Angles α and β are the two acute angles in a right triangle, where α = 5x 3 + 20 and β = 2x 3 + 14. Find α. A) 24° B) 30° C) 60° D) 66°

Answer:

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

Step-by-step explanation:

Air containing 0.04% carbon dioxide

V, volume of room is 6000 ft3.

Q, rate of air 2000 ft3/min,

initial concentration of 0.4% carbon dioxide,

determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes?

firstly, we find the time taken for air to completely filled the room

Q = V/t

t = V/Q = 6000/2000 = 3min

so, its take 3mins for air to be completely filled in the room and for exhaust air to move out.

there is  an initial concentration of 0.4% carbon dioxide, and the air pump in is 0.04%.

therefore,

3mins = 0.04% of CO2

3*60 =180sec = 0.04%

1sec = 0.04/180 = 0.00022%/sec

so at any time the concentration of CO2 is 0.4 + 0.00022 =0.40022%/sec

What is the concentration at 10 minute

the concentration at 10minutes = the concentration for 1minute because at every minutes, the concentration moves in is moves out. = concentration for 2000ft3.

for 0.04% = 6000ft3

   ?          = 2000ft3

              = 2000* 0.04)/6000 =0.0133%

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

Answer: this corresponds to a chi-square test for Goodness of Fit.

Step-by-step explanation:

Goodness of fit Chi-squared test gives confirmation on whether a separately observed frequency distribution differs from a theoretical or general distribution.

In this case, the researcher studies a fraction of income earners - low income earners. He is interested in knowing whether the distribution of pay of low income earners is different from the distribution of pay of ALL income earners.

This survey hence corresponds to a chi-square test for goodness of fit.

Answer:

Let [tex]A = (a_1, ..., a_n)[/tex] and [tex]B = (b_1, ..., b_n)[/tex] bases of V. The matrix of change from A to B is the matrix n×n whose columns are vectors columns of the coordinates of vectors [tex]b_1, ..., b_n[/tex] at base A.

The, we case correspond to find the coordinates of vectors of C,

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

at base B.

1. We need to find [tex]a,b,c\in\mathbb{R}[/tex] such that

[tex]\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right][/tex]

Then we find these values solving the linear system

[tex]\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&-1\\0&-1&1&-1\end{array}\right][/tex]

Using rows operation we obtain the echelon form of the matrix

[tex]\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&1\end{array}\right][/tex]

now we use backward substitution

[tex]c=1\\-b+c=-1,\; b=2\\a-2b+2c=2,\; a=4[/tex]

Then the coordinate vector of [tex]\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right][/tex] is [tex]\left[\begin{array}{ccc}4\\2\\1\end{array}\right][/tex]

2. We need to find [tex]a,b,c\in\mathbb{R}[/tex] such that

[tex]\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right][/tex]

Then we find these values solving the linear system

[tex]\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&0\\0&-1&1&-1\end{array}\right][/tex]

Using rows operation we obtain the echelon form of the matrix

[tex]\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&2\end{array}\right][/tex]

now we use backward substitution[tex]c=2\\-b+c=-1,\; b=3\\a-2b+2c=2,\; a=4[/tex]

Then the coordinate vector of [tex]\left[\begin{array}{ccc}2\\0\\-1\end{array}\right][/tex] is [tex]\left[\begin{array}{ccc}4\\3\\2\end{array}\right][/tex]

3. We need to find [tex]a,b,c\in\mathbb{R}[/tex] such that

[tex]\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right][/tex]

Then we find these values solving the linear system

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

Using rows operation we obtain the echelon form of the matrix

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

now we use backward substitution[tex]c=-2\\-b+c=2,\; b=-4\\a-2b+2c=2,\; a=-2[/tex]

Then the coordinate vector of [tex]\left[\begin{array}{ccc}-3\\1\\2\end{array}\right][/tex] is [tex]\left[\begin{array}{ccc}-2\\-4\\-2\end{array}\right][/tex]

Then the change of basis matrix from B to C is

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

To find the change of basis matrix from basis B to basis C in R3, invert basis B, multiply it by basis C, and the resulting matrix transforms coordinates from B to C: [[1 2 -2], [2 3 -4], [4 4 -7]].

Here's how to go about finding the change of basis matrix from basis B to basis C in R3:

1. Write down the vector coordinates of interest. These coordinates are given by basis B and basis C:
   basis B : {⟨2,−1,1⟩,⟨−2,2,−1⟩,⟨1,−1,0⟩}
   basis C : {⟨2,−1,−1⟩,⟨2,0,−1⟩,⟨−3,1,2⟩}

2. Find the inverse of basis B. The inverse of a matrix is such that if you multiply the original matrix by its inverse you get the identity matrix — a simple "1, 0" matrix. This step effectively reverses the transformation provided by basis B.

3. Then, calculate the product of basis B's inverse and basis C. This essentially re-projects the coordinates of basis B onto basis C.

4. The resulting matrix is your Change of Basis matrix from B to C. In our calculation, this comes out as:

Change of Basis from B to C:
   [[ 1.  2. -2.]
    [ 2.  3. -4.]
    [ 4.  4. -7.]]

This matrix will transform any vector in coordinates relative to basis B into coordinates relative to basis C. The first row indicates how much of each vector in B is needed to form the first vector in C, the second row for the second vector in C, and so on.

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Answer:

[tex]1.75*10^{-27}[/tex]

Step-by-step explanation:

If, collectively, 15 people weigh more than 3500 pounds, that means each person must weigh more than 3500/15 = 233.33 pounds.

If the distribution for population weights is normal at mean = 180 and standard deviation = 25 lbs, that means the probability for 1 person to weigh higher than 233 lbs is

[tex]1 - P(x > 233, \mu = 180, \sigma = 25) = 1 - 0.984 = 0.016[/tex]

For all 15 people to have higher weigh than that then the probability is

[tex]0.016^{15} = 1.75*10^{-27}[/tex]

This is indeed very unlikely to happen

Final answer:

It would be highly unusual for 15 people in an elevator to have a combined weight of more than 3,500 lbs; the z-score of 8.27 reflects an extremely small probability, pointing to a rare event.

Explanation:

To determine how unusual it would be for 15 people in an elevator to have a combined weight of more than 3,500 lbs, we use the Central Limit Theorem. Given that each person's weight is a random variable X that is normally distributed with a mean (μ) of 180 lbs and a standard deviation (σ) of 25 lbs, the sum of the weights of 15 people will also be normally distributed with a mean (μtotal) of 15 * 180 lbs and a standard deviation (σtotal) of √15 * 25 lbs, due to the Central Limit Theorem.

The next step is to calculate these values:

Now we calculate the z-score to determine how many standard deviations away 3,500 lbs is from the mean:

Z = (X - μtotal) / σtotal

Z = (3500 - 2700) / 96.82 ≈ 8.27

Using standard normal distribution tables or a calculator, we find that the probability of a z-score of 8.27 is extremely small, indicating that it would be highly unusual for 15 people to weigh more than 3,500 lbs in total.

Answer: Yes, this sample provide evidence that people living in rural communities live longer than 77 years.

Step-by-step explanation:

Since we have given that

Average lifespan in the USA = 77 years

We need to check whether the people living in rural communities live longer than 77 years.

So, Hypothesis would be

[tex]H_0:\mu=77\\\\H_a:\mu>77[/tex]

Since n = 12

[tex]\bar{x}=81.03\\\\s=1.53[/tex]

since n <30 so, we will use t test.

So, the test statistic value is given by

[tex]t=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\\\t=\dfrac{81.03-77}{\dfrac{1.53}{\sqrt{12}}}\\\\\\t=\dfrac{4.03}{0.4416}\\\\t=9.125[/tex]

degrees of freedom = df = n-1 = 12-1 =11

At 95% significance level , t = 1.796

Since, 1.796< 9.125

So, we will reject the null hypothesis.

Hence, Yes, this sample provide evidence that people living in rural communities live longer than 77 years.

The null hypothesis is that the average lifespan of rural Idaho residents is 77 years, and the alternative hypothesis is that it's greater than 77 years. The calculated t-value is approximately 9.974, which indicates that the rural Idaho sample’s lifespan is significantly longer than the national average of 77 years.

The null and alternative hypotheses are as follows:

You can calculate the t-statistic using the formula: t = (x_bar - mu) / (s / sqrt(n)), where x_bar is the sample mean, mu is the population mean, s is the sample standard deviation, and n is the sample size. Plugging in the given values, we get:

t = (81.03 - 77) / (1.53 / sqrt(12))  = 9.974, approx.

The resulting t-value indicates that the rural Idaho sample’s lifespan is significantly above the national average of 77 years. Hence, the sample provides evidence that people living in rural Idaho communities live longer than the standard American lifespan of 77 years.

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Answer:

a) 6.3 minutes

Step-by-step explanation:

Population mean (μ) = 7.2 minutes  

Standard deviation (σ) = 0.56 minutes

The z-score for any running time 'X' is given by:

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

In this scenario, the company is looking for the top 5% runners, that is, runners at and below the 5-th percentile of the normal distribution. The equivalent z-score for the 5-th percentile is 1.645.

Therefore, the minimum speed, X, a runner needs to achieve in order to be sponsored is:

[tex]-1.645=\frac{X-7.2}{0.56}\\X= 6.3\ minutes[/tex]

Answer: B. 461 acres

Step-by-step explanation:

Given : In an agricultural study, the average amount of corn yield is normally distributed with a mean of 185.2 bushels of corn per acre, with a standard deviation of 23.5 bushels of corn.

i.e. [tex]\mu=185.2\ \ , \ \sigma=23.5[/tex]

Let x denotes the amount of corn yield.

Now, the probability that the amount of corn yield is more than 190 bushels of corn per acre.

[tex]P(x>190)=P(\dfrac{x-\mu}{\sigma}>\dfrac{190-185.2}{23.5})[/tex]

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

[tex]=P(z>0.2043)=1-P(z<0.2043)[/tex]  [∵ P(Z>z)=1-P(Z<z)]

[tex]1-0.5809405[/tex]   [using z-value calculator or table]

[tex]=0.4190595[/tex]

Now, If a study included 1100 acres then the expected number to yield more than 190 bushels of corn per acre :-

[tex]0.4190595\times1100=460.96545\approx461\text{ acres}[/tex]

hence, the correct answer is B. 461 acres .

Answer:

0.6563 or 65.63% of brook trout caught will be between 12 and 18 inches

Step-by-step explanation:

Mean trout length (μ) = 14 inches

Standard deviation (σ) = 3 inches

The z-score for any given trout length 'X' is defined as:  

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

For a length of X =12 inches:

[tex]z=\frac{12-14}{3}\\z=-0.6667[/tex]

According to a z-score table, a score of -0.6667 is equivalent to the 25.25th percentile of the distribution.

For a length of X =18 inches:

[tex]z=\frac{18-14}{3}\\z=1.333[/tex]

According to a z-score table, a score of 1.333 is equivalent to the 90.88th percentile of the distribution.

The proportion of trout caught between 12 and 18 inches, assuming a normal distribution, is the interval between the equivalent percentile of each length:

[tex]P(12\leq X\leq 18) = 90.88\% - 25.25\%\\P(12\leq X\leq 18) = 65.63\%[/tex]

To find the proportion of brook trout caught between 12 and 18 inches, calculate the z-scores for these values and find the area between them on a standard normal distribution curve.

To find the proportion of brook trout caught between 12 and 18 inches in length, we need to calculate the z-scores for these values and then find the area between the z-scores on a standard normal distribution curve.

First, we calculate the z-score for 12 inches: z = (12 - 14) / 3 = -2/3.

Second, we calculate the z-score for 18 inches: z = (18 - 14) / 3 = 4/3.

Using a z-table or a calculator, we can find the area to the left of -2/3 and the area to the left of 4/3. Subtracting these two areas will give us the proportion of brook trout caught between 12 and 18 inches.

Answer:

add all of it

Step-by-step explanation:

The solution is x = -5

Step-by-step explanation:

Given equation is:

[tex]3^{(x-1)} = 9^{(x+2)}[/tex]

In order to solve the eponnetial equations, we have to equate the bases of both sides so that the exponents can be put equal

So,

Replacing 9 with 3^2

[tex]3^{(x-1)} = (3^2)^{(x+2)}[/tex]

When there are exponents on exponents, both are multiplied so,

[tex]3^{(x-1)} = 3^{(2x+4)}[/tex]

As the bases on both sides are same, the exponents can be put equal

So,

[tex]x-1 = 2x+4[/tex]

Adding 1 on both sides

[tex]x-1+1 = 2x+4+1\\x = 2x+5[/tex]

Subtracting 2x from both sides

[tex]x-2x = 2x-2x+5\\-x = 5\\x = -5[/tex]

Hence,

The solution is x = -5

Keywords: Exponents, Equations

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In light diffraction, fringe separation is inversely proportional to the slit width. When the slit width in an experiment is doubled, it should result in the separation between the dark fringes on the screen being halved. Therefore, if the original fringe separation was 10 mm, the new separation should be 5 mm.

The subject question pertains to the topic of light diffraction, particularly through varying slit widths. The scenario described is a single slit diffraction experiment involving an argon laser with a specific wavelength. In diffraction, the fringe separation is inversely proportional to the slit width. This is because the angle of diffraction is determined by the wavelength of light divided by the slit width, according to the formula sinθ = λ/D, where θ is the diffraction angle, λ is the wavelength, and D is the slit width.

Therefore, when the slit width is doubled, the diffraction angles for the minima (dark fringes) will be halved assuming all other conditions remain the same. Consequently, the separation between the dark fringes on the screen will also be halved. So, if the original separation was 10 mm, the new separation when the slit width is doubled should be 5 mm.

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Answer:

The sampling distribution of the sample mean has approximately a normal distribution because of

c) the central limit theorem.

Step-by-step explanation:

Given that the average exam score of students of a large class is 70 with a standard deviation of 10.

From the above students a sample of 36 students is selected, and the mean score of these students is computed.

As per central limit theorem we have when samples are drawn at random from population, with sample size sufficiently large to represent the population then sample mean follows a normal distribution.

Here population size N = 70 and sample size n =36

we can say sample size is greater than 30 and sufficiently large to represent the population. Also we can assume that these are randomly drawn.

So the answer would be

The sampling distribution of the sample mean has approximately a normal distribution because of

c) the central limit theorem.

Using the tangent function of trigonometry, the satellite's distance from Angeles is 890.53 miles.

Distance between Phoenix and Los Angeles = 340 miles

1 mile = 5,280 feet

340 miles = 1,795,200 feet (340 miles * 5280 feet/mile)

Angle of elevation at Phoenix = 60°

Angle of elevation at Los Angeles = 75°

Tangent Equation:

tan(75°) = x/d

Solving the equation for x:

x = d * tan(75°)

Substituting the distance value into this equation for x:

x = 1,795,200 * tan(75°) ≈ 4,702,000 feet ≈ 890.53 miles (4,702,000/5,280)

Thus, we can conclude that the distance of the satellite at Los Angeles is approximately, 4,702,000 feet or 890.53 miles.

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The distance between the satellite and Los Angeles is approximately 462.215 miles.

To find the distance between the satellite and Los Angeles, we can use trigonometry. Let's assume that the satellite is at point A between Phoenix and Los Angeles, and point B is Los Angeles.

We know that the angle of elevation at Phoenix (angle α) is 60°, and the angle of elevation at Los Angeles (angle β) is 75°.

We can use the tangent function to find the distance from Los Angeles to the satellite:

tan(β) = opposite/adjacent

We know that the opposite side is the distance between Phoenix and Los Angeles, which is 340 miles. So we can write:

tan(75°) = 340/AB

Rearranging the equation, we get:

AB = 340/tan(75°)

Using a calculator, we find that AB ≈ 462.215 miles.

Therefore, the satellite is approximately 462.215 miles away from Los Angeles.

Answer:

a) P(A graduate is offered fewer than two jobs) = 0.53.

b) P(A graduate is offered more than one job) = 0.47.

Step-by-step explanation:

Let X be a random variable denoting the number of jobs offers that a university graduate gets within a month of graduation.

The probability that a university graduate will be offered no jobs within a month of graduation is estimated to be 10% i.e. [tex]P(X=0)=0.10[/tex]

The probability of receiving one job offers has similarly been estimated to be 43% i.e. [tex]P(X=1)=0.43[/tex]

The probability of receiving two job offers has similarly been estimated to be 34% i.e. [tex]P(X=2)=0.34[/tex]

The probability of receiving three job offers has similarly been estimated to be 13% i.e. [tex]P(X=3)=0.13[/tex]

a) P (A graduate is offered fewer than two jobs) i.e. P(X<2)

So, [tex]P(X<2)=P(X=0)+P(X=1)[/tex]

[tex]P(X<2)=0.10+0.43[/tex]

[tex]P(X<2)=0.53[/tex]

P(A graduate is offered fewer than two jobs) = 0.53.

b) P (A graduate is offered more than one job) i.e. P(X>1)

So, [tex]P(X>1)=P(X=2)+P(X=3)[/tex]

[tex]P(X>1)=0.34+0.13[/tex]

[tex]P(X>1)=0.47[/tex]

P(A graduate is offered more than one job) = 0.47.

The probability that a graduate is offered fewer than two jobs is 53%, and the probability that a graduate is offered more than one job is 47%.

In the scenario given, you want to find two probabilities: A) the probability that a graduate is offered fewer than two jobs, and B) the probability that a graduate is offered more than one job. The total probability should add up to 100%, or a probability of 1.

For part A), 'fewer than two jobs' could mean either no job offers or one job offer. We know from the information given that the probability of no job offer is 10% (or 0.10) and the possibility of one job offer is 43% (or 0.43). So you add these two probabilities together: 0.10 + 0.43 = 0.53. Therefore, the probability that a graduate is offered fewer than two jobs is 53%.

For part B), 'more than one job' could mean either two job offers or three job offers. From the information given, we can find that the probability of receiving two job offers is 34% (or 0.34) and the probability for three job offers is 13% (or 0.13). Adding these two probabilities gives 0.34 + 0.13 = 0.47. Hence, the probability that a graduate is offered more than one job is 47%.

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Answer:

1.50 marks per pound

Step-by-step explanation:

Data provided in the question:

Selling price of three-wheel cars made in North Edsel = 5000 pounds

Selling price of four-wheel cars made in south Edsel = 10,000 marks

Real exchange rate between North and South Edsel

= four​ three-wheel cars for three​ four-wheel cars

i.e

⇒ 4 × 5000 pounds  = 3 × 10,000 marks

or

1 pounds = [ ( 3 × 10,000 ) ÷ ( 4 × 5,000) ]

or

1 pound = 30,000 ÷ 20,000

or

1 pound = 1.50 marks

Hence,

The nominal exchange rate = 1.50 marks per pound

Final answer:

The nominal exchange rate between North and South Edsel is 3750 pounds.

Explanation:

The real exchange rate is the ratio at which goods and services of one country can be exchanged for those of another country. In this case, the real exchange rate between North and South Edsel is four three-wheel cars for three four-wheel cars. So, for every four four-wheel cars from South Edsel, you can exchange them for three three-wheel cars from North Edsel.

Since the price of the three-wheel cars from North Edsel is 5000 pounds, and the ratio is four three-wheel cars for three four-wheel cars, the nominal exchange rate between the two countries would be:

5000 pounds * (3 four-wheel cars / 4 three-wheel cars) = 3750 pounds.

Therefore, the nominal exchange rate between the two countries is 3750 pounds.

Answer:

a) [tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]

b) [tex]P(X\leq 4)=0.6+0.24+0.096+0.0384=0.9744[/tex]

c) [tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[0.6+0.24+0.096]=0.064[/tex]

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

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

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

[tex]X\sim Geo (1-p)[/tex]

Part a

For this case we want this probability

[tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]

Part b

For this case we want this probability:

[tex]P(X\leq 4)=P(X=1)+P(X=2)+P(X=3)+P(X=4)[/tex]

If we find the individual probabilities we got:

[tex]P(X=1)=(1-0.6)^{1-1} 0.6 = 0.6[/tex]

[tex]P(X=2)=(1-0.6)^{2-1} 0.6 = 0.24[/tex]

[tex]P(X=3)=(1-0.6)^{3-1} 0.6 = 0.096[/tex]

[tex]P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384[/tex]

And replacing we have:

[tex]P(X\leq 4)=0.6+0.24+0.096+0.0384=0.9744[/tex]

Part c

For this case at least 4 trials means that the random variable X needs to be 4 or more

[tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[P(X=1)+P(X=2)+P(X=3)][/tex]

And we found already the probabilities P(X=1),P(X=2) and P(X=3) so we just need to replace:

[tex]P(X\geq 4)=1-P(X<4)=1-P(X\leq 3)=1-[0.6+0.24+0.096]=0.064[/tex]

The three scenarios are calculated using a combination of geometric and binomial probability distribution. For the first success to happen on the 4th trial is 0.0384. For a success to happen within the first four trials is 0.8847, and for the first success to need at least four trials is 0.3600.

This question has to do with the concept of probability distribution in math, and specifically with the geometric distribution and binomial probability distribution. Given the probability of a successful optical alignment p = 0.6, we're being asked about different scenarios involving success on specific trials.

(a) For the first successful alignment to happen on the 4th trial, the first 3 trials have to be failures and the 4th one, a success. The probability would be (0.4)^3 * 0.6 = 0.0384.

(b) This means we want a successful alignment on the 1st, 2nd, 3rd or 4th trials. For this we sum up the probabilities of each case: (0.6) + (0.4*0.6) + (0.4)^2*0.6 + (0.4)^3*0.6 = 0.8847.

(c) For a successful alignment to occur in at least 4 trials, we have to subtract the probability of success within the first 3 trials from 1. Therefore, 1 - ((0.6) + (0.4*0.6) + (0.4)^2*0.6) = 0.3600.

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Answer:

A. False

B. True

C. False

Step-by-step explanation:

A. Angles are not congruent. (CDA is bigger than AED)

B. Both angles are on opposite sides; therefore they are congruent. (They are the same measurement.)

C. BC is shorter than AB. Not congruent.

Final answer:

To reject the null hypothesis (H0) if the calculated t-value is less than -1.9842.

Explanation:

To establish the appropriate rejection region for a significance level of 0.05, one compares the test statistic (denoted as 't') with a critical value. In this case, the correct rejection region dictates that the null hypothesis (H0) should be rejected if the calculated 't' is less than -1.9842.

Simply put, if the derived 't'-value falls below -1.9842, it implies rejecting the null hypothesis. This decision leads to the conclusion that there is enough evidence to support the alternative hypothesis (Ha: μ < 60). This meticulous comparison adheres to statistical principles, ensuring a reliable interpretation of the test outcomes and providing a solid foundation for decision-making in hypothesis testing.

Answer:

a)  [tex]\bar X=9.07[/tex]

b) The 95% confidence interval is given by (8.197;9.943)  

c) [tex]m=2.14 \frac{1.580}{\sqrt{15}}=0.873[/tex]

d)  3 possible ways

1) Increasing the sample size n.  

2) Reducing the variability. If we have more data probably we will have less variation.

3) Lower the confidence level. Because if we have lower confidence then the quantile from the t distribution would belower and tthe margin of error too.

Step-by-step explanation:

Notation and definitions  

n=15 represent the sample size  

[tex]\bar X= 9.07[/tex] represent the sample mean  

[tex]s=1.580[/tex] represent the sample standard deviation  

m represent the margin of error  

Confidence =95% or 0.95

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Part a: Find the point estimate of the population mean.

The point of estimate for the population mean [tex]\mu[/tex] is given by:

[tex]\bar X =\frac{\sum_{i=1}^{n} x_i}{n}[/tex]

The mean obteained after add all the data and divide by 15 is [tex]\bar X=9.07[/tex]

Calculate the critical value tc  

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. The degrees of freedom are given by:  

[tex]df=n-1=15-1=14[/tex]  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.025,14)" for [tex]t_{\alpha/2}=-2.14[/tex]  

"=T.INV(1-0.025,14)" for [tex]t_{1-\alpha/2}=2.14[/tex]  

The critical value [tex]tc=\pm 2.14[/tex]  

Part c: Calculate the margin of error (m)  

First we need to calculate the standard deviation given by this formula:

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex]

s=1.580

The margin of error for the sample mean is given by this formula:  

[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]  

[tex]m=2.14 \frac{1.580}{\sqrt{15}}=0.873[/tex]  

Part b: Calculate the confidence interval  

The interval for the mean is given by this formula:  

[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]  

And calculating the limits we got:  

[tex]9.07 - 2.14 \frac{1.580}{\sqrt{15}}=8.197[/tex]  

[tex]9.07 + 2.14 \frac{1.580}{\sqrt{15}}=9.943[/tex]  

The 95% confidence interval is given by (8.197;9.943)  

Part d: How can we reduce the margin of error?

We can reduce the margin of error on the following ways:

1) Increasing the sample size n.  

2) Reducing the variability. If we have more data probably we will have less variation.

3) Lower the confidence level. Because if we have lower confidence then the quantile from the t distribution would belower and tthe margin of error too.

The publisher's point estimate for average reading time is 9.2 minutes. The 95% confidence interval for this estimate is between 8.18 and 10.22 minutes. The margin of error is approximately 1.04 minutes, which can be reduced by increasing the sample size.

The subject matter of this question is statistics, specifically dealing with the calculation and interpretation of point estimates, confidence intervals, and margins of error. Here are the steps to solve your question:

Point estimate of the population mean: This is the estimated population mean, which you find by taking the average of your sample. If you sum up all the time spent and divide by the number of people (15), you'll get the point estimate, which ends up being 9.2 minutes.

95% confidence interval for the population mean: This is computed using the sample mean, the standard deviation of the sample, and the value from a t-distribution table for a specific confidence level (95% or 0.05 significance level in this case). The calculations, based on the standard deviation, result in a 95% confidence interval of about 8.18 to 10.22.

Margin of error: The margin of error can be calculated as the difference between the sample mean and the extreme end of the confidence interval, which is about 1.04 in this case.

Reducing the margin of error: This can be achieved by increasing the sample size, which will decrease the standard error.

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Answer:

z=13.36 (Statistic)

[tex]p_v =P(Z>13.36)\approx 0[/tex]

The p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of men with red/green color blindness is significant higher than the proportion of female with red/green color blindness .

Step-by-step explanation:

1) Data given and notation

[tex]X_{MCB}=56[/tex] represent the number of men with red/green color blindness

[tex]X_{WCB}=5[/tex] represent the number of women with red/green color blindness

[tex]n_{MCB}=600[/tex] sample of male selected

[tex]n_{WCB}=600[/tex] sample of demale selected

[tex]p_{MCB}=\frac{56}{600}=0.093[/tex] represent the proportion of men with red/green color blindness

[tex]p_{WCB}=\frac{5}{2150}=0.0023[/tex] represent the proportion of women with red/green color blindness

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the value for the test (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion for men with red/green color blindness is a higher than the rate for women  , the system of hypothesis would be:

Null hypothesis:[tex]p_{MCB} \leq p_{WCB}[/tex]

Alternative hypothesis:[tex]p_{MCB} > \mu_{WCB}[/tex]

We need to apply a z test to compare proportions, and the statistic is given by:

[tex]t=\frac{p_{MCB}-p_{WCB}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{MCB}}+\frac{1}{n_{WCB}})}}[/tex]   (1)

Where [tex]\hat p=\frac{X_{MCB}+X_{WCB}}{n_{MCB}+n_{WCB}}=\frac{56+5}{600+2150}=0.0221[/tex]

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

[tex]z=\frac{0.093-0.0023}{\sqrt{0.0221(1-0.0221)(\frac{1}{600}+\frac{1}{2150})}}=13.36[/tex]  

4) Statistical decision

For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test.  

Since is a one side test the p value would be:

[tex]p_v =P(Z>13.36)\approx 0[/tex]

So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of men with red/green color blindness is significant higher than the proportion of female with red/green color blindness .

To test the claim that men have a higher rate of red/green color blindness, we need to compare the proportions of color blindness in men and women using a chi-square test of independence.

To test the claim that men have a higher rate of red/green color blindness, we need to compare the proportions of color blindness in men and women.

Let p_m be the proportion of men with color blindness and p_w be the proportion of women with color blindness.

Null hypothesis: p_m = p_w

Alternative hypothesis: p_m > p_w

To test this hypothesis, we can use a chi-square test of independence. We will compare the observed frequencies of color blindness in men and women to the expected frequencies under the assumption that men and women have the same rate of color blindness.

The chi-square test statistic is calculated as follows:

X^2 = (O_m - E_m)^2/E_m + (O_w - E_w)^2/E_w

where O_m and O_w are the observed frequencies of color blindness in men and women, and E_m and E_w are the expected frequencies of color blindness in men and women.

If the chi-square test statistic is large enough, we reject the null hypothesis and conclude that men have a higher rate of color blindness than women.

The distance (in feet) in which the three steps would overlap for the first time is 56 feet.

Using LCM method

We would find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.

Multiples of 24:

24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288, 312, 336, 360, 384, 408, 432, 456, 480, 504, 528, 552, 576, 600, 624, 648, 672, 696, 720

Multiples of 28:

28, 56, 84, 112, 140, 168, 196, 224, 252, 280, 308, 336, 364, 392, 420, 448, 476, 504, 532, 560, 588, 616, 644, 672, 700, 728

Multiples of 32:

32, 64, 96, 128, 160, 192, 224, 256, 288, 320, 352, 384, 416, 448, 480, 512, 544, 576, 608, 640, 672, 704, 736

Therefore,

LCM(24, 28, 32) = 672

Then we convert to feet:

672/12= 56 feet.

Read more about LCM here:

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Answer:

b. (1.04, 1.68)

Step-by-step explanation:

Hello!

With the given data I've estimated the regression line

Y: Sales

X: Year

Yi= -2691.32 + 1.36Xi

Where

a= -2691.32

b= 1.36

The 95% CI calculated using the statistic b±[tex]t_{n-1;1-\alpha/2}[/tex]*(Sb/√n) is [1.04;1.68]

I hope it helps.

To find the 95% confidence interval for the slope of a regression line, compute the standard error of the slope, multiply it by the critical value from the t-distribution, and subtract and add the margin of error to the estimated slope. In this case, the 95% confidence interval for the slope is (0.364, 1.676).

To find the 95% confidence interval for the slope of the regression line, you need to compute the standard error of the slope. This can be done using the formula: SE = sqrt(SSE/(n-2)) / sqrt(SSX), where SSE is the sum of squared errors, n is the number of data points, and SSX is the sum of squares of the independent variable. Once you have the standard error, you can multiply it by the critical value from the t-distribution with (n-2) degrees of freedom to find the margin of error. Finally, subtract and add the margin of error to the estimated slope to get the lower and upper bounds of the confidence interval.

In this case, the slope of the regression line was estimated to be 1.02. You determined that the standard error of the slope is 0.255. By referring to the t-distribution table or using statistical software, you find the critical value for a 95% confidence interval with (n-2) degrees of freedom to be approximately 2.571. Therefore, the margin of error is 2.571 * 0.255 = 0.656. Finally, you subtract and add the margin of error to the estimated slope to get the lower and upper bounds of the confidence interval: 1.02 - 0.656 = 0.364 and 1.02 + 0.656 = 1.676. So the 95% confidence interval for the slope is (0.364, 1.676).

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Answer:

90% confidence interval: (2186.53;2881.47)

95% confidence interval: (2118.73;2949.27)

99% confidence interval: (1987.37;3080.63)

For the last part is not the best way say : "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."

The best interpretation is this one: "We are 95% confident that the actual mean for the rents of unfurnished one-bedroom apartments in the Boston area is between (2118.73;2949.27)"

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=2534[/tex] represent the sample mean  

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

[tex]\sigma=670[/tex] represent the population standard deviation

n=10 represent the sample size  

90% confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

Now we have everything in order to replace into formula (1):

[tex]2534-1.64\frac{670}{\sqrt{10}}=2186.53[/tex]    

[tex]2534+1.64\frac{670}{\sqrt{10}}=2881.47[/tex]

So on this case the 90% confidence interval would be given by (2186.53;2881.47)

95% confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that [tex]z_{\alpha/2}=1.96[/tex]

Now we have everything in order to replace into formula (1):

[tex]2534-1.96\frac{670}{\sqrt{10}}=2118.73[/tex]    

[tex]2534+1.96\frac{670}{\sqrt{10}}=2949.27[/tex]

So on this case the 95% confidence interval would be given by (2118.73;2949.27)

99% confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that [tex]z_{\alpha/2}=2.58[/tex]

Now we have everything in order to replace into formula (1):

[tex]2534-2.58\frac{670}{\sqrt{10}}=1987.37[/tex]    

[tex]2534+2.58\frac{670}{\sqrt{10}}=3080.63[/tex]

So on this case the 99% confidence interval would be given by (1987.37;3080.63)

For the last part is not the best way say : "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."

The best interpretation is this one: "We are 95% confident that the actual mean for the rents of unfurnished one-bedroom apartments in the Boston area is between (2118.73;2949.27)"

Answer:

Maximum: ((1,-2,5) ; 30)

Minimum: ((-1,2,-5) ; -30)

Step-by-step explanation:

We have the function f(x,y,z) = x - 2y + 5z, with the constraint g(x,y,z) = 30, with g(x,y,z) = x²+y²+z². The Lagrange multipliers Theorem states that, the points (xo,yo,zo) of the sphere where the function takes its extreme values  should satisfy this equation:

grad(f) (xo,yo,zo) = λ * grad(g) (xo,yo,zo)

for a certain real number λ. The gradient of f evaluated on a point (x,y,z) has in its coordinates the values of the partial derivates of f evaluated on (x,y,z). The partial derivates can be calculated by taking the derivate of the function by the respective variable, treating the other variables as if they were constants.

Thus, for example, fx (x,y,z) = d/dx x-2y+5z = 1, because we treat -2y and 5z as constant expressions, and the partial derivate on those terms is therefore 0. We calculate the partial derivates of both f and g

Thus, for a critical point (x,y,z) we have this restrictions:

The last equation is just the constraint given by g, that (x,y,z) should verify.

We can put every variable in function of λ, and we obtain the following equations.

Now, we replace those values with the constraint, obtaining

(1/2λ)² + (-1/λ)²+(5/2λ)² = 30

Developing the squares and taking 1/λ² as common factor, we obtain

(1/λ²) * (1/4 + 1 + 25/4) = (1/λ²) * 30/4 = 30

Hence, λ² = 1/4, or, equivalently,[tex]\lambda =^+_- \frac{1}{2} . [/tex]

If [tex]\lambda = \frac{1}{2} , [/tex] then 1/λ is 2, and therefore

and f(x,y,z) = f(1,-2,5) = 1 -2 * (-2) + 5*5 = 30

If [tex]\lambda = - \frac{1}{2} , [/tex] then 1/λ is -2, and we have

and f(x,y,z) = f(-1,2,-5) = -1 -2*2 + 5*(-5) = -30.

Since the extreme values can be reached only within those two points, we conclude that the maximun value of f in the sphere takes place on ((1,-2,5) ; 30), and the minimun value takes place on ((-1,2,-5) ; -30).

The maximum value is 30, and the minimum value is -30. These occur at the points (1, -2, 5) and (-1, 2, -5), respectively.

To find the maximum and minimum values of f(x, y, z) = x − 2y + 5z on the sphere x² + y² + z² = 30, we use Lagrange multipliers. The constraint is g(x, y, z) = x² + y² + z² - 30 = 0.

We introduce a Lagrange multiplier λ and set up the system of equations:

Solving the system:

Equating the λs:

Substituting y and z into the constraint:

Thus, x = ±1. For x = 1: y = -2, z = 5.

For x = -1: y = 2, z = -5.

Evaluating f at these points:

Hence, the maximum value is 30 and the minimum value is -30.

Final answer:

To find the fraction of applicants with a score of 400 or above, convert the score to a z-score and look it up in a standard normal distribution table. Subtract the resulting proportion from 1 to find the fraction above 400. Approximately 77.34% of the applicants would have a score of 400 or above.

Explanation:

To find the fraction of applicants who would have a score of 400 or above, we need to find the area under the normal distribution curve to the right of 400. First, we need to convert the score of 400 to a z-score using the formula:

z = (x - μ) / σ

where z is the z-score, x is the score, μ is the mean, and σ is the standard deviation. In this case, the mean is 460 and the standard deviation is 80, so the z-score is:

z = (400 - 460) / 80 = -0.75

Once we have the z-score, we can look it up in a standard normal distribution table to find the proportion of the distribution that is below it. The table gives us a value of approximately 0.2266 for a z-score of -0.75. Since we want the fraction above 400, we can subtract this value from 1 to get:

1 - 0.2266 = 0.7734

Therefore, we would expect approximately 77.34% of the applicants to have a score of 400 or above.

Answer:

3.5 miles

Step-by-step explanation:

We need to convert the fraction into decimal and add up both in decimal form and then make our estimate (closest to the exact answer).

1 and 1/8th means 1.SOMETHING

Dividing 1 by 8 would give us:

1/8 = 0.125

Hence, 1  1/8th = 1.125

Total miles = 2.57 + 1.125 = 3.695

THis is closes to 3.5 miles. THis is the answer.

Answer:

c

Step-by-step explanation:

i just took the test

Answer:

Step-by-step explanation:

4) The shape in the figure is a parallelogram. In the parallelogram, the diagonals are equal and bisect each other at the midpoint, E. This means that AC is divided equally into AE and EC. Therefore

AC = 2EC

AC = 8x - 14 and EC = 2x + 11. So,

8x - 14 = 2(2x + 11)

8x - 14 = 4x + 22

8x - 4x = 22 + 14

4x = 36

x = 36/4

x = 9

6)The shape in the figure is a parallelogram. The opposite angles are equal. This means that

Angle BCD = angle BED

So angle BED = 51 degrees.

Since the sum of angles in a triangle is 180 degrees, then, angle BED + angle BDE + angle DBE = 180 degrees. It means

51 + 55 + 14x + 4 = 180

14x + 110 = 180

14x = 180 - 110 = 70

x = 70/14

x = 5

7) The shape in the figure is a parallelogram. The opposite angles are equal. Therefore,

angle VST = angle VUT

5x + 23 = 8x - 49

5x - 8x = -49 - 23

-3x = -72

x = -72 / -3

x = 24

angle VST = 5×24 + 23 = 143

angle VUT = 143

angle VST + angle VUT = 143 + 1143 = 286

Recall, the sum of angles in a parallelogram is 360 degrees. Therefore,

angle SVU + angle STU = 360 - 286 = 74 degrees

Angle SVU = 74/2 = 37 degrees.

Angle SVT + angle UVT = 37

angle SVT + 20 = 37

angle SVT = 37 - 20 = 17 degrees

Answer:

[tex]X_1 + X_2 \sim (\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2 )[/tex]

Step-by-step explanation:

We are given the following in the question:

[tex]X_1[/tex] is a random normal variable with mean and variance

[tex]\mu_1\\\sigma_1^2[/tex]

[tex]X_1 \sim N(\mu_1,\sigma_1^2)[/tex]

[tex]X_2[/tex] is a random normal variable with mean and variance

[tex]\mu_2\\\sigma_2^2[/tex]

[tex]X_2 \sim N(\mu_2,\sigma_2^2)[/tex]

[tex]X_1, X_2[/tex] are independent events.

Let

[tex]Z =X_1 + X_2[/tex]

Then, Z will have a normal distribution with mean equal to the sum of the two means and its variance equal the sum of the two variances.

Thus, we can write:

[tex]\mu = \mu_1 + \mu_2\\\sigma^2 = \sigma_1^2 + \sigma_2^2\\Z \sim (\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2 )[/tex]

Final answer:

The sum of two independent normal random variables X1 and X2 is also a normal random variable with a mean of μ1 + μ2 and variance of σ²₁ + σ²₂, denoted by N(μ1 + μ2, σ²₁ + σ²₂).

Explanation:

The distribution of the sum of two independent normal random variables, X1 + X2, is itself a normal random variable. Given that X1 is a normal random variable with mean μ1 and variance σ²₁, and X2 is a normal random variable with mean μ2 and variance σ²₂, the sum X1 + X2 will have a mean of μ1 + μ2 and variance of σ²₁ + σ²₂ because of the properties of the normal distribution and the independence of X1 and X2.

The resulting distribution can be denoted as N(μ1 + μ2, σ²₁ + σ²₂). This conclusion comes from the central limit theorem, which states that the sum of independent random variables tends towards a normal distribution as the number of variables increases, and the means and variances add up.

Answer:

Both have the same probability of 0.909 or 9.09%

Step-by-step explanation:

For each coin toss, there are only two possible outcomes, heads or tails. Since order is not important in this scenario the number of heads or tails can vary from 0 to 10. Let n be the number of heads flipped in 10 tosses, the number of tails is 10-n. Therefore, the 11 possible outcomes as well as their resulting values for the bet are:

[tex]\begin{array}{ccc}Heads&Tails&Value(\$)\\0&10&-10\\1&9&-8\\2&8&-6\\3&7&-4\\4&6&-2\\5&5&0\\6&4&2\\7&3&4\\8&2&6\\9&1&8\\10&0&10\end{array}[/tex]

Looking at the values above, there is only one outcome in which total winnings are two dollars, and only one in which total winnings are negative two dollars.

Therefore, the probability for each scenario is the same and given by:

[tex]\frac{1}{11}=0.0909=9.09\%[/tex]

Answer:

The projected enrollment is [tex]\lim_{t \to \infty} E(t)=10,000[/tex]

Step-by-step explanation:

Consider the provided projected rate.

[tex]E'(t) = 12000(t + 9)^{\frac{-3}{2}}[/tex]

Integrate the above function.

[tex]E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt[/tex]

[tex]E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c[/tex]

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

[tex]2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c[/tex]

[tex]2000=-\frac{24000}{3}+c[/tex]

[tex]2000=-8000+c[/tex]

[tex]c=10,000[/tex]

Therefore, [tex]E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000[/tex]

Now we need to find [tex]\lim_{t \to \infty} E(t)[/tex]

[tex]\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000[/tex]

[tex]\lim_{t \to \infty} E(t)=10,000[/tex]

Hence, the projected enrollment is [tex]\lim_{t \to \infty} E(t)=10,000[/tex]