step explanation:

The interpretation of 90% confidence interval says that a person can be 90% confident that the true population mean lies in it.

Given : A 90% confidence interval that results from examining 653 customers in one fast food chains drive through has a lower bound of 178.2 seconds and an upper bound of 181.6 seconds.

i.e.  90% confidence interval for population mean drive through service time of fast food restaurants is between 178.2 seconds and 181.6 seconds.

It means a person can be 90% confident that the mean drive through service time of this fast food chain is between 178.2 seconds and 181.6 seconds.

Hence, the correct answer is option d) .one can be 90% confident that the mean drive through service time of this fast food chain is between 178.2 seconds and 181.6 seconds.

Answer:

-$0.90

Step-by-step explanation:

There are only two possible outcomes, winning $23 (W) or losing $15 (L). Therefore:

[tex]P(W) + P(L) = 1[/tex]

The probability of the player making his next 3 free throws (P(W)) is:

[tex]P(W) = \frac{217}{302}*\frac{217}{302}*\frac{217}{302}\\P(W) = 0.37098[/tex]

The probability of the player NOT making his next 3 free throws (P(L)) is:

[tex]P(L) = 1 - P(W) = 1 - 0.37098\\P(L) = 0.62902[/tex]

Expected value (EV) is given by the payoff of each outcome multiplied by its probability:

[tex]EV = (23*0.37098) -(15*0.62902)\\EV = -\$0.90[/tex]

The expected value of the proposition is -$0.90

Answer:

The 95% confidence interval for the variance in the pounds of impurities would be [tex] 3.829 \leq \sigma^2 \leq 14.121[/tex].

Step-by-step explanation:

1) Data given and notation

[tex]s^2 =6.62[/tex] represent the sample variance

s=2.573 represent the sample standard deviation

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

n=20 the sample size

Confidence=95% or 0.95

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 \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 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,19)" "=CHISQ.INV(0.975,19)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=32.852[/tex]

[tex]\chi^2_{1- \alpha/2}=8.907[/tex]

And replacing into the formula for the interval we got:

[tex]\frac{(19)(6.62)}{32.852} \leq \sigma^2 \frac{(19)(6.62)}{8.907}[/tex]

[tex] 3.829 \leq \sigma^2 \leq 14.121[/tex]

So the 95% confidence interval for the variance in the pounds of impurities would be [tex] 3.829 \leq \sigma^2 \leq 14.121[/tex].

The confidence interval for the population variance in the pounds of impurities is calculated using the sample variance, the degrees of freedom, and the critical values from a chi-squared distribution. The values are then plugged into the formulas to give the 95% confidence interval for the population variance.

The subject of this question is in the field of statistics and it asks about the calculation of the 95% confidence interval for the population variance. In this case, we are trying to estimate the variance in the pounds of impurities in fertilizer bags based on a sample of twenty 100-pound bags with a variance of 6.62.

To solve this, we make the assumption that the population from which the bags are sampled follows a normal distribution. The confidence interval for the variance of a normal distribution can be found using the chi-squared distribution. The formula for this assumes that our samples are independent and identically distributed.

For a 95% confidence interval and d.f. (degrees of freedom) = n - 1 = 20 - 1 = 19, the upper and lower critical values (in this case for a chi-squared distribution, rounded to three decimal places) are 8.231 and 32.852 respectively. We can then use these values in the following formulas:

Lower Limit = sample variance * (d.f. / upper critical value) = 6.62*(19/32.852)

Upper Limit = sample variance * (d.f. / lower critical value) = 6.62*(19/8.231)

Finally, evaluate these to get our confidence interval.

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Answer:

  If lines have the same slope, then they are parallel.

Step-by-step explanation:

The first part of the given statement sets up the condition for the second part to be true. A conditional statement makes that explicit.

  (lines described by first part) are (lines described by second part) ⇒

  if (lines are described by first part), then (lines are described by second part)

Your conditional statement can be written ...

  If lines have the same slope, then they are parallel.

This is a one-sample z-test in statistics. You need to set null and alternative hypotheses, calculate the z-score, and compare it to the critical value to test if the mean score of second grade students in the school district significantly exceeds the nationwide average.

The subject matter here falls within Statistics, a branch of Mathematics. You want to test whether the mean score of second grade students in the school district is significantly greater than the nationwide average. To solve this, we perform a one-sample z-test, because we know the population standard deviation (σ = 15).

To start, we set the null hypothesis that the mean score for the district is equal to the nationwide average of μ0 = 50. The alternative hypothesis, according to the superintendent's belief, is that the mean score for the district is greater than 50.

You then calculate the z-score, which is the ratio of the difference between the sample mean and the population mean to the standard deviation of the population divided by the square root of the sample size (z = (X-bar - μ0) / (σ / √n). Here, X-bar is 52, n is 60 and σ is 15. After calculating the z-score, you would compare it to the critical z-value for the selected level of significance (typically 0.05 for a two-tailed test). If the calculated z score is larger than the critical z value, you would reject the null hypothesis and accept the alternative hypothesis, thus concluding that the second graders in the district have greater math skills than the nationwide average.

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Answer:

[tex]P(\bar x>60)=P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]

Is a very improbable event.

Step-by-step explanation:

We want to calculate the probability that the total weight exceeds the limit when the average weight x exceeds 6000/100=60.

If we analyze the situation we this:

If [tex]x_1,x_2,\dots,x_100[/tex] represent the 100 random beggage weights for the n=100 passengers . We assume that for each [tex]i=1,2,3,\dots,100[/tex] for each [tex]x_i[/tex] the distribution assumed is normal with the following parameters [tex]\mu=49, \sigma=18[/tex].

Another important assumption is that the each one of the random variables are independent.

1) First way to solve the problem

The random variable S who represent the sum of the 100 weight is given by:

[tex]S=x_1 +x_2 +\dots +x_100 =\sum_{i=1}^{100} x_i[/tex]

The mean for this random variable is given by:

[tex]E(S)=\sum_{i=1}^{100} E(x_i)=100\mu = 100*49=4900[/tex]

And the variance is given by:

[tex]Var(S)=\sum_{i=1}^{100} Var(x_i)=100(\sigma)^2 = 100*(18)^2[/tex]

And the deviation:

[tex]Sd(S)=\sqrt{100(\sigma)^2} = 10*(18)=180[/tex]

So we have this distribution for S

[tex]S \sim (4900,180)[/tex]

On this case we are working with the total so we can find the probability on this way:

[tex]P(S>6000)=P(z>\frac{6000-4900}{180})=P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]

2) Second way to solve the problem

We know that the sample mean have the following distribution:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}}[/tex]

If we are interested on the probability that the population mean would be higher than 60 we can find this probability like this:

[tex]P(\bar x >60)=P(\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}>\frac{60-49}{\frac{18}{\sqrt{100}}})[/tex]

[tex]P(z>6.11)=1-P(z<6.11)=4.98x10^{-10}[/tex]

And with both methods we got the same probability. So it's very improbable that the limit would be exceeded for this case.

By using reference angle, the exact value of csc (-1020)° is [tex]\frac {-2\sqrt3}{3}[/tex].

How to find the exact value

To find the exact value of csc (-1020)° using the reference angle, first determine the reference angle for -1020°.

To find the reference angle, add or subtract multiples of 360° to bring the angle into the range of 0° to 360°.

-1020° + 360° = -660°

Since -660° is still negative,  add another 360°:

-660° + 360° = -300°

Now we have a reference angle of 300°.

The cosecant function (csc) is the reciprocal of the sine function. We know that the sine function is positive in the first and second quadrants, so the cosecant function will be positive in those quadrants.

The exact value of csc(300°) can be found by taking the reciprocal of the sine of 300°:

csc(300°) = 1/sin(300°)

To find the sine of 300°, use the periodicity of the sine function:

sin(300°) = sin(300° - 360°) = sin(-60°)

The sine of -60° is the same as the sine of 60°, but with a negative sign:

sin(-60°) = -sin(60°) = [tex]\frac {-\sqrt3}{2}[/tex]

Now, find the reciprocal:

csc(300°) = 1/(-√3/2) = [tex]\frac {-2}{\sqrt3}[/tex]

To rationalize the denominator, multiply the numerator and denominator by √3:

csc(300°) = [tex]\frac {-2}{\sqrt3}[/tex] * [tex]\frac {\sqrt3}{\sqrt3}[/tex] =  [tex]\frac {-2\sqrt3}{3}[/tex]

Therefore, the exact value of csc (-1020)° is [tex]\frac {-2\sqrt3}{3}[/tex].

Complete question

Use reference angle to find the exact value of csc (–1020)°.

Answer:

dont escape lol

Step-by-step explanation:

the answer is 2

Answer:    

z=1.59

If we compare the p value and the significance level given [tex]\alpha=0.025[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we fail to reject the null hypothesis, so we can conclude that the mean repellency of the new bug repellent is greater than 89% at 0.025 of signficance.    

Step-by-step explanation:    

1) Data given and notation    

[tex]\bar X=0.92[/tex] represent the mean effectiveness of a new bug repellent for the sample    

[tex]\sigma=0.08[/tex] represent the population standard deviation for the sample    

[tex]n=18[/tex] sample size    

[tex]\mu_o =0.89[/tex] represent the value that we want to test  

[tex]\alpha=0.025[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)    

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

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean repellency of the new bug repellent is greater than 89% or 0.89, the system of hypothesis would be:    

Null hypothesis:[tex]\mu \leq 0.89[/tex]    

Alternative hypothesis:[tex]\mu > 0.89[/tex]    

We don't know the population deviation, and the sample size <30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic    

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

[tex]t=\frac{0.92-0.89}{\frac{0.08}{\sqrt{18}}}=1.59[/tex]    

Calculate the P-value    

First we need to calculate the degrees of freedom given by:

[tex]df=n-1=18-1=17[/tex]

Since is a one-side upper test the p value would be:    

[tex]p_v =P(t_{17}>1.59)=0.065[/tex]

In Excel we can use the following formula to find the p value "=1-T.DIST(1.59;17;TRUE)"  

Conclusion    

If we compare the p value and the significance level given [tex]\alpha=0.025[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we fail to reject the null hypothesis, so we can conclude that the mean repellency of the new bug repellent is greater than 89% at 0.025 of signficance.    

The amount of water silo can hold in terms of π is 324π cubic feet

Given that cylindrical shaped silo has a diameter of 18 feet and a height of 4 feet

To find: Amount of water silo can hold in terms of π

This means we are asked to find the volume of cylindrical shaped silo

The volume of cylinder is given as:

[tex]\text {volume of cylinder }=\pi \mathrm{r}^{2} \mathrm{h}[/tex]

Where "r" is the radius of cylinder

"h" is the height of cylinder

π is the constant has a value 3.14

Given diameter = 18 feet

[tex]radius = \frac{diameter}{2} = \frac{18}{2} = 9[/tex]

Substituting the values in formula, we get

[tex]\text {volume of cylinder }=\pi \times 9^{2} \times 4[/tex]

[tex]\text {volume of cylinder }=\pi \times 81 \times 4=324 \pi[/tex]

Thus amount of water silo can hold in terms of π is 324π cubic feet

Answer:324π

Step-by-step explanation: