Determine the horizontal change of a line with an x intercept at (3,0) and a y intercept at (0,2)

Answer:

a) Statement is true

b) Statement incorrect. Company goal is to attend customer at 190 second at the most. So test should be one tail test  (right)

c) We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error

Step-by-step explanation:

a)  as the significance level is = 0,02  that means  α = 0,02 (the chance of error type I )  and the β (chance of type error II   is

1  -  0.02  =  0,98

b) The company establishe in 190 seconds time for customer at its ticket counter (if this time is smaller is excellent ) company is concerned about bigger time because that could be an issue for customers. Therefore the test should be a one test-tail to the right

c) test statistic

Hypothesis test should be:

null hypothesis                H₀   =  190

alternative hypothesis    H₀   >  190

t(s)  =  ( μ  -  μ₀ ) / s/√n      ⇒   t(s)  = ( 202 - 190 )/(28/√100 )

t(s)  = 12*10/28

t(s)  = 4.286

That value is far away of any of the values found for 99 degree of fredom and between  α  ( 0,025 and 0,01 ). We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error

If we look table  t-student we will find that for 99 degree of freedom and α = 0.02.

Answer:

(0,-6)

Step-by-step explanation:

recall that the y-intercept is simply the point where the line crosses the y-axis at x = 0.

to find the y intercept, we simply substitute x=0 into the equation and solve for y

12x - 10y = 60, when x = 0,

12(0) - 10y = 60

-10y = 60

y = 60 / (-10)

y = -6

hence the coordinate of the y - intercept is (0,-6)

Answer:

give me a second to research the answer

Step-by-step explanation:

Answer:

a) The 99% confidence interval would be given by [tex]-15.277 \leq \mu_1 -\mu_2 \leq 27.277[/tex]

b) No, since the confidence interval contains the 0 we don't have enough evidence to conclude that we have significant differences between the two means.

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 means 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.  

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".  

[tex]\bar X_1 =140.2[/tex] represent the sample mean 1

[tex]\bar X_2 =134.2[/tex] represent the sample mean 2

n1=7 represent the sample 1 size  

n2=13 represent the sample 2 size  

[tex]s_1 =17[/tex] sample standard deviation for sample 1

[tex]s_2 =15.1[/tex] sample standard deviation for sample 2

[tex]\mu_1 -\mu_2[/tex] parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}[/tex] (1)  

And the pooled variance can be founded with the following formula:

[tex]s^2_p=\frac{(n_x -1)s_x^2 +(n_y-1)s_y^2}{n_x +n_y -2}[/tex]

[tex]s^2_p=\frac{(7 -1)17^2 +(13-1)15.1^2}{7 +13 -2}=248.34[/tex]

[tex]S_p =15.759[/tex] the pooled deviation

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:

[tex]\bar X_1 -\bar X_2 =140.2-134.2=6[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:  

[tex]df=n_1 +n_2 -1=7+13-2=18[/tex]  

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,18)".And we see that [tex]t_{\alpha/2}=2.88[/tex]  

The standard error is given by the following formula:

[tex]SE=S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}[/tex]

And replacing we have:

[tex]SE=15.759\sqrt{\frac{1}{7}+\frac{1}{13}}=7.388[/tex]

Part a Confidence interval

Now we have everything in order to replace into formula (1):  

[tex]6-2.88(15.759)\sqrt{\frac{1}{7}+\frac{1}{13}}=-15.277[/tex]  

[tex]6+2.88(15.759)\sqrt{\frac{1}{7}+\frac{1}{13}}=27.277[/tex]  

So on this case the 99% confidence interval would be given by [tex]-15.277 \leq \mu_1 -\mu_2 \leq 27.277[/tex]  

Part b Does the interval suggest that there is a difference in the mean counts of the two researchers?

No, since the confidence interval contains the 0 we don't have enough evidence to conclude that we have significant differences between the two means.

To fill out the normal distribution curve, use the average weight and standard deviation. To find the percentage of people who would receive a bag with a weight greater than a certain amount, calculate the Z-score and use a Z-table.

To fill out the normal distribution curve for this situation, we can use the given information. The bags have an average weight of 1,040 grams with a standard deviation of 25 grams. This means that the mean of the distribution is 1,040 and the standard deviation is 25. We can plot the normal distribution curve using these values.

To find the percentage of people who would receive a bag that had a weight greater than 1,115 grams, we need to find the area under the normal distribution curve to the right of 1,115. We can calculate this using a Z-score calculator or a Z-table. The Z-score for 1,115 is (1,115 - 1,040) / 25 = 3. From the Z-table, we can find that the area to the right of Z-score 3 is approximately 0.0013. This means that approximately 0.13% of people would receive a bag that weighs more than 1,115 grams.

Answer: a)  0.5328   b) 0.9772

Step-by-step explanation:

Given : Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms.

[tex]\mu=200[/tex]   and   [tex]\sigma=10[/tex]

We assume that the resistance in circuits are normally distributed.

a) Let x denotes the average resistance of the circuit.

Sample size : n= 25

Then, the probability that the average resistance for the 25 resistors is between 199 and 202 ohms :-

[tex]P(199<x<200)=P(\dfrac{199-200}{\dfrac{10}{\sqrt{25}}}<\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{202-200}{\dfrac{10}{\sqrt{25}}})\\\\=P(-0.5<z<1)\ \ [\because z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(z<1)-P(z<-0.5)\\\\=P(z<1)-(1-P(z<0.5))\ \ [\because\ P(Z<-z)=1-P(Z<z)]\\\\=0.8413-(1-0.6915)\ \ [\text{By z-table}]\\\\=0.5328[/tex]

b) Total resistors = 25

Let Z be the total resistance of 25 resistors.

To find P(Z≤5100 ohms) , first we find the mean and variance for Z.

Mean= E(Y) = E(25 X)=25 E(X)=25(200)= 5000 ohm

[tex]Var(Y)=Var(25\ X)=25^2(\dfrac{\sigma^2}{n})=25^2\dfrac{(10)^2}{25}=2500[/tex]

The probability that the total resistance does not exceed 5100 ohms will be :

[tex]P(Y\leq5000)=P(\dfrac{Y-\mu}{\sqrt{Var(Y)}}<\dfrac{5100-5000}{\sqrt{2500}})\\\\=P(z\leq2)=0.9772\ \ [\text{By z-table}][/tex]

Hence, the probability that the total resistance does not exceed 5100 ohms = 0.9772

Answer:

Total number of  ways 6 red marbles can be chosen=541549008

Step-by-step explanation:

16 marbles are chosen in which 6 are red marbles and remaining marbles ,which are 10, are blue marbles.

In order to find in how many ways 6 red marbles can be chosen we will proceed as:

Out of 17 red marbles 6 are chosen and out of 18 blue marbles 10 are chosen.

Total number of  ways 6 red marbles can be chosen= [tex]17_{C_6} * 18_{C_1_0}[/tex]

Total number of  ways 6 red marbles can be chosen=[tex]\frac{17!}{6!*(17-6)!} * \frac{18!}{10!*(18-10)!}[/tex]

Total number of  ways 6 red marbles can be chosen= 12376*43758

Total number of  ways 6 red marbles can be chosen=541549008

Answer: N = 541,549,008

Therefore the number of ways to select 6 red marbles is 541,549,008

Step-by-step explanation:

Given;

Number of red marbles total = 17

Number of blue marbles total = 18

Number of red marbles to be selected = 6

Number of blue marbles to be selected = 16 - 6 = 10

To determine the number of ways 6 red marbles can be selected N.

N = number of ways 6 red marbles can be selected from 17 red marbles × number of ways 10 blue marbles can be selected from 18 blue marbles

N = 17C6 × 18C10

N = 17!/(6! × (17-6)!) × 18!/(10! × (18-10)!)

N = 17!/(6! × 11!) × 18!/(10! × 8!)

N = 541,549,008

Therefore the number of ways to select 6 red marbles is 541,549,008

Answer:

7th term = 1.

Step-by-step explanation:

Given that, first term of increasing geometric progression is 9-4√5.

each term (starting with the second) is equal to the difference of the term following it and the term preceding it.

let first term of geometric progression be a and the increasing ratio be r.

⇒ The geometric progression is   a , ar , ar² , ar³, ....... so on.

Given, each term (starting with the second) is equal to the difference of the term following it and the term preceding it.

⇒ second term = (third term - first term)

⇒ ar = (ar² - a)

⇒ r = r² - 1

⇒ r² - r -1 =0

⇒ roots of this equation is r = [tex]\frac{1+\sqrt{5} }{2}[/tex]  , [tex]\frac{1-\sqrt{5} }{2}[/tex]

 (roots of ax²+bx+c are [tex]\frac{-b+\sqrt{b^{2} -4ac} }{2a}[/tex] and [tex]\frac{-b-\sqrt{b^{2} -4ac} }{2a}[/tex])

and it is given, increasing geometric progression

⇒ r > 0.

⇒ r = [tex]\frac{1+\sqrt{5} }{2}[/tex].

Now, nth term in geometric progression is arⁿ⁻¹.

⇒ 7th term = ar⁷⁻¹ = ar⁶.

     = (9-4√5)([tex]\frac{1+\sqrt{5} }{2}[/tex])⁶

     = (0.05572809)(17.94427191)  =  1

⇒ 7th term = 1.

Answer:

a)P(X=2) = (2C2)(0.91)^2 (1-0.91)^{2-2}=0.8281[/tex]

b) P(X=8)=(8C8)(0.91)^8 (1-0.91)^{8-8}=0.4703[/tex]

c) [tex]P(X \geq 1)=1-P(X<1)=1-P(X=0)=1-0.4703=0.5297[/tex]

Step-by-step explanation:

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]  

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: [tex]P(A)+P(A') =1[/tex]

a. Two computers are chosen at random. What is the probability that both computers are ancient?

[tex]P(X=2)=(2C2)(0.91)^2 (1-0.91)^{2-2}=0.8281[/tex]

b. Eight computers are chosen at random. What is the probability that all eight computers are ancient​?

On this case we are looking for this probability:

[tex]P(X=8)=(8C8)(0.91)^8 (1-0.91)^{8-8}=0.4703[/tex]

c. What is the probability that at least one of eight randomly selected computers is cutting-edge​? Would it be unusual that at least one of eighteight randomly selected computers is cutting-edge​?

Since we are interested on the cutting edge class the new probability of success would be p=1-0.91=0.09. And we want to find this probability:

[tex]P(X \geq 1)=1-P(X<1)=1-[P(X=0)][/tex]

And we can find the indiviudal probabilitiy like this:

[tex]P(X=0)=(8C0)(0.09)^0 (1-0.09)^{8-0}=0.4703[/tex]

And if we replace we got:

[tex]P(X \geq 1)=1-P(X<1)=1-P(X=0)=1-0.4703=0.5297[/tex]

The probability of both computers being ancient is about 82.81%, all eight being ancient is about 43.05%, and at least one of eight being cutting-edge is roughly 56.95%, which is not unusual.

Probability of Selecting Ancient Computers

When dealing with probabilities of independent events, such as selecting computers at random, we can calculate the probability of multiple events occurring in sequence by multiplying the probabilities of each individual event.

Note that these calculations assume that each computer's classification as ancient or cutting-edge is independent of the others.

Answer:Angle AEC is 139 degrees.

Step-by-step explanation:

Since line EC bisects angle BED, it divides angle BED equally into 2. This means that

Angle BEC = angle CED

If angle CED = 4x + 1

Therefore,

Angle BED = 2 × angle CED

= 2(4x + 1) = 8x + 2

The sum of the angles in a straight line is 180 degrees. Therefore

Angle AEB + angle BED = 180

Angle AEB = 11x - 12. Therefore

11x - 12 + 8x + 2 = 180

19x - 10 = 180

19x = 180 + 10 = 190

x = 190/19 = 10

Angle BEC = 4x + 1 = 4×10 + 1 = 41

Angle AEB = 11x - 12 = 11×10 - 12 = 98

Angle AEC = 41 + 98 = 139

Answer: AEC = 139

Step-by-step explanation:

You first have to find x.

To find x, we need to add all the angles together to get 180°.

Since EC bisects BED, we know angle BEC and CED equal the same measure.

AEB + BEC + CED = 180°

(11x - 12) + (4x + 1) + (4x + 1) = 180°

Add like terms and solve

19x - 10 = 180

19x = 190

x = 10

Now, we substitute 'x' in AEB and BEC

AEB = 11x - 12

11 (10) - 12

AEB = 98

BEC = 4x + 1

4 (10) + 1

BEC = 41

98 + 41 = 139

AEC = 139

Answer:

12 6/8 or 12 3/4

Step-by-step explanation:

Answer:the total amount that they recycled is 12 3/4 boxes

Step-by-step explanation:

Maria's class recycled 2 2/8 boxes of paper in a month. Converting 2 2/8 boxes of paper in a month to improper fraction, it becomes

18/8 boxes of paper in a month.

They recycled another 10 4/8 boxes the next month. Converting 10 4/8 boxes to improper fraction, it becomes 84/8 boxes.

Therefore, the total amount of boxes that they recycled would be

18/8 + 84/8 = 102/8 boxes

Converting to whole number, it becomes

12 3/4 boxes

Final answer:

The degrees of freedom for a chi-square test of independence with two variables each having three categories and a sample size of n = 75 is 4.

Explanation:

The question at hand involves determining the degrees of freedom (df) for a chi-square test of independence where two variables each have three categories, and the sample size is n = 75. To calculate the degrees of freedom for this scenario, you use the formula df = (r - 1)(c - 1) where r represents the number of rows (categories of one variable) and c represents the number of columns (categories of the other variable).

In this case, with both variables having three categories, we have r = 3 and c = 3, which gives us:

df = (3 - 1)(3 - 1) = (2)(2) = 4.

Therefore, the correct answer is a. 4 degrees of freedom for the chi-square distribution when testing for independence with a sample size of n = 75.

Answer:

The value of the test statistic and degrees of freedom is 2.148 and 11 respectively.

Step-by-step explanation:

Consider the provided information.

The mean annual tuition and fees for a sample of 12 private colleges was 36,800 with a standard deviation of 5,000 .

Thus, n = 12, [tex]\bar x=36800[/tex] σ = 5000

degrees of freedom = n-1 = 12-1 = 11

[tex]H_0: \mu = 33700\ and\ H_a: \mu \neq 33700[/tex]

Formula to find the value of z is: [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

Where [tex]\bar x[/tex] is mean of sample, μ is the mean of population, σ is the standard deviation of population and n is number of observations.

[tex]z=\frac{36800-33700}{\frac{5000}{\sqrt{12}}}[/tex]

[tex]z=2.148[/tex]

Hence, the value of the test statistic and degrees of freedom is 2.148 and 11 respectively.

It's a six sided polygon.  For any polygon the external angles add to 360 degrees.  The internal angles shown are the supplements of the external angles.   We have

(180 - θ₁) + (180 - θ₂) + ... + (180 - θ₆) = 360

6(180) - 360 = θ₁ + θ₂ + θ₃ + θ₄ + θ₅ + θ₆

720 = θ₁ + θ₂ + θ₃ + θ₄ + θ₅ + θ₆

The six angles add up to 720 degrees, and five of them add to

126+101+135+147+96=605

So y = 720 - 605 = 115

The degree sign is external to y so not part of the answer:

Answer: 115

Answer:y = 115 degrees

Step-by-step explanation:

The given polygon has 6 sides. It is a hexagon. The sum of the interior angles of a polygon is

180(n - 2)

Where

n is the number of sides that the polygon has. This means that n = 6

Therefore, the sum of the interior angles would be

180(6 - 2) = 720 degrees

Therefore,

126 + 101 + 135 + 147 + 96 + y = 720

605 + y = 720

Subtracting 605 from both sides of the equation, it becomes

y = 720 - 605 = 115 degrees

Step-by-step explanation:

Since we have given that

Sample size = 25

Mean = 46 years

Standard deviation = 5 years

We need to find the margin of error of a 98% confidence interval for the true mean client age.

So, Margin of error is given by

[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=2.33\times \dfrac{5}{\sqrt{25}}\\\\=2.33\times \dfrac{5}{5}\\\\=2.33[/tex]

Hence, margin of error is 2.33.

Answer:

A. Different volumes of water are moved different distances.

Step-by-step explanation:

Integration is used to find work required to pump water out of a tank because different volumes of water are moved different distances and to sum it all we the tool required is integration. Moreover, work done is a path function or an inexact differential. It does depend upon the path followed by the process.

hence the correct answer is A.

Answer:

[tex]p_v =P(t_{74}<-1.227)=0.112[/tex]    

If we compare the p value and a significance level for example [tex]\alpha=0.1[/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, so we can conclude that the true mean it's not significantly less than 20 min.

Step-by-step explanation:

Data given and notation    

[tex]\bar X=18.81[/tex] represent the average lateral recumbency for the sample    

[tex]s=8.4[/tex] represent the sample standard deviation    

[tex]n=75[/tex] sample size    

[tex]\mu_o =20[/tex] represent the value that we want to test    

[tex]\alpha[/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 apply a left tailed  test.  

What are H0 and Ha for this study?    

Null hypothesis:  [tex]\mu \geq 20[/tex]  

Alternative hypothesis :[tex]\mu < 20[/tex]  

Compute the test statistic  

The statistic for this case 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{18.81-20}{\frac{8.4}{\sqrt{75}}}=-1.227[/tex]

The degrees of freedom are given by:

[tex]df=n-1=75-1=74[/tex]    

Give the appropriate conclusion for the test  

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

[tex]p_v =P(t_{74}<-1.227)=0.112[/tex]    

Conclusion    

If we compare the p value and a significance level for example [tex]\alpha=0.1[/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, so we can conclude that the true mean it's not significantly less than 20 min.

Answer:

Yes, this equation has a solution. According to Intermediate Value Theorem at least one solution for [0,2]

Step-by-step explanation:

Hi there!

1) Remember a definition.

Intermediate Value Theorem:

If [tex]f[/tex] is continuous on a given closed interval [a,b], and f(a)≠f(b) and f(a)<k<f(b) then there has to be at least one number 'c' between 'a' and 'b', such that f(c)=k

----

(Check the first graph as an example)

2) The Intermediate Value Theorem can be applied to determine whether there is a solution on a given interval.

Let's choose the interval [tex][0,2][/tex]

[tex]f(x)=x^{3}-3x-1\\f(0)=(0)^{3}-3(0)-1\\f(0)=-1\\f(0)<0\\[/tex]

Proceed to the other point: 2

[tex]f(x)=x^{3}-3x-1\\f(2)=(2)^{3}-3(2)-1\\f(2)=1\\f(2)>0\\[/tex]

3) Check the 2nd Graph for a the Visual answer, of it.  And the 3rd graph for all solutions of this equation.

Answer:

a. none of these answers

Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.281)

Step-by-step explanation:

Data given and notation  

[tex]\bar X=13.5[/tex] represent the mean height for the sample  

[tex]s=3.2[/tex] represent the sample standard deviation for the sample  

[tex]n=9[/tex] sample size  

[tex]\mu_o =10[/tex] represent the value that we want to test

[tex]\alpha=0.01[/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 check if the mean is higher than 10 years, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 10[/tex]  

Alternative hypothesis:[tex]\mu > 10[/tex]  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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{13.5-10}{\frac{3.2}{\sqrt{9}}}=3.28[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=9-1=8[/tex]  

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

[tex]p_v =P(t_{(8)}>3.281)=0.00558[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that we have a mean higher than 10 years at 1% of significance.  

a. none of these answers

Answer:  The required ordered pair (p, q) is (-7, -12).

Step-by-step explanation:  Given that (x+2)(x-1) divides the following polynomial f(x) :

[tex]f(x)=x^5-x^4+x^3-px^2+qx+4.[/tex]

We are to find the ordered pair (p,q).

We have the following theorem :

Factor theorem : If (x-a) divides a polynomial h(x), then h(a) = 0.

According to the given information, we can say that (x+2) divides f(x). So, we get

[tex]f(-2)=0\\\\\Rightarrow (-2)^5-(-2)^4+(-2)^3-p(-2)^2+q(-2)+4=0\\\\\Rightarrow -32-16-8-4p-2q+4=0\\\\\Rightarrow 2p+q=-26~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

Also, (x-1) is a factor of f(x). So,

[tex]f(1)=0\\\\\Rightarrow (1)^5-(1)^4+(1)^3-p(1)^2+q(1)+4=0\\\\\Rightarrow 1-1+1-p+q+4=0\\\\\Rightarrow p-q=5~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]

Adding equations (i) and (ii), we get

[tex]3p=-21\\\\\Rightarrow p=-7.[/tex]

From equation (ii), we get

[tex]-7-q=5\\\\\Rightarrow q=-12.[/tex]

Thus, the required ordered pair (p, q) is (-7, -12).

The point that does not satisfy the inequality is

A) (4, 0)

To find points satisfying an inequality in a two-dimensional space:

Graph the Inequality: Plot the boundary line as if it were an equation (solid or dashed based on inclusion/exclusion).

Shade the Region: Determine the side of the line representing the solution set by testing a point; shade the satisfying side.

Identify Points: All points within the shaded region, including the boundary if included, satisfy the inequality.

Read more on inequality here https://brainly.com/question/24372553

#SPJ2

Answer:

Step-by-step explanation:

Given

mean [tex]\mu =12.9 Pounds[/tex]

Standard deviation [tex]\sigma =1\ Pound[/tex]

[tex]P(x<11)=P\left ( \frac{x-\mu }{\sigma }< \frac{11-12.9}{1}\right )[/tex]

[tex]P(x<11)=P\left ( z< -1.9\right )[/tex]

From z table

[tex]P(x<11)=0.0289\approx 0.029[/tex]

Answer:

Null hypothesis:[tex]\mu = 100[/tex]  

Alternative hypothesis:[tex]\mu \neq 100[/tex]  

[tex]t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921[/tex]  

[tex]p_v =2*P(t_{11}<-0.921)=0.377[/tex]  

Step-by-step explanation:

1) Data given and notation  

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]

[tex]s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}[/tex]

The results obtained are:

[tex]\bar X=98.375[/tex] represent the sample mean  

[tex]s=0.6.109[/tex] represent the sample standard deviation  

[tex]n=12[/tex] sample size  

[tex]\mu_o =100[/tex] represent the value that we want to test  

[tex]\alpha[/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 check if the mean is equal to 100pCL/L, the system of hypothesis are :  

Null hypothesis:[tex]\mu = 100[/tex]  

Alternative hypothesis:[tex]\mu \neq 100[/tex]  

Since we don't know the population deviation, 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".  

3) Calculate the statistic  

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

[tex]t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921[/tex]  

4) P-value  

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

[tex]df=n-1=12-1=11[/tex]

Since is a two sided test the p value would given by:  

[tex]p_v =2*P(t_{11}<-0.921)=0.377[/tex]  

5) Conclusion  

If we compare the p value and the significance level assumed [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, so we can conclude that the true mean is not significant different from 100 at 5% of significance.  

The question asked is about setting a hypothesis test to see if the mean reading from radon detectors differs from 100 pCI/L. The null and alternative hypotheses should be formed, calculations should be performed and the p-value compared with the significance level to determine if the null hypothesis should be retained or rejected.

The subject of this question is setting up a hypothesis test for determining if the mean reading from the radon detectors differs from 100 pCI/L. The first step is to set up the null and alternate hypothesis. The null hypothesis (H0) would be that the population mean reading is 100 pCI/L (µ = 100), whereas the alternative hypothesis (H1) would propose that the mean reading differs from 100 pCI/L (µ ≠ 100).

Next, we would calculate the sample mean and sample standard deviation. Using these calculations, you can perform a t-test to compare the sample mean to the proposed population mean of 100 pCI/L. The decision of rejecting or not rejecting the null hypothesis relies on the comparison of the p-value obtained from the test statistic with the significance level.

Without the actual calculations, it is not possible to conclude whether the data suggests that the population mean reading under these conditions differs from 100 pCI/L or not. However, if the p-value is less than the significance level (commonly 0.05), we would reject the null hypothesis and conclude that the data provides enough evidence to suggest that the population mean differs from 100.

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Answer:

68.26% corresponds to the life span of fruit flies that are between 29 days and 37 days

Step-by-step explanation:

Given that the life span of a species of fruit fly have a bell-shaped distribution, with a mean of 33 days and a standard deviation of 4 days.

Let X be the life span

X is N(33,4)

we can convert X into Z standard normal variate by

[tex]z=\frac{x-33}{4}[/tex]

To find the percentage  corresponds to the life span of fruit flies that are between 29 days and 37 days

For this let us find probability for x lying between 29 and 37 days

[tex]29\leq x\leq 37\\=-1\leq z\leq 1[/tex]

Probability = P(|z|<1) = 0.6826

Convert to percent

68.26% corresponds to the life span of fruit flies that are between 29 days and 37 days

To find the percentage that corresponds to the life span of fruit flies between 29 days and 37 days, we need to calculate the z-scores for both values and then use the standard normal distribution table to find the area under the curve between those z-scores. The percentage is approximately 68%.

To find the percentage that corresponds to the life span of fruit flies between 29 days and 37 days, we need to calculate the z-scores for both values and then use the standard normal distribution table to find the area under the curve between those z-scores.

To calculate the z-scores, we use the formula: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.

For 29 days: z = (29 - 33) / 4 = -1

For 37 days: z = (37 - 33) / 4 = 1

Looking at the standard normal distribution table or using a calculator, we find that the area between -1 and 1 is approximately 68%. Therefore, the percentage that corresponds to the life span of fruit flies between 29 days and 37 days is 68%.

Answer:

Consider the following calculations

Step-by-step explanation:

The complete R snippet is as follows

install.packages("Sleuth3")

library("Sleuth3")

attach(case0502)

data(case0502)

## plot

# plots

boxplot(Percent~ Judge, data=case0502,ylab="Values",

main="Boxplots of the Data",col=c(2:7,8),horizontal=TRUE)

# perform anova analysis

a<- aov(lm(Percent~ Judge,data=case0502))

#summarise the results

summary(a)

### we can use the independent sample t test here

sp<-case0502[which(case0502$Judge=="Spock's"),]

nsp<-case0502[which(case0502$Judge!="Spock's"),]

## perform the test    

t.test(sp$Percent,nsp$Percent)

The results are CHECK THE IMAGE ATTACHED

b)

> summary(a)

Df Sum Sq Mean Sq F value Pr(>F)

Judge 6 1927 321.2 6.718 6.1e-05 *** as the p value is less than 0.05 , hence there is a significant difference in the percent of women included in the 6 judges’ venires who aren’t Spock’s judge

Residuals 39 1864 47.8

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

c)

t.test(sp$Percent,nsp$Percent)

  Welch Two Sample t-test

data: sp$Percent and nsp$Percent

t = -7.1597, df = 17.608, p-value = 1.303e-06 ## as the p value is less than 0.05 , hence we reject the null hypothesis in favor of alternate hypothesis and conclude that there is a significant difference in the percent of women incuded in Spock’s venires versus the percent included in the other judges’ venires combined

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-19.23999 -10.49935

sample estimates:

mean of x mean of y

14.62222 29.49189

Answer:

n=3394

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 means 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.  

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 population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Solution to the problem

The confidence interval for the mean is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by [tex]\alpha=1-0.98=0.02[/tex] and [tex]\alpha/2 =0.01[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-2.33, z_{1-\alpha/2}=2.33[/tex]

The margin of error for the proportion interval is given by this formula:  

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]    (a)  

And on this case we have that [tex]ME =\pm 0.02[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  

We can use as estimation of [tex]\hat p=0.5[/tex] since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.02}{2.33})^2}=3393.06[/tex]  

And rounded up we have that n=3394

Answer:

B

Step-by-step explanation:

Type I error is basically rejection of the null hypothesis when the null hypothesis is true. In the given scenario the null hypothesis consists of the percentage of students who own cars is 35%. Hence the type I error would be rejection of null hypothesis that the percentage of students who own cars is 35% while the percentage is 35%.

Final answer:

A Type I error is rejecting the null hypothesis when it's true, and a Type II error is failing to reject the null hypothesis when it's false. In this case, Type I error is option B and Type II error is option C.

Explanation:

When performing hypothesis testing, there is potential to make two types of errors: Type I error and Type II error. A Type I error occurs when we reject the null hypothesis even though the null hypothesis is actually true. In the context of the question, the Type I error would correspond to option B: Reject the null hypothesis that the percentage of college students who own cars is equal to 35% when that percentage is actually equal to 35%.

In contrast, a Type II error happens when we fail to reject the null hypothesis whereas the null hypothesis is false. It is correctly identified by option C: Fail to reject the null hypothesis that the percentage of college students who own cars is equal to 35% when that percentage is actually different from 35%.

Answer:

[tex] df = n-1=10-1=9[/tex]

a. 9

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".  

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".  

We assume that we have the following system of hypothesis:

H0: The data follows the distribution proposed

H1: The data not follows the distribution proposed

The statistic to check the hypothesis is given by:  

[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]  

This statistic have a Chi Square distribution distribution with k-1 degrees of freedom, where n represent the number of categories on this case k=10. And if we find the degrees of freedom we got:

[tex] df = k-1=10-1=9[/tex]

a. 9

Final answer:

The correct answer is a. 9.

The degrees of freedom for a chi-square goodness-of-fit test with 10 categories is 9, which is calculated by subtracting one from the number of categories.

Explanation:

The degrees of freedom for a chi-square goodness-of-fit test are calculated as the number of categories minus one. In the case of having 10 categories, the degrees of freedom would be 10 - 1 = 9.

Therefore, the correct answer is a. 9.

Remember, the chi-square test statistic helps us determine how well an observed distribution fits an expected distribution, and degrees of freedom are essential for determining the critical values of the test from the chi-square distribution.

For a chi-square distribution, as the degrees of freedom increase, the curve becomes more symmetrical.

Answer:

So, the interval is : (7.206,7.794)

Step-by-step explanation:

The mean μ = 7.5

Standard deviation σ=3

n = 400

At 95% confidence interval, the z score is 1.96

[tex]7.5+1.96(\frac{3}{\sqrt{400} } )[/tex]

And [tex]7.5-1.96(\frac{3}{\sqrt{400} } )[/tex]

7.5+0.294 and 7.5-0.294

So, the interval is : (7.206,7.794)

Answer:

a) If we compare the p value and the significance level given [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, so we can conclude that the true mean is not significantly different from 8.17 at 5% of signficance.  

b) Since is a two sided test the p value would be:  

[tex]p_v =2*P(t_{(49)}<-1.525)=0.067[/tex]  

Step-by-step explanation:

1) Data given and notation  

[tex]\bar X=8.159[/tex] represent the mean weight for the sample  

[tex]s=0.051[/tex] represent the sample standard deviation

[tex]n=50[/tex] sample size  

[tex]\mu_o =8.17[/tex] represent the value that we want to test

[tex]\alpha=0.05[/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)  

Part a

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 8.17, the system of hypothesis would be:  

Null hypothesis:[tex]\mu = 8.57[/tex]  

Alternative hypothesis:[tex]\mu \neq 8.57[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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{8.159-8.17}{\frac{0.051}{\sqrt{50}}}=-1.525[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=50-1=49[/tex]  

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

[tex]p_v =2*P(t_{(49)}<-1.525)=0.067[/tex]  

Conclusion  

If we compare the p value and the significance level given [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, so we can conclude that the true mean is not significantly different from 8.17 at 5% of signficance.