The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the distance of a focus from the center. Because a>c, it follows that e<1. Write a brief paragraph about the general shape of each of the following ellipses. Be sure to justify your conclusions. (a) Eccentricity close to 0 (b) Eccentricity=5 (c) Eccentricity close to 1

Answers

Answer 1

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as [tex]e=\frac{c}{a}[/tex]

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

[tex]\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)[/tex]

b) Eccentricity =5

[tex]5=\frac{c}{a} \:c=5a[/tex]

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

[tex]If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0[/tex]

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

[tex]a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)[/tex]

[tex]Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1[/tex]

The Eccentricity E Of An Ellipse Is Defined As The Number C/a, Where A Is The Distance Of A Vertex From

Related Questions

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00. The 95% confidence interval for the average hourly wage (in $) of all information system managers is

Answers

Answer:

The 95% confidence interval for the average hourly wage of all information system managers is (39.14,42.36)

Step-by-step explanation:

In order to calculate the 95% confidence interval for the average hourly wage we would have to calculate first the margin of error as follows:

ME=t0.05/2,n-1s/√n

for n=75, t0.025,74=1.993

Therefore, ME=1.993*7/√75

ME=1.61

Therefore, the 95% confidence interval for the average hourly income of all syatem manager would be as follows:

(X-ME,X+ME)=(40.75-1.61,40.75+1.61)

=(39.14,42.36)

Pet Place sells pet food and supplies including a popular bailed hay for horses. When the stock of this hay drops to 20 bails, a replenishment order is placed. The store manager is concerned that sales are being lost due to stock outs while waiting for a replenishment order. It has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails. The manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 bails

Answers

Answer:

The probability that demand during lead-time will exceed 20 bails is 0.2033.

Step-by-step explanation:

We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.

Let X = demand during the lead-time

So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

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

where, [tex]\mu=[/tex] population mean demand = 15 bails

           [tex]\sigma[/tex] = standard deviation = 6 bails

Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)

       P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)

                                                             = 1 - 0.7967 = 0.2033

HELP ASAP GIVING BRANLIST!!

Answers

Answer:

Question 1: 3 - 5 hours.

Question 2: 0 - 1 hour

Step-by-step explanation:

Question 1: As you can see in the diagram, the guy is moving really slowly and is almost stuck, therefore, it is 3 - 5  hours.

Question 2:  In hours 0 - 1, you can see that the graph is the closest to vertical as it gets.

The production department has installed a new spray machine to paint automobile doors. As is common with most spray guns, unsightly blemishes often appear because of improper mixtures or other problems. A worker counted the number of blemishes on each door. Most doors had no blemishes; a few had one; a very few had two; and so on. The average number was 0.5 per door. The distribution of blemishes followed the Poisson distribution. Out of 10,000 doors painted, about how many would have no blemishes

Answers

Answer:

The numbers of doors that will have no blemishes will be about 6065 doors

Step-by-step explanation:

Let the number of counts by the  worker of each blemishes on the door be (X)

The distribution of blemishes followed the Poisson distribution with parameter  [tex]\lambda =0.5[/tex] / door

The probability mass function on of a poisson distribution Is:

[tex]P(X=x) = \dfrac{e^{- \lambda } \lambda ^x}{x!}[/tex]

[tex]P(X=x) = \dfrac{e^{- \ 0.5 }( 0.5)^ x}{x!}[/tex]

The probability that no blemishes occur is :

[tex]P(X=0) = \dfrac{e^{- \ 0.5 }( 0.5)^ 0}{0!}[/tex]

[tex]P(X=0) = 0.60653[/tex]

P(X=0) = 0.6065

Assume the number of paints on the door by q = 10000

Hence; the number of doors that have no blemishes  is = qp

=10,000(0.6065)

= 6065

Suppose that four microchips in a production run of sixty are defective. A sample of six is to be selected to be checked for defects. (a) How many different samples can be chosen

Answers

Answer:

50,063,860 different samples can be chosen

Step-by-step explanation:

The order in which the microchips are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

[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]

How many different samples can be chosen

We choose 6 microchips from a set of 60. So

[tex]C_{60,6} = \frac{60!}{6!(60-6)!} = 50063860[/tex]

50,063,860 different samples can be chosen

6. Factor the expression.
9b2 + 48b + 64
A (3b + 8)2
B (-3b + 8)2
C (-3b - 82
D (3b - 8)2
70%​

Answers

Answer:

A. [tex](3b+8)^2[/tex]

Step-by-step explanation:

[tex]9b^2+48b +64\\=(3b)^2 + 2\times 3b\times 8 +(8)^2\\=(3b+8)^2[/tex]

Lara’s Inc. is currently an unlevered firm with 450,000 shares of stock outstanding, with a market price of $15 a share. The company has earnings before interest and taxes of $314,000. Lara's met with his bankers, Warne Incorporated and agreed to borrow $825,000, at 5 percent. You are an ardent investor and you currently own 20,000 shares of Lara's stock. If you seek to unlevered your position; how many shares of Lara's stock will you continue to own, if you can loan out funds at 5 percent interest? Ignore taxes in your deliberations. Kindly show all workings.

Answers

Answer:

tiStep-by-step explanaon:

EASY MIDDLE SCHOOL MATH WITH INSTANT BRAINIEST AND MANY POINTS, PLZ HELP!!!

Answers

Answer:

Second option is the correct choice. See the explanation below.

Step-by-step explanation:

[tex]A=\frac{bh}{2}\\\\\mathrm{Switch\:sides}:\\\\\frac{bh}{2}=A\\\\\mathrm{Multiply\:both\:sides\:by\:}2\\\\\frac{2bh}{2}=2A\\\\hb=2A\\\\\mathrm{Divide\:both\:sides\:by\:}b;\\\\\frac{hb}{b}=\frac{2A}{b}\\\\h=\frac{2A}{b}[/tex]

Best Regards!

Answer:

[tex]h = \frac{2A}{b} [/tex]

Option B is the right option.

Solution,

[tex]a = \frac{bh}{2} \\ 2a = bh(cross \: multiplication) \\ 2a = b \times h \\ h = \frac{2A}{b} [/tex]

hope this helps...

Good luck on your assignment..

NEED GEOMETRY HELP ASAP (12 POINTS)

Answers

Answer:

HJ > PK

Step-by-step explanation:

Notice that the side PL in one triangle has the same length as side GJ in the other, and side GH has the same size as side LK of the other triangle. Now what is different is the angle subtended between these sides in the case of the triangle on the lower left, the subtended angle is [tex]90^o[/tex] , which is larger angle than that subtended between equal sides on the other triangle ([tex]85^o[/tex])

Therefore, if the angle subtended by the equivalent sides in the triangle on the left is larger than the angle subtended on the right hand side triangle, then the sides associated with such angle aperture must keep the inequality. That is:

Since [tex]\angle\,G\,\,\,>\,\,\,\angle \,L[/tex], then   HJ > PK

Consider random samples selected from the population of all female college soccer players in the United States. Assume the mean height of female college soccer players in the United States is 66 inches and the standard deviation is 3.5 inches. Which do you expect to have less variability (spread): a sampling distribution with sample size n

Answers

Answer:

Option C is correct.

The sampling distribution with sample size n=100 will have less variability.

Step-by-step explanation:

Complete Question

Consider random samples selected from the population of all female college soccer players in the United States. Assume the mean height of female college soccer players in the United States is 66 inches and the standard deviation is 3.5 inches. Which do you expect to have less variability (spread): a sampling distribution with sample size n = 100 or a sample size of n = 20.

A. Both sampling distributions will have the same variability.

B.The sampling distribution with sample size n=20 will have less variability

C. The sampling distribution with sample size n =100 will have less variability

Solution

The central limit theorem allows us to say that as long as

- the sample is randomly selected from the population distribution with each variable independent of each other and with the sample having an adequate enough sample size.

- the random sample is normal or almost normal which is guaranteed if the population distribution that the random sample was extracted from is normal or approximately normal,

1) The mean of sampling distribution (μₓ) is approximately equal to the population mean (μ)

μₓ = μ = 66 inches

2) The standard deviation of the sampling distribution or the standard error of the sample mean is related to the population standard deviation through

σₓ = (σ/√N)

where σ = population standard deviation = 3.5 inches

N = Sample size

And the measure of variability for a sampling distribution is the magnitude of the standard deviation of the sampling distribution.

For sampling distribution with sample size n = 100

σₓ = (3.5/√100) = 0.35 inch

For sampling distribution with sample size n = 20

σₓ = (3.5/√20) = 0.7826 inch

The standard deviation of the sampling distribution with sample size n = 20 is more than double that of the sampling distribution with sample size n = 100, hence, it is evident that the bigger the sample size, the lesser the standard deviation of the sampling distribution and the lesser the variability that the sampling distribution shows.

Hope this Helps!!!

There is more than one integer greater than 1 which leaves a reminder of1 when divided by each of the four smallest primes

Answers

Answer:

210

Complete question found at brainly(ID): 18678557 is stated below.

There is more than one integer, greater than 1, which leaves a remainder of 1 when divided by each of the four smallest primes. What is the difference between the two smallest such integers?

Step-by-step explanation:

Prime numbers are numbers that can only be divided by itself and 1

The smallest of the prime numbers we have = 2, 3, 5, 7

Since the integers greater than 1 are said to be divided by the four smallest prime numbers, we would assume the number of integers are 4 in total.

Let the integers be T

From the question:

Integer/(prime number) = quotient + (remainder/prime number)

Integer/(prime number) = Q + R/P

Let the different quotients derived from all 4 prime number = w, x, y, z

For prime 2:

T/2 = w + 1/2

T/2 - 1/2 = w

(T-1)/2 = w

T = 2w + 1

T-1 = 2w

Following the above solution

For prime 3:

T = 3x + 1

T-1 = 3x

For prime 5:

T = 5y + 1

T-1 = 5y

For prime 7:

T = 7z + 1

T-1 = 7z

T-1= T-1 = T-1 = T-1

2w = 3x = 5y = 7z

T-1 = LCM of all the prime numbers

T- 1 = 2×3×5×7

T-1 = 210

T = 210+1 = 211

T = 211

The smallest of the integer greater than 1 that leaves a remainder of 1 = 1(T-1) + 1 = 211

The next after the smallest number: 2(T-1) +1= 2(210) + 1 = 421

The two smallest number = 1(T-1) + 1 and 2(T-1) +1 respectively

The difference between the two smallest such integers = 421-211 = 210

Two thousand dollars is deposited into a savings account at 8.5​% interest compounded continuously. ​(a) What is the formula for​ A(t), the balance after t​ years? ​(b) What differential equation is satisfied by​ A(t), the balance after t​ years? ​(c) How much money will be in the account after 5 ​years? ​(d) When will the balance reach ​$3000​? ​(e) How fast is the balance growing when it reaches ​$3000​?

Answers

Answer:

a)[tex]A(t)=2000e^{0.085t}[/tex]

b)[tex]A'(t)=170e^{0.085t}[/tex]

c)$3059.1808

d)t=4.77 years

e) The balance growing is $254.99/year

Step-by-step explanation:

We are given that Two thousand dollars is deposited into a savings account at 8.5​% interest compounded continuously.

Principal = $2000

Rate of interest = 8.5%

a) What is the formula for​ A(t), the balance after t​ years? ​

Formula [tex]A(t)=Pe^{rt}[/tex]

So,[tex]A(t)=2000e^{0.085t}[/tex]

B)What differential equation is satisfied by​ A(t), the balance after t​ years?

So, [tex]A'(t)=2000 \times 0.085 e^{0.085t}[/tex]

[tex]A'(t)=170e^{0.085t}[/tex]

c)How much money will be in the account after 5 ​years? ​

Substitute t = 5 in the formula "

[tex]A(t)=2000e^{0.085t}\\A(5)=2000e^{0.085(5)}\\A(5)=3059.1808[/tex]

d)When will the balance reach ​$3000​?

Substitute A(t)=3000

So, [tex]3000=2000e^{0.085t}[/tex]

t=4.77

The balance reach $3000 in 4.77 years

e)How fast is the balance growing when it reaches ​$3000​?

Substitute the value of t = 4.77 in derivative formula :

[tex]A'(t)=170e^{0.085t}\\A'(t)=170e^{0.085 \times 4.77}\\A'(t)=254.99[/tex]

Hence the balance growing is $254.99/year

Find the following measure for this figure.
Volume =

Answers

Answer:

91 2/3 pi cubic units

Step-by-step explanation:

The formula for the volume of cone is [tex]\dfrac{1}{3}\pi r^2 h[/tex]. Plugging in the given numbers, you get:

[tex]\dfrac{1}{3}\cdot \pi \cdot 5^2 \cdot 11= 91 \ 2/3 \pi[/tex]

Hope this helps!

Answer:

[tex]Volume=\frac{1}{3} \,275\,\pi[/tex] cubic units

Notice that this answer doesn't agree with any of the first three in the list provided via the screenshot

Step-by-step explanation:

Recall the formula for the volume of a cone:

[tex]Volume=\frac{1}{3} Base\,*\,Height[/tex]

In this case the Height is 11 units, and they also give us the radius of the circular base (5 units) from which we can find the circle's base area:

[tex]Area_{circle} = \pi\,R^2\\Area_{circle}=\pi\,(5)^2\\Area_{circle}=25 \pi[/tex]

therefore the total volume becomes:

[tex]Volume=\frac{1}{3} Base\,*\,Height\\Volume=\frac{1}{3} 25\,\pi\,*\,11\\\\Volume=\frac{1}{3} \,275\,\pi[/tex]

Determine the present value P that must be invested to have the future value A at simple interest rate r after time t. A​ = ​$7000.00​, r​ = 9.5​%, t​ = 9 months

Answers

Answer:

  $6534.42

Step-by-step explanation:

Put the given values into the simple interest formula and solve for the remaining variable.

  A = P(1 +rt)

where P is the principal invested, r is the annual rate, and t is the number of years.

  $7000 = P(1 +0.095(9/12)) = 1.07125P

  $7000/1.07125 = P ≈ $6534.42

The value that must be invested is $6534.42.

The World Issues club has decided to donate 60% of all their fundraising activities this year to Stephen Lewis Foundation. This foundation was created to help ease the pain of HIV/AIDS in Africa. Lewis, a Canadian, works for the United Nations trying to determine ways to stop the spread of this deadly disease from crippling an entire continent. Choose a variable to represent the money earned during fundraising activities and the revenue generated for the foundation Use these variables to create an equation that will determine the amount of money the foundation will receive. In their latest bake sale, the club raised $72. Calculate the amount the foundation will receive. At the end of the year, the World Issues Club mailed a cheque to the foundation for $850. How much money did they fundraise in total?

Answers

Answer:

$43.20$1416.67

Step-by-step explanation:

Let the money earned during fundraising activities =x

Since the World Issues club has decided to donate 60% of all their fundraising activities this year to Stephen Lewis Foundation.

The amount of money the foundation will receive

=60% of x

= 0.6x

In the bake sale, the club raised $72.

Therefore, the amount the foundation will receive =0.6*72=$43.20

At the end of the year, the World Issues Club mailed a cheque to the foundation for $850.

Therefore:

0.6x=850

x=850/0.6

x=$1416.67

The total amount of money the club raised is $1416.67.

Find the area of a triangle that has the base of 5 inches and a height of 3 3/4 inches

Answers

Answer:

9.375 in^2

Step-by-step explanation:

Find the area of the smaller sector.
A
6 in
030°
Area = [? ]in?
B
Round your answer to the nearest hundredth.

Answers

Answer:

9.42 in²

Step-by-step explanation:

The area of whole circle S=pi*R²    , where pi is appr. 3.14,  R= 6 in

S= 3.14*6² =113.04 in²

The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²

The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.

To find the area of a sector, you can use the formula:

Area of sector = (θ/360) × π × r²

where θ is the central angle in degrees, r is the radius of the sector.

The central angle is 30 degrees and the radius is 6 inches.

Plugging these values into the formula:

Area of sector = (30/360) × π × 6²

= (1/12) × π × 36

= (1/12) × 3.14159 × 36

= 9.42478 square inches

To learn more on Area of sector click:

https://brainly.com/question/29055300

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pls helppppp with my math​

Answers

Answer:

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

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]\frac{5}{6}-\frac{2}{3}=\frac{5}{6}-\frac{2*2}{3*2}\\\\=\frac{5}{6}-\frac{4}{6}\\\\=\frac{5-4}{6}\\\\=\frac{1}{6}[/tex]

Someone flips five coins, but you don’t see the outcome. The person reports that no tails are showing. What is the probability that the person flipped 5 heads?

Answers

Answer:

0.03125 = 3.125% probability that the person flipped 5 heads

Step-by-step explanation:

For each coin, there are only two possible outcomes. Either it was heads, or it was tails. The result of a coin toss is independent of other coin tosses. 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.

Five coins:

This means that n = 5.

Fair coin:

Equally as likely to be heads or tails, so p = 0.5.

What is the probability that the person flipped 5 heads?

This is P(X = 5).

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

[tex]P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125[/tex]

0.03125 = 3.125% probability that the person flipped 5 heads

The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.5 ppm and standard deviation 1.4 ppm. 18 randomly selected large cities are studied. Round all answers to two decimal places.
A. xBar~ N( ____) (____)
B. For the 18 cities, find the probability that the average amount of pollutants is more than 9 ppm.
C. What is the probability that one randomly selected city's waterway will have more than 9 ppm pollutants?
D. Find the IQR for the average of 18 cities.Q1 =
Q3 =
IQR:
2. X ~ N(30,10). Suppose that you form random samples with sample size 4 from this distribution. Let xBar be the random variable of averages. Let ΣX be the random variable of sums. Round all answers to two decimal places.
A. xBar~ N(___) (____)
B. P(xBar<30) =
C. Find the 95th percentile for the xBar distribution.
D. P(xBar > 36)=
E. Q3 for the xBar distribution =

Answers

Answer:

1)

A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552

C) P(X> 9)= 0.36317

D) IQR= 0.4422

2)

A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)

B) P( [tex]\frac{}{X}[/tex]<30)= 0.50

C) P₉₅= 32.60

D) P( [tex]\frac{}{X}[/tex]>36)= 0

E) Q₃: 31.0586

Step-by-step explanation:

Hello!

1)

The variable of interest is

X: pollutants found in waterways near a large city. (ppm)

This variable has a normal distribution:

X~N(μ;σ²)

μ= 8.5 ppm

σ= 1.4 ppm

A sample of 18 large cities were studied.

A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

The population mean is the same as the mean of the variable

μ= 8.5 ppm

The standard deviation is

σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108

So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B)

P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)

To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.

Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]

Then using the Z table you'll find the probability of

P(Z≤1.51)= 0.93448

Then

1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552

C)

In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:

P(X> 9)= 1 - P(X ≤ 9)

Z= (X-μ)/δ= (9-8.5)/1.44

Z= 0.347= 0.35

P(Z≤0.35)= 0.63683

Then

P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317

D)

The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:

Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25

Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:

P(Z≤z₁)= 0.25

Using the table you have to identify the value of Z that accumulates 0.25 of probability:

z₁= -0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)

z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)

[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5=  8.2789 ppm

The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

Using the table you have to identify the value of Z that accumulates 0.75 of probability:

z₃= 0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5=  8.7211 ppm

IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422

2)

A)

X ~ N(30,10)

For n=4

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

Population mean μ= 30

Population variance σ²/n= 10/4= 2.5

Population standard deviation σ/√n= √2.5= 1.58

[tex]\frac{}{X}[/tex] ~ N(30;2.5)

B)

P( [tex]\frac{}{X}[/tex]<30)

First you have to standardize the value and then look for the probability:

Z=  ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0

P(Z<0)= 0.50

Then

P( [tex]\frac{}{X}[/tex]<30)= 0.50

Which is no surprise since 30 y the value of the mean of the distribution.

C)

P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95

P( Z≤ z₀)= 0.95

z₀= 1.645

Now you have to reverse the standardization:

z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)

z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)

[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60

P₉₅= 32.60

D)

P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0

E)

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

z₃= 0.67

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586

Q₃: 31.0586

PLEASE ANSWER FAST !!!
What is the range of the function g for given domain ?

Answers

Answer:

The answer is B

Step-by-step explanation:

Hope this helps.. if not im sorry :(

The assembly time for a product is uniformly distributed between 8 and 12 minutes.The mean and the variance of the assembly time are: a.4 minutes and 16 (minute)2 b.8 minutes and 12 (minute)2 c.12 minutes and 1.33 (minute)2 d.10 minutes and 1.33 (minute)2

Answers

Answer:

d. 10 minutes and 1.33 minutes.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean of the uniform distribution is:

[tex]M = \frac{a + b}{2}[/tex]

The variance of the uniform distribution is given by:

[tex]V = \frac{(b-a)^{2}}{12}[/tex]

The assembly time for a product is uniformly distributed between 8 and 12 minutes.

This means that [tex]a = 8, b = 12[/tex].

Mean:

[tex]M = \frac{8 + 12}{2} = 10[/tex]

Variance:

[tex]V = \frac{(12-8)^{2}}{12} = 1.33[/tex]

So the correct answer is:

d. 10 minutes and 1.33 minutes.

A consumer group was interested in comparing the operating time of cordless toothbrushes manufactured by two different companies. Group members took a random sample of 18 toothbrushes from Company A and 15 from Company B. Each was charged overnight and the number of hours of use before needing to be recharged was recorded. Company A toothbrushes operated for an average of 119.7 hours with a standard deviation of 1.74 hours; Company B toothbrushes operated for an average of 120.6 hours with a standard deviation of 1.72 hours. The 90% confidence interval is (-1.93, 0.13). The correct interpretation is:__________
A. We are 90% confident that, on average, there is no difference in operating hours between toothbrushes from Company A compared to those from Company B.
B. We are 90% confident that, on average, there is a difference in operating hours between toothbrushes from Company A compared to those from Company B.
C. We are 90% confident that, on average, the toothbrushes from Company B operate longer before needing to be recharged than the toothbrushes from Company A.
D. We are 90% confident that, on average, the toothbrushes from Company A operate longer before needing to be recharged than the toothbrushes from Company B.

Answers

Answer:

The degrees of freedom are given by:

[tex] df = n_A +n_B -2 = 18 +15-2= 31[/tex]

And the 90% confidence interval for this case is:

[tex] -1.90 \leq \mu_A -\mu_B \leq 0.13[/tex]

And for this case since the confidence interval contains the value 0 we can conclude that:

A. We are 90% confident that, on average, there is no difference in operating hours between toothbrushes from Company A compared to those from Company B.

Step-by-step explanation:

We know the following info given:

[tex] \bar X_A= 119.7[/tex] sample mean for A

[tex] s_A = 1.74[/tex] sample deviation for A

[tex] n_A = 18[/tex] sample size from A

[tex] \bar X_B= 120.6[/tex] sample mean for B

[tex] s_B = 1.72[/tex] sample deviation for B

[tex] n_B = 15[/tex] sample size from B

The degrees of freedom are given by:

[tex] df = n_A +n_B -2 = 18 +15-2= 31[/tex]

And the 90% confidence interval for this case is:

[tex] -1.90 \leq \mu_A -\mu_B \leq 0.13[/tex]

And for this case since the confidence interval contains the value 0 we can conclude that:

A. We are 90% confident that, on average, there is no difference in operating hours between toothbrushes from Company A compared to those from Company B.

What is the slope of the line with the two
points A(-4, 8) and B(-9, 12)?

Answers

Answer:

slope = -4/5

Step-by-step explanation:

A line passes two points (x1, y1) and (x2, y2).

The slope of this line can be calculate by the formula:

s = (y2 - y1)/(x2 - x1)

=>The line that passes A(-4, 8) and B(-9, 12) has the slope:

s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5

Hope this helps!

Please answer this correctly

Answers

The correct answer to this is 1. Trust me

1. A random sample of 64 customers at a drive-through bank window is observed, and it is found that the teller spends an average of 2.8 minutes with each customer, with a standard deviation of 1.2 minutes. Is there sufficient evidence to conclude that the teller spends less than 3 minutes with each customer

Answers

Answer:

The 95% confidence interval is 2.5 < u <3.1.

Step-by-step explanation:

The provided sample mean is X = 2.8 and the sample standard deviation is s = 1.2, and the sample size is n = 64.

1. Null and Alternative Hypotheses:

The following null and alternative hypotheses need to be tested:

H0  u = 3

Ha: u < 3

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

2. Rejection Region Based

on the information provided, the significance level is alpha = 0.05, and the critical value for a left-tailed test is t c = -1.669.

The rejection region for this left-tailed test is R = t : t < -1.669

3. Test Statistics

The t-statistic is computed as follows:

t = (X - uo)/[s/n^(1/2)] =

replacing

t = (2.8 - 3)/ [1.2/64 ^(1/2)]

t =-1.333  

4. Decision about the null hypothesis

Since it is observed that t = -1.333 > t c = -1.669, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.0936, and since p= 0.0936 => 0.05, it is concluded that the null hypothesis is not rejected.

5. Conclusion It is concluded that the null hypothesis H0 is not rejected. Therefore, there is not enough evidence to claim that the population mean u is less than 3, at the 0.05 significance level.

Confidence Interval

The 95% confidence interval is 2.5 < u <3.1.

What is the result of −18⋅16 2/3? Enter the result as an improper fraction and as a mixed number.

Answers

Answer:

-30000/100

300 0/1

Step-by-step explanation:

We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.

16 2/3 = 16.67

We do the multiplication:

−18⋅16 2/3 = -300

We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:

-30000/100

How mixed number would it be:

300 0/1

Please answer this question I give brainliest thank you! Number 8

Answers

Answer:

The third options

Step-by-step explanation:

Counting we can see that 10 students went to two or less states, and 10 went to three or more

Answer: half of the students has visited 2 or fewer states.

Explanation: if we count the total amount of students in the survey we can determine that there was a total of 20 students. Half of 20 is 10. Option three is true because if we count the amount of students who have visited 2 or fewer a states it adds up to 10 making that statement true

ali's typing rate between 8:00 am and noon is 48 words per minute . after lunch a lunch break, Ali's typing rate between 1:00 pm and 4:00 pm is 2,040 words per hour . what is Ali's average typing rate per minute for the whole time she works?

Answers

Answer:

41 word/min

Step-by-step explanation:

Before noon Ali works:

4 hours= 4*60 min= 240 min

She types:

240*48= 11520 words

After lunch she works:

4 hours

She types:

4*2040= 8160 words

Total Ali works= 4+4= 8 hours= 480 min

Total Ali types= 11520+8160= 19680 words

Average typing rate= 19680 words/480 min= 41 word/min

A research organization keeps track of what citizens think is the most important problem facing the country today. They randomly sampled a number of people in 2003 and again in 2009 using a different random sample of people in 2009 than in 2003 and asked them to choose the most important problem facing the country today from the following​ choices, war,​ economy, health​ care, or other. Which of the following is the correct test to use to determine if the distribution of​ "problem facing this country​today" is different between the two different​ years?

A.

Use a​ chi-square test of homogeneity.

B.

Use a paired​ t-test.

C.

Use a​ two-sample z-test for proportions.

D.

Use a​ chi-square goodness-of-fit test.

Answers

Answer:

Step-by-step explanation:

From the information given, the population is divided into sub groups. Each group would consist of citizens picking a particular choice as the most important problem facing the country. The choices are the different categories. In this case, the null hypothesis would state that the distribution of proportions for all categories is the same in each population. The alternative hypothesis would state that the distributions is different. Therefore, the correct test to use to determine if the distribution of​ "problem facing this country ​today" is different between the two different​ years is

A) Use a​ chi-square test of homogeneity.

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