Water is being pumped continuously from a pool at a rate proportional to the amount of water left in the pool. Initially there was 15,000 gallons of water in the pool; six minutes later there was 13,800 gallons.

At what rate was the amount of water in the pool decreasing when there were 14,000 gallons remaining and when will there be 5,000 gallons remaining?

Please show all steps.

Answer:

As the distance between the y-intercepts of f(x) and g(x) is 3.

So, the y-intercept of f(x) is 3 units below g(x), as shown in attached figure a.

Step-by-step explanation:

Let us consider the equation of straight line in the form of slope-intercept form such as:

[tex]y = mx + b[/tex]      

where m is the slope of the line and b is the y-intercept.

Determining the y-intercept of f(x) = x - 1

As the given function f(x) = x-1 is in the form of [tex]y = mx + b[/tex].

So, the y-intercept will be -1.

y-intercept can also be computed by putting x = 0, as at y-intercept the value of x is zero.

f(x) = x - 1

y = 0 - 1 ⇒ y = -1

Determining the y-intercept of g(x) = x + 2

As the given function g(x) = x + 2 is in the form of [tex]y = mx + b[/tex].

So, the y-intercept will be 2.

y-intercept can also be computed by putting x = 0, as at y-intercept the value of x is zero.

g(x) = x + 2

y = 0 + 2 ⇒ y = 2

Determining the distance between the y - intercepts of f(x) and g(x)

So, the distance between the y - intercepts of f(x) and g(x): 2 - (-1) = 3

So, the y-intercept of f(x) is 3 units below g(x), as shown in attached figure a.

Keywords: y-intercept, graph, function

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Answer:

0.278

Step-by-step explanation:

Given that Nuri joins a game for a car. The rule is that Nuri pick one key from box either A, B, or C. A box has two keys but only one can be used. B box has three keys but only one can be used. C box has two keys but none of them can be used.

Each box is equally likely to be selected.

In other words

[tex]P(A) = P(B) = P(C)=\frac{1}{3}[/tex]

If A is selected then probability of winning is using the correct key out of two keys i.e. 0.5

If B is selected then probability of winning is using the correct key out of three keys i.e. 0.333

If c is selected then probability of winning is using the correct key out of two keys i.e. 0.00

So the probability that Nuri can win the car

= [tex]\frac{1}{3} *0.5+\frac{1}{3} *0.333+\frac{1}{3} *0\\= 0.278[/tex]

Answer:

0.5 or 50%

Step-by-step explanation:

For any given value of 'x' representing the time between arrivals of two customers. If 0 < x <120, then the cumulative distribution function is:

[tex]\frac{x-0}{120-0}=\frac{x}{120}[/tex]

Therefore, the probability that the time between the arrivals of two customers will be more than 60 seconds is determined by:

[tex]P(X>60) = 1 -\frac{60}{120}\\P(X>60) = 0.5[/tex]

The probability is 0.5 or 50%.

Answer:

Therefore, Michael concludes option C)

C)[tex](DA)^{2}=(DG)^{2}+(AG)^{2}[/tex]

Step-by-step explanation:

Given:

1. DG = 3 and the area of square DEFG is 9.

2. AG = 4 and the area of square GHIA is 16.

3. DA = 5 and the area of square ABCD is 25.

So we have,

[tex](DG)^{2}=3^{2}=9\\ \\(AG)^{2}=4^{2}=16\\\\(DA)^{2}=5^{2}=25\\[/tex]

Now Add DG² and AG² we get

[tex](DG)^{2}+(AG)^{2}=9+16=25=(DA)^{2}[/tex]

Which is also called as  Pythagoras theorem i.e

[tex](\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}[/tex]

Therefore, Michael concludes option C)

C)[tex](DA)^{2}=(DG)^{2}+(AG)^{2}[/tex]

Answer:

Step-by-step explanation:

a)  While the mean is below 4 the standard deviation tells us that there is a pretty high chance for the value to be above 4.  One standard deviation away is 1.79 - 5.65.  We are concerned with things over 4 so we'll look at the upper half, which is 3.72 - 5.65.  Being one standard deviation to the right means there is a 34.1% chance of the readings being in this range.  And there is even chances of it being slightly higher, though that is comparatively low.  But even as low as 10% is usually considered too high a chance to risk.  If you don't understand how the standard deviation got me those percents let me know.

b) alternative hypothesis is always the option where we want to prove it.  So we want to prove the concentration is 4 or above.  So the null is less than 4 and the alternative is greater than or equal to 4.  Do you know the correct symbols?  if not I can get those written out.  As for the p value we need the confidence level for the question, do you have that?

The two hypothesis:

And the reasoning of the inspector is incorrect because it ignores the large standard deviation.

We know that the mean radon concentration must be smaller than 4.0 pC/L.

In this particular house, the mean is 3.72 pC/L with a really large standard deviation of 1.93 pC/L.

And the inspector says that the radon abatement is not necessary, as the mean is smaller than 4.0 pC/L.

Now, as you can see, the standard deviation is really large. This means that over a given period of time, the mean concentration per liter may be larger than 4.0 pC (and then decreases). But this would imply that the exposure over large periods of times could be really large. This is why the reasoning is incorrect.

b) The null hypothesis is what we want to prove. In this case, is that the mean concentration is smaller than 4.0 pC/L.

The alternative hypothesis is the other option, in this case, that the concentration is equal or larger than 4.0 pC/L.

using the correct notation and defining C as the concentration we can write:

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Answer:

Sample size should be atleast 60025

Step-by-step explanation:

Given that a school board wishes to know the current mean reading level of 6th graders throughout a very large school district. Past experience shows that the standard deviation of such reading scores is about 2.5

For 95% confidence we have significance level = 5%

Margin of error = 0.04

Margin of error should be less than 0.04

i.e. Z critical value for 95% *std error <0.04

[tex]1.96*\frac{2.5}{\sqrt{n} } <0.02\\\\n>60025[/tex]

Sample size should be atleast 60025

The school board needs to sample 151 6th graders to be 95% confident that the sample mean reading level is within 0.4 of the true mean.

The school board is interested in determining the required sample size needed to estimate the mean reading level of 6th graders. Given the standard deviation of 2.5 and the margin of error of 0.4 for a 95% confidence interval, we must use the formula for sample size in estimating a mean. The formula is:

n = (z*s/E)^2

where n is the sample size, z is the z-score corresponding to the desired confidence level, s is the population standard deviation, and E is the margin of error. For a 95% confidence level, the z-score is 1.96.

Therefore, the sample size n would be:

n = (1.96*2.5/0.4)^2

n = (4.9/0.4)^2

n = (12.25)^2

n = 150.0625

Since we cannot have a fraction of a sample, we round up to the nearest whole number. Thus, the school board needs to sample 151 6th graders to be 95% confident that the sample mean reading level is within 0.4 of the true mean.

Answer:

There is no effect of the good intentions i.e. to restrain their diet out of concern about their weight, have on their eating habits.

Step-by-step explanation:

The question is incomplete as some information is not provided, please refer below the remaining information of the question.

Here are the data on grams of potato chips consumed (note that the study report gave the standard error of the mean rather than the standard deviation):

Group          n     bar{x}         s

Unrestrained  9     63.83      24.72

Restrained  11     34      39.8

a. The standard error of the difference between sample means  ( ± 0.0001 )  is:

b. The critical value  ( ± 0.001 )  from the t distribution for confidence interval 80 % using the conservative degrees of freedom is:

c. Give a 80 % confidence interval   ( ± 0.01 )  that describes the effect of restraint:

Answer:

a) The standard error of the difference between sample means:

S E  =  √ s 2 1 /  n 1  +  s 2 2 /n 2

    =  √ s e 2 1 +  s e 2 2

    =  √ 24.72 2 +  39.8 2

    =  46.85

b) The degrees of freedom:

D f  =  n 1  + n 2  − 2

=  9  +  11  −  2

= 18

The confidence level  =  0.80

The significance level,  α  =  0.20

The critical value from the t distribution for confidence interval 80%:

t c r i t i c a l  =  t α / 2 ,  d f  =  t 0.10 , 18  =  ±  1.33

c) The 80% confidence interval:

( μ 1 − μ 2 ) = ( ¯ x 1 − ¯ x 2 )  ±  t ⋅ S E

= ( 63.83 − 34 ) ±  1.33 × 46.85  

= 29.83 ± 62.31

− 32.48 < ( μ 1 − μ 2 ) <  92.14

As the interval contains the zero. So, it can be concluded that there is no effect of the good intentions i.e. to restrain their diet out of concern about their weight, have on their eating habits.

Good intentions regarding diet may impact eating habits among college students, but environmental influences often override intentions.

The effect of good intentions on eating habits, especially among those who are trying to restrain their diet due to concerns about weight, can vary greatly. Research shows that consumption patterns, such as an increase in eating fast food and sugary beverages, coupled with a decrease in physical activity, contribute to significant weight gain among college students. Implementing healthier eating and physical activity habits is crucial, but it is also important to acknowledge that neither changing diets nor exercise are effective on their own for preventing health issues. Both elements must be combined to reduce the risk of chronic diseases such as cardiovascular disease and cancers.

Good intentions may have some impact on eating habits, but the susceptibility to environmental influences, such as availability of unhealthy snacks, can override these intentions. Therefore, colleges and universities are actively pursuing comprehensive approaches to encourage both healthy eating and increased physical activity. For individuals concerned about their diet and weight, it is essential to engage in consistent healthy behaviors, rather than relying purely on intentions.

Answer:

B

Step-by-step explanation:

Absolute value is essentially just every number, but positive. (like -2's absolute value is 2). This is also a number's distance from zero. 2 is 2 away from zero, and -2 is 2 away from zero.

Absolute value returns the positive value of a numerical expression.

Absolute value is used to calculate distance on a coordinate plane because (a) Absolute value is the distance between 2 points on a number​ line, so it gives the distance between any 2 points.

The distance between points A and B on the coordinate plane is:

[tex]\mathbf{Distance = |A - B|}[/tex]

or

[tex]\mathbf{Distance = |B - A|}[/tex]

The above equations mean that:

Absolute values will calculate the distance between any two points, irrespective of whether the point is 0 or not.

Hence, the correct option is: (a)

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Answer:

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)

The equation of the given line is

y=2x+2

Comparing with the slope intercept form, slope = 2

If two lines are parallel, it means that they have the same slope. Therefore, the slope of the line passing through (2,-1) is 2

To determine the intercept, we would substitute m = 2, x = 2 and y = -1 into y = mx + c. It becomes

- 1 = 2×2 + c = 4 + c

c = - 1 - 4 = - 5

The equation becomes

y = 2x - 5

Answer:

1. Plane False

2. Sphere False

3. Ellipsoid  False

4. Circular cylinder  True

Step-by-step explanation:

For this case we have the following curve [tex]C(t) = (cos t , sin t , t[/tex]

And we can express like this the terms for the curve or each component:

[tex] x= cos t, y= sin t , z =t[/tex]

1. Plane False

The general equation for a plane is given by:

a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0.

For this case we don't satisfy this since have sinusoidal functions and this equation is never satisfied.

2. Sphere False

The general equation for a sphere is given by:

(x - a)² + (y - b)² + (z - c)² = r²

And for this case if we see our parametric equation again that is not satisfied since we have two cosenoidal functions. And another function z=t

3. Ellipsoid  False

The general equation for an ellipsoid is given by:

x^2/a2 + y^2/b2 + z^2/c2 = 1

And for this case again that's not satisfied since we have

[tex]\frac{cos^2 t}{a^2} + \frac{sin^2 t}{b^2}+\frac{t^2}{c^2} \neq 1[/tex]

4. Circular cylinder  True

The general equation for a circular cylinder is given by:

[tex]x^2 +y^2 = r^2[/tex]

And if we replace the equations that we have we got:

[tex] cos^2 t + sin^2 t = 1[/tex] from the fundamental trigonometry property.

So then we see that our function satisfy the condition and is the most appropiate option.

Answer:

C. The parent population must be normally distributed

Step-by-step explanation:

The parent population must follow a normal distribution in order to use a t procedure

Answer:

y= -3x +10

Step-by-step explanation:

you already have your slop so plug it in and your y intercept is 10

In this context, cos(θ)=0.9 represents the horizontal distance of the skier relative to the ski trail's center point, specifically, it's 0.9 times the length of the trail's radius to the east. Correct answer is - The skier is 0.88 radius lengths to the East of the center of the ski trail.

In this question, you're being asked about the meaning of cos(θ) = 0.9 in a real-world context. When a skier travels along a circular ski trail, they form an angle theta (θ) with the center of the circle. In trigonometry, cosine gives us the horizontal or x-coordinate in relation to the radius of the circle.

In this case, it means that the skier is 0.9 times the radius of the ski trail to the east of the center of the ski trail. As the radius of the ski trail is 1.25 km, this puts the skier 0.9 x 1.25 = 1.125 km to the east. So, the correct answer would be 'The skier is 0.88 radius lengths to the East of the center of the ski trail'.

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Answer:

Attached as image png.

Step-by-step explanation:

The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. An article gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions. 62 51 53 58 42 54 56 61 59 64 51 53 64 62 50 68 54 55 57 50 55 50 56 55 46 56 54 55 53 47 48 56 58 48 63 58 57 56 54 59 54 52 50 55 60 51 56 59 Construct a boxplot of the data.

The boxplot is a method to represent a series of numerical data through their quartiles. In this way, the box diagram shows at a glance the median and quartiles of the data.

Answer:

Hence safely 9 watermelons can be placed in a single container.

Step-by-step explanation:

Given that the  weight (in pounds) of ``medium-size'' watermelons is normally distributed with mean 15 and variance 4.

X = weight in pounds of medium size watermelons is N(15, 2)

Let the water melons stored be n

Then the sample of n has a mean of (15) and std error = [tex]\frac{2}{\sqrt{n} }[/tex]

Capacity = 140

Hence we can say either 8 or 9

If n=9, we have weight = 15*9+1.96*2/3 = 136.40

Hence safely 9 watermelons can be placed in a single container.

Answer:

Step-by-step explanation:

given that an article in the Sacramento Bee (March 8, 1984, p. A1) reported on a study finding that "men who drank 500 ounces or more of beer a month (about 16 ounces a day) were three times more likely to develop cancer of the rectum than non-drinkers."

To support this some data collected should be given.  The data should be from a sample of a large size randomly drawn from 2 groups one men who drank 500 oz or more of beer and other group not drank that much.  These people medical history with cancer patients number also should have been given.

Final answer:

The Sacramento Bee article is missing key numerical details such as the baseline rate of rectal cancer in non-drinkers, the number of study participants, the total number of cancer cases, the study duration, and potential confounding factors. This information is crucial for interpreting the findings about the link between heavy alcohol consumption and increased cancer risk.

Explanation:

The report from the Sacramento Bee on a study finding that men who consumed a significant amount of beer had a higher risk of developing rectal cancer is missing several crucial pieces of numerical information. Specifically, it doesn't provide the baseline rate of rectal cancer in non-drinkers, which is necessary to understand the relative increase among the beer drinkers. Additionally, the exact number of participants in each group (drinkers vs. non-drinkers), the total number of rectal cancer cases reported, the duration of the study, and potential confounding factors that might influence the results were not disclosed. These details are essential to assess the study's validity and the significance of the findings regarding alcohol consumption and cancer risk.

Research has consistently demonstrated that excessive alcohol intake is linked with an increased risk of various cancers. Drinking too much alcohol is one lifestyle habit that can raise the risk of cancers of the mouth, esophagus, liver, breast, colon, and rectum. Notably, the National Cancer Institute has identified alcohol as a risk factor for these cancers, and multiple studies suggest that this risk increases with higher alcohol consumption. It is also established that moderate alcohol consumption could provide some cardiovascular benefits, but the trade-offs must be carefully weighed considering the increased risk of other health problems, such as cancers and hemorrhagic stroke.

Answer:

Half half for both since all three are fair coins.

Step-by-step explanation:

Final answer:

The cumulative probability of getting a certain number of heads in 4 coin flips can be calculated with the pbinom() function in R, which uses the binomial distribution, suitable for independent Bernoulli trials with two outcomes and a constant success probability.

Explanation:

Binomial Probability Distribution in R

To calculate the cumulative probability of getting a certain number of heads in multiple coin flips using binomial distribution, the pbinom() function in R can be used. We are looking at a scenario where a coin is flipped 4 times, and we want to find the cumulative probability of getting 0, 1, 2, 3, or 4 heads.

The binomial distribution is appropriate here because coin flipping is a Bernoulli trial: there are two possible outcomes (heads or tails), the probability of success (head in this case) is constant, and each flip is independent of others.

The cumulative probability can be calculated as follows:

Probability of no more than 2 heads: pbinom(2, 4, 0.5)

Probability of no more than 4 heads (which is certain): pbinom(4, 4, 0.5) equals to 1

You will get results that match options A through E provided in the question.

Answer:5.25

Step-by-step explanation:

Answer:

The sample proportion is 8.4%.

Step-by-step explanation:

The sample proportion f people in the United States who earn more than $75,000 each year is the number of people who reported earning more than $75,000 each year divided by the size of the sample.

The proportion is:

[tex]P = \frac{168}{2000} = 0.084[/tex]

The sample proportion is 8.4%.

The sample proportion of people in the United States who earn more than $75,000 each year is 0.084.

Number of people who reported earning more than $75,000: 168

Total number of people in the sample: 2,000

[tex]\[ \hat{p} = \frac{168}{2000} \][/tex]

[tex]\[ \hat{p} = 0.084 \][/tex]

 To three decimal places, the sample proportion is 0.084

Answer:

The degree measure of the slice representing the mint patties rounded to the nearest degree is 69°.

Step-by-step explanation:

Consider the provided information.

A two-pound bag of assorted candy contained 100 caramels,

83 mint patties,

93 chocolate squares,

80 nut clusters,

79 peanut butter taffy pieces.

Now find the degree measure of the slice representing the mint patties rounded to the nearest degree as shown:

[tex]\frac{83}{100+83+93+80+79} \times 360^0[/tex]

[tex]\frac{83}{435} \times 360^0[/tex]

[tex]68.69\approx69^0[/tex]

Hence, the degree measure of the slice representing the mint patties rounded to the nearest degree is 69°.

Answer:

A sample size of at least 737 specimens is required.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]

Now, find the width M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

In this problem, we have that:

[tex]M = 3, \sigma = 31.62[/tex]

So:

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]3 = 2.575*\frac{31.62}{\sqrt{n}}[/tex]

[tex]3\sqrt{n} = 81.4215[/tex]

[tex]\sqrt{n} = 27.1405[/tex]

[tex]n = 736.60[/tex]

A sample size of at least 737 specimens is required.

Answer:

The minimum sample size is N=1537.

Step-by-step explanation:

We want to know the sample size to estimate the proportion of voters with a margin of error of 0.05, with a 95% of confidence.

The margin of error can be defined as:

[tex]e=UL-LL=(p+z*\sqrt{\frac{p(1-p)}{N}}) -(p-z*\sqrt{\frac{p(1-p)}{N}})=2z\sqrt{\frac{p(1-p)}{N}}[/tex]

We can calculate N from this

[tex]e=2z\sqrt{\frac{p(1-p)}{N}}\\\\\frac{p(1-p)}{N}=(\frac{e}{2z})^2\\\\N= (\frac{2z}{e})^2*p(1-p)[/tex]

The sample size will be calculated for p=0.5, which is the proportion that requires the larger sample size (maximum variance).

The z-value for a 95% CI is z=1.96.

The minimum sample size is then

[tex]N= (\frac{2z}{e})^2*p(1-p)=(\frac{2*1.96}{0.05})^2*0.5(1-0.5)=6146.56*0.5*0.5=1536.64[/tex]

The minimum sample size is N=1537.

Answer:

Equation:

250 + 132.5y = x

Solution for solving for x:

x = 250 + 132.5y

Solution for solving for y:

y = (x - 250)/132.5

In this situation, the solutions mean that the equation is linear or first grade because the solutions can be calculated in any of the following four forms:

1. Adding the same number to both sides

2. Subtracting the same number from both sides

3. Multiplying both sides by the same number

4. Dividing both sides by the same number

Completing the question and statement correctly:

The equation  represents the relationship between the quantities in this situation, where x is the weight, in grams, of the filled box and y the number of shirts in the box.

Step-by-step explanation:

1. Let's review all the information provided to us to answer the question correctly:

Weight of an empty box = 250 grams

Weight of each T-shirt = 132.5 grams

2. Name two possible solutions to the equation that represent the relationship between the quantities in this situation, where x  is the weight, in grams, of the filled box and y the number of shirts in the box. What do the solutions mean in this situation?

x = weight, in grams, of the filled box

y = number of shirts in the box

Equation:

250 + 132.5y = x

Solution for solving for x:

x = 250 + 132.5y

Solution for solving for y:

y = (x - 250)/132.5

In this situation, the solutions mean that the equation is linear or first grade because the solutions can be calculated in any of the following four forms:

1. Adding the same number to both sides

2. Subtracting the same number from both sides

3. Multiplying both sides by the same number

4. Dividing both sides by the same number

Step-by-step explanation:

The two possible solution are 647.5 gm and 912.5 gm.

Let's use the following equation:

Weight of the filled box (W) = Weight of the empty box (250 grams) + (Number of T-shirts) * (Weight of each T-shirt, 132.5 grams)

So, the equation would be:

W = 250 + 132.5 * N

Where:

- W is the weight of the filled box in grams.

- N is the number of T-shirts in the box.

Now, let's find two possible solutions:

Solution 1:

Suppose the box is filled with 3 T-shirts.

W = 250 + 132.5 * 3

W = 250 + 397.5

W = 647.5 grams

Solution 2:

Suppose the box is filled with 5 T-shirts.

W = 250 + 132.5 * 5

W = 250 + 662.5

W = 912.5 grams

- Solution 1: If the box is filled with 3 T-shirts, it would weigh 647.5 grams in total.

This means that the combined weight of the 3 T-shirts added to the empty box's weight is 647.5 grams.

- Solution 2: If the box is filled with 5 T-shirts, it would weigh 912.5 grams in total.

This means that the combined weight of the 5 T-shirts added to the empty box's weight is 912.5 grams.

In general, the equation allows you to calculate the weight of the filled box based on the number of T-shirts you put in it. Each additional T-shirt adds 132.5 grams to the total weight, considering the empty box's weight of 250 grams.

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Answer:

[tex]\chi^2_{\alpha/2}=43.195[/tex]

[tex]\chi^2_{1- \alpha/2}=14.573[/tex]

b. 14.573 and 43.195.

Step-by-step explanation:

Previous concepts

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) Solution to the problem

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]

We need on this case to calculate the critical values. First we need to calculate the degrees of freedom given by:

[tex]df=n-1=28-1=27[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table for the Chis square distribution with 27 degrees of freedom to find the critical values. We need a value that accumulates 0.025 of the area on the left tail and 0.025 of the area on the right tail.  

The excel commands would be: "=CHISQ.INV(0.025,27)" "=CHISQ.INV(0.975,27)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=43.195[/tex]

[tex]\chi^2_{1- \alpha/2}=14.573[/tex]

b. 14.573 and 43.195.

Answer:I am not very sure but I think [tex]\frac{1}{51.5}[/tex]

Step-by-step explanation:

The probability of picking the fake coin given it lands heads [tex]\(n\)[/tex] times is [tex]\( \frac{1}{100 \times 2^n} \).[/tex]

To find the probability that you have picked the fake coin given that it lands heads [tex]\(n\)[/tex] times in a row, we can use Bayes' theorem.

Let [tex]\(F\)[/tex] be the event of picking the fake coin, and [tex]\(H\)[/tex] be the event of getting heads [tex]\(n\)[/tex] times in a row.

We are asked to find [tex]\(P(F|H)\)[/tex], the probability of picking the fake coin given that [tex]\(n\)[/tex] heads are observed in a row.

By Bayes' theorem:

[tex]\[ P(F|H) = \frac{P(H|F) \times P(F)}{P(H)} \][/tex]

We know:

[tex]- \(P(H|F) = 1\)[/tex] (since the fake coin has two heads).

[tex]- \(P(F) = \frac{1}{100}\)[/tex] (since there is only one fake coin out of 100).

[tex]- \(P(H)\)[/tex] is the probability of getting [tex]\(n\)[/tex] heads in a row, which is the same for both the fake and normal coins. This is [tex]\(0.5^n\).[/tex]

Now, substituting the values:

[tex]\[ P(F|H) = \frac{1 \times \frac{1}{100}}{0.5^n} \][/tex]

[tex]\[ P(F|H) = \frac{1}{100 \times 0.5^n} \][/tex]

[tex]\[ P(F|H) = \frac{1}{100 \times 2^n} \][/tex]

So, the probability of picking the fake coin given that it lands heads [tex]\(n\)[/tex] times in a row is [tex]\( \frac{1}{100 \times 2^n} \).[/tex]

Answer: 0.0668

Step-by-step explanation:

Given : The volume of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.45 ounces and a standard deviation of 0.30 ounce.

i.e. [tex]\mu=12.45[/tex] and [tex]\sigma=0.30[/tex]

Let x denotes the volume of soda a dispensing machine pours into a 12-ounce can.

Then, the proportion of the soda cans contain less than the advertised 12 ounces of soda will be :-

[tex]P(x<12)=P(\dfrac{x-\mu}{\sigma}<\dfrac{12-12.45}{0.30})\\\\=P(z<-1.5)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-P(z<1.5)\ \ [\because P(Z<-z)=1-P(Z<z)]\\\\=1-0.9332\ [\text{By z-table}]\\\\=0.0668[/tex]

Hence, the proportion of the soda cans contain less than the advertised 12 ounces of soda = 0.0668

Final answer:

The proportion of 12-ounce soda cans that contain less than the advertised amount of soda is approximately 6.68%, which is found by calculating the z-score of 12 ounces and looking up the corresponding proportion in the standard normal distribution table.

Explanation:

To find the proportion of soda cans that contain less than the advertised 12 ounces of soda, we need to calculate the z-score for 12 ounces using the mean and standard deviation of the soda volumes. The z-score tells us how many standard deviations away from the mean a certain value is.

The z-score formula is:

Z = (X - μ) / σ

Where X is the value (12 ounces), μ is the mean (12.45 ounces), and σ is the standard deviation (0.30 ounces).

Plugging in the values we get:

Z = (12 - 12.45) / 0.30
Z = -0.45 / 0.30
Z = -1.5

Once we have the z-score, we can use the standard normal distribution table to find the proportion of values that lie below this z-score. The table gives us the proportion of the distribution that is to the left of the z-score. A z-score of -1.5 corresponds to a proportion of approximately 0.0668.

Therefore, the proportion of soda cans containing less than 12 ounces is approximately 0.0668, or 6.68%.

Answer:

[tex]t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974[/tex]    

[tex]p_v =2*P(t_{39}<-1.8974)=0.0652[/tex]

If we compare the p value with a significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.    

Step-by-step explanation:

Assuming the following problem: "In Hamilton County, Ohio the mean number of days needed to sell a home is 86 days (Cincinnati Multiple Listing Service, April, 2012). Data for the sale 40 of homes in a nearby county showed a sample mean of 80 days with a sample standard deviation of 20 days. Conduct a hypotheses test to determine whether the mean number of days until a home is sold is different than the Hamilton county mean of 86 days in the nearby county. "

1) Data given and notation    

[tex]\bar X=80[/tex] represent the sample mean

[tex]s=20[/tex] represent the sample standard deviation

[tex]n=40[/tex] sample size    

[tex]\mu_o =86[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)    

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

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean is different from 86, the system of hypothesis would be:    

Null hypothesis:[tex]\mu = 86[/tex]    

Alternative hypothesis:[tex]\mu \neq 86[/tex]    

We don't know the population deviation, so for this case we can use the 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".

3) Calculate the statistic    

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

[tex]t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974[/tex]    

4) Calculate the P-value    

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

[tex]df=n-1=40-1=39[/tex]

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

[tex]p_v =2*P(t_{39}<-1.8974)=0.0652[/tex]

In Excel we can use the following formula to find the p value "=2*T.DIST(-1.8974,39,TRUE)"  

5) Conclusion    

If we compare the p value with a significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.    

Answer:

[tex]t=\pm 2.95[/tex]

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The t distribution or Student’s t-distribution is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.  

Data given

Confidence =0.99 or 99%

[tex]\alpha=1-0.99=0.01[/tex] represent the significance level

n =16 represent the sample size

We don't know the population deviation [tex]\sigma[/tex]

Solution for the problem

For this case since we don't know the population deviation and our sample size is <30 we can't use the normal distribution. We neeed to use on this case the t distribution, first we need to calculate the degrees of freedom given by:

[tex]df=n-1=16-1=15[/tex]

We know that [tex]\alpha=0.01[/tex] so then [tex]\alpha/2=0.005[/tex] and we can find on the t distribution with 15 degrees of freedom a value that accumulates 0.005 of the area on the left tail. We can use the following excel code to find it:

"=T.INV(0.005;15)" and we got [tex]t_{\alpha/2}=-2.95[/tex] on this case since the distribution is symmetric we know that the other critical value is [tex]t_{\alpha/2}=2.95[/tex]

Answer:

Since p >0.05 at 5% level we find that there is no evidence    of a difference in the proportion of defective parts produced by the two suppliers

Step-by-step explanation:

Given that a manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective; an SRS of 100 parts from supplier 2 finds 10 defective.

Let p 1 p1 and p 2 p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective.

[tex]H 0 : p 1 = p 2 \\H a : p 1 \neq  p 2 .[/tex]

(two tailed test )

Sample             I            II          total

N                     400      100        500

x                        20         10          30

p                       0.05    0.10      0.06

[tex]Std error =\sqrt{\bar p(1- \bar p)(\frac{1}{n_1} +\frac{1}{n_2} )}  \\=0.0266[/tex]

Z= -1.8831

p value = 0.0601

Since p >0.05 at 5% level we find that there is no evidence    of a difference in the proportion of defective parts produced by the two suppliers

Option 2 is correct. The  P-value of the test is 0.06.

The question requires testing if there is a significant difference between the proportions of defective parts from two suppliers. We test this using a two-proportion z-test and set up our hypotheses as H0: P₁ = P₂ and Ha: P₁ ≠ P₂.

Collect the Sample Data:

Sample 1: n₁ = 400, x₁ = 20

Sample 2: n₂ = 100, x₂ = 10

Calculate the Sample Proportions:

P₁ = x₁/n₁ = 20/ 400 = 0.05

P₂ = x₂/n₂ = 10/100 = 0.10

Calculate the Pooled Proportion:

[tex]\hat p[/tex] = (x₁ + x₂) / ( n₁ + n₂) = (20 + 10) / (400 + 100) = 30 / 500 = 0.06

Calculate the Standard Error (SE) of the Difference in Proportions:

[tex]SE &= \sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} \\ SE &= \sqrt{0.06 \times 0.94 \left( \frac{1}{400} + \frac{1}{100} \right)} \\ SE &= \sqrt{0.06 \times 0.94 \left( 0.0025 + 0.01 \right)} \\ SE &= \sqrt{0.06 \times 0.94 \times 0.0125} \\ SE &= \sqrt{0.000705} \\ SE &\approx 0.02655[/tex]

Calculate the Test Statistic z:

[tex]z &= \frac{\hat{p}_1 - \hat{p}_2}{SE} \\ z &= \frac{0.05 - 0.10}{0.02655} \\ z &\approx -1.886[/tex]

Find the P-value:

Since this is a two-tailed test, we need to find the probability of obtaining a test statistic as extreme as $z = -1.886$. The P-value is found using the standard normal distribution.

The P-value for $z = -1.886$ is approximately $0.06$.

Complete question:

A manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective; an SRS of 100 parts from supplier 2 finds 10 defective. Let P₁ and P₂ be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. The manufacturer wants to know if there is evidence of a difference in the proportion of defective parts produced by the two suppliers. To make this determination, you test the hypotheses H0 : P₁ = P₂ and Ha : P₁ ≠ P₂.

The P-value of your test is:

0.03

0.06

0.116

a) There is a mass of 25 kilograms of salt in the tank after 3.5 hours.

b) The concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute.

Salt concentration (c(t)), in kilograms per liter, is usually modelled by first order non-homogeneous linear differential equation, whose form is derived from principle of mass conservation and described below:

[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V} \cdot c(t) = \frac{\dot V\cdot c_{o}}{V}[/tex]   (1)

Where:

The solution of this differential equation is:

[tex]c(t) = \left(c_{i}-c_{o}\right)\cdot e^{-\frac{\dot V}{V}\cdot t }+c_{o}[/tex]   (2)

Where:

[tex]c_{i} = \frac{m_{s}}{V}[/tex]   (3)

Where [tex]m_{s}[/tex] is the initial salt mass in the tank, in kilograms and t is the time, in minutes.

a) The amount of salt is found by multiplying the volume occupied by the water and the salt concentration in the tank. If we know that [tex]m_{s} = 50\,kg[/tex], [tex]V = 2000\,L[/tex], [tex]\dot V = 7\,\frac{L}{min}[/tex], [tex]c_{o} = 0.0125\,\frac{kg}{L}[/tex] and [tex]t = 3.5\,h[/tex], then the amount of salt in the tank:

[tex]c_{i} = \frac{50\,kg}{2000\,L}[/tex]

[tex]c_{i} = 0.025\,\frac{kg}{L}[/tex]

[tex]c(12600) = (0.025-0.0125)\cdot e^{-\frac{7}{2000}\cdot (12600)}+0.0125[/tex]

[tex]c(12600) = 0.0125[/tex]

And the salt mass in the tank is:

[tex]m = \left(0.0125\,\frac{kg}{L} \right)\cdot \left(2000\,L\right)[/tex]

[tex]m = 25\,kg[/tex]

There is a mass of 25 kilograms of salt in the tank after 3.5 hours. [tex]\blacksquare[/tex]

b) For [tex]t \to +\infty[/tex], [tex]c(t) \to c_{o}[/tex]. Therefore, the concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute. [tex]\blacksquare[/tex]

To learn more on concentration processes, we kindly invite to check this verified question: https://brainly.com/question/14162783

a) The amount of salt in the tank after 3.5hr is 25Kg

b) The concentration of the salt as time approaches infinity is 0.125Kg/min.

Let y represents the total amount of salt at time(t)

dy/dt = rate of solution in - rate of solution out of the tank

But, concentration = amount of salt at time t/ volume of water at time t

= y/ 2000

dy/dt = 0.0875 - 7y/2000dy/dt

dy/dt = (175 - 7y/2000)dy/dt

separating variables

(1/7y-175)dy = - 1/200dtIn7y - 175

Integration both sides

y = 175 + e^(–1/2000t) + k/7

when the beginning y = 50, t = 0, k = 175

substituting the values

At t = 3.5 hrs

y = 25kg

When t approaches to infinity

Concentration is 0.125kg/min

Learn more about rate of flow of solution here:https://brainly.com/question/14309748

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