Montoya (2007) asked 56 men and 82 women to rate 21 different body parts on a scale of 1 (no opinion) to 5 (very desirable). They found that men and women rated the eyes similarly, with an average rating of about 3.77 ± 1.23 (M ± SD). Assuming these data are normally distributed, answer the following questions. (Round your answers to two decimal places.) (a) What percentage of participants rated the eyes at least a 5 (very desirable)? % (b) What percentage rated the eyes at most a 1 (no opinion)? %

Answer:

(P(t)) = P₀/(1 - P₀(kt)) was proved below.

Step-by-step explanation:

From the question, since β and δ are both proportional to P, we can deduce the following equation ;

dP/dt = k(M-P)P

dP/dt = (P^(2))(A-B)

If k = (A-B);

dP/dt = (P^(2))k

Thus, we obtain;

dP/(P^(2)) = k dt

((P(t), P₀)∫)dS/(S^(2)) = k∫dt

Thus; [(-1)/P(t)] + (1/P₀) = kt

Simplifying,

1/(P(t)) = (1/P₀) - kt

Multiply each term by (P(t)) to get ;

1 = (P(t))/P₀) - (P(t))(kt)

Multiply each term by (P₀) to give ;

P₀ = (P(t))[1 - P₀(kt)]

Divide both sides by (1-kt),

Thus; (P(t)) = P₀/(1 - P₀(kt))

(P(t)) = P₀/(1 - P₀(kt))

According to the, since β and also δ are both proportional to P, we can deduce the following equation ;

Then dP/dt = k(M-P)P

Then dP/dt = (P^(2))(A-B)

Now, If k = (A-B);

After that dP/dt = (P^(2))k

Thus, we obtain;

Now dP/(P^(2)) = k dt

((P(t), P₀)∫)dS/(S^(2)) = k∫dt

Thus; [(-1)/P(t)] + (1/P₀) = kt

Simplifying,

Then 1/(P(t)) = (1/P₀) - kt

Multiply each term by (P(t)) to get ;

After that 1 = (P(t))/P₀) - (P(t))(kt)

Multiply each term by (P₀) to give ;

Now P₀ = (P(t))[1 - P₀(kt)]

Then Divide both sides by (1-kt),

Thus; (P(t)) = P₀/(1 - P₀(kt))

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Answer:

Probability that a randomly selected individual who was unemployed in 2000 would have been unemployed between 12 and 13 weeks is 0.83144 .

Step-by-step explanation:

We are given that according to the Bureau of Labor Statistics, in calendar year 2000, the average duration of unemployment was 12.7 weeks, with a standard deviation of 0.3 weeks.

Let X = randomly selected individual who was unemployed in 2000

Since distribution is approximately normal, so X ~ N([tex]\mu=12.7,\sigma^{2} = 0.3^{2}[/tex])

The z score probability distribution is given by;

                 Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 12.7 weeks

            [tex]\sigma[/tex] = population standard deviation = 0.3 weeks

So, the probability that a randomly selected individual who was unemployed in 2000 would have been unemployed between 12 and 13 weeks is given by = P(12 < X < 13) = P(X < 13) - P(X <= 12)

P(X < 13) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{13-12.7}{0.3}[/tex] ) = P(Z < 1) = 0.84134

P(X <= 12) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{12-12.7}{0.3}[/tex] ) = P(Z < -2.33) = 1 - P(Z <= 2.33) = 1 - 0.99010

                                                                                                   = 0.0099

Therefore, P(12 < X < 13) = 0.84134 - 0.0099 = 0.83144 .

Answer:

Interval within 1 standard deviation: (16.3,19.7)

Interval within 2 standard deviation: (14.6, 21.4)

Interval within 3 standard deviation: (12.9, 23.1)

Step-by-step explanation:

We are given the following in the question:

Population mean, [tex]\mu[/tex] = 18

Standard deviation, [tex]\sigma[/tex] = 1.7

We have to find the following intervals:

Interval within 1 standard deviation:

[tex]\mu \pm 1\sigma\\=18 \pm 1.7\\=(16.3, 19.7)[/tex]

Interval within 2 standard deviation:

[tex]\mu \pm 2\sigma\\=18 \pm 2(1.7)\\= 18 \pm 3.4\\=(14.6,21.4)[/tex]

Interval within 3 standard deviation:

[tex]\mu \pm 3\sigma\\=18 \pm 2(1.7)\\= 18 \pm 5.1\\=(12.9,23.1)[/tex]

Answer: a) 2 miles

b) 4 miles

Step-by-step explanation:

There are two right angle triangles formed in the rectangle.

Taking 30 degrees as the reference angle, the length of the side walk, h represents the hypotenuse of the right angle triangle.

The width, w of the park represents the opposite side of the right angle triangle.

The length of the park represents the adjacent side of the right angle triangle.

a) to determine the width of the park w, we would apply

the tangent trigonometric ratio.

Tan θ, = opposite side/adjacent side. Therefore,

Tan 30 = w/2√3

1/√3 = w/2√3

w = 1/√3 × 2√3

w = 2

b) to determine the the length of the side walk h, we would apply

the Cosine trigonometric ratio.

Cos θ, = adjacent side/hypotenuse. Therefore,

Cos 30 = 2√3/h

√3/2 = 2√3/h

h = 2√3 × 2/√3

h = 4

Answer:

22%

Step-by-step explanation:

In the event of changing a test answer there are three possible outcomes, which should add up to 100%: changing from incorrect to correct (51%),  changing from correct to incorrect (27%) and  changing from incorrect to incorrect (X). Therefore, the percent of changes from incorrect to​ incorrect was:

[tex]100\% = 51\%+27\%+X\\X= 22\%[/tex]

22% of the changes were  from incorrect to​ incorrect.

Answer:

The column height of air after the inversion is 7 cm.

Step-by-step explanation:

The initial pressure balance is given as

P_1+20 cm=76 cm of Hg

P_1=76-20 cm of Hg

P_1=56 cm of Hg

The initial volume of the air with cross sectional area as 12 cm2 and the length of air column as 12 is given as V_1=12 cm *1 =12 cm3

After the inversion

P_2=20 cm+76 cm of Hg

P_2=96 cm of Hg

The volume of the air after the inversion with cross sectional area as 1 cm2 and the length of air column as x is given as V_2=x *1 =x cm3

Now as temperature is constant and the cross sectional area is also constant so

[tex]P_1V_1=P_2V_2\\56\times 12=96\times x\\x=\dfrac{56\times 12}{96}\\x=7 cm[/tex]

So the column height of air after the inversion is 7 cm.

Answer:

7cm

Step-by-step explanation:

P1 =H-h =76-20=56; P2 =H+h =96

P1V1 =P2V2

56x12=96xV2

V2 =56x12/96 = 7 cm

Answer:

Step-by-step explanation:

If two triangles are equal, it means that the ratio of the length of each side of one triangle to the length of the corresponding side of the other triangle is constant. Also, corresponding angles are congruent.

1) Triangle TUV is similar to triangle SQR because

Angle Q is congruent to angle U

TU/SQ = UV/QR = 2

2) Triangle ABC is not similar to triangle DEF because the ratio of AB to DF is not constant.

The total cost of the loan, including down payment, PMI, mortgage payments, and yearly taxes and insurance over 30 years, for purchasing a house at $138,000 with specific conditions rounds to approximately $333,400.00.

To calculate the total cost of the loan for purchasing a house at $138,000.00 with a 10% down payment, an interest rate of 4.875%, PMI payments of $25.88 for 77 months, yearly taxes of $2400.00, and insurance of $750.00 per year, the following steps are undertaken:

Rounding to the nearest hundred gives us a total cost of approximately $333,400.00.

Answer:

a) P=0.2861

b) P=0.0954

c) P=0.3815

d) P=0.6185

Step-by-step explanation:

The question is incomplete:

Among the N=16 students taking a graduate statistics class, A=10 are master students and the other N-A=6 are doctorial students. A random sample of n=5 students is going to be selected to work on a class project. Use X to denote the number of master students in the sample. Keep at least 4 decimal digits if the result has more decimal digits.

a) The probability that exactly 4 master students are in the sample is closest to?

b) The probability that all 5 students in the sample are master students is closest to?

c) The probability that at least 4 students in the sample are master students is closest to?

d) The probability that at most 3 students in the sample are master students is closest to?

We use a binomial distribution with n=5, with p=10/16=0.625 (proportion of master students).

a)

[tex]P(k=4)=\binom{5}{4}p^4q^1=5*0.625^4*0.375=5*0.1526*0.3750\\\\P(k=4)=0.2861[/tex]

b)

[tex]P(k=5)=\binom{5}{5}p^5q^0=1*0.625^5*1=\\\\P(k=4)=0.0954[/tex]

c)

[tex]P(k\geq4)=P(k=4)+P(k=5)=0.2861+0.0954=0.3815[/tex]

d)

[tex]P(k\leq3)=1-P(x\geq4)=1-0.3815=0.6185[/tex]

Answer:

[tex]P(X<4)=P(\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=0.2[/tex]  

[tex]P(z<\frac{4-\mu}{\sigma})=0.2[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-0.842<\frac{4-\mu}{1.1}[/tex]

And if we solve for the mean we got

[tex]\mu =4 +0.842*1.1=4.926[/tex]

Step-by-step explanation:

Assuming this question "Suppose that the lifetimes of TV tubes are normally distributed with a standard deviation of 1.1 years. Suppose also that exactly 20% of the tubes die before 4 years. Find the mean lifetime of TV tubes. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place. ?

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the lifetimes of TV tubes of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,1.1)[/tex]  

Where [tex]\mu[/tex] and [tex]\sigma=1.1[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

For this part we know the following condition:

[tex]P(X>4)=0.8[/tex]   (a)

[tex]P(X<4)=0.2[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(z>-0.842)=0.8

If we use condition (b) from previous we have this:

[tex]P(X<4)=P(\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=0.2[/tex]  

[tex]P(z<\frac{4-\mu}{\sigma})=0.2[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-0.842<\frac{4-\mu}{1.1}[/tex]

And if we solve for the mean we got

[tex]\mu =4 +0.842*1.1=4.926[/tex]

Mean lifetime is 4.9 years. Found using z-score of -0.8416 with 20% dying before 4 years, [tex]\(\sigma = 1.1\)[/tex].

To find the mean lifetime of TV tubes, we will use the properties of the normal distribution and the given information. Here's the step-by-step solution:

1. Identify the given values:

  - Standard deviation [tex](\(\sigma\))[/tex]: 1.1 years

  - Percentage of tubes that die before 4 years: 20% (or 0.20)

2. Find the z-score corresponding to the given percentage:

  - Since the percentage is 20%, we need to find the z-score for which 20% of the area under the normal curve lies to the left.

  - Using a standard normal distribution table or a calculator, the z-score for 0.20 is approximately -0.8416.

3. Set up the z-score formula:

  The z-score formula is given by:

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

where [tex]\(X\)[/tex] is the value (4 years in this case), [tex]\(\mu\)[/tex] is the mean lifetime we want to find, and [tex]\(\sigma\)[/tex] is the standard deviation.

4. Substitute the known values into the z-score formula and solve for [tex]\(\mu\)[/tex]:

  [tex]\[ -0.8416 = \frac{4 - \mu}{1.1} \][/tex]

  Rearrange to solve for [tex]\(\mu\)[/tex]:

  [tex]\[ -0.8416 \times 1.1 = 4 - \mu \][/tex]

  [tex]\[ -0.92576 = 4 - \mu \][/tex]

  [tex]\[ \mu = 4 + 0.92576 \][/tex]

  [tex]\[ \mu \approx 4.9258 \][/tex]

5. Round the mean lifetime to one decimal place:

  [tex]\[ \mu \approx 4.9 \][/tex]

So, the mean lifetime of the TV tubes is approximately 4.9 years.

Answer:

Step-by-step explanation:

Hello!

You have the variable:

X: Weight of a small bag of candies (ounces)

n= 12 small bags

i) Sample mean, X[bar]= 3 ounces

This is the expected weight of the small bags of candies.

ii) Population Standard deviation, σ= 0.1 ounces

The population standard deviation is a measurement of variability, it shows you how dispersed are the population values from the population mean.

iii) Sample standard deviation, S= 0.13 ounces

The sample standard deviation is the point estimate of the population standard deviation, it is a measure of variability, it shows the dispersion of the data around the sample mean.

I hope it helps!

Answer:

Step-by-step explanation:

Hello!

X: Cholesterol level of a woman aged 30-39. (mg/dl)

This variable has an approximately normal distribution with mean μ= 190.14 mg/dl

1. You need to find the corresponding Z-value that corresponds to the top 9.3% of the distribution, i.e. is the value of the standard normal distribution that has above it 0.093 of the distribution and below it is 0.907, symbolically:

P(Z≥z₀)= 0.093

-*or*-

P(Z≤z₀)= 0.907

Since the Z-table shows accumulative probabilities P(Z<Z₁₋α) I'll work with the second expression:

P(Z≤z₀)= 0.907

Now all you have to do is look for the given probability in the body of the table and reach the margins to obtain the corresponding Z value. The first column gives you the integer and first decimal value and the first row gives you the second decimal value:

z₀= 1.323

2.

Using the Z value from 1., the mean Cholesterol level (μ= 190.14 mg/dl) and the Medical guideline that indicates that 9.3% of the women have levels above 240 mg/dl you can clear the standard deviation of the distribution from the Z-formula:

Z= (X- μ)/δ ~N(0;1)

Z= (X- μ)/δ

Z*δ= X- μ

δ=(X- μ)/Z

δ=(240-190.14)/1.323

δ= 37.687 ≅ 37.7 mg/dl

I hope it helps!

Answer:

Calculate the Macaulay duration of Annuity B at the time of purchase is 1.369.

Step-by-step explanation:

First, we use 0.93 to calculate the v which equals 1/(1+i).

[tex]\frac{0+1*v+2v^{2} }{1+v+v^{2} }[/tex] = 0.93

After rearranging the equation, we get 1.07[tex]v^{2}[/tex] + 0.07v - 0.93=0

So, v=0.9

Mac D: [tex]\frac{0+1*v+2*v^{2}+ 3*v^{2} }{1+v+v^{2}+ v^{3} }[/tex]

After substituting the value of v, we get Mac D = 1.369.

Answer:

So the probability is P=0.00256.

Step-by-step explanation:

We have 26 letters, so the probability that the first letter in the name is the same is 1/26.

The probability that the second letter in the name is the same is 1/26 and the probability that the third letter in the name is the same is 1/26.

Out of ten people we choose 2.

So the probability is:

[tex]P=C_2^{10}\left(\frac{1}{26}\right)^3\\\\P=\frac{10!}{2!(10-2)!}\cdot \left(\frac{1}{26}\right)^3\\\\P=\frac{45}{17576}\\\\P=0.00256\\[/tex]

So the probability is P=0.00256.

Hypothesis:

The ratio of ladies giving multiple birth to total number of women for any race will be the same.

Test:

Ratio of white women giving multiple births = 94 / 3132 = 0.0300

Ratio of black women giving multiple births = 20 / 606 = 0.0330

Conclusion:

There is no racial difference in the likelihood of multiple births. Although we do see a difference in the ratios calculated above, the difference is small enough to be due to sample size difference of white and black women. The smaller number of total black women makes the ratio calculated from this sample have a higher probability to deviate from what is expected. This deviation will account for the difference in probability between both races.

We can see the effects of this small sample size by increasing or decreases the numerator by 1 for black women:

21 / 606 = 0.0347

19 / 606 = 0.0313

This change in the data of one woman produces a very large percentage change in our ratio for black women (5%). Thus despite inaccuracy due to small sample size, our hypothesis is correct.

Final answer:

A hypothesis test for the difference in proportions can be used to assess if there is a racial difference in the likelihood of multiple births between white and black women, based on the given data. The null hypothesis is no difference, and if the test statistic is significant, it may indicate a racial difference.

Explanation:

To evaluate any racial differences in the likelihood of multiple births between white women and black women based on the given data, we can perform a hypothesis test. Specifically, this would be a test for the difference between two proportions.

For white women:

Multiples: 94
Total births: 3132

For black women:

Multiples: 20
Total births: 606

Using a chi-squared distribution table or calculator, at α=0.05 and 1 degree of freedom, the critical value is approximately 3.841.

Since our calculated chi-squared value (0.2094) is less than the critical value (3.841), we fail to reject the null hypothesis.

Therefore, there is no significant evidence to conclude that there is a racial difference in the likelihood of multiple births among white and black women in this county.

Answer:

[tex]63.44^{0}[/tex]

Step-by-step explanation:

Tom's resultant speed is calculated as [tex]\sqrt{1^{2}+2^{2} }[/tex]

= [tex]\sqrt{5}[/tex]

The distance tom swim is the hypotenuse of the right angle triangle.

sin = opposite/hypotenuse = [tex]\frac{1}{\sqrt{5} }[/tex]  = 1/5 = 0.2

arcsin (0.2) = [tex]26.56^{0}[/tex]

upstream angle =

[tex]90^{0} - 26.56^{0}[/tex] =

[tex]63.44^{0}[/tex]

Answer:

there is no stagnation point

Step-by-step explanation:

for the velocity field V→(u,v)= (0.46 + 2.1x)i→ + (−2.8 - 2.1y)j , the stagnation point is found when the velocity vectors converge in one point ( thus also stays in that place when the point is reached). Thus the stagnation point can be found when the divergence of the velocity field is <0 ( thus it does not diverge , but converges)

div(V) = ∇*V= d/dx (0.46 + 2.1x) + d/dy (−2.8 - 2.1y) = 0

2.1 - 2.1 = 0

since div(V) can never be  <0 , there is no stagnation point

Answer:

(a) The average value of the given function is 12/π

(b) c = 1.238 or 2.808

Step-by-step explanation:

The average value of a function on a given interval [a, b] is given as

f(c) = (1/(b - a))∫f(x)dx;

from x = b to a

Now, given the function

f(x) = 6sin(x) - 3sin(2x), on [0, π]

The average value of the function is

1/(π-0) ∫(6sinx - 3sin2x)dx

from x = 0 to π

= (1/π) [-6cosx + (3/2)cos2x]

from 0 to π

= (1/π) [-6cosπ + (3/2)cos 2π - (-6cos0 + (3/2)cos0)]

= (1/π)(6 + (3/2) - (-6 + 3/2) )

= (1/π)(12) = 12/π

f(c) = 12/π

b) if f_(ave) = f(c), then

6sinx - 3sin2x = 12/π

2sinx - sin2x = 4/π

But sin2x = 2sinxcosx, so

2sinx - 2sinxcosx = 4/π

sinx - sinxcosx = 2/π

sinx(1 - cosx) = 2/π

This equation can only be estimated to be x = 1.238 or 2.808

Solving sin(x)(1 - cos(x)) = 2/π on [0, π] yields x = π/2 as the only solution. This corresponds to approximately 1.5708, satisfying the given equation.

To solve the equation sin(x)(1 - cos(x)) = 2/π, we can use the double-angle identity for sine, which states that sin(2x) = 2sin(x)cos(x). Rewrite the equation in terms of sin(2x):

sin(x)(1 - cos(x)) = 2/π

sin(x) - sin(x)cos(x) = 2/π

Now, substitute sin(x) = 2/π into the equation:

(2/π) - (2/π)cos(x) = 2/π

Multiply both sides by π to simplify:

2 - 2cos(x) = 2

Subtract 2 from both sides:

-2cos(x) = 0

Divide by -2:

cos(x) = 0

Now, find the values of x where cos(x) = 0, which occurs at x = π/2 and 3π/2. Since we are looking for solutions in the interval [0, π], x = π/2 is the only valid solution.

So, the solution to the equation sin(x)(1 - cos(x)) = 2/π on the interval [0, π] is x = π/2, which is approximately 1.5708.

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Answer:

a) 0.011

b) -0.0624

Step-by-step explanation:

See attached pictures.

Final answer:

Uncertainty in the FTIR spectrometer measurements can be estimated as the standard deviation of the measurements, yielding 0.0109 ppm. Bias is estimated as the difference between the mean of the measurements (1.0823 ppm) and the true value (1.1447 ppm), resulting in a bias of -0.0624 ppm.

Explanation:

To address the student's question regarding the calibration of an FTIR spectrometer and the estimation of uncertainty and bias in measurements, we shall consider the given data.

Uncertainty in measurements can be estimated using the standard deviation of the measurements, which provides an indication of the spread of the data around the mean. To calculate the uncertainty:

Using the provided measurements of carbon content, we calculate the uncertainty as follows:

This standard deviation represents the uncertainty in the measurements.

Estimation of Bias

Bias in the measurements can be estimated as the difference between the mean of the measurements and the true value. Thus, the bias is calculated by subtracting the true carbon content from the mean measurement:

Bias = Mean - True value = 1.0823 ppm - 1.1447 ppm = -0.0624 ppm

The negative sign indicates that the measurements are, on average, lower than the true value.

Answer:

a) Poisson distribution

b) 99.33% probability that in any one minute at least one purchase is​ made

c) 0.05% probability that seven people make a purchase in the next four ​minutes

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given time interval.

5 people per minute check out from their shopping cart and make an online purchase.

This means that [tex]\mu = 5[/tex]

a) What model might you suggest to model the number of purchases per​ minute? ​

The only information that we have is the mean number of an event(purchases) in a time interval. Each event is also independent fro each other. So you should suggest the Poisson distribution to model the number of purchases per​ minute.

b) What is the probability that in any one minute at least one purchase is​ made? ​

Either no purchases are made, or at least one is. The sum of the probabilities of these events is 1. So

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

We want to find [tex]P(X \geq 1)[/tex]

So

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

In which

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067[/tex]

1 - 0.0067 = 0.9933.

99.33% probability that in any one minute at least one purchase is​ made

c) What is the probability that seven people make a purchase in the next four ​minutes?

The mean is 5 purchases in a minute. So, for 4 minutes

[tex]\mu = 4*5 = 20[/tex]

We have to find P(X = 7).

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-20}*(20)^{7}}{(7)!} = 0.0005[/tex]

0.05% probability that seven people make a purchase in the next four ​minutes

The Poisson distribution model is used when the data consist of counts of occurrences.

a) The Poisson distribution model is used when the data consist of counts of occurrences.

b) Given that: λ (mean number of occurrence) = 5 people per minute, hence:

[tex]P(X\ge 1)=1-P(X=0)=1-\frac{e^{-\lambda }\lambda^x}{x!}= 1-\frac{e^{-5 }5^0}{5!}=0.9999[/tex]

The probability that in any one minute at least one purchase is​ made is 0.9999.

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The equation of the quadratic function is [tex]y=(x-7)(x+3)[/tex]

Explanation:

The vertex form of the quadratic function is given by

[tex]y=a(x-h)^{2}+k[/tex]

It is given that the quadratic function has a vertex at [tex](2,-25)[/tex]

The vertex is represented by the coordinate [tex](h,k)[/tex]

Hence, substituting [tex](h,k)=(2,-25)[/tex] in the vertex form, we get,

[tex]y=a(x-2)^{2}-25[/tex]

Now, substituting the x - intercept [tex](7,0)[/tex] , we have,

[tex]0=a(7-2)^{2}-25[/tex]

[tex]0=a(5)^{2}-25[/tex]

[tex]25=a(25)[/tex]

 [tex]1=a[/tex]

Thus, the value of a is 1.

Hence, substituting [tex]a=1[/tex], [tex](h,k)=(2,-25)[/tex] in the vertex form [tex]y=a(x-h)^{2}+k[/tex] , we get,

[tex]y=1(x-2)^{2}-25[/tex]

[tex]y=(x-2)^{2}-25[/tex]

[tex]y=x^2-2x+4-25[/tex]

[tex]y=x^2-2x-21[/tex]

[tex]y=(x-7)(x+3)[/tex]

Thus, the equation of the quadratic function is [tex]y=(x-7)(x+3)[/tex]

Answer:

the answer is d

Step-by-step explanation:

The surface area of the triangular prism is 1664 square inches.

Explanation:

Given that the triangular prism has a length of 20 inches and has a triangular face with a base of 24 inches and a height of 16 inches.

The other two sides of the triangle are 20 inches each.

We need to determine the surface area of the triangular prism.

The surface area of the triangular prism can be determined using the formula,

[tex]SA= bh+pl[/tex]

where b is the base, h is the height, p is the perimeter and l is the length

From the given the measurements of b, h, p and l are given by

[tex]b=24[/tex] , [tex]h= 16[/tex] , [tex]l=20[/tex] and

[tex]p=20+20+24=64[/tex]

Hence, substituting these values in the above formula, we get,

[tex]SA= (24\times16)+(64\times20)[/tex]

Simplifying the terms, we get,

[tex]SA=384+1280[/tex]

Adding the terms, we have,

[tex]SA=1664 \ square \ inches[/tex]

Thus, the surface area of the triangular prism is 1664 square inches.

Answer:

0.344 = 34.4% probability that at least 1 of them have blue eyes.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have blue eyes, or they have not. The probability of a person having blue eyes is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

A survey reports that the probability a person has blue eyes is 0.10.

This means that [tex]p = 0.1[/tex]

4 people are randomly selected at Miramar College

This means that [tex]n = 4[/tex]

Find the probability that at least 1 of them have blue eyes.

Either none of them have blue eyes, or at least one do. The sum of the probabilities of these events is decimal 1. So

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

We want [tex]P(X \geq 1)[/tex]. So

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

In which

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

[tex]P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.656[/tex]

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.656 = 0.344[/tex]

0.344 = 34.4% probability that at least 1 of them have blue eyes.

To find the probability that at least one out of four people has blue eyes at Miramar College, we use the complement rule and calculate that the probability is 0.344 after rounding to three decimal places.

The question asks us to calculate the probability that at least one person out of four randomly selected people at Miramar College will have blue eyes, given that the probability a person has blue eyes is 0.10. To find the probability of at least one person having blue eyes, we can use the complement rule. The complement of at least one person having blue eyes is that no person has blue eyes.

The probability that a single person does not have blue eyes is 1 - 0.10 = 0.90. For four independent selections, the probability that none of them have blue eyes is 0.90 raised to the fourth power. Therefore, the probability that at least one out of four people has blue eyes is 1 minus this result.

Calculation:

P(no one has blue eyes) = 0.90 ^ 4

P(no one has blue eyes) = 0.6561

P(at least one has blue eyes) = 1 - P(no one has blue eyes)

P(at least one has blue eyes) = 1 - 0.6561

P(at least one has blue eyes) = 0.3439

Thus, the probability that at least one person out of the four has blue eyes, rounded to three decimal places, is 0.344.

Answer:

[tex]P(9 < x < 12)=P(X=10)+P(X=11)[/tex]

[tex]P(X=10)=(19C10)(0.2)^{10} (1-0.2)^{19-10}=0.00127[/tex]

[tex]P(X=11)=(19C11)(0.2)^{11} (1-0.2)^{19-11}=0.00026[/tex]

[tex]P(9 < X < 12)=0.00127 +0.00026=0.0015[/tex]

Step-by-step explanation:

Previous concepts

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

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=19, p=0.2)[/tex]

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

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

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

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

And we want to find this probability:

[tex]P(9 < x < 12)=P(X=10)+P(X=11)[/tex]

[tex]P(X=10)=(19C10)(0.2)^{10} (1-0.2)^{19-10}=0.00127[/tex]

[tex]P(X=11)=(19C11)(0.2)^{11} (1-0.2)^{19-11}=0.00026[/tex]

[tex]P(9 < X < 12)=0.00127 +0.00026=0.0015[/tex]

The probability that less than 12 but more than 9 bulbs from the sample are defective is approximately 1.0230 (rounded to four decimal places).

We want to find the probability of having between 10 and 11 defective bulbs out of a sample of 19, with a 20% defect rate.

Probability of a single bulb being defective (p): 20% or 0.20.

Probability of a single bulb not being defective (q): 1 - p = 1 - 0.20 = 0.80.

Now, we'll calculate the probabilities for k = 10 and k = 11 using the binomial probability formula:

P(X = k) = (n choose k) * p^k * q^(n-k),

where:

n is the total number of trials (sample size), which is 19 in this case.

k is the number of successful trials (defective bulbs) we want.

For k = 10:

P(X = 10) = (19 choose 10) * (0.20)^10 * (0.80)^(19-10)

P(X = 10) = 92,378.49 * 0.0000001024 * 0.1073741824

P(X = 10) ≈ 0.9899 (rounded to four decimal places)

For k = 11:

P(X = 11) = (19 choose 11) * (0.20)^11 * (0.80)^(19-11)

P(X = 11) = 61,967.28 * 0.000002048 * 0.0262144

P(X = 11) ≈ 0.0331 (rounded to four decimal places)

Now, sum these probabilities to find the overall probability of having between 10 and 11 defective bulbs:

P(10 ≤ X ≤ 11) = P(X = 10) + P(X = 11)

P(10 ≤ X ≤ 11) ≈ 0.9899 + 0.0331

P(10 ≤ X ≤ 11) ≈ 1.0230 (rounded to four decimal places)

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Answer:

Both the answers are as in the solution.

Step-by-step explanation:

As the given matrix is not in the readable form, a similar question is found online and the solution of which is attached herewith.

Part a:

Given matrix is : A = [tex]\left[\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right][/tex]

Here,

[tex]det(A) =\left|\begin{array}{ccc}0&3&4\\1&2&3\\-3&-7&8\end{array}\right| = -55 \neq 0.[/tex]

Then, A is non-singular matrix.

Here, A₁₁= 0.

If we write A as LU with L lower triangular matrix and U upper triangular matrix, then A₁₁=L₁₁U₁₁.

So, As

A₁₁ = 0 gives L₁₁U₁₁= 0 ,

This indicates that either L₁₁= 0 or U₁₁ = 0.

If L₁₁= 0 or U₁₁ = 0, this would made the corresponding matrix singular, which contradicts the condition as  A is non-singular.

Therefore, A has no LU decomposition.

Part b:

By the implementation of the various row operations

interchange R1 and R2

[tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\-3&-7&8\end{array}\right][/tex]

R3+3R1=R3

[tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&-1&17\end{array}\right][/tex]

R3+(1/3)R2 = R3

[tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right][/tex]

Therefore, U = [tex]\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right][/tex].

Here, LP = E₁₂=E₃₁=-3 &E₃₂=-1/3

[tex]LP=\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-3&0&1\end{array}\right]\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&-1/3&1\end{array}\right][/tex]

[tex]LP=\left[\begin{array}{ccc}0&1&0\\1&0&0\\-3&-1/3&1\end{array}\right][/tex]

[tex]LP=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-1/3&-3&1\end{array}\right]\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right][/tex]

So now U is given as

[tex]U=\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&55/3\end{array}\right]\\L=\left[\begin{array}{ccc}1&0&0\\0&1&0\\-1/3&-3&1\end{array}\right]\\P=\left[\begin{array}{ccc}0&1&0\\1&0&0\\0&0&1\end{array}\right]\\[/tex]

Answer:

The required sample size is 350 to estimate the proportion with a margin of error of 0.05          

Step-by-step explanation:

We are given the following in the question:

[tex]\hat{p} = 0.65[/tex]

Confidence level = 95%

Margin of error = 0.05

Confidence interval:

[tex]\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

Margin of error =

[tex]z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

Putting values, we get,

[tex]0.05 = 1.96\sqrt{\dfrac{(0.65)(1-0.65)}{n}}\\\\n = \dfrac{(1.96)^2(0.65)(1-0.65)}{(0.05)^2}\\\\n = 349.5856\\n \approx 350[/tex]

Thus, the required sample size is 350 to estimate the proportion with a margin of error of 0.05

The proportion of students who do not change their passwords with a 5% margin of error using a preliminary estimate of 65%, a sample size of approximately 369 students is needed.

To determine the sample size required to estimate the proportion of students who do not change their password with a margin of error of 0.05, using a preliminary estimate of 0.65, we use the formula for determining sample size in a proportion survey:

n = (Z² × p(1 - p)) / E²

Where:

n = required sample size

Z = Z-score associated with the desired confidence level (For a 95% confidence level, Z = 1.96)

p = preliminary estimate of the proportion (0.65 in this case)

E = margin of error (0.05 as specified)

Substituting the values into the formula, we get:

n = (1.96² × 0.65(1 - 0.65)) / 0.05²

  = 369

Hence, to estimate the proportion of students who do not change their password with a 5% margin of error, a sample size of approximately 369 students is required.

This result helps ensure that research or surveys conducted on certain populations achieve a high level of accuracy within the specified confidence level and margin of error.

Answer:

B. 0.4542 to 0.5105

Step-by-step explanation:

A 90% confidence interval for p is calculated as:

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

This apply if n*p≥5 and n*(1-p)≥5

Where p is the proportion of sample, n is the size of the sample and [tex]z_{\alpha /2}[/tex] is equal to 1.645 for a 90% confidence.

Then, in this case p, n*p and n*(1-p) are calculated as:

[tex]p=\frac{410}{850} =0.4824[/tex]

n*p = (850)(0.4824) = 410

n*(1-p) = (850)(1-0.4824) = 440

So, replacing values we get:

[tex]0.4824-1.645\sqrt{\frac{0.4824(1-0.4824)}{850} }\leq p\leq 0.4824+1.645\sqrt{\frac{0.4824(1-0.4824)}{850} }[/tex]

[tex]0.4824-0.0282\leq p\leq 0.4824+0.0282[/tex]

[tex]0.4542\leq p\leq 0.5105[/tex]

It means that a  90% confidence interval for p is 0.4542 to 0.5105

Answer:

The correct answer in the option is;

B. 0.4542 to 0.5105.

Step-by-step explanation:

To solve the question, we note that

Total number of residents, n = 850

Number supporting property tax levy = 410

Proportion supporting tax levy, p = [tex]\frac{410}{850}[/tex] = 0.48235

The formula for confidence interval is

[tex]p +/-z*\sqrt{\frac{p(1-p)}{n} }[/tex]

Where

z = z value

The z value from the tables at 90 % = 1.64

Therefore we have

The confidence interval given as

[tex]0.48235 +/-1.64*\sqrt{\frac{0.48235(1-0.48235)}{850} }[/tex] = 0.48235 ± 2.811 × 10⁻²

= 0.4542 to 0.5105

The confidence interval is 0.4542 to 0.5105.

Answer:

D

Step-by-step explanation:

Surface area of this rectangular prism is A=2(1x10+8x10+8x1) so the area for this one is 196 ft

Answer:

a. data(wages)

plot(wages, type='o', ylab='wages per hour')

Step-by-step explanation:

a.  Display and interpret the time series plot for these data.

#take data samples from wages

data(wages)

plot(wages, type='o', ylab='wages per hour')

see others below

b. Use least squares to fit a linear time trend to this time series. Interpret the regression output. Save the standardized residuals from the fit for further analysis.

#linear model

wages.lm = lm(wages~time(wages))

summary(wages.lm) #r square is correct

##  

## Call:

## lm(formula = wages ~ time(wages))

##  

## Residuals:

##      Min       1Q   Median       3Q      Max  

## -0.23828 -0.04981  0.01942  0.05845  0.13136  

##  

## Coefficients:

##               Estimate Std. Error t value Pr(>|t|)    

## (Intercept) -5.490e+02  1.115e+01  -49.24   <2e-16 ***

## time(wages)  2.811e-01  5.618e-03   50.03   <2e-16 ***

## ---

## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

##  

## Residual standard error: 0.08257 on 70 degrees of freedom

## Multiple R-squared:  0.9728, Adjusted R-squared:  0.9724  

## F-statistic:  2503 on 1 and 70 DF,  p-value: < 2.2e-16

c. plot(y=rstandard(wages.lm), x=as.vector(time(wages)), type = 'o')

d. #we find Quadratic model trend

wages.qm = lm(wages ~ time(wages) + I(time(wages)^2))

summary(wages.qm)

##  

## Call:

## lm(formula = wages ~ time(wages) + I(time(wages)^2))

##  

## Residuals:

##       Min        1Q    Median        3Q       Max  

## -0.148318 -0.041440  0.001563  0.050089  0.139839  

##  

## Coefficients:

##                    Estimate Std. Error t value Pr(>|t|)    

## (Intercept)      -8.495e+04  1.019e+04  -8.336 4.87e-12 ***

## time(wages)       8.534e+01  1.027e+01   8.309 5.44e-12 ***

## I(time(wages)^2) -2.143e-02  2.588e-03  -8.282 6.10e-12 ***

## ---

## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

##  

## Residual standard error: 0.05889 on 69 degrees of freedom

## Multiple R-squared:  0.9864, Adjusted R-squared:  0.986  

## F-statistic:  2494 on 2 and 69 DF,  p-value: < 2.2e-16

#time series plot of the standardized residuals

plot(y=rstandard(wages.qm), x=as.vector(time(wages)), type = 'o')

wages.qm = lm(wages ~ time(wages) + I(time(wages)^2))

summary(wages.qm)

##  

## Call:

## lm(formula = wages ~ time(wages) + I(time(wages)^2))

##  

## Residuals:

##       Min        1Q    Median        3Q       Max  

## -0.148318 -0.041440  0.001563  0.050089  0.139839  

##  

## Coefficients:

##                    Estimate Std. Error t value Pr(>|t|)    

## (Intercept)      -8.495e+04  1.019e+04  -8.336 4.87e-12 ***

## time(wages)       8.534e+01  1.027e+01   8.309 5.44e-12 ***

## I(time(wages)^2) -2.143e-02  2.588e-03  -8.282 6.10e-12 ***

## ---

## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

##  

## Residual standard error: 0.05889 on 69 degrees of freedom

## Multiple R-squared:  0.9864, Adjusted R-squared:  0.986  

## F-statistic:  2494 on 2 and 69 DF,  p-value: < 2.2e-16

e. #time series plot of the standardized residuals

plot(y=rstandard(wages.qm), x=as.vector(time(wages)), type = 'o')