The distribution of cholesterol levels in teenage boys is approximately normal withnbsp mu equals 170 and sigma equals 30 ​(Source: U.S. National Center for Health​ Statistics). Levels above 200 warrant attention. Find the probability that a teenage boy has a cholesterol level greater than 225.

Answers

Answer 1

Answer: 0.0336

Step-by-step explanation:

Given : The distribution of cholesterol levels in teenage boys is approximately normal with mean :[tex]\mu= 170[/tex]

Standard deviation : [tex]\sigma= 30[/tex]

The formula for z-score is given by :-

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

For x=225

[tex]z=\dfrac{225-170}{30}=1.83[/tex]

The p-value =[tex]P(z>1.83)=1-P(z<1.83)[/tex]

[tex]=1-0.966375=0.033625\approx0.0336[/tex]

The probability that a teenage boy has a cholesterol level greater than 225 =0.0336


Related Questions

The probability of buying a movie ticket with a popcorn coupon is 0.629 and without a popcorn coupon is 0.371. If you buy 29 movie tickets, we want to know the probability that more than 16 of the tickets have popcorn coupons. Consider tickets with popcorn coupons as successes in the binomial distribution. Give the numerical value of the parameter p in this binomial distribution scenario.

Answers

Answer:

The probability[tex]0.75095[/tex] and the parameter [tex]p=0.629[/tex]

Step-by-step explanation:

The formula for probability  in a binomial distribution is:

[tex]b(x;n,p)=\frac{n!}{x!(n-x)!}\ast p^{x}\ast(1-p)^{n-x}[/tex]

where p is the probability of success (ticket with popcorn coupon), n is the number of trials (tickets bought) and x the number of successes desired. In this case p=0.629 (probability of buying a movie ticket with coupon), n=29,  and x=17,18,19, ...29.

[tex]b(17;29,0.629)=\frac{29!}{17!(29-17)!}\ast0.629^{17}\ast(1-0.629)^{29-17}=0.133\,25\\ b(18;29,0.629)=\frac{29!}{18!(29-18)!}\ast0.629^{18}\ast(1-0.629)^{29-18}=0.150\,61[/tex]

[tex]b(19;29,0.629)=\frac{29!}{19!(29-19)!}\ast0.629^{19}\ast(1-0.629)^{29-19}=0.147\,84\\ b(20;29,0.629)=\frac{29!}{20!(29-20)!}\ast0.629^{20}\ast(1-0.629)^{29-20}=0.125\,32 \\ b(21;29,0.629)=\frac{29!}{21!(29-21)!}\ast0.629^{21}\ast(1-0.629)^{29-21}=0.091\,06 \\ b(22;29,0.629)=\frac{29!}{22!(29-22)!}\ast0.629^{22}\ast(1-0.629)^{29-22}=0.056\,14[/tex]

[tex]b(23;29,0.629)=\frac{29!}{23!(29-23)!}\ast0.629^{23}\ast(1-0.629)^{29-23}=2.896\,8\times10^{-2} \\ b(24;29,0.629)=\frac{29!}{24!(29-24)!}\ast0.629^{24}\ast(1-0.629)^{29-24}=1.227\,8\times10^{-2}\\ b(25;29,0.629)=\frac{29!}{25!(29-25)!}\ast0.629^{25}\ast(1-0.629)^{29-25}=4.163\,4\times10^{-3} \\ b(26;29,0.629)=\frac{29!}{26!(29-26)!}\ast0.629^{26}\ast(1-0.629)^{29-26}=1.085\,9\times10^{-3} \\ b(27;29,0.629)=\frac{29!}{27!(29-27)!}\ast0.629^{27}\ast(1-0.629)^{29-27}=2.045\,7\times10^{-4}[/tex]

[tex]b(28;29,0.629)=\frac{29!}{28!(29-28)!}\ast0.629^{28}\ast(1-0.629)^{29-28}=2.477\,4\times10^{-5} \\ b(29;29,0.629)=\frac{29!}{29!(29-29)!}\ast0.629^{29}\ast(1-0.629)^{29-29}=1.448\,3\times10^{-6}[/tex]

The probability of more than 16 is equal to the sum of the probability of x=17, 17,18,19, ...29.

[tex]b(x>16;29,0.629)=0.13325+0.15061+0.14784+0.12532+0.09106+0.05614+2.8968\times10^{-2}+1.2278\times10^{-2}+4.1634\times10^{-3}+1.0859\times10^{-3}+2.0457\times10^{-4}+2.4774\times10^{-5}+1.4483\times10^{-6}=0.75095[/tex]

Final answer:

The numerical value of the parameter p in this binocular distribution scenario, where getting a popcorn coupon is defined as a success, is 0.629.

Explanation:

In this binomial distribution scenario, we are considering buying a movie ticket with a popcorn coupon as a success. The probability of success (p) is given as 0.629. Therefore, in this context, the numerical value of the parameter p for the binomial distribution is 0.629.

This means that each time a ticket is bought, there is a 0.629 chance (or 62.9%) of getting a popcorn coupon with the ticket. This probability stays constant with each new ticket purchase. In other words, the purchase of one ticket does not influence the likelihood of the outcome of the next ticket. This makes the scenario suitable to be modeled using a binomial distribution.

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Last​ year, a person wrote 128 checks. Let the random variable x represent the number of checks he wrote in one​ day, and assume that it has a Poisson distribution. What is the mean number of checks written per​ day? What is the standard​ deviation? What is the​ variance?

Answers

Answer: The mean number of checks written per​ day =  [tex]\lambda=0.3507[/tex]

[tex]\text{Variance}(\sigma^2)=\lambda=0.3507[/tex]

[tex]\text{Standard deviation}=0.5922[/tex]

Step-by-step explanation:

Given : A person wrote 128 checks in last year.

Consider , the last year is a no-leap year.

The number of days in ;last year = 365 days

Let X be the number of checks in one day.

Then , [tex]X=\dfrac{128}{365}=0.350684931507\approx0.3507[/tex]

The mean number of checks written per​ day =  [tex]\lambda=0.3507[/tex]

Now X follows Poisson distribution with parameter [tex]\lambda=0.3507[/tex].

Then , [tex]\text{Variance}(\sigma^2)=\lambda=0.3507[/tex]

[tex]\Rightarrow\sigma=\sqrt{\lambda}=\sqrt{0.3507}=0.59219929078\approx0.5922[/tex]

PLEASE HELP ME GET THESE FINISHED

Answers

Answer:

g(-0.5) = -1

g(0.2) = 0

g(0.5) = 1

Step-by-step explanation:

We are given the value of g(x) for which the x is defined.

Solving

g(-0.5) = -1

As given g(x) = -1 if -1.5 ≤ x ≤ 0.5

g(0.2) = 0

As given g(x) = 0 if -0.5 < x < 0.5

g(0.5) = 1

As given g(x) = 1 if 0.5 ≤ x < 1.5

The mean per capita income is 15,451 dollars per annum with a variance of 298,116. What is the probability that the sample mean would differ from the true mean by less than 22 dollars if a sample of 350 persons is randomly selected? Round your answer to four decimal places.

Answers

Answer: 0.5467

Step-by-step explanation:

Let X be the random variable that represents the income (in dollars) of a randomly selected person.

Given : [tex]\mu=15451[/tex]

[tex]\sigma^2=298116\\\\\Rightarrow\ \sigma=\sqrt{298116}=546[/tex]

Sample size : n=350

z-score : [tex]\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

To find the probability that the sample mean would differ from the true mean by less than 22 dollars, the interval will be

[tex]\mu-22,\ \mu+22\\\\=15,451 -22,\ 15,451 +22\\\\=15429,\ 15473[/tex]

For x=15429

[tex]z=\dfrac{15429-15451}{\dfrac{546}{\sqrt{350}}}\approx-0.75[/tex]

For x=15473

[tex]z=\dfrac{15473-15451}{\dfrac{546}{\sqrt{350}}}\approx0.75[/tex]

The required probability :-

[tex]P(15429<X<15473)=P(-0.75<z<0.75)\\\\=1-2(P(z<0.75))=1-2(0.2266274)=0.5467452\approx0.5467[/tex]

Hence, the required probability is 0.5467.

17. Prove the following statement: Let n e Z. If n is odd, then n2 is odd. Proof 6 pts. 2 VÎ±Î¶Ï there fore, n and n Please See tue classnotes

Answers

Answer with explanation:

It is given that , n is Odd integer.

If , n is odd, then it can be Written as with the help of Euclid division lemma

 → n= 2 p +1, as 2 p is even , and adding 1 to it converts it into Odd.

As Euclid lemma states that for any three integers, a ,b and c ,when a is divided by b, gives quotient c and remainder r , then it can be written as:

 a= b c+r, →→0≤r<b

Now, n² will be of the form

(2 p +1)²=(2 p)²+2 × 2 p×1+ (1)²

             =4 p²+4 p +1

⇒Multiplying any positive or negative Integer by 4, gives Even integer and sum  or Difference of two even integer is always even.

So, 4 p²+4 p, will be an even term.But Adding , 1 to it converts it into Odd Integer.

Hence, if n is an Odd number then , n² will be also odd.

Does the mean of a normal distribution is always positive? How about the standard deviation?

Answers

Answer:

Mean of a Normal Distribution:

The mean of normal distribution is not always positive, it is equally distributed around mean,mode and median and can be any value from ranging from negative to positive to infinity.

Standard Deviation:

For the standard deviation, it can not be negative. It can only be equal to any positive values i.e., values [tex]\geq 0[/tex].

Find the point on the plane 4x+3y+z=10 that is nearest to​ (2,0,1). What are the values of​ x, y, and z for the​ point? x= 28 / 13 y = 3 / 26 z= 27 / 26 ​(Type integers or simplified​ fractions.)

Answers

To find the point on the plane that is nearest to (2,0,1), we minimize the squared distance between the two points using partial derivatives and set them equal to 0. The values of x, y, and z for the point are x = 28/13, y = 3/26, and z = 27/26.

To find the point on the plane that is nearest to (2,0,1), we need to find the coordinates that satisfy the equation 4x+3y+z=10 and minimize the distance between the point and (2,0,1).

This can be done by minimizing the squared distance between the two points. Using the formula for distance, we get the squared distance as:

d^2 = (x-2)^2 + y^2 + (z-1)^2

To minimize the squared distance, we can find the partial derivatives with respect to x, y, and z and set them equal to 0.

Solving these equations, we find that x = 28/13, y = 3/26, and z = 27/26.

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A sample of ????=25 diners at a local restaurant had a mean lunch bill of $16 with a standard deviation of ????=$4 . We obtain a 95% confidence interval as (14.43,17.57) . Which action will not reduce the margin of error?

Answers

Answer:

"Decreasing the sample size."

Step-by-step explanation:

The margin of error is :

[tex]ME=z\times \frac{\sigma }{\sqrt{n}}[/tex]

Therefore, as n decreases, margin of error increases.

In the question we are asked " Which action will not reduce the margin of error."

Therefore, the correct option for which action will not reduce the margin of error is "Decreasing the sample size."

The probability that Mary will win a game is 0.02, so the probability that she will not win is 0.98. If Mary wins, she will be given $160; if she loses, she must pay $16. If X = amount of money Mary wins (or loses), what is the expected value of X? (Round your answer to the nearest cent.)

Answers

Given:

Probability of winning, P(X) = 0.02

Probability of losing, P([tex]\bar{X}[/tex]) = 0.98

Wining amount = $160

Losing amount = $16

Step-by-step explanation:

Let the expected amount of money win be  'X'

Expected value of X, E(X) = Probability of winning, P(X).Probability of winning, P(X)  - Probability of losing, P([tex]\bar{X}[/tex]).Losing amount

Now,

E(X) = ([tex]0.02\times 160 - 0.98\times 16[/tex])

E(X) = -12.48

Expected value of X = -12.48

Expected loss value = $12.48 loss

Since 2007, a particular fund returned 13.3% compounded monthly. How much would a $4000 investment in this fund have been worth after 2 years? (Round your answer to the nearest cent.)

Answers

Answer:

$5,211.30

Step-by-step explanation:

We have to calculate compound interest with the formula [tex]A=P(1+\frac{r}{n})^{nt}[/tex]

Where A =  Amount after maturity

           P = Principal amount ( $4,000)

           r = Rate of interest 13.3% in decimal ( 0.133)

           n = number of compounding period, monthly ( 12 )

            t = Time in years ( 2 )

Now we put the values in the formula

[tex]A=4,000(1+\frac{0.133}{12})^{(12\times 2)}[/tex]

[tex]A=4,000(1+0.0110833)^{(24)}[/tex]

[tex]A=4,000\times 1.0110833^{24}[/tex]

[tex]A=4,000\times 1.3028262297[/tex]

A =  $5,211.30

After 2 years investment would be $5,211.30.

4> Solve by using Laplace transform: y'+5y'+4y=0; y(0)=3 y'(o)=o

Answers

Answer:

[tex]y=3e^{-4t}[/tex]

Step-by-step explanation:

[tex]y''+5y'+4y=0[/tex]

Applying the Laplace transform:

[tex]\mathcal{L}[y'']+5\mathcal{L}[y']+4\mathcal{L}[y']=0[/tex]

With the formulas:

[tex]\mathcal{L}[y'']=s^2\mathcal{L}[y]-y(0)s-y'(0)[/tex]

[tex]\mathcal{L}[y']=s\mathcal{L}[y]-y(0)[/tex]

[tex]\mathcal{L}[x]=L[/tex]

[tex]s^2L-3s+5sL-3+4L=0[/tex]

Solving for [tex]L[/tex]

[tex]L(s^2+5s+4)=3s+3[/tex]

[tex]L=\frac{3s+3}{s^2+5s+4}[/tex]

[tex]L=\frac{3(s+1)}{(s+1)(s+4)}[/tex]

[tex]L=\frac3{s+4}[/tex]

Apply the inverse Laplace transform with this formula:

[tex]\mathcal{L}^{-1}[\frac1{s-a}]=e^{at}[/tex]

[tex]y=3\mathcal{L}^{-1}[\frac1{s+4}]=3e^{-4t}[/tex]

A study was conducted to measure the effectiveness of a diet program that claims to help manage weight. Subjects were randomly selected to participate. Before beginning the program, each participant was given a score based on his or her fitness level. After six months of following the diet, each participant received another score. The study wanted to test whether there was a difference between before and after scores. What is the correct alternative hypothesis for this analysis?

Answers

Answer: u (sub d) is inequal to zero

Step-by-step explanation: Because this is a paired t-test, our alternative hypothesis would be u(sub d) is inequal to zero.

Final answer:

The alternative hypothesis for a study on the effectiveness of a diet program would express the expectation of a statistically significant change in fitness level scores, likely a decrease if lower scores indicate better fitness, after participating in the program compared to before.

Explanation:

The correct alternative hypothesis for a study that aims to measure the effectiveness of a diet program in managing weight would look at whether there is a statistically significant difference in the fitness level scores of the participants before and after following the program. As the question suggests evaluating the effectiveness of a diet program, we are particularly interested in seeing an improvement, which would mean expecting a lower score after the program if the score represents a measure where lower is better.

Thus, the alternative hypothesis (H1) should reflect the expectation of improvement. If the fitness score is such that a lower score indicates better fitness, the alternative hypothesis would be:

H1: The mean fitness level score after the program is lower than the mean fitness level score before the program.

This implies that the diet program is effective in improving fitness levels. On the contrary, if a higher fitness score indicates better fitness, the alternative hypothesis would be framed to reflect an expected increase in the score after the program.

Which equation correctly describes the relationship between the measures of the angles and arcs formed by the intersecting secants?


m∠1=1/2(mAB−mEF)

m∠1=1/2(mAB+mEF)

m∠1=1/2mAB

m∠1=mAB+mEF

Answers

Answer:

  m∠1=1/2(mAB+mEF)

Step-by-step explanation:

The measure of the angle is half the sum of the intercepted arcs.

Answer:

[tex]m\angle 1 = \frac{1}{2}(m\angle AB+m\angle EF)[/tex]

Step-by-step explanation:

When two chords intersect each other inside a circle, the measure of the angle formed is one half the sum of the measure of the intercepted arcs.

Here, the chords FA and BE intersected each other inside the circle,

Also, angle 1 is the angle formed by the intersection,

Thus, from the above statement,

[tex]m\angle 1 = \frac{1}{2}(m\angle AB+m\angle EF)[/tex]

Second option is correct.

My Notes OAsk Your Tea The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. If 1 gram of this isotope is present initially, how long will it take for 85% of the lead to decay? (Round your answel to two decimal places.) hr

Answers

Answer:

10.96 hours will take for 85% of the lead to decay.

Step-by-step explanation:

Suppose A represents the amount of Pb-209 at time t,

According to the question,

[tex]\frac{dA}{dt}\propto A[/tex]

[tex]\implies \frac{dA}{dt}=kA[/tex]

[tex]\int \frac{dA}{A}=\int kdt[/tex]

[tex]ln|A|=kt+C_1[/tex]

[tex]A=e^{kt+C_1}[/tex]

[tex]A=e^{C_1} e^{kt}[/tex]

[tex]\implies A=C e^{kt}[/tex]

Let [tex]A_0[/tex] be the initial amount,

[tex]A_0=C e^{0} = C[/tex]

[tex]\implies A=A_0 e^{kt}[/tex]

Since, the half-life of 3.3 hours.

[tex]\implies \frac{A_0}{2}=A_0 e^{3.3k}\implies e^{3.3k}=0.5\implies k=-0.21004[/tex]

[tex]\implies A=A_0 e^{-0.21004t}[/tex]

Here, [tex]A_0=1\text{ gram}[/tex]

[tex]A=(100-85)\% \text{ of }A_0=15\%\text{ of }A_0=0.15A_0[/tex]

By substituting the values,

[tex]0.15A_0=A_0 e^{-0.21004t}[/tex]

[tex]0.15=e^{-0.21004t}[/tex]

[tex]\implies t\approx 10.96\text{ hour}[/tex]

Use the Newton-Raphson method to determine the solution of the simultaneous nonlinear equations: y=−x2+x+0.75 y+5xy=x2 Use the initial guesses of x = y = 1.2, and iterate until the 4th iteration. (Round the final answers to five decimal places.) The values of x and y are as follows: iterationxy01.21.21 0.0290321.39412 3 0.239294

Answers

Answer:

Step-by-step explanation:

Let's solve for y.

−x2+x+0.75y+5xy=x2

Step 1: Add x^2 to both sides.

−x2+5xy+x+0.75y+x2=x2+x2

5xy+x+0.75y=2x2

Step 2: Add -x to both sides.

5xy+x+0.75y+−x=2x2+−x

5xy+0.75y=2x2−x

Step 3: Factor out variable y.

y(5x+0.75)=2x2−x

Step 4: Divide both sides by 5x+0.75.

y(5x+0.75)

5x+0.75

=

2x2−x

5x+0.75

y=

2x2−x

5x+0.75

Answer:

y=

2x2−x

5x+0.75

Final answer:

The Newton-Raphson method is a numerical technique used to find roots of nonlinear equations. Through iterations, the method identifies a value that satisfies the equations provided. Doing this for four iterations, we find that x=0.239294 and y=1.39412.

Explanation:

The Newton-Raphson method uses iterations to find the solution of nonlinear equations. Given the equations y = -x^2+x+0.75 and y+5xy=x^2 and the initial guesses x=y=1.2, we'll use the Newton-Raphson method to find the roots.

We can represent these equations as f(x,y) = -x^2+x+0.75 - y and g(x,y) = x^2 - y - 5xy = 0. The Newton-Raphson method works by using the Jacobian matrix, comprised of the partial derivatives of the equations with respect to x and y, to estimate new values for x and y with each iteration. Starting with our initial values of x and y, we then repeatedly apply the formula to calculate new values of x and y until we've reached the desired number of iterations.

Using this method, you end up with the following values: x=0.239294 and y=1.39412 for iteration 4. Remember, this method uses an iterative approach, and different starting values might yield slightly different results.

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Find a simplified weighted voting system which is equivalent to

[8: 9, 3, 2, 1] and

[20: 8, 6, 3, 2, 1].

Answers

Answer: The explanation is as follows:

Step-by-step explanation:

(a) [8: 9, 3, 2, 1]

q = 8

Here, coalition is as follows:

[P1, P2, P3, P4] = [9, 3, 2, 1]

for the above coalition, the combined weight is

[P1, P2, P3, P4] = 9+3+2+1 = 15 ⇒ combined weight

For simplified weighted voting system;

q = combined weight ⇒ both the terms have to be equal for a simplified weighted voting system.

But, here 8 ≠ 15

It is not a simplified weighted voting system.

(b) [20: 8, 6, 3, 2, 1]

q = 20

Here, coalition is as follows:

[P1, P2, P3, P4, P5] = [8, 6, 3, 2, 1]

for the above coalition, the combined weight is

[P1, P2, P3, P4, P5] = 8+6+3+2+1 = 20 ⇒ combined weight

For simplified weighted voting system;

q = combined weight

Since,  20 = 20

It is a simplified weighted voting system.


I need the answer to this math question.

1) Divide 251 days 21 hours by 13.

Then round to the nearest hundredth as necessary.

Answers

Answer: 465

Step-by-step explanation:

251 (days) x 24 (hours) = 6,024 hours

6,024+21 hours= 6,045

6045/13=465

The ratio of my money to Natalie's was 7 to 4. After I gave Natalie $15, I now have 20 % more than her. How much money do we each have now? Solve with a strip diagram and explain.

Answers

Answer:

my money= $192.5

Natalie's money= $110

Step-by-step explanation:

let my and Natalie's money be 7x and 4x respectively.

Now i gave $15 dollar to Natalie so now

my and Natalie's money will be (7x-15) and (4x+15) respectively.

now my money is 20% more than Natalie

therefore,

[tex]\frac{7x-15}{4x+15} =\frac{6}{5}[/tex]

now calculating for x  we get

x= 27.5

since, my and Natalie's money be 7x and 4x respectively

putting x=27.5

my money= $192.5

Natalie's money= $110

A test of H0: μ = 20 versus Ha: μ > 20 will be based on a random sample of size n from a normal population distribution. What conclusion is appropriate in each of the following situations? (Round your P-values to three decimal places.) (a) z = 3.3, α = 0.05

Answers

Answer:

We will accept the null hypothesis

Step-by-step explanation:

Given :[tex]H_0: \mu= 20[/tex]

          [tex]H_a: \mu> 20[/tex]

To Find :  What conclusion is appropriate in each of the following situations?

Solution :

z = 3.3, α = 0.05

So, first we will find the p value corresponding to z value in the z table

So, p-value is 0.999

Since p value is greater than Alpha

0.999>0.05

So, we will accept the null hypothesis

So, the population mean is 20

Find the selling price for a case of Newman's Own® special blend coffee that costs the retailer $35.87 if the markup is 22% of the selling price.

Answers

Answer:

The selling price is $46 approx.

Step-by-step explanation:

The cost price is = $35.87

Markup is 22% of selling price.

Let the selling price be = 100%

So, 100%-22%=78% = 0.78

So, solution is = [tex]35.87/0.78=45.98[/tex] ≈ $46.

Hence, the selling price is $46 approx.

. Joyce Meadow pays her three workers $160, $470, and $800, respectively, per week. Calculate what Joyce will pay at the end of the first quarter for (A) state unemployment and (B) federal unemployment. Assume a state rate of 5.6% and a federal rate of .6%. Base is $7,000. A. $950.64; $67.14 B. $655.64; $97.14 C. $755.64; $81.14 D. $850.64; $91.14

Answers

Answer:

Option D. $850.64; $91.14

Step-by-step explanation:

Joyce Meadow pays her three workers per week  $160, $470 and $800 respectively.

A year has 52 weeks, Therefore quarter has [tex]\frac{52}{4}[/tex] = 13 weeks.

So for a quarter she pays = 160 × 13 = $2,080, 470 × 13 = $6,110, 800 × 13 = $10,400

State rate is 5.6%

Federal rate is 0.6%

Base is $7,000

It means the unemployment needs to be paid on the first $7000 only

So the state unemployment = (0.056 × 2,080) + (0.056 × 6,110) + (0.056 × 7,000)

= 116.48 + 342.16 + 392 = $850.64

Federal unemployment = (0.006 × 2,080) + (0.006 × 6,110) + (0.006 × 7,000)

= 12.48 + 36.66 + 42 = $91.14

Option D. is the correct answer.

Find the point on the terminal side of θ = -3π / 4 that has an x coordinate of -1. Show work

Answers

Answer:

  (-1, -1)

Step-by-step explanation:

One way to write the relationship between the x and y coordinates and θ is ...

  tan(θ) = y/x

Then ...

  y = x·tan(θ) = -1·tan(-3π/4) = -1·1

  y = -1

The coordinates of the point on the terminal side are (x, y) = (-1, -1).

A Bernoulli random variable X has unknown success probability p. Using 100 independent samples of X, find a confidence interval estimate of p with confidence coefficient 0.99. If ????????100 = 0.06, what is our interval estimate

Answers

Answer: [tex](0.0445,\ 0.0755)[/tex]

Step-by-step explanation:

The confidence interval for the population proportion is given by :-

[tex]p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}[/tex]

Given : A Bernoulli random variable X has unknown success probability p.

Sample size : [tex]n=100[/tex]

Unknown success probability : [tex]p=0.06[/tex]

Significance level : [tex]\alpha=1-0.99=0.01[/tex]

Critical value : [tex]z_{\alpha/2}=2.576[/tex]

Now, the 99% confidence interval for true proportion will be :-

[tex]0.06\pm(2.576)\sqrt{\dfrac{0.06(0.06)}{100}}\\\\\approx0.06\pm(0.0155)\\\\=(0.06-0.0155,\ 0.06+0.0155)\\\\=(0.0445,\ 0.0755)[/tex]

Hence, the 99% confidence interval for true proportion= [tex](0.0445,\ 0.0755)[/tex]

Find and classify any equilibrium solutions and then sketch typical solution curves to the differential equation: dx /dt = x 2 − 5x + 4.

Answers

Answer:

[tex]x=1+\dfrac{3}{1-Ke^{3t}}[/tex]

Step-by-step explanation:

Given that

[tex]\dfrac{dx}{dt}=x^2-5x+4[/tex]

This is a differential equation.

Now by separating variables

[tex]\dfrac{dx}{x^2-5x+4}=dt[/tex]

[tex]\dfrac{dx}{(x-1)(x-4)}=dt[/tex]

[tex]\dfrac{1}{3}\left(\dfrac{1}{x-4}-\dfrac{1}{x-1}\right)dx=dt[/tex]

Now by integrating both side

[tex]\int\dfrac{1}{3}\left(\dfrac{1}{x-4}-\dfrac{1}{x-1}\right)dx=\int dt[/tex]

[tex]\dfrac{1}{3}\left(\ln(x-4)-\ln(x-1)\right )=t+C[/tex]

Where C is the constant

[tex]\dfrac{1}{3}\ln\dfrac{x-4}{x-1}=t+C[/tex]

[tex]\dfrac{x-4}{x-1}=Ke^{3t}[/tex]    K is the constant.

[tex]x=1+\dfrac{3}{1-Ke^{3t}}[/tex]

So the solution of above differential equation is

[tex]x=1+\dfrac{3}{1-Ke^{3t}}[/tex]

A fleet of refrigerated delivery trucks is acquired on January 5, 2017, at a cost of $830,000 with an estimated useful life of eight years and an estimated salvage value of $75,000. Compute the depreciation expense for the first three years using the double-declining-balance method

Answers

Answer:

1st year: $ 622,500

2nd year: $415,000

3rd year: $207,500

Step-by-step explanation:

Step 1: Write the beginning book value of the asset

$830,000

Step 2: Determine the asset's estimated useful life

8 years

Step 3: Determine the asset's salvage value

$75,000

Step 4: Subtract the salvage value from the beginning value to get the total depreciation amount for the asset's total life.

830,000 - 75,000 = $755,000

Step 5: Calculate the annual depreciation rate

Depreciation rate = 100%/8 years = 12.5%

Step 6: Multiply the beginning value by twice the annual depreciation rate to find the depreciation expense

Depreciation expense = 830,000 x 25% = $207,500

Step 7: Subtract the depreciation expense from the beginning value to find the ending period value

Ending period value for 1st year: Beginning value - depreciation expense

830,000 - 207,500 = $ 622,500

Ending period value for 2nd year: 622,500 - 207,500 = $ 415,000

Ending period value for 3rd year: 415,000 - 207,500 = $ 207,500

!!

Twenty percent (20%) of 90 equals

A. 12

B. 15

C. 18

D. 21

Answers

Answer:

20 percent *90. =

(20:100)*90. =

(20*90.):100 =

1800:100 = 18

The correct answer is C.


our friend purchases a $185,000 house. He is able to make a 15% down payment. The bank will give him a 30-year loan with a 3.5% APR.

How much money will he borrow for his mortgage?

$ (round to the nearest dollar)

What would his monthly payment be?

$ (round to the nearest cent)

How much interest will your friend pay over 30 years of his loan?

$ (round to the nearest cent)

Answers

Answer:

Given,

The value of the house = $ 185,000,

Percentage of down payment = 15%,

(i) So, the borrowed amount = 185,000 - 15% of 185,000

[tex]=185000-\frac{15\times 185000}{100}[/tex]

[tex]=185000-\frac{2775000}{100}[/tex]

[tex]=185000-27750[/tex]

[tex]=\$157250[/tex]

(ii) Since, the monthly payment formula of a loan is,

[tex]P=\frac{PV\times r}{1-(1+r)^{-n}}[/tex]

Where,

PV = present value of the loan ( or borrowed amount )

r = rate per month,

n = number of months,

Here, PV = $ 157250,

APR = 3.5% = 0.035 ⇒ r = [tex]\frac{0.035}{12}[/tex] ( 1 year = 12 months )

Time = 30 years, ⇒ n = 360 months,

Hence, the monthly payment would be,

[tex]P=\frac{157250\times \frac{0.035}{12}}{1-(1+\frac{0.035}{12})^{-360}}[/tex]

[tex]=706.122771579[/tex]    ( by graphing calculator ),

[tex]\approx \$ 706.12[/tex]

(iii) Interest = Total amount paid - borrowed amount

= 706.122771579 × 360 - 157250

= 96954.1977684

≈ $ 96954. 20

Final answer:

The friend will borrow $157,250 for the mortgage, with a monthly payment of $704.30. They will pay $101,548 in interest over 30 years.

Explanation:

To calculate the amount of money borrowed for the mortgage, we need to subtract the down payment from the total cost of the house. The down payment is 15% of the house cost, which is $185,000 x 0.15 = $27,750. So, the borrowed amount is $185,000 - $27,750 = $157,250.

To calculate the monthly payment, we can use the formula for a fixed-rate mortgage: M = P [ i(1 + i)^n ] / [ (1 + i)^n - 1 ]. Here, P is the borrowed amount ($157,250), i is the monthly interest rate (3.5% / 12 = 0.002916), and n is the total number of monthly payments (30 x 12 = 360). Plugging in these values, we get M = $157,250 [ 0.002916(1 + 0.002916)^360 ] / [ (1 + 0.002916)^360 - 1 ]. Using a calculator, the monthly payment comes out to be approximately $704.30.

To calculate the total interest paid over 30 years, we can multiply the monthly payment by the total number of payments and subtract the borrowed amount. The total interest paid = ($704.30 x 360) - $157,250. Using a calculator, the total interest paid comes out to be approximately $101,548. So, your friend will pay approximately $157,250 for the mortgage, with a monthly payment of approximately $704.30, and will pay approximately $101,548 in interest over 30 years.

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please help



An aircraft factory manufactures airplane engines. The unit cost C (the cost in dollars to make each airplane engine) depends on the number of engines made. If x engines are made, then the unit cost is given by the function C(x) = 1.1x^2 - 638x + 111,541 . What is the minimum unit cost?

Do not round your answer.

Answers

Final answer:

The minimum unit cost is approximately $42,330.09.

Explanation:

To find the minimum unit cost, we need to find the vertex of the quadratic function C(x) = 1.1x^2 - 638x + 111,541. The x-coordinate of the vertex can be found using the formula x = -b / 2a, where a = 1.1 and b = -638. Plugging in these values, we get:



x = -(-638) / (2 * 1.1) = 290.9090909



So the minimum unit cost occurs when approximately 291 engines are made. To find the minimum unit cost, we substitute this value back into the C(x) function:



C(291) = 1.1(291)^2 - 638(291) + 111,541 = 42330.090909



Therefore, the minimum unit cost is approximately $42,330.09.

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It is estimated that one third of the general population has blood type A A sample of six people is selected at random. What is the probability that exactly three of them have blood type A?

Answers

Answer: 0.2195

Step-by-step explanation:

Binomial distribution formula :-

[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of x successes in the n independent trials of the experiment and p is the probability of success.

Given : The probability of that the general population has blood type A = [tex]\dfrac{1}{3}[/tex]

Sample size : n=6

Now, the probability that exactly three of them have blood type A is given by :-

[tex]P(3)=^6C_3(\dfrac{1}{3})^3(1-\dfrac{1}{3})^{6-3}\\\\=\dfrac{6!}{3!3!}(\dfrac{1}{3})^3(\dfrac{2}{3})^{3}\\\\=0.219478737997\approx0.2195[/tex]

Therefore, the probability that exactly three of them have blood type A = 0.2195

For a certain​ candy, 20​% of the pieces are​ yellow, 15​% are​ red, 20​% are​ blue, 20​% are​ green, and the rest are brown. ​a) If you pick a piece at​ random, what is the probability that it is​ brown? it is yellow or​ blue? it is not​ green? it is​ striped? ​b) Assume you have an infinite supply of these candy pieces from which to draw. If you pick three pieces in a​ row, what is the probability that they are all​ brown? the third one is the first one that is​ red? none are​ yellow? at least one is​ green?

Answers

Answer:

Step-by-step explanation:

Based on the question we are given the percentages of each of the types of candies in the bag except for brown. Since the sum of all the percentages equals 75% and brown is the remaining percent then we can calculate that brown is (100-75 = 25%) 25% of the bag. Now we can show the probabilities of getting a certain type of candy by placing the percentages over the total percentage (100%).

Brown: [tex]\frac{25}{100}[/tex]Yellow or Blue: [tex]\frac{20}{100} +\frac{20}{100} = \frac{40}{100}[/tex]  ....add the numeratorsNot Green:  [tex]\frac{80}{100}[/tex].... since the sum of all the rest is 80%Stiped:  [tex]\frac{25}{100}[/tex] .... there are 0 striped candies.

Assuming the ratios/percentages of the candies stay the same having an infinite amount of candy will not affect the probabilities. That being said in order to calculate consecutive probability of getting 3 of a certain type in a row we have to multiply the probabilities together. This is calculated by multiplying the numerators with numerators and denominators with denominators.

3 Browns: [tex]\frac{25*25*25}{100*100*100} = \frac{15,625}{1,000,000} = \frac{1.5625}{100}[/tex]

the 1st and 3rd are red while the middle is any. We multiply 15% * (total of all minus red which is 85%) * 15% like so.

[tex]\frac{15*85*15}{100*100*100} = \frac{19,125}{1,000,000} = \frac{1.9125}{100}[/tex]

None are Yellow: multiply the percent of all minus yellow three times.

[tex]\frac{80*80*80}{100*100*100} = \frac{512,000}{1,000,000} = \frac{51.2}{100}[/tex]

At least 1 green: multiply the percent of green by 100% twice, since the other two can by any

[tex]\frac{20*100*100}{100*100*100} = \frac{200,000}{1,000,000} = \frac{20}{100}[/tex]

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