A password is required to be 12 to 16 characters in length. Characters can be digits (0-9), upper or lower-case letters (A-Z, a-z) or special characters. There are 10 permitted special characters. There is an additional rule that not all characters can be letters (i.e. there has to be at least one digit or one special character.) How many permitted passwords are there? Give your answer in un-evaluated/un-simpli ed form and explain it fully.

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:

[tex](-\infty,3)[/tex]

Step-by-step explanation:

We are given that

[tex](x-3)y''+4y=x[/tex]

[tex]y''+\frac{4}{x-3}y=\frac{x}{x-3}[/tex]

y(0)=0

y'(0)=1

By comparing with

[tex]y''+p(x)y'+q(x)y=g(x)[/tex]

We get

[tex]p(x)=\frac{4}{x-3}[/tex]

[tex]g(x)=\frac{x}{x-3}[/tex]

q(x)=0

p(x),q(x) and g(x) are continuous for all real values of x except 3.

Interval on which p(x),q(x) and g(x) are continuous

[tex](-\infty,3)[/tex]and (3,[tex]\infty)[/tex]

By unique existence theorem

Largest interval which contains 0=[tex](-\infty,3)[/tex]

Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution=[tex](-\infty,3)[/tex]

The largest interval on which includes x=0 for which given initial-value problem has unique solution is [tex](-\infty, 3)[/tex]

The given parameters are:

(x − 3)y'' + 4y = x,

y(0) = 0

y'(0) = 1

Divide the equation (x − 3)y'' + 4y = x through by (x - 3)

[tex]y'' + \frac{4y}{x - 3} = \frac{x}{x - 3}[/tex]

Compare the above equation to the following equation

y" + p(x) y' +  q(x)y = g(x)

Then, we have:

[tex]p(x) = \frac{4y}{x - 3}[/tex]

[tex]q(x) = 0[/tex]

[tex]g(x) = \frac x{x - 3}[/tex]

The domains of functions p(x) and g(x) are all set of real values except 3

This is represented as:

[tex](-\infty, 3)\ u\ (3,\infty)[/tex]

Using the unique existence theorem, we have:

The largest interval that contains x = 0 is [tex](-\infty, 3)[/tex]

Hence, the largest interval on which includes x=0 for which given initial-value problem has unique solution is [tex](-\infty, 3)[/tex]

Read more about intervals at:

https://brainly.com/question/1399055

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.

https://brainly.com/question/33433874

#SPJ3

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.

https://brainly.com/question/23091711

#SPJ12

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:

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.

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)

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

Answer:

A. No

City driving mean requires less miles than Highway driving mean.

B. No

City driving median requires less miles than Highway driving median.

C. No

City driving mode requires less miles than Highway driving mode

D.

The driving conditions in the City is better than the driving conditions in the Highway because all measures of centers are lesser than those of Highway.

Step-by-step explanation:

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:

The interval is approximately a 98% confidence interval

Step-by-step explanation:

From the question :  Error, E= 3% = 0.03, Total population, n=1000,  number of people that are concerned that their taxes will be audited, p = 19% = 0.19

E^{2}=z_{\alpha/2}^{2}\cdot \frac{p(1-p)}{n}

0.03^{2}=z_{\alpha/2}^{2}\cdot \frac{0.19(1-0.19)}{1000}

z_{\alpha/2}^{2}=5.848

z_{\alpha/2}=2.418

Area right to 2.418 is 0.0078. So

\alpha/2=0.0078

Therefore \alpha=0.0156\approx 0.02

Thus, the interval is approximately a 98% confidence interval.

Answer:

98 percent

Step-by-step explanation:

plato

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

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.

https://brainly.com/question/32117953

#SPJ3

Answer:

Requirements are computed below.

Step-by-step explanation:

For ABC analysis:

Step1: Calculate the total value of each item.

SKU Number of Items Value ($/item) Value of the product ($)

A 220 67 14740

B 225 3 675

C 245 21 5145

D 145 7 1015

E 230 26 5980

F 240 83 19920

Step2: Reorganize the table on the basis of total value in descending order, & calculate % share of total value of investment and cumulative % share of total value of investment.

SKU Number of Items Value ($/item) Value of the product ($) % of Total value Cumulative % of total value Segment

SKU Number of Items Value ($/item) Value of the product ($) % of Total value Cumulative % of total value Segment

F 240 83 19920 41.96 41.96 A

A 220 67 14740 31.05 73.01 A

E 230 26 5980 12.60 85.60 B

C 245 21 5145 10.84 96.44 B

D 145 7 1015    2.14   98.58 C

B 225 3 675         1.42         100.00 C

----------

% of total number of items in segment B = (230+245)/1305 = 36.4%

% of total value of inventory in segment B = (12.6 + 10.4)/100 = 23.43%

% of the total weight of inventory in segment B = (9660+14700)/67795 = 35.93%

SKU Number of Items Value ($/item) Weight (lb/item) Total weight

A 220 67 52 11440

B 225 3 59 13275

C 245 21 60 14700

D 145 7 48 6960

E 230 26 42 9660

F 240 83 49 11760

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.

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:

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.

https://brainly.com/question/33168340

#SPJ3

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

Find out more on exponential function at: https://brainly.com/question/12940982

Answer: 2a + b - 3cd

Step-by-step explanation:

a 2 + b - cd 3

2a + b - 3cd

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:

(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:

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:

0.7262

Step-by-step explanation:

Let random variable = X

Probability of getting a cookie with at least 55 chips:

P(55≤ X ≤78), where X1 = 55 and X2 = 78

Let mean, U = 66 and  standard deviation, S.D. = 10

Let Z = X - U/S.D; Z1 = X1 - U/S.D and Z2 = X2 - U/S.D

Z1 = 55 - 66/10 = - 1.1

Z2 = 78 - 66/10 = 1.2

P(X1≤ X ≤ X2) = P(X1 -U/S.D ≤ X - U/S.D ≤ X2 - U/S.D)

P(Z1 ≤ Z ≤ Z2) = P(0 ≤ Z ≤ Z1) + P(0 ≤ Z ≤ Z2)

Hence, (55 ≤ X ≤ 78) = P( - 1.1 ≤ Z ≤ 1.2)

∴ P(-1.1 ≤ Z ≤ 1.2) = P(0 ≤ Z ≤ 1.1) + (0 ≤ Z ≤ 1.2)

=   0.3413 + 0.3849 = 0.7262

Answer:

Step-by-step explanation:

You should try going too jeeska

Answer:

Step-by-step explanation:

Elise’s garden is rectangular. The formula for determining the area of a rectangle is expressed as

Area = length × width

From the information given,

Length of garden = 6 feet

Width of garden = 12.6 feet

Area of garden = 6 × 12.6 = 75.6 square feet

Tomato plants require 2 square feet of space. This means that the number of tomato plants that Elise can fit in her garden is

75.6/2 = 37.8

Since the number of tomato plants must be whole number, then the number of tomato plants that Elise can fit in her garden is 37

Answer:

a) {4,5,6,7,8,9,10}

b) {1,2,3,4,5}

c) {4,5,6}

Answer:

P(A) = 3/20

Step-by-step explanation:

P(A)=P(blue)P(head)=(3/10)(1/2)=3/20

as there are 10 cards in total, out of which 3 are blue so the probability to get the blue card is, P(blue) = 3/10. and the probability of getting a head when a coin is tossed is P(head) = 1/2.

So in total

P(A) = P(blue)*p(head) = (3/10)*(1/2) = 3/20 = 0.15

Final answer:

P(A) = 0.15.

Explanation:

To find P(A), the probability of event A happening, we need to consider two separate events: picking a blue card and tossing a head on a coin.

First, the probability of picking a blue card from the special deck is 3 out of 10, because there are 3 blue cards among the 10 total cards.

Next, the probability of tossing a head on a coin is 1 out of 2, since a coin has two sides and both outcomes (heads and tails) are equally likely.

To find the combined probability P(A) of the two independent events, we multiply the probabilities:

[tex]P(A) = P(Blue Card) \(\times\) P(Head Coin Toss)[/tex]

Thus:

[tex]P(A) = (3/10) \(\times\) (1/2) = 3/20 = 0.15[/tex]

Rounded to two decimal places, P(A) = 0.15.

Answer:

[tex]\frac{x-2}{-1} =\frac{y-4}{1} =\frac{z-4}{-4} =t[/tex]

Step-by-step explanation:

Given that a line passes through P(2,4,4)

Also the line is perpendicular to the plane

[tex]-1x+1y-4z=1.[/tex]

From the equation of the plane we can say that normal to the plane has direction ratios as (-1,1,-4)

Since the required line is also perpendicular to the plane, the direction ratios of the required line is

(-1,1,4)

It passes through (2,4,4)

If Q(x,y,z) are general points on the line then

Direction ratios of PQ are = (x-2, y-4, z-4)

These are proportional to (-1,1,4)

So parametric form of the line is

[tex]\frac{x-2}{-1} =\frac{y-4}{1} =\frac{z-4}{-4} =t[/tex]

Whem t=0 we get the point P.

Answer:

22.96% probability that a hotel room costs between $250.00 and $285.00

Step-by-step explanation:

Problems of normally distributed samples can be 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 = 244, \sigma = 55[/tex]

What is the probability that a hotel room costs between $250.00 and $285.00?

This is the pvalue of Z when X = 285 subtracted by the pvalue of Z when X = 250. So

X = 285

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

[tex]Z = \frac{285 - 244}{55}[/tex]

[tex]Z = 0.75[/tex]

[tex]Z = 0.75[/tex] has a pvalue of 0.7734

X = 250

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

[tex]Z = \frac{250 - 244}{55}[/tex]

[tex]Z = 0.11[/tex]

[tex]Z = 0.11[/tex] has a pvalue of 0.5438

0.7734 - 0.5438 = 0.2296

22.96% probability that a hotel room costs between $250.00 and $285.00

Answer:

0.436

Step-by-step explanation:

Given that a professor planned to give an examination in a large class on the Monday before Thanksgiving vacation

Some students asked whether he could change the date because so many of their classmates had at least one other exam on that date.

They speculated that at least 40% of the class had this problem.

The professor agreed to poll the class, and if there was convincing evidence that the proportion with at least one other exam on that date was greater than .40, he would change the date.

No of students in total = 250

Reported they had another exam = 109

proportion of the class reported that they had another exam on the date

= [tex]\frac{109}{250} =0.436[/tex]

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