A probability distribution showing the probability of success does not change from trial to trial, is termed a:
-uniform probability distribution
-binomial probability distribution
-hypergeometric probability distribution
-normal probability distribution

The answer to this problem involves applying integrals, norms, and concepts of angles between vectors to the functions f(x) and g(x). The INNER PRODUCT is the integral of the products of the two functions, the norms are the square roots of the inner products of the functions with themselves, and the angle between the functions is calculated using the dot product and norms.

To find the inner product 〈f,g〉, the norms ∥f∥ and ∥g∥, and the angle αf,g between the functions f(x)=−10x2−6 and g(x)=−9x−4, we'll apply concepts from vector calculus. The inner product (also known as the dot product) is the integral from 0 to 1 of the products of the two functions. The norm of a function is the square root of the inner product of the function with itself. The angle between two vectors in a Vector Space, in this case the space of continuous functions C0[0,1], is given by cos(α) = 〈f,g〉/( ∥f∥∙ ∥g∥). Integrating and solving these equations will give us the desired values.

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These are the corrected values for the inner product, norms, and angle between [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]in the vector space [tex]\(C^0[0,1]\).[/tex]

1. The inner product \(\langle f,g \rangle\) is given by the integral [tex]\(\int_{1}^{0} f(x)g(x) \, dx\).[/tex] We will compute this integral with[tex]\(f(x) = -10x^2 - 6\) and \(g(x) = -9x - 4\).[/tex]

[tex]\[ \langle f,g \rangle = \int_{1}^{0} (-10x^2 - 6)(-9x - 4) \, dx \][/tex]

 Expanding the integrand, we get:

[tex]\[ \int_{1}^{0} (90x^3 + 40x^2 + 54x + 24) \, dx \][/tex]

Integrating term by term, we have:

[tex]\[ \left[ \frac{90}{4}x^4 + \frac{40}{3}x^3 + \frac{54}{2}x^2 + 24x \right]_{1}^{0} \][/tex]

[tex]\[ = \left( \frac{90}{4} \cdot 0^4 + \frac{40}{3} \cdot 0^3 + \frac{54}{2} \cdot 0^2 + 24 \cdot 0 \right) - \left( \frac{90}{4} \cdot 1^4 + \frac{40}{3} \cdot 1^3 + \frac{54}{2} \cdot 1^2 + 24 \cdot 1 \right) \][/tex]

[tex]\[ = - \left( \frac{90}{4} + \frac{40}{3} + \frac{54}{2} + 24 \right) \][/tex]

[tex]\[ = - \left( 22.5 + 13.333 + 27 + 24 \right) \][/tex]

[tex]\[ = -87.833 \][/tex]

So, [tex]\(\langle f,g \rangle = -87.833\).[/tex]

2. The norm of f is given by [tex]\(\|f\|[/tex] = [tex]\sqrt{\langle f,f \rangle}\)[/tex], where:

[tex]\[ \langle f,f \rangle = \int_{1}^{0} (-10x^2 - 6)^2 \, dx \][/tex]

[tex]\[ = \int_{1}^{0} (100x^4 + 120x^2 + 36) \, dx \][/tex]

[tex]\[ = \left[ \frac{100}{5}x^5 + \frac{120}{3}x^3 + 36x \right]_{1}^{0} \][/tex]

[tex]\[ = - \left( \frac{100}{5} + \frac{120}{3} + 36 \right) \][/tex]

[tex]\[ = - \left( 20 + 40 + 36 \right) \][/tex]

[tex]\[ = -96 \][/tex]

[tex]\[ \|f\| = \sqrt{96} \][/tex]

3. Similarly, the norm of g is given by [tex]\(\|g\|[/tex] = [tex]\sqrt{\langle g,g \rangle}\)[/tex], where:

[tex]\[ \langle g,g \rangle = \int_{1}^{0} (-9x - 4)^2 \, dx \][/tex]

[tex]\[ = \int_{1}^{0} (81x^2 + 72x + 16) \, dx \][/tex]

[tex]\[ = \left[ \frac{81}{3}x^3 + 36x^2 + 16x \right]_{1}^{0} \][/tex]

[tex]\[ = - \left( \frac{81}{3} + 36 + 16 \right) \][/tex]

[tex]\[ = - \left( 27 + 36 + 16 \right) \][/tex]

[tex]\[ = -79 \][/tex]

[tex]\[ \|g\| = \sqrt{79} \][/tex]

4. The angle \[tex](\alpha_{f,g}\)[/tex] between [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex] is found using the formula:

[tex]\[ \cos(\alpha_{f,g}) = \frac{\langle f,g \rangle}{\|f\| \cdot \|g\|} \][/tex]

[tex]\[ \alpha_{f,g} = \cos^{-1}\left(\frac{-87.833}{\sqrt{96} \cdot \sqrt{79}}\right) \][/tex]

[tex]\[ \langle f,g \rangle = -87.833, \quad \|f\| = \sqrt{96}, \quad \|g\| = \sqrt{79}, \quad \alpha_{f,g} = \cos^{-1}\left(\frac{-87.833}{\sqrt{96} \cdot \sqrt{79}}\right) \][/tex]

The correct calculations for the inner product and norms should be as follows:

[tex]\[ \langle f,g \rangle = \int_{0}^{1} (90x^3 + 40x^2 + 54x + 24) \, dx \][/tex]

[tex]\[ \|f\| = \sqrt{\int_{0}^{1} (100x^4 + 120x^2 + 36) \, dx} \][/tex]

[tex]\[ \|g\| = \sqrt{\int_{0}^{1} (81x^2 + 72x + 16) \, dx} \][/tex]

The angle calculation remains the same. Let's correct the integrals:

[tex]\[ \langle f,g \rangle = \left[ \frac{90}{4}x^4 + \frac{40}{3}x^3 + \frac{54}{2}x^2 + 24x \right]_{0}^{1} \][/tex]

[tex]\[ = \frac{90}{4} + \frac{40}{3} + \frac{54}{2} + 24 \][/tex]

[tex]\[ = \frac{90}{4} + \frac{40}{3} + \frac{54}{2} + 24 \][/tex]

[tex]\[ = 87.833 \][/tex]

So,[tex]\(\langle f,g \rangle = 87.833\).[/tex]

[tex]\[ \|f\| = \sqrt{\left[ \frac{100}{5}x^5 + \frac{120}{3}x^3 + 36x \right]_{0}^{1}} \][/tex]

[tex]\[ = \sqrt{20 + 40 + 36} \][/tex]

[tex]\[ = \sqrt{96} \][/tex]

[tex]\[ \|g\| = \sqrt{\left[ \frac{81}{3}x^3 + 36x^2 + 16x \right]_{0}^{1}} \][/tex]

[tex]\[ = \sqrt{27 + 36 + 16} \][/tex]

[tex]\[ = \sqrt{79} \][/tex]

The corrected values are:

[tex]\[ \langle f,g \rangle = 87.833, \quad \|f\| = \sqrt{96}, \quad \|g\| = \sqrt{79} \][/tex]

The angle calculation remains the same:

[tex]\[ \alpha_{f,g} = \cos^{-1}\left(\frac{87.833}{\sqrt{96} \cdot \sqrt{79}}\right) \][/tex]

Answer:

A

Step-by-step explanation:

Having an inexpensive wedding helps young couples avoid financial burdens that may strain their marriage. This is true as it will ensure the young couples spend wisely. They understand that if they splurge, the returns will not be as great as that from a long-lasting marriage.

Answer:

Step-by-step explanation:

Given

Paraboloid [tex]z=x^2+y^2[/tex] and Plane [tex]x+y=1[/tex] are bounded to get the area

In first octant

value of x varies from [tex]0\leq x\leq 1[/tex]

value of y varies from [tex]0\leq y\leq 1[/tex]

Volume [tex]V=\int_{0}^{1}\int_{0}^{1-x}\left ( x^2+y^2\right )dydx[/tex]

[tex]V=\int_{0}^{1}\left [ x^2y+\frac{y^3}{3}\right ]_{0}^{1-x}[/tex]

[tex]V=\int_{0}^{1}\left [ x^2(1-x)+\frac{(1-x)^3}{3}\right ]dx[/tex]

[tex]V=\int_{0}^{1}\left ( \frac{-4x^3}{3}+2x^2-x+\frac{1}{3}\right )dx[/tex]

[tex]V=\frac{1}{6}[/tex]

The volume of the region bounded by the paraboloid z=x2+y2 and the plane x+y=1 in the first octant is 1/6

The paraboloid and the plane are given as:

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

[tex]x+y = 1[/tex]

Make y the subject

[tex]y = 1 - x[/tex]

The plane is in the first octant.

So, the volume is represented as:

[tex]V = \int\limits^1_0 { \int\limits^y_0 {z^2} \, dy } \, dx[/tex]

The integral becomes

[tex]V = \int\limits^1_0 { \int\limits^{1-x}_0 {x^2 + y^2} \, dy } \, dx[/tex]

Integrate with respect to y

[tex]V = \int\limits^1_0 [x^2y + \frac 13y^3]\limits^{1-x}_0 } \, dx[/tex]

Expand

[tex]V = \int\limits^1_0 [x^2(1-x) + \frac 13(1-x)^3]} \, dx[/tex]

Expand

[tex]V = \int\limits^1_0 [x^2-x^3 + \frac 13(1 -x + x^2 - x^3)]} \, dx[/tex]

Expand

[tex]V = \int\limits^1_0 [x^2-x^3 + \frac 13 -x + x^2 - \frac {x^3}3]} \, dx[/tex]

Evaluate the like terms

[tex]V = \int\limits^1_0 [ \frac 13 -x + 2x^2 - \frac {4x^3}3]} \, dx[/tex]

Integrate

[tex]V = [\frac 13x -\frac 12x^2 + \frac 23x^3 - \frac 13x^4]\limits^1_0[/tex]

Expand

[tex]V = \frac 13(1) -\frac 12(1)^2 + \frac 23(1)^3 - \frac 13(1)[/tex]

[tex]V = -\frac 12(1)^2 + \frac 23(1)^3[/tex]

Evaluate the exponents

[tex]V = -\frac 12 + \frac 23[/tex]

Add the fractions

[tex]V = \frac{-3 + 4}{6}[/tex]

[tex]V = \frac{1}{6}[/tex]

Hence, the volume of the region is 1/6

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Answer:

Step-by-step explanation:

Given

Balloon velocity is 14 ft/s upwards  

Distance between balloon and cyclist is 140 ft

Velocity of motor cycle is 88 ft/s

After 10 sec  

motorcyclist traveled a distance of [tex]d_c=88\times 10=880 ft[/tex]

Distance traveled by balloon in 10 s

[tex]d_b=14\times 10=140 ft[/tex]

net height of balloon from ground =140+140=280 ft[/tex]

at [tex]t=10 s[/tex]

distance between cyclist and balloon is [tex]z=\sqrt{280^2+880^2}[/tex]

[tex]z=923.47 ft[/tex]

now suppose at any time t motorcyclist cover a distance of x m and balloon is at a height of h m

thus [tex]z^2=(88t)^2+(14t+140)^2[/tex]

differentiating w.r.t time

[tex]\Rightarrow z\frac{\mathrm{d} z}{\mathrm{d} t}=14\cdot \left ( 14t+140\right )+88\cdot \left ( 88t\right )[/tex]

[tex]\Rightarrow at\ t=10 s[/tex]

[tex]\Rightarrow \frac{\mathrm{d} z}{\mathrm{d} t}=\frac{14\cdot \left ( 14t+140\right )+88\cdot \left ( 88t\right )}{z}[/tex]

[tex]\Rightarrow \frac{\mathrm{d} z}{\mathrm{d} t}=\frac{14\times 280+88^2\times 10}{923.471}[/tex]

[tex]\Rightarrow \frac{\mathrm{d} z}{\mathrm{d} t}=\frac{81,360}{923.471}=88.102\ ft/s[/tex]

The sample of 100 horses is too small to obtain an unbiased estimate.

Applied research frequently focuses more on finding solutions to particular issues that directly impact people in the present. As an illustration, a social psychologist conducting fundamental research on violence would consider the various potential causes of violence in general.

Given, A study is being planned by a group of veterinary researchers to determine how many enteroliths are present on average in affected horses. When the study is financed, the researchers will sign up the first 100 horses they observe.

A study plan, which is considered to be the most important piece of information describing the projected research sample of an investigator, is the main part of an application. An important analyst is also given the opportunity to speak on the planned study, highlighting its benefits and explaining the implementation and conduct of the research. Because it covers your whole research project from start to finish and identifies and explains your emphasis, technique, and goals, a research plan is crucial.

Therefore, The sample size of 100 horses is insufficient to provide a fair assessment.

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Answer:

[tex]\mu_{p_E -p_H} = 0.4-0.6 =-0.2[/tex]

[tex]SE_{p_E -p_H}=\sqrt{\frac{0.4(1-0.4)}{20} +\frac{0.6 (1-0.6)}{20}}=0.1549[/tex]

b. Normal with mean μ= -0.2 and standard deviation σ=0.1549

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".  

[tex]p_E[/tex] represent the real population proportion for Engineering

[tex]\hat p_E =\frac{8}{20}=0.4[/tex] represent the estimated proportion for Engineering

[tex]n_E=20[/tex] is the sample size required for Engineering

[tex]p_H[/tex] represent the real population proportion for Humanities  

[tex]\hat p_H =\frac{12}{20}=0.6[/tex] represent the estimated proportion for Humanities

[tex]n_H=20[/tex] is the sample size required for Humanities

We assume that the population proportions follows a normal distribution since we satisfy these conditions:

[tex]np\geq 5[/tex] ,[tex]n(1-p)\geq 5[/tex]

[tex]20*0.4= 8\geq 5 , 20(1-0.4)=12\geq 5[/tex]

The population proportion have the following distribution  

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

And we are interested on the distribution for [tex]p_E-p_H[/tex]

For this case we know that the distribution for the differences of proportions is also normal and given by:

[tex]p_E -p_H \sim N(\hat p_E -\hat p_H, \sqrt{\frac{\hat p_E(1-\hat p_E)}{n_E} +\frac{\hat p_H (1-\hat p_H)}{n_H}}[/tex]

So then we can find the mean and the deviation like this:

[tex]\mu_{p_E -p_H} = 0.4-0.6 =-0.2[/tex]

[tex]SE_{p_E -p_H}=\sqrt{\frac{0.4(1-0.4)}{20} +\frac{0.6 (1-0.6)}{20}}=0.1549[/tex]

So then the best option is :

b. Normal with mean μ= -0.2 and standard deviation σ=0.1549

The shape is unknown since there is no knowledge about it and the representative sample is limited.

Total number of interview in engineering and humanities = 20

Number of interview in engineering = 8

Number of interview in humanities = 12

⇒ Probability of interview in engineering - Probability of interview in humanities

⇒ 8/20 - 12/20

⇒ -0.20

The shape is unknown since there is no knowledge about it and the representative sample is limited.

Therefore, Option D is the right answer.

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Answer:Bradley practices for 328500 minutes more than Jonas

Step-by-step explanation:

Jonas practices guitar 2 hours a day for 2 years. 24 hours make a day. Assuming there are 365 days in a year.

If he plays 2 hours for 2 days

Number of hours for which he practiced for 365 days would be (365×2)/2 = 365 hours.

Over the course of 3 years, number of hours for which he practiced would be 365×3 = 1095 hours

Bradley practices the guitar 5 hours a day more then Jonas. Jonas practiced for 1 hour a day. This means that Bradley practices the guitar for 5+1 = 6 hours a day. In a year, it would be 6×365 = 2190 hours. In 3 years, it would be 3×2190 = 6570 hours

Difference in number of hours between Bradley and Jonas is

6570 - 1095 = 5475 hours. Since 60 minutes = 1 hour,

5475 hours would be

5475×60 = 328500 minutes.

Answer:

Step-by-step explanation:

Given that the television show Degenerate Housewives has been successful for many years. That show recently had a share of 18, meaning that among the TV sets in use, 18% were tuned to Degenerate Housewives.

Each household is independent of the other and there two outcomes.

X no of households that are tuned to Degenerate Housewives. is Bin (12, 0.18)

the probability that none of the households are tuned to Degenerate Housewives. P(none) =_______0.0924_____

the probability that at least one household is tuned to Degenerate Housewives. P(at least one) = ____1-0.0924 = 0.9076______

the probability that at most one household is tuned to Degenerate Housewives. P(at most one) = _____0.3359______

If at most one household is tuned to Degenerate Housewives, does it appear that the 18% share value is wrong? (Hint: Is the occurrence of at most one household tuned to Degenerate Housewives unusual?)

No 33% is not unusual.

Answer:

Option A is right

Step-by-step explanation:

Given that an  article in an educational journal discussing a non-conventional high school reading program reports a 95% confidence interval of (520 , 560) for the average score on the reading portion of the SAT.

Let X be the  score on the reading portion of the SAT.

By central limit theorem x bar, the sample mean will follow a normal.

95% confidence interval shows that mean

= average of the two limits

= 540

Margin of error = upper bound of confidence interval - Mean

=[tex]560-540=20[/tex]

Option A is right

Answer:

The different single-scoop ice-cream cones can you buy from this vendor=12

Step-by-step explanation:

Given that a  vendor sells ice cream from a cart on the boardwalk. He offers vanilla, chocolate, strawberry, blueberry, and pistachio ice cream, served on either a waffle, sugar, or plain cone.

Different types of icecreams are 4

Types of cones are 3

For each ice cream say we can have either waffle, sugar or plain cone.

i.e. for each type of ice cream, we have 3 different cones.

Hence for 4 types of ice creams we have 3*4=12 different single scoop ice creamcones.

There are 15 different single-scoop ice-cream cones that can be bought from the vendor.

To calculate the number of different single-scoop ice-cream cones that can be bought from the vendor, we need to multiply the number of ice cream flavors with the number of cone types. The vendor offers 5 flavors and 3 cone types, so the total number of different single-scoop ice-cream cones that can be bought is 5 x 3 = 15.

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We can be 95% confident that the true proportion of all voters in the state who favor approval lies between 0.438 < p < 0.505. Option B is the correct answer.

Here's a breakdown of the steps involved in constructing the confidence interval:

Calculate the sample proportion (P):

P = 408 (voters favoring approval) / 865 (total voters surveyed) = 0.4714

Determine the critical value (z):

For a 95% confidence interval, the critical value from the standard normal distribution is 1.96.

Calculate the margin of error:

Margin of error = z * √(P * (1 - P) / n)

Margin of error = 1.96 * √(0.4714 * (1 - 0.4714) / 865) = 0.0335

Construct the confidence interval:

Lower bound = P - margin of error = 0.4714 - 0.0335 = 0.4379

Upper bound = P + margin of error = 0.4714 + 0.0335 = 0.5049

Round the bounds to three decimal places:

0.438 < p < 0.505

Therefore, we can be 95% confident that the true proportion of all voters in the state who favor approval lies between 0.438 and 0.505.

Answer:

ok

Step-by-step explanation:Jawo

A market research group is interested in comparing the mean weight loss for two different popular diets. And this indicates that Diet A is better than Diet B.

An assumption or concept is given as a hypothesis for the purpose of debating it and testing if it might be true.

Given:

For diet A :

Let n₁ = 10 and X₁ = 1.5, σ₁ = 0.4

For diet B:

Let n₂ = 12 and X₂ = 1.2, σ₂ = 0.6

And α = 0.05

Hypothesis:

H₀: μ₁ = μ₂

H₁ : μ₁ > μ₂

The test:

t =   (x₁ - x₂)  / S√(1/n₁ + 1/n₂)

Where S = √(n₁σ₁² + n₂σ₂²)/(n₁+n₂-2)

S = √(1.6+4.32)/(20)

S= √(0.296)

S = 0.544

Now substituting the value of S to the test.

t =  (x₁ - x₂) / S√(1/n₁ + 1/n₂)

t = (1.5-1.2)/ (0.544)√(1/10 + 1/12)

t = (0.3) / (0.544)(0.428)

t = (0.3) / (0.2329)

t = 1.29

So the p-value is 0.212463888

Here,  α = 0.05 > 0.212463888

If P value < α  then reject H₀ at % level of significance, accept otherwise.

H₁ : μ₁ > μ₂

So, Diet A is better than Diet B

Therefore, diet A is better than diet B

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Answer:

c. y? =+4.198 x

Step-by-step explanation:

Hello!

Using the given data you need to estimate the equation of linear regression.

Dependent variable:

Y: Annual percentage change in stock price for a company.

Independent variable:

X: Annual percentage change in profits for the company.

The population regression line equation is:

[tex]Y_i= \alpha + \beta X_i + E_i[/tex]

To estimate the equation you need to find the point estimator for α and β.

The following formulas are the ones to use:

α ⇒ a= y[bar] - bX[bar]

β ⇒ b= (∑xy - [(∑x)(∑y)]/n)/(∑X²-(∑x)²/n)

As you can see you need to make several summatories before calculating the values of a and b:

n= 7

∑x= 36.30

∑x²= 667.09

∑y= 51.60

∑y²= 747.98

∑xy= 560.74

Sample mean of de dependent variable Y[bar]= ∑y/n= (51.60/7)= 7.37

Sample mean of the independent variable X[bar]= ∑x/n= (36.30/7)= 5.19

b= [tex]\frac{560.74-\frac{(36.3*51.6)}{7} }{667.09-\frac{(51.60)^2}{7} }[/tex]

b= 0.6122

a= [tex]7.37-(0.61*5.19)[/tex]

a= 4.196

The estimated regression equation is:

Y= 4.196 + 0.6122x

I hope it helps!

Answer:

Step-by-step explanation:

Given that researchers are studying the relationship between honesty, age, and self-control conducted an experiment on 160 children between the ages of 5 and 15.

a) The population is the  children between the ages of 5 and 15 in general

D. All children between the ages of 5 and 15

b) Sample is

D. 160 children between the ages of 5 and 15

c) Yes becauB. If the sample is randomly selected and representative of the entire population, then the results can be generalized to the target population. Furthermore, since this study is experimental, the findings can be used to infer causal relationships.se randomly selected and also sample size is above 30

The population of interest is all children aged 5 to 15; the study's sample consists of the 160 participating children. Results can be generalized if the sample is representative, and as an experimental study, it can establish causal relationships.

The population of interest in the study is D. All children between the ages of 5 and 15 because the researchers are attempting to understand a broader phenomenon that could be applicable to all children within this age range, not just those in the study. The sample for the study is D. 160 children between the ages of 5 and 15 because these are the actual participants that researchers have collected data from within this experiment.

Regarding the generalizability and establishment of causal relationships of the study, the correct option is B. If the sample is randomly selected and representative of the entire population, then the results can be generalized to the target population. Furthermore, since this study reports the manipulation of an independent variable (the instruction given to one group and not to the other), it qualifies as an experimental study, and its results can inform us about causal relationships.

Final answer:

The probability of getting a red card or a face card from a deck of playing cards is 8/13.

Explanation:

To determine P(E1 or E2), we need to find the probability of either event E1 (outcome is a red card) or event E2 (outcome is a face card) occurring.

To find P(E1), we can count the number of red cards in the deck, which is 26 (there are 2 red suits and each suit has 13 cards).

To find P(E2), we can count the number of face cards in the deck, which is 12 (there are 3 face cards in each suit and 4 suits).

Since E1 and E2 are not mutually exclusive (some cards are both red and face cards), we need to subtract the number of cards that satisfy both events to avoid double counting. In this case, there are 6 red face cards in the deck (2 red suits each with 3 face cards).

Therefore, P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2) = 26/52 + 12/52 - 6/52 = 32/52 = 8/13.

Final answer:

The probability of drawing either a red card or a face card from a standard deck of 52 playing cards is 8/13.

Explanation:

To calculate P(E1 or E2), where E1 is the event of drawing a red card and E2 is the event of drawing a face card from a standard deck of 52 playing cards, we can use the formula for the probability of the union of two events: P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2). Since there are 13 red cards in each of the two red suits (hearts and diamonds), P(E1) = 26/52. There are 3 face cards (jack, queen, king) in each of the four suits, so P(E2) = 12/52. However, among these face cards, 6 are red (3 in hearts and 3 in diamonds), so P(E1 and E2) = 6/52. Thus, the probability of drawing either a red card or a face card is P(E1 or E2) = 26/52 + 12/52 - 6/52.

In conclusion: P(E1 or E2) = (26 + 12 - 6) / 52

Simplifying that, we get: P(E1 or E2) = 32/52, which can be reduced to P(E1 or E2) = 8/13.

Answer:

a)For this case the best estimator for the true mean is the sample mean [tex]\hat \mu =\bar X[/tex] because:

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

And if we assume that each observation [tex]X_1 , X_2,...,X_n [/tex] follows a normal distribution [tex]X_i \sim N(\mu,\sigma)[/tex] then we have:

[tex]E(\bar X)=\frac{1}{n} n\mu =\mu[/tex]

So then yes the [tex]\bar X[/tex[ is a unbiased estimator for the true mean.

b) [tex]z_{\alpha/2}=1.96[/tex]

c) [tex] ME= 1.96 \frac{0.4}{\sqrt{50}}=0.1109[/tex]

Step-by-step explanation:

Previous concepts

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".  

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".

[tex]\bar X=4.1[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)  

[tex]\sigma=0.4[/tex] represent the population standard deviation  

n=50 represent the sample size  

Assuming the X follows a normal distribution  

[tex]X \sim N(\mu, \sigma=0.4[/tex]

The sample mean [tex]\bar X[/tex] is distributed on this way:

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

A. What is the point estimate in this scenario? Is it an unbiased estimator?

For this case the best estimator for the true mean is the sample mean [tex]\hat \mu =\bar X[/tex] because:

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

And if we assume that each observation [tex]X_1 , X_2,...,X_n [/tex] follows a normal distribution [tex]X_i \sim N(\mu,\sigma)[/tex] then we have:

[tex]E(\bar X)=\frac{1}{n} n\mu =\mu[/tex]

So then yes the [tex]\bar X[/tex[ is a unbiased estimator for the true mean.

B. What is the critical z-value for a 95% interval?

The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex]  

Using the normal standard table, excel or a calculator we see that:  

[tex]z_{\alpha/2}=1.96[/tex]

C. What is the margin of error for a 95% interval in this scenario? Show your work.

The margin of error is given by:

[tex] ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

If we replace the values that we have we got:

[tex] ME= 1.96 \frac{0.4}{\sqrt{50}}=0.1109[/tex]

The confidence interval on this case is given by:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

Since we have all the values we can replace:

[tex]4.1 - 1.96\frac{0.4}{\sqrt{50}}=3.989[/tex]  

[tex]4.1 + 1.96\frac{0.4}{\sqrt{50}}=4.211[/tex]  

So on this case the 95% confidence interval would be given by (3.989;4.211)  

Answer:

a) We would expect to see 500*0.88=440

b) [tex]z=\frac{0.9 -0.88}{\sqrt{\frac{0.88(1-0.88)}{500}}}=1.376[/tex]  

[tex]p_v =2*P(Z>1.376)=0.167[/tex]  

So the p value obtained was a very high value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion is not significant different from 0.9.

The p value is a criterion to decide if we reject or not the null hypothesis, when [tex]p_v <\alpha[/tex] we reject the null hypothesis in other case we FAIL to reject the null hypothesis. And represent the "probability of obtaining the observed results of a test, assuming that the null hypothesis is correct".  

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=450 represent the people that have the seat belt fastened

[tex]\hat p=\frac{450}{500}=0.9[/tex] estimated proportion of people that have the seat belt fastened

[tex]p_o=0.88[/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)  

Part a

We would expect to see 500*0.88=440

Part b

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion changes fro m 0.88.:  

Null hypothesis:[tex]p=0.88[/tex]  

Alternative hypothesis:[tex]p \neq 0.88[/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].

Calculate the statistic  

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

[tex]z=\frac{0.9 -0.88}{\sqrt{\frac{0.88(1-0.88)}{500}}}=1.376[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about 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.  

The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(Z>1.376)=0.167[/tex]  

So the p value obtained was a very high value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion is not significant different from 0.9.

The p value is a criterion to decide if we reject or not the null hypothesis, when [tex]p_v <\alpha[/tex] we reject the null hypothesis in other case we FAIL to reject the null hypothesis. And represent the "probability of obtaining the observed results of a test, assuming that the null hypothesis is correct".  

Answer:

a) The 90% confidence interval would be given by (230.212;233.128)

b) For this case if we analyze the confidence interval we see that not contains the value of 235. So we can't support the claim that the true mean is higher than 235 Hz at 10% of significance.

Step-by-step explanation:

Previous concepts  and data given

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".

Data: 230.66, 233.05, 232.58, 229.48, 232.58

We can calculate the mean and the deviation from these 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]

[tex]\bar X=231.67[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s=1.531 represent the sample standard deviation

n=5 represent the sample size  

Part a) Confidence interval

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

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

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=5-1=4[/tex]

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,4)".And we see that [tex]t_{\alpha/2}=2.13[/tex]

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

[tex]231.67-2.13\frac{1.531}{\sqrt{5}}=230.212[/tex]    

[tex]231.67+2.13\frac{1.531}{\sqrt{5}}=233.128[/tex]

So on this case the 90% confidence interval would be given by (230.212;233.128)

Part b)  Do the results of your calculations support the claim that mean natural frequency is 235 Hz?  

For this case if we analyze the confidence interval we see that not contains the value of 235. So we can't support the claim that the true mean is higher than 235 Hz at 10% of significance.

To determine the 90% confidence interval for mean natural frequency, calculate the mean, standard deviation, find the t-value, and compute the margin of error and CI. The claim that the mean frequency is 235 Hz is assessed by checking if this value is within the calculated CI.

To find the 90% two-sided confidence interval (CI) on mean natural frequency for the five delaminated beams, we first calculate the mean (\( \bar{x} \)) and the standard deviation (s) of the given frequencies. Then, we use the t-distribution, as the sample size is small (n=5), to find the t-value for a 90% CI, which corresponds to a significance level (alpha) of 0.10 and degrees of freedom (df) of n-1. Finally, we apply the formula for the 90% CI as follows:

(b) To assess whether the mean natural frequency is 235 Hz, we would check if this value lies within the 90% CI. If it does not, the results do not support the claim that the mean natural frequency is 235 Hz.

Answer:

a) 0.3731

b) 0.9241

c) Mean = 1.9

Standard Deviation = 0.9987

Step-by-step explanation:

We are given the following information:

P(chocolate chip cookie) = 47.5% = 0.475

Then the number of children follows a binomial distribution, where

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

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 4

a) x = 2

We have to evaluate:

[tex]P(x =2)\\= \binom{4}{2}(0.475)^2(1-0.475)^{4-2}\\= 0.3731[/tex]

b) x = 1

We have to evaluate

[tex]P(x \geq 1) = 1 - P(x = 0)\\= 1 - \binom{4}{0}(0.475)^0(1-0.475)^{4-0}\\= 1 - 0.0759\\= 0.9241[/tex]

c) Mean and standard deviation of distribution

[tex]\text{Mean} = np = 4\times 0.475 = 1.9\\\text{Standard deviation} = \sqrt{np(1-p)} = \sqrt{4\times 0.475(1-0.475)} = 0.9987[/tex]

Answer:

Option e -  0.5597

Step-by-step explanation:

Given : Suppose that a random sample of size 49 is to be selected from a population with mean 49 and standard deviation 6.

To find :  What is the approximate probability that will be more than 0.5 away from the population mean?

Solution :  

Applying z-score formula,

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

Here, [tex]\mu=49,\sigma=6, n=49[/tex]

As, probability that will be more than 0.5 away from the population mean

i.e. x=49.5,

[tex]z=\frac{49.5-49}{\frac{6}{\sqrt{49}}}[/tex]

[tex]z=\frac{0.5}{0.857}[/tex]

[tex]z=0.58[/tex]

In z-score table the value of z at 0.58 is 0.7190.

x=48.5,

[tex]z=\frac{48.5-49}{\frac{6}{\sqrt{49}}}[/tex]

[tex]z=\frac{-0.5}{0.857}[/tex]

[tex]z=-0.58[/tex]

In z-score table the value of z at -0.58 is 0.2810.

Now, probability that will be more than 0.5 away from the population mean is given by,

[tex]P=P(X>49.5)+P(X<48.5)[/tex]

[tex]P=1-P(X<49.5)+P(X<48.5)[/tex]

[tex]P=1-0.7190+0.2810[/tex]

[tex]P=0.562[/tex]

Therefore, option e is correct.

Answer: The maximum error = $105.76.

Step-by-step explanation:

Formula to find the maximum error:

[tex]E= z^*\dfrac{\sigma}{\sqrt{n}}[/tex]

, where n= sample size.

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

z*= Critical value(two-tailed).

As per given  , we have

[tex]\overline{x}=1438[/tex]

n= 35

[tex]\sigma=269[/tex]

For 98% confidence  , the significance level = [tex]1-0.98=0.02[/tex]

By z-table , the critical value (two -tailed) =[tex]z^* = z_{\alpha/2}=z_{0.01}=2.326[/tex]

Now , the maximum error = [tex]E= (2.326)\dfrac{269}{\sqrt{35}}[/tex]

[tex]E= (2.326)\dfrac{269}{5.9160797831}[/tex]

[tex]E= (2.326)\times45.4692989044=105.761589252\pprox105.76[/tex]

Hence, With 98% confidence level , the maximum error = $105.76.

To estimate the difference between the proportion of elementary school teachers and high school teachers who are satisfied, we need to calculate a 95% confidence interval. The CI is (0.0181, 0.1589).

To estimate the difference between the proportion of all elementary school teachers who are satisfied and all high school teachers who are satisfied, we need to calculate a 95% confidence interval (CI). First, we calculate the sample proportions for both groups. The sample proportion of elementary school teachers who are very satisfied is 226/397 = 0.5698, and the sample proportion of high school teachers who are very satisfied is 129/268 = 0.4813. Next, we calculate the standard error using the formula: √(phat1*(1-phat1)/n1 + phat2*(1-phat2)/n2), where phat1 and phat2 are the sample proportions and n1 and n2 are the sample sizes. Substituting the values, we get √((0.5698*(1-0.5698)/397 + 0.4813*(1-0.4813)/268)). This gives us a standard error of 0.0359.

Next, we need to calculate the margin of error (ME) by multiplying the standard error by 1.96 (which corresponds to a 95% confidence level). ME = 1.96 * 0.0359 = 0.0704.

Finally, we can calculate the confidence interval by subtracting the margin of error from the difference in sample proportions and adding the margin of error to the difference in sample proportions. The difference in sample proportions is 0.5698 - 0.4813 = 0.0885. So, the 95% confidence interval is (0.0885 - 0.0704, 0.0885 + 0.0704), which simplifies to (0.0181, 0.1589).

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Answer:

We are confident at 99% that the difference between the two proportions is between [tex]0.380 \leq p_{Republicans} -p_{Democrats} \leq 0.420[/tex]

Step-by-step explanation:

Part a

Data given and notation  

[tex]X_{D}=3266[/tex] represent the number people registered as Democrats

[tex]X_{R}=2137[/tex] represent the number of people registered as Republicans

[tex]n=7525[/tex] sampleselcted

[tex]\hat p_{D}=\frac{3266}{7525}=0.434[/tex] represent the proportion of people registered as Democrats

[tex]\hat p_{R}=\frac{2137}{7525}=0.284[/tex] represent the proportion of people registered as Republicans

The standard error is given by this formula:

[tex]SE=\sqrt{\frac{\hat p_D (1-\hat p_D)}{n_{D}}+\frac{\hat p_R (1-\hat p_R)}{n_{R}}}[/tex]

And the standard error estimated given by the problem is 0.008

Part b

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]p_A[/tex] represent the real population proportion of Democrats that approve of the way the California Legislature is handling its job  

[tex]\hat p_A =\frac{1894}{3266}=0.580[/tex] represent the estimated proportion of Democrats that approve of the way the California Legislature is handling its job  

[tex]n_A=3266[/tex] is the sample size for Democrats

[tex]p_B[/tex] represent the real population proportion of Republicans that approve of the way the California Legislature is handling its job  

[tex]\hat p_B =\frac{385}{2137}=0.180[/tex] represent the estimated proportion of Republicans that approve of the way the California Legislature is handling its job

[tex]n_B=2137[/tex] is the sample for Republicans

[tex]z[/tex] represent the critical value for the margin of error  

The population proportion have the following distribution  

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

The confidence interval for the difference of two proportions would be given by this formula  

[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]  

For the 90% confidence interval the value of [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.64[/tex]  

And replacing into the confidence interval formula we got:  

[tex](0.580-0.180) - 1.64 \sqrt{\frac{0.580(1-0.580)}{3266} +\frac{0.180(1-0.180)}{2137}}=0.380[/tex]  

[tex](0.580-0.180) + 1.64 \sqrt{\frac{0.580(1-0.580)}{3266} +\frac{0.180(1-0.180)}{2137}}=0.420[/tex]  

And the 99% confidence interval would be given (0.380;0.420).  

We are confident at 99% that the difference between the two proportions is between [tex]0.380 \leq p_{Republicans} -p_{Democrats} \leq 0.420[/tex]

Answer:

Part (A) [tex]H_0:\mu=2.5[/tex] and [tex]H_a:\mu\neq2.5[/tex]

Part (B) [tex]H_0:\mu=16.35[/tex] and [tex]H_a:\mu<16.35[/tex]

Step-by-step explanation:

Consider the provided information.

Null hypothesis refers the population parameter is equal to the claimed value.

If there is no statistical significance in the test then it is know as the null which is denoted by [tex]H_0[/tex], otherwise it is known as alternative hypothesis denoted by [tex]H_a[/tex].

Part (A): The diameter of a spindle in a small motor is supposed to be 2.5 millimeters (mm) if the spindle is either too small or too large, the motor will not work properly.

Thus, the required null and alternative hypothesis are:

[tex]H_0:\mu=2.5[/tex] and [tex]H_a:\mu\neq2.5[/tex]

Part (B) The nationwide average price of a certain brand of laundry detergent is $16.35 and Harry thinks that prices in Caldwell are lower than the rest of the country.

Thus, the required null and alternative hypothesis are:

[tex]H_0:\mu=16.35[/tex] and [tex]H_a:\mu<16.35[/tex]

The null and alternative hypotheses for the given situations are as follows: a) H0: μ = 2.5 mm, Ha: μ ≠ 2.5 mm. b) H0: μ = $16.35, Ha: μ < $16.35.

a) H0: The mean diameter of the spindles is 2.5 mm (μ = 2.5 mm).

Ha: The mean diameter of the spindles is not equal to 2.5 mm (μ ≠ 2.5 mm).

b) H0: The mean price of the laundry detergent in Caldwell is $16.35 (μ = $16.35).

Ha: The mean price of the laundry detergent in Caldwell is less than $16.35 (μ < $16.35).

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Answer:

The cost of producing those 50 blu-rays is given by, 1337.5 unit and it is 26.75 unit for each blu-ray.

Step-by-step explanation:

The cost of producing x blu-rays is given by,

B(x) = 1250 + 1.75x unit -----------------------------(1)

So, cost of producing 50 blu-rays is given by,

B(50) = [tex]1250 + 50 \times (1.75)[/tex] unit

         = (1250 + 87.5) unit

         = 1337.5 unit

So, for each blu-ray, it is,

[tex]\frac {1337.5}{50}[/tex] unit

= 26.75 unit

Answer:

At least 81 observations should be made to be 95% confident that the observed data is within a tolerance of ±10 percent of the population data.

Step-by-step explanation:

The following equation is used to compute the minimum sample size required to estimate the population proportion  within the required margin of error:

n≥ p×(1-p) × [tex](\frac{z}{ME} )^2[/tex] where

Then, n≥ 0.30×0.70 × [tex](\frac{1.96}{0.10} )^2[/tex] ≈ 80.67

At least 81 observations should be made to be 95% confident that the observed data is within a tolerance of ±10 percent of the population data.

Random observations should include different employees, and sampling time should also be random.

The probability of drawing a 2 and then drawing another 2 in the board game is 1/72.

To calculate the probability of drawing a 2 and then drawing another 2 in the board game, we first need to understand the concept of probability. Probability represents the likelihood of an event occurring, ranging from 0 (impossible) to 1 (certain). In this case, we're interested in the probability of two specific events happening sequentially: first drawing a 2, and then drawing another 2.

In the board game scenario, there are 9 numbers in total. So, the probability of drawing a 2 on the first draw is 1 out of 9 since there's one 2 among the nine numbers. After drawing the first number and not replacing it, there are now 8 numbers left. If the first draw was a 2, then there's one less 2 available among the remaining 8 numbers. So, the probability of drawing another 2 on the second draw is 1 out of 8.

To find the overall probability of both events happening in sequence (drawing a 2 and then drawing another 2), we multiply the probabilities of each event. Thus, the probability of drawing a 2 and then drawing another 2 is (1/9) * (1/8) = 1/72.

Therefore, the probability of drawing a 2 and then drawing another 2 in the board game is indeed 1/72. This means that out of all the possible sequences of two draws without replacement, only 1 out of 72 sequences will result in drawing a 2 followed by another 2.

Complete question :- In a board game, students draw a number do not replace it and then draw a second number

Determine the probability of each event occurring,

1

6

6

9

21

62

What is the probability of drawing a 2, then drawing another 2?

Answer:

n=601

Step-by-step explanation:

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{\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 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

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]

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 0.04[/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)  

Since we don't have a prior estimation for the proportion we can use 0.5 as estimation. And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25[/tex]  

And rounded up we have that n=601

Answer:

91.3

Step-by-step explanation:

To find the temperature of the tea after 5 minutes, use the given function and the information given in the question. By solving the function for the constant C, you can then find the temperature of the tea after 5 minutes.

To find the temperature of the tea after 5 minutes, we need to use the given function and the information given in the question.

First, we can use the given function f(t)=Ce(−kt)+70 to find the value of C. Since we know that after 3 minutes the temperature of the tea is 100 degrees, we can substitute t=3 and T=100 into the function to get:

100 = Ce(−3k)+70

Next, we can use the same function to find the temperature of the tea after 5 minutes. Substituting t=5 into the function and using the value of C that we found earlier, we get:

T = Ce(−5k)+70

Rounding to the nearest tenth, the temperature of the tea after 5 minutes will be approximately 89.4 degrees Fahrenheit.

Answer:

We conclude that the population mean is not equal to 17.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 17

Sample mean, [tex]\bar{x}[/tex] = 14.12

Sample size, n = 40

Alpha, α = 0.05

Population standard deviation, σ = 4

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 17\\H_A: \mu \neq 17[/tex]

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

a) Formula:

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

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{14.12 - 17}{\frac{4}{\sqrt{40}} } = -4.5536[/tex]

b) P-value can be calculated from the standard z-table.

P-value = 0.0000

c) Since the p-value is less than the significance level, we reject the null hypothesis and accept the alternate hypothesis. Thus, the population mean is not equal to 17

d) Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]

e) Rejection Rule:

We reject the null hypothesis if it is less than lower critical value and greater than the upper critical value

If the z-statistic lies outside the acceptance region which is from -1.96 to +1.96, we reject the null hypothesis.

f) Since the calculated z-stat lies outside the acceptance region, we reject the null hypothesis and accept the alternate hypothesis. Thus, the population mean is not equal to 17.

The test statistic is -1.78 and the p-value is 0.0761, indicating that we fail to reject the null hypothesis. Therefore, it cannot be concluded that the population mean is not equal to 17.

The test statistic can be calculated using the formula:

test statistic = (sample mean - population mean) / (population standard deviation / sqrt(sample size))

Plugging in the given values, we get:

test statistic = (14.12 - 17) / (4 / sqrt(40))

Calculating this gives us a test statistic value of -1.78.

The p-value can be calculated using the test statistic. We need to find the probability that a test statistic at least as extreme as -1.78 would occur assuming the null hypothesis is true. Using a standard normal distribution table or software, we find the p-value to be approximately 0.0761.

Since the p-value is greater than the significance level (alpha = 0.05), we fail to reject the null hypothesis. Therefore, we can conclude that there is not enough evidence to suggest that the population mean is not equal to 17.

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