A sorority has 38 members, 28 of whom are full members and 10 are pledges. Two persons are selected at random from the membership list of the sorority. Find the requested probabilities. (Enter the probabilities as fractions.)

To find the probability that a 14 square foot metal sheet has at least 10 defects, we can use the Poisson distribution formula.

To find the probability that a 14 square foot metal sheet has at least 10 defects, we can use the Poisson distribution formula.

The Poisson distribution formula is given by: P(x) = (e^-λ * λ^x) / x!

Where:

To find P(x ≥ 10), we need to calculate the probabilities for x = 10, 11, 12, 13, 14 and sum them up.

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Answer: C.$10,000

Step-by-step explanation:

Assuming the interest was compounded annually, then we would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

A = $31066

r = 6.5% = 6.5/100 = 0.065

n = 1 because it was compounded once in a year.

t = 18 years

Therefore,

31066 = P(1+ 0.065/1)^1 × 18

31066 = P(1.065)^18

31066 = 3.12P

P = 31066/3.12

P = $9957.1

Approximately $10000 to the nearest dollar

Answer:

(a) The probability that exactly 5 of the 6 consumers recognize the brand name is 0.0369.

(b) The probability that all of the selected consumers recognize the brand name is 0.0041.

(c) The probability that at least 5 of the selected consumers recognize the brand name is 0.041.

(d) The events of 5 customers recognizing the brand name is unusual.

Step-by-step explanation:

Let X = number of consumer's who recognize the brand.

The probability of the random variable X is, P (X) = p = 0.40.

A random sample of size, n = 6 consumers are selected.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...[/tex]

(a)

Compute the value of P (X = 5) as follows:

[tex]P(X=5)={6\choose 5}0.40^{5}(1-0.40)^{6-5}\\=6\times0.01024\times0.60\\=0.036864\\\approx0.0369[/tex]

Thus, the probability that exactly 5 of the 6 consumers recognize the brand name is 0.0369.

(b)

Compute the value of P (X = 6) as follows:

[tex]P(X=6)={6\choose 6}0.40^{6}(1-0.40)^{6-6}\\=1\times 0.004096\times1\\=0.004096\\\approx0.0041[/tex]

Thus, the probability that all of the selected consumers recognize the brand name is 0.0041.

(c)

Compute the value of P (X ≥ 5) as follows:

P (X ≥ 5) = P (X = 5) + P (X = 6)

              = 0.0369 + 0.0041

              = 0.041

Thus, the probability that at least 5 of the selected consumers recognize the brand name is 0.041.

(d)

An event is considered unusual if the probability of its occurrence is less than 0.05.

The probability of 5 customers recognizing the brand name is 0.0369.

This probability value is less than 0.05.

Thus, the events of 5 customers recognizing the brand name is unusual.

Answer:

Expected Number=71

Variance =4959

Standard Deviation= 70.42

Step-by-step explanation:

Expected Number = E (x) = np

Here p = 0.71 and q= 1- 0.71= 0.29 n= 100

Expected Number = E (x) = np = 0.71*100= 71

b) The Variance and Standard Deviation for the number of these tweets with no reaction

Suppose the number of tweets are hundred then the number of tweets with no reaction would be 71 .

Variance = E(x²) - [E(x)]² =  (100)²- (71)²= 10000- 5041= 4959

Standard Deviation= √4959= 70.42

Answer:

The forklift truck can raise maximum of 43 boxes.

Step-by-step explanation:

We are given the following in the question:

Maximum limit of  forklift truck =

1750 kilograms

Let x be the number of 40 kilogram boxes that this forklift truck can raise.

Thus, we can write the following inequality that describes the maximum number of boxes raised.

[tex]40x \leq 1750\\\\x \leq \dfrac{1750}{40}\\\\x \leq 43.75\\Max(x) = 43[/tex]

Thus, the forklift truck can raise maximum of 43 boxes.

Answer:

a) [tex]\frac{dm}{dt} = -k\cdot m[/tex], b) [tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex], c) [tex]m(t) = 10\cdot e^{-\frac{t}{2438.155} }[/tex], d) [tex]m(300) \approx 8.842\,g[/tex]

Step-by-step explanation:

a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:

[tex]\frac{dm}{dt} = -k\cdot m[/tex]

b) The general solution is found after separating variables and integrating each sides:

[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]

Where [tex]\tau[/tex] is the time constant and [tex]k = \frac{1}{\tau}[/tex]

c) The time constant is:

[tex]\tau = \frac{1690\,yr}{\ln 2}[/tex]

[tex]\tau = 2438.155\,yr[/tex]

The particular solution of the differential equation is:

[tex]m(t) = 10\cdot e^{-\frac{t}{2438.155} }[/tex]

d) The amount of radium after 300 years is:

[tex]m(300) \approx 8.842\,g[/tex]

Answer:

a) -dm/m = k*dt

b) m=m₀*e^(-k*t)

c) m=10 g*e^(-4.1* 10⁻⁴ years⁻¹*t)

d) m= 8.842 gr

( same results obtained previously by Xero099)

Step-by-step explanation:

Since the amount of radium is proportional to its quantity . Denoting as m the mass of Radium , and t as time. we have:

a) -dm/dt = k*m , where k is the proportionality constant

b) solving the differential equation;

-dm/m = k*dt

∫dm/m = -∫k*dt

ln m = -k*t + C

for t=0 , we have m=m₀ (initial mass)

then

ln m₀ = 0 + C → C=ln m₀

therefore

ln (m/m₀) = -k*t

m=m₀*e^(-k*t)

c) for t= 1690 years , the quantity of radium is half , then m=m₀/2 , thus

ln (m₀/2/m₀) = -k*t

ln (1/2) =   -k*t

(ln 2 )/t = k

k =  ln 2  / 1690 years = 4.1* 10⁻⁴ years⁻¹

then for m₀= 10 g

m=10 g*e^(-4.1* 10⁻⁴ years⁻¹*t)

d) at t= 300 years

m=10 g*e^(-4.1* 10⁻⁴ years⁻¹*300 years) = 8.842 gr

then the mass is m= 8.842 gr

Answer:

10

Step-by-step explanation:

I assume it's a rectangle, therefore:

P = 50

a = 15 ft

b = ?

P = 2a + 2b

50 = 2 * 15 + 2b

50 = 30 + 2b

2b = 50 - 30

2b = 20

b = 10

A box with a base width four times its length, built from 350 m^2 material, will have maximum volume when its dimensions are: length = sqrt(35) m, width = 4sqrt(35) m, and height = 350/(12*sqrt(35)) m.

We know that the box's area, which is the sum of all the faces, is given by 2lw+2lh+2wh = 350 square meters, where l is the length, w is the width, and h is the height. Since the base width is four times the base length, we can say w=4l. Substituting this into the equation, we solve for h, h = (350 - 2l^2) / (6l).

The volume of the box is given by V = lwh. Substituting the value of h from the previous equation, we get V = l * 4l * (350 - 2l^2) / (6l) = 4/3 *l^2 *(175 - l^2). To maximize the volume, we need to find the value of l when the derivative of the volume equation is 0. Differentiating, setting the derivative equal to 0, and solving for l, we find l = sqrt(35), and substituting this into the earlier equation, we find h = 350/(12*sqrt(35)). The dimensions of the box that will maximize volume are, length = sqrt(35) m, width = 4sqrt(35) m, and height = 350/(12*sqrt(35)) m.

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Answer:

True, this is the meaning of the z-score. Z = 0.65 means that the raw score is 0.65 standard deviations above the mean.

Step-by-step explanation:

The Z-score measures how many standard deviations a raw score is from the mean.

For example, a z-score of -2 means that the raw score is 2 standard deviations below the mean.

Another example, a z-score of 2 means that the raw score is 2 standard deviations above the mean.

If z= 0.65, then the raw score is 0.65 standard deviations above the mean.

True, this is the meaning of the z-score. Z = 0.65 means that the raw score is 0.65 standard deviations above the mean.

Answer:

Step-by-step explanation:

Given a rectangular tank with dimension (10ft by 20ft by 16ft)

Then the volume of the tank is

Volume =length × breadth ×height

Volume=10×20×16=3200ft³

V=3200ft³

Then,

∆F= weight density × volume

∆F= 62.4×3200

∆F= 199,680 lb

Then,

Let the 0-point on the x-axis be at the bottom of the tank, so the level of the water ranges from x = 0 to x = 16ft. (It would just as well to let 0 be ground level and let x range from x = −26ft to x − 0.) Then a slice of water at level x is raised (26− x)ft

Then ∆x=(26-x)ft

Work is given as

W= -∫F∆xdx. From 0 to 16

W= -∫199,680(26-x)dx From 0 to 16

W=-199,680∫26-x dx From 0 to 16

W=-199,680 (26x-x²/2). From 0 to 16

W=-199,680(26×16-0.5×16² -0-0)

W=-199,680(288)

W=-57,507,840J

The work done to empty the tank is 57,507,840J

Answer:

2 X 10∧6 ft.lb

Step-by-step explanation:

Volume of rectangular tank =  Base area X Height = (10 X 20 X 16)ft³ = 3200ft³

Mass of tank filled with water = 3200ft³ X 62.4 lb/ft³ = 199680lb or 90706.37kg

Hence, work done due to gravity,Δg = m * g * h, where, m = mass of tank, g = gravity = 9.8m/s² and h = height = 10ft or 3.05 meter(m)

∴ Δg = 90706.37 * 9.8 * 3.05 =  2.711 X10∧6 J or 2 X 10∧6 ft.lb

Answer:

The lowest 10 percent of the citizens would need at least 52.8 minutes to complete the form.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 40, \sigma = 10[/tex]

The slowest 10 percent of the citizens would need at least how many minutes to complete the form

This is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.

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

[tex]1.28 = \frac{X - 40}{10}[/tex]

[tex]X - 40 = 1.28*10[/tex]

[tex]X = 52.8[/tex]

The lowest 10 percent of the citizens would need at least 52.8 minutes to complete the form.

Final answer:

To find the time needed for the lowest 10 percent of citizens to complete the 2010 U.S. Census 'long' form, we use the z-score formula to calculate the value. The lowest 10 percent of citizens would need at least 26.8 minutes to complete the form.

Explanation:

To find how many minutes would be needed for the lowest 10 percent of citizens to complete the 2010 U.S. Census 'long' form, we need to use the z-score formula.

First, we calculate the z-score using the formula: z = (x - mean) / standard deviation, where x is the value we want to find the z-score for, mean is the mean of the distribution, and the standard deviation is the standard deviation of the distribution.

Next, we use a z-table or a calculator to find the z-score that corresponds to a cumulative probability of 0.1. The z-table gives us a z-score of -1.28. We can rearrange the z-score formula to solve for x: x = (z * standard deviation) + mean. Plugging in the values, we get: x = (-1.28 * 10) + 40 = 26.8. Therefore, the lowest 10 percent of citizens would need at least 26.8 minutes to complete the form.

Answer:

The proportion of the women that have blood pressures lower than 68 is 0.1038 or 10.38%.

Step-by-step explanation:

Given:

Blood pressure required (x) = 68

Mean blood pressure (μ) = 80.7

Standard deviation (σ) = 10.1

The distribution is normally distributed.

So, first, we will find the z-score of the distribution using the formula:

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

Plug in the values and solve for 'z'. This gives,

[tex]z=\frac{68-80.7}{10.1}=-1.26[/tex]

So, the z-score of the distribution is -1.26.

Now, we need the probability [tex]P(x\leq 68 )=P(z\leq -1.26)[/tex].

From the normal distribution table for z-score equal to -1.26, the value of the probability is 0.1038. This is the area to the left of the curve or less than z-score value which is what we need.

Therefore, the proportion of the women that have blood pressures lower than 68 is 0.1038 or 10.38%.

Answer:

The 95% confidence interval for population mean is (72, 80).

Step-by-step explanation:

The (1 - α) % confidence interval for population mean (μ) is:

[tex]CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}[/tex]

The Margin of error for this confidence interval is:

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

The confidence interval for μ can also be written as:

[tex]CI=\bar x\pm MOE[/tex]

Given:

[tex]\bar x=76\\MOE=4[/tex]

Compute the 95% confidence interval for population mean as follows:

[tex]CI=\bar x\pm MOE\\=76\pm4\\=(72, 80)[/tex]

Thus, the 95% confidence interval for population mean is (72, 80).

Answer:

[tex] \bar X \pm ME[/tex]

And if we find the limits we got:

[tex] 76-4 = 72[/tex]

[tex] 76 + 4 = 80[/tex]

So then the 95% confidence interval would be (72.0,80.0)

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 = 76[/tex] represent the sample mean for the sample  

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

s represent the sample standard deviation  

n=49 represent the original sample size  

Confidence =95% or 0.95

ME=4 represent the margin of error.

Solution to the problem

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)  

The margin of error is defined as:

[tex] ME= t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]

The formula for the confidence interval is equivalent to:

[tex] \bar X \pm ME[/tex]

And if we find the limits we got:

[tex] 76-4 = 72[/tex]

[tex] 76 + 4 = 80[/tex]

So then the 95% confidence interval would be (72.0,80.0)

Answer:

(a) Initial concentration of salt = 0.07 kg/L

(b) [tex]\therefore Y'(t)= 0.28 -\frac{y}{125}[/tex]

(c) [tex]Y(t)=35 +35e^{-\frac{1}{125} t}[/tex]

(d)Therefore the amount of salt after 5 hours is  =38.18 kg

(e) The concentration of salt in the solution in the tank as time approaches infinity is = 0.035 kg/L

Step-by-step explanation:

Given that,

A tank contains 70 kg of salt and 1000 L of water.

(a)

[tex]\textrm{Concentration of salt }=\frac{\textrm{mass of salt}}{\textrm{volume of water}}[/tex]

                                 [tex]=\frac{70}{1000} kg/L[/tex]

                                =0.07 kg/L

(b)

Let Y(t) be the amount of salt at any instant time t.

Therefore

[tex]Y'(t)= Y_{in}-Y_{out}[/tex]

[tex]Y_{in}=\textrm{concentration of salt} \times \textrm{rate of enter}[/tex]

     =(8×0.035) kg/min

     =0.28 kg/min

Since the rate of water in and out are same , the amount of solution remain constant.

[tex]Y_{out}= (\frac{y}{1000}\times 8) kg/min[/tex]

       [tex]=\frac{y}{125} kg/L[/tex]

[tex]\therefore Y'(t)= 0.28 -\frac{y}{125}[/tex]

(c)

The above equation can be rewrite as

[tex]Y'(t) +\frac{y}{125} =0.28[/tex]

The coefficient of y is p(t) [tex]=\frac{1}{125}[/tex]

The integrating factor of the D.E is[tex]=e^{\int p(t) dt=[/tex] [tex]e^{\int \frac{1}{125} dt[/tex]  [tex]=e^{\frac{1}{125} t[/tex]

Multiplying the integrating factor of both sides of D.E

[tex]e^{\frac{1}{125} t} \ \frac{dY}{dt} +e^{\frac{1}{125} t} .\frac{1}{125} Y=0.28 \ e^{\frac{1}{125} t}[/tex]

Integrating both sides

[tex]\int e^{\frac{1}{125} t} \ \ dY+\int e^{\frac{1}{125} t} .\frac{1}{125} Y \ dt=\int0.28 \ e^{\frac{1}{125} t}\ dt[/tex]

[tex]\Rightarrow e^{\frac{1}{125} t} \ Y= \frac{0.28e^{\frac{1}{125}t} }{\frac{1}{125}} +C[/tex]

[tex]\Rightarrow Y=35+Ce^{-\frac{1}{125}t}[/tex]

At initial when t= 0, y =70

Therefore

[tex]70=35+Ce^0[/tex]

⇒C= 70-35

⇒C=35

Therefore

[tex]Y(t)=35 +35e^{-\frac{1}{125} t}[/tex]

(d)

When t= 5 hour = 300 min

To find the amount of salt after 5 hour , we need to put the value of t in the general solution of D.E

Therefore the amount of salt after 5 hours is =  [tex]Y(300)=35+35e^{-\frac{300}{125} }[/tex]

                                                                                          = 38.18 Kg

(e)

When t=∞

[tex]Y(\infty )= 35 +e^{-\infty}[/tex]

         = 35 Kg   [tex][e^{-\infty}=0][/tex]

Since the amount of water is remain same i.e 1000 L

Therefore the concentration of salt is [tex]=\frac{35}{1000}kg/L[/tex]

                                                                 =0.035 kg/L

This problem involves using differential equations to model the concentration of salt in a tank over time. The initial concentration is determined, an initial value problem is set up, the initial value problem is solved, the amount of salt after a specific time is found, and the approaching concentration as time goes to infinity is identified.

(a) The initial concentration of the solution can be found by dividing the amount of salt by the volume of water. So, 70 kg / 1000 L = 0.07 kg/L.

(b) Let y be the amount of salt in the tank at time t. Then, dy/dt = 0.035 kg/L * 8 L/min - 0.07 kg/L * 8 L/min since salt is being added at a rate of 0.035 kg/L *8 L/min and leaving at a concentration of 0.07 kg/L * 8 L/min. Thus, dy/dt = 0.28 kg/min - 0.56 kg/min = -0.28 kg/min. The initial condition is y(0) = 70 kg.

(c) To solve this differential equation, we use the method of separation of variables. We get y(t) = 70 - 0.28t in kg.

(d) After 5 hours, or 300 minutes, the amount of salt in the tank is y(300) = 70 - 0.28*300 = 16 kg.

(e) As time approaches infinity, the concentration of salt will approach the incoming salt concentration, which is 0.035 kg/L since the solution is mixed and leaves at that rate.

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Answer:

Therefore Rotten Tomato ratings explain about 75% of the variation in Metascore ratings is the correct answer here.

Step-by-step explanation:

The R2 value for regression or the coefficient of determination represents the proportion of variation in dependent variable that is explained by regression ( the rest of it is residual variation )

The given question is about the analysis of a linear regression model to examine the relationship between Rotten Tomatoes ratings and Metascore ratings for popular movies.

The given question pertains to linear regression analysis in statistics. It involves examining the relationship between Rotten Tomatoes ratings and Metascore ratings for a sample of 75 popular movies. The r-squared value of 0.75 indicates that approximately 75% of the variation in Metascore ratings can be explained by the linear relationship with Rotten Tomatoes ratings. Additionally, for every one-point increase in Rotten Tomatoes ratings, the regression model predicts a 0.75-point increase in Metascore ratings.

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Answer:

Step-by-step explanation:

Hello!

X: amount spent in a supermarket impulse buying in a 10 min unplanned shopping interval by one customer.

It is known that the mean of this variable is μ= $48 and its standard deviation is δ=$7

a.

The Central limit theorem states that if there is a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

X[bar]≈N(μ;σ²/n)

The mean of the sampling distribution is μ= $48

The standard deviation of the sampling distribution is σ/√n= $7/√60= $0.90

It is not necessary to make any assumption about the distribution of X since n=60 is considered large enough, you can directly approximate the sampling distribution to normal regardless of the distribution of X.

b.

To calculate this probability you have to use the approximation of the sampling distribution:

Z= (X[bar]-μ)/(σ/√n)≈N(0;1)

μ= $48

σ/√n= $0.90

P(46≤X[bar]≤50)= P(X[bar]≤50) - P(X[bar]≤46)

P(Z≤(50-48)/0.90) - P(Z≤(46-48)/0.90)

P(Z≤2.22) - P(Z≤-2.22)= 0.987 - 0.013= 0.974

c.

If we assume that X has an approximately normal distribution, then you will use it's a mean and standard deviation to reach the asked probability.

Z= (X-μ)/δ≈N(0;1)

μ= $48

δ= $7

P(46≤X≤50)= P(X≤50) - P(X≤46)

P(Z≤(50-48)/7) - P(Z≤(46-48)/7)

P(Z≤0.29) - P(Z≤-0.29)= 0.61409 - 0.38591= 0.22818≅ 0.2282

I hope it helps!

The Central Limit Theorem ensures that the average amount spent by a sample of 60 customers due to impulse buying is normally distributed with a mean of $48 and a standard error of approximately $0.90.

Regarding part (a) of the question, by applying the Central Limit Theorem (CLT), we can state that for the sample of n = 60 customers, the sampling distribution of the sample mean x will be approximately normal due to the large sample size, even if the distribution of x is not normal. The mean of the distribution μx remains the same at $48, and the standard deviation (often referred to as the standard error, σx) can be calculated by dividing the population standard deviation by the square root of the sample size (n), which is σx = $7/√60 ≈ $0.90. Thus, the correct statement is: 'The sampling distribution of x is approximately normal with mean μx = $48 and standard error σx = $0.90.'

Answer:

Option b. should not be rejected

Step-by-step explanation:

We are given that the contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces.

We have to test whether the variance of the population is significantly more than 0.003, i.e.;

  Null Hypothesis, [tex]H_0[/tex] : [tex]\sigma[/tex] = [tex]\sqrt{0.003}[/tex]

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\sigma > \sqrt{0.003}[/tex]

The test statistics used here for testing variance is;

          T.S. = [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2}__n_-_1[/tex]

where, s = sample standard deviation = 0.06

           n = sample size = 26 cans

So, Test statistics = [tex]\frac{(26-1)0.06^{2} }{0.003 }[/tex] ~ [tex]\chi^{2}__2_5[/tex]

                            = 30

So, at 5% level of significance chi square table gives critical value of 37.65 at 25 degree of freedom. Since our test statistics is less than the critical so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that null hypothesis should not be rejected and variance of population is 0.003.

Answer:

In this case, the 30% represents the proportion of the sample. It is a statistic that can be used to estimate a parameter of the population.

Step-by-step explanation:

In this case, the 30% represents the proportion of this specific sample (survey taken by the magazine).

It is a statistic that can be used to estimate a parameter of the population. In this case, it may be used to estimate the true proportion of "people in New York who believe that the Yankees will miss the playoffs this year".

If a new sample is taken, a new statistic will be calculated that may or may not be equal to 30%.

The 30% in the article represents the proportion of people in New York City who believe that the Yankees will miss the playoffs this year. The article provides a confidence interval of 25% to 35% for this proportion, indicating a 92% confidence level.

The 30% in the article represents the proportion of people in New York City who believe that the Yankees will miss the playoffs this year. The article states that the survey found that 30% of the population had this belief. Additionally, the article provides a confidence interval of 25% to 35% for this proportion, implying that there is a 92% confidence that the true proportion lies within this range.

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Answer:

There is enough evidence to support the claim the mean nicotine content is actually higher than advertised.          

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 1.5

Sample mean, [tex]\bar{x}[/tex] = 1.53

Sample size, n = 100

Alpha, α = 0.05

Population standard deviation, σ = 0.17

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 1.5\\H_A: \mu > 1.5[/tex]

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

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{1.53 - 1.5}{\frac{0.17}{\sqrt{100}} } = 1.7647[/tex]

Now, we calculate the p-value from the standard normal table.

P-value = 0.038807

Since the p-value is less than the significance level we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim the mean nicotine content is actually higher than advertised.

Answer:

Yes, the mean nicotine content is actually higher than advertised.

Step-by-step explanation:

We are given that the nicotine content in cigarettes of a certain brand is known to be right-skewed with mean (in milligrams) μ and a known standard deviation σ = 0.17.

Also, the brand advertises that the mean nicotine content of its cigarettes, [tex]\mu[/tex] is 1.5, but measurements on a random sample of 100 cigarettes of this brand give a sample mean of x bar = 1.53.

Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 1.5

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 1.5

The test statistics used here will be;

     T.S. = [tex]\frac{xbar - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)

Here, n = sample size = 100

So, test statistics = [tex]\frac{1.53 - 1.5}{\frac{0.17}{\sqrt{100} } }[/tex] = 1.765

Now, at 0.05 level of significance, the standard z table gives critical value of 1.6449. Since our test statistics is more than the critical value as 1.765 > 1.6449 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region.

Therefore, we conclude that the mean nicotine content is actually higher than advertised.

Answer:

The probability that a student has passed this class with at least a C is 0.73

Step-by-step explanation:

A = 4 B = 3 C = 2 D = 1 E = 0

P(0) = 0.1

P(1) = 0.17

P(2) = 0.21

P(3) = 0.32

P(4) = 0.2

Probability that student passed this class with at least a C mean the score must be more than C

P(X\geq 2) = P(2) + P(3) + P(4)\\

P(X\geq 2) = 0.21 + 0.32 + 0.2\\

P(X\geq 2) = 0.73

Final answer:

The probability that a student has passed an introductory statistics class with at least a C grade is found by summing the probabilities for grades C and above, resulting in a probability of 0.73 or 73%.

Explanation:

The question asks for the probability that a student has passed an introductory statistics class with at least a C grade, which is represented as a grade of at least 2 when A = 4, B = 3, and so forth. The distribution of grades and their probabilities are given as follows: X represents the grade, and P(X) represents the probability of obtaining that grade. Therefore, to find the probability of passing the class with at least a C, we need to sum the probabilities of getting a grade of 2 or higher.

P(X=2) = 0.21

P(X=3) = 0.32

P(X=4) = 0.2

Adding these probabilities together:

P(X≥2) = 0.21 + 0.32 + 0.2 = 0.73

Therefore, the probability that a student has passed this class with at least a C is 0.73 or 73%.

Step-by-step explanation:

sec² x − 2 = 0

sec² x = 2

cos² x = ½

cos x = ±√½

x = π/4, 3π/4, 5π/4, 7π/4

The answer is B. It is valid for Dr. Aronson to use the independent-measures t-test to investigate the effect of stereotype threat on math test performance among Caucasian engineering students. The t-test, which compares the means of two groups, is appropriate as long as the assumptions of independent measures, normal distribution, and homogeneity of variances are met.

The answer is B. The independent-measures t-test is a statistical test used to compare the means of two independent groups to determine if the differences between them are statistically significant. It is valid for Dr. Aronson to use this test because he has two groups of participants (control and experimental) who are separate and independently assigned. The large sample size also makes it more likely that the distribution of scores follows a normal pattern, even if this is not guaranteed.

The assumptions underlying the independent-measures t-test include the independence of observations, normal distribution of the population, and homogeneity of variance (equal variances). In this scenario, nothing suggests these assumptions would be violated. The students are independently assigned to the groups and their scores would not affect one another, satisfying the independence of observations.

Although we do not know for sure if the populations from which the samples come are normally distributed or if the variances within both samples are equal, these are assumptions of the test and not data dependent, therefore we can't definitively argue they are reasons not to use the test.

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Answer:

I Expect 6.4167 persons arriving during a 55.minute period.

Step-by-step explanation:

If the mean in one hour (60 minutes) is 7, then in a 55 minute period you can expect, by using a rule of three, 7*55/60 = 6.4167 persons. This is because the amount of expected people is always proportional to the amount of time spent waiting for poisson distributions.

Final answer:

To find the margin of error and 95% confidence interval for the proportion of companies likely to require higher health care contributions, we use a sample proportion of 0.52 and a sample size of 1000 to calculate a standard error of 0.0158. Multiplying this by the z-value of 1.96 gives a margin of error of 0.0310. The 95% confidence interval is thus (0.4890, 0.5510).

Explanation:

The question asks us to calculate the margin of error and a 95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage, based on a sample of 1,000 companies where 52% are likely to require higher contributions.

We start by identifying the sample proportion (p-hat) and the sample size (n):
p-hat = 0.52
n = 1000

The formula to calculate the standard error (SE) of the proportion is:
[tex]SE = \sqrt{((p-hat * (1 - p-hat)) / n)[/tex]

Using this formula, we calculate the SE:
[tex]SE = \sqrt{((0.52 * (1 - 0.52)) / 1000)[/tex]
[tex]SE = \sqrt{((0.52 * 0.48) / 1000)[/tex]
[tex]SE = \sqrt{(0.2496 / 1000)[/tex]
[tex]SE = \sqrt{(0.0002496)[/tex]
SE = 0.0158

Now we determine the z-value for a 95% confidence interval, which is commonly 1.96.

The margin of error (ME) is calculated by multiplying the z-value by the SE:
ME = z * SE
ME = 1.96 * 0.0158
ME = 0.0310

Finally, we calculate the 95% confidence interval with the formula:
(p-hat - ME, p-hat + ME)
(0.52 - 0.0310, 0.52 + 0.0310)
(0.4890, 0.5510)

Therefore, the margin of error is 0.0310 and the 95% confidence interval for the proportion of companies is (0.4890, 0.5510).

Answer:

The probability that a person with at least 55 years is more likely to buy the product will be 31.12%

Step-by-step explanation:

The exercise gives a condition which decreases the table to particular rows. The condition is "more likely" to buy a product, we will only work with the row with the label is "more likely". Therefore, the denominator of our equation will be 1301.

The exercise asks for the probability that a person with at least 55 years given the condition. So, we have:

Probability = (405/1301) x100 = 31.12%

Answer:

0.727

Step-by-step explanation:

The probability of an event can be estimated by taking the ratio of the number of times the events occur and the total number of possible outcomes. In the problem, the condition given is that the person is at least 55 years of age. Considering the column for 55+, the total number of people in the age bracket is 557 and 405 people are more likely to purchase the product. Thus, the probability is: 405/557 = 0.727.

Answer:

Therefore the rate of change of area [tex]25t^4+16t^3[/tex] square inches/ s

Step-by-step explanation:

Given that,

A rectangle is growing such that the length of rectangle is(5t+4) and its height is t⁴.

Where t is in second and dimensions are in inches.

The area of a rectangle is = length× height

Therefore the area of the rectangle is

A(t) = (5t+4) t⁴

⇒A(t) = t⁴(5t+4)

To find the rate change of area we need to find out the first order derivative of the area.

Rules:

[tex](1)\frac{dx^n}{dx} = nx^{n-1}[/tex]

[tex](2) \frac{d}{dx}(f(x) .g(x))= f'(x)g(x)+f(x)g'(x)[/tex]

A(t) = t⁴(5t+4)

Differentiate with respect to t

[tex]\frac{d}{dt} A(t)=\frac{d}{dt} [t^4(5t+4][/tex]

[tex]\Rightarrow \frac{dA(t)}{dt} =(5t+4)\frac{dt^4}{dt} +t^4\frac{d}{dt} (5t+4)[/tex]

[tex]\Rightarrow \frac{dA(t)}{dt} = (5t+4)4t^{4-1}+t^4.5[/tex]

[tex]\Rightarrow \frac{dA(t)}{dt} =4t^3(5t+4)+5t^4[/tex]

[tex]\Rightarrow \frac{dA(t)}{dt} = 20t^4+16t^3+5t^4[/tex]

[tex]\Rightarrow \frac{dA(t)}{dt} = 25t^4+16t^3[/tex]

Therefore the rate of change of area [tex]25t^4+16t^3[/tex] square inches/ s

The rate of change of the rectangle's area, with respect to time, is obtained by differentiating the expression for the area with respect to time. Using the product rule, the derivative is found to be (5t4 + 4t3)(5+4t) inches per second.

The subject of this question is primarily calculus, focusing on the concept of derivatives and rates of change. The area, A, of the rectangle can be found by multiplying the length by the height, hence A = (5t+4)t4. To find the rate of change of area with respect to time, we differentiate A with respect to time, t. This gives us the derivative dA/dt = d/dt [(5t+4)t4]. Using the product rule for differentiation (uv)' = u'v + uv', we find dA/dt = (5t4 + 4t3)(5+4t). Hence, the rate of change of the area with respect to time is (5t4 + 4t3)(5+4t) inches per second.

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The question is not complete and the complete question is;

On April 1, the unpaid balance in an account was $218. A payment of $30 was made on April 11. On April 21, a $40 purchase was made. The finance charge rate was 15% per month of the average daily balance. Find the new balance at the end of April.

Answer:

Average daily balance = $211.33

Step-by-step explanation:

From the question, we know the following;

On April 1, the unpaid balance was $218.

On April 11, a payment of $30 was made.

On April 21, a $40 purchase was made.

Now, from april 1st till 10th, the unpaid balance was $218.

Since a $30 deposit was paid on the 11th,it means that from april 11th till 20th, the unpaid balance was $218 - $30 = $188

Also since a $40 purchase was made on April 21st,it means that from april 21st till 30th, the unpaid balance was $188 + $40 = 228

Now, since we want to find the average daily balance for the month, let's find the total daily balance and divide by the number of days in the month.

Total balance was:

10 days from april 1st till 10th = 10 x $218 = $2180.

10 days from april 11th till 20th = 10 x $188 = $1880.

10 days from april 21st till 30th = 10 x 228 = $2280.

Adding up, the total =$2180 + $1880+ $2280 = $6340

Now the average daily balance for the month which has 30 days = $6340/30 = $211.33

Answer:

Population of bacteria four hours later [tex]A=616.5187[/tex]  or nearest whole number

[tex]A=616[/tex]

Step-by-step explanation:

Given,

Let amount of bacteria four hours later is=A

Rate of increase of bacteria per hour  [tex]=\frac{\partial x}{\partial t}=4.0565e^{1.3t}[/tex]

Initial population of bacteria is =54

Time 4 hours

Find the population of bacteria 4 hours later

Solution,

[tex]\int_{54}^{A}dx=\int_{0}^{4}4.0565e^{1.3t}dt[/tex]

[tex]\left [ x \right ]_{54}^{A}=4.0565/1.3\left [e^{1.3t} \right ]_{0}^{4}[/tex]

[tex]A-54=4.0565/1.3\left ( e^{1.3\times 4} -1\right )[/tex]

[tex]A-54=562.5187301[/tex]

[tex]A=562.5187301+54[/tex]

[tex]A=616.5187[/tex]

Population of bacteria 4 hours later is[tex]A=616[/tex]

The population of the bacteria four hours later was 39708

An exponential function is in the form:

y = abˣ

Where a is the initial value of y and b is the multiplication factor.

Given that the  initial population is 54 bacteria and it increases in size at a rate of 4.0565e^(1.3t) bacteria per hour. In 4 hours:

[tex]Bacteria\ population=54 * 4.0565e^{1.3*4}=39708[/tex]

The population of the bacteria four hours later was 39708

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Answer: at 50 mph, she drove for 4 hours.

at 45 mph, she drove for 2 hours.

Step-by-step explanation:

Let t represent the time that she spent driving at 50 miles per hour.

Taylor took 6 hours to drive home from college for Thanksgiving break. This means that the time that she spent driving at 45 miles per hour is (6 - t) hours.

Distance = speed × time

Distance covered while driving 50 miles per hour is

50t

Distance covered while driving 45 miles per hour is

45(6 - t)

Since the total distance that she drove is 290 miles, it means that

50t + 45(6 - t) = 290

50t + 270 - 45t = 290

50t - 45t = 290 - 270

5t = 20

t = 20/5 = 4

At 45 miles per hour, she drove at

6 - 4 = 2 hours

Answer:

Since Timothy scored on the higher percentile, he performed better with respect to the rest of the students in the class.

Step-by-step explanation:

When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.

So, when comparing scores, whoever scored on the higher percentile scored more than a higher percentage of students, so had the better relative score.

In this problem, we have that

Timothy scored on the 72nd percentile.

Clayton scored on the 70th percentile.

Since Timothy scored on the higher percentile, he performed better with respect to the rest of the students in the class.