Suppose that the NY state total population remains relatively fixed 20Mil, with 8.4Mil of the people living in the city and remaining are in the suburbs. Each year 3.5% of the people living in the city move to the suburbs, and 1.7% of the suburban population moves to the city. What is the long-term distribution of population, after 100 years (what is the population in the city and in the suburbs)? Plot population of city and suburbs over period of 100 years. Submit, 1) answer(s), 2) Matlab code, 3) graph(s)

Answer:

x=6

Step-by-step explanation:

x+2=8

subtract 2 from both sides

x=6

Answer:

Option B will be your answer

Step-by-step explanation:

V=2 1/2×1/2×2

=2.5 cube

volume of 1/2 cubic units =)(1/2×1/2×1/2=0.125cube^3

2.5/0.125=20

Hope it helps...

v = 2½

number of half unit cubes = 20

Answer:

1.

Step-by-step explanation:

base formula

is length × height so that would be 11×4=44

base= 44

height=4

volume= 132

2.

base formula

since its a cube it would be length× hight = 36

base= 36

hight=6

volume =216

3.

base formula

pie×radius^2 so that would be 3.14×1.5^2= 14.80

base 14.80

hight 4

volume= 59.2

4

Answer: No. Pennant Sports Shop did not sell the same ratio of defense sticks to total lacrosse sticks on both days.

Step-by-step explanation:

On Friday, there were 25 lacrosse sticks sold, 10 of which were defense sticks. The ratio of defense sticks to lacrosse will be:

= 10/25 = 2/5 = 2:5

On Saturday, there were 50 lacrosse sticks sold, 15 of which were defense stick. The ratio of defense sticks to lacrosse will be:

= 15/50 = 3/10 = 3:10

Based on the calculation above, Pennant Sports Shop did not sell the same ratio of defense sticks to total lacrosse sticks on both days

The probability P(Z < -1.06) is approximately 0.142. The probability P(Z > 2.386) is about 0.008. The probability P(-0.777 < Z < 0.777) is approximately 0.456.

The probability P(X < 9.8) ≈ 0.211. The probability P(X > 10.2) =  = 0.212. The probability P(X < 9.8 or X > 10.2) = 0.423.

To locate the distribution of X and the indicated possibilities for the given instances, we need to use the residences of the everyday distribution. Given that the populace has a median (μ) of 10 and a widespread deviation (σ) of 2.5, we will continue as follows:

a. N = 7, P(X < 9):

For a pattern size of seven, the distribution of X follows a normal distribution with the equal mean (10) however a trendy deviation of σ/sqrt(n) = 2.5/[tex]\sqrt{7}[/tex] ≈ 0.944.

To discover P(X < nine), we need to standardize the cost of 9 with the use of the Z-rating formula: Z = (X - μ) / σ.

Substituting the values, we get Z = (9 - 10) / 0.944 ≈ -1.06.

Using a standard regular distribution table or calculator, we are able to locate that the chance P(Z < -1.06) is approximately 0.142.

B. N = 12, P(X > 11.5):

For a sample length of 12, the distribution of X follows a regular distribution with the same suggestion (10) but a well-known deviation of σ/[tex]\sqrt{n}[/tex] = 2.5/[tex]\sqrt{12}[/tex] ≈ 0.7217.

To discover P(X > 11.5), we standardize the value of 11.5 for the usage of the Z-rating method: Z = (X - μ) / σ.

Substituting the values, we get Z = (11.5 - 10) / 0.7217 ≈ 2.386.

Using a trendy everyday distribution table or calculator, we will locate that the chance P(Z > 2.386) is about 0.008.

C. N = 15, P(9.5 < X < 10.25):

For a sample size of 15, the distribution of X follows a normal distribution with identical implies (10) however a popular deviation of σ/sqrt(n) = 2.5/[tex]\sqrt{15}[/tex]≈ 0.6455.

To discover P(9.5 < X < 10.25), we need to standardize the values using the Z-score components.

Z1 = (9.5 - 10) / 0.6455 ≈ -0.777, and Z2 = (10.25 - 10) / 0.6455 ≈ 0.777.

Using a widespread ordinary distribution desk or calculator, we can locate that P(-0.777 < Z < 0.777) is approximately 0.456.

D. N = 100, P(X < 9.8 or X > 10.2):

For a sample size of 100, the distribution of X follows a regular distribution with the equal implies (10) however a general deviation of σ/sqrt(n) = 2.5/[tex]\sqrt{100}[/tex] = 0.25.

To find P(X < 9.8 or X > 10.2), we need to calculate the probabilities for each person's case and subtract them from 1.

P(X < 9.8) = P(Z < (9.8 - 10) / 0.25) ≈ P(Z < -0.8) ≈ 0.211.

P(X > 10.2) = P(Z > (10.2 - 10) / 0.25) ≈ P(Z < -0.8) ≈ 1 - P(Z < 0.8) ≈ 1 - 0.788 = 0.212.

Therefore, P(X < 9.8 or X > 10.2) ≈ P(X < 9.8) + P(X > 10.2) ≈ 0.211 + 0.212 = 0.423.

Remember to consult a trendy everyday distribution desk or use a calculator to locate the possibilities associated with the Z-scores.

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Answer: A: y=5x+7

Step-by-step explanation: Hope this help :D

The average number of full-time students in samples of size 16 is B) 3.5.

Because of the extraordinarily huge population, this can be regarded a binomial distribution if all students globally are considered. A normal distribution is commonly used to approximate the binomial distribution. As a result, the mean equals the expectation:

E[x] = np = (16)(0.22) = 3.52

μ = 3.52

The likelihood of success raised to the power of the number of successes is multiplied by the probability of failure raised to the power of the difference between the number of successes and the number of trials. The product is then multiplied by the sum of the number of trials and successes.

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Answer:

1x + 15?

Step-by-step explanation:

Answer: Z = 83

X = 7

Step-by-step explanation:

Z = 83 because the opposite side is also 83 and they are the same angle. X = 7 because the whole thing equals 360 and 83 +83 = 166 and 360 - 166 = 194 and 194 divided by 2 equals 97 and 15 x 7 - 8 = 97 and the other side is also the same because the angles are the same

Answer:

$85.2

Step-by-step explanation:

Answer:

85.2 dollar

Step-by-step explanation:

Tax is 5.2 because 80 times 0.065 is that, then you add them up

Answer:2 11/24

Step-by-step explanation:

Answer:

2.4583

Step-by-step explanation:

I helped you.

First

5÷6 and 7÷8 and again similarly 3÷4 and sum the all outcome come from these eqations

The value of the test statistic is  x² = 3.841.

Therefore option B is correct.

Expected frequency for "Health insurance" and "Men"

= (80 * 70) / 110= 50.909

Expected frequency for "Health insurance" and "Women"

= (30 * 70) / 110 =  19.091

Expected frequency for "No health insurance" and "Men"

= (80 * 40) / 110

= 29.091

Expected frequency for "No health insurance" and "Women"

= (30 * 40) / 110 =  10.909

We find the chi-square statistic

χ² = Σ [(O - E)² / E]

χ² = [(50 - 50.909)² / 50.909] + [(20 - 19.091)² / 19.091] + [(30 - 29.091)² / 29.091] + [(10 - 10.909)² / 10.909]

= (0.081² / 50.909) + (0.909² / 19.091) + (0.909² / 29.091) + (0.081² / 10.909)

=  0.00131 + 0.04537 + 0.02784 + 0.00604

= 0.08056

We find  the critical value for a 0.05 significance level and 1 degree of freedom = 3.841.

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complete question:

A survey conducted in a small business yielded the results shown in the table. Test the claim that health care coverage is independent of gender. Use a 0.05 significance level. What is the value of the test statistic?

                                    Men          Women

Health insurance           50              20

No health insurance      30               10

OA. x^2=0.1637

OB. x^2=8.263

OC. x^2 =3.841

OD. x^2 =5.821

Answer:

1st one = 4th option

2nd one = 3rd option

Step-by-step explanation:

Answer:

10.5 cm

Step-by-step explanation:

A line segment is defined as a part of a line. It is a straight line having two end points. The only difference between a line and line segment is that a line is continuous with indefinite length whereas a line segment has a definite length to it between the two points.

In the figure, the length of the line segment is 10.5 cm (rounded to the nearest tenth).

Answer:

10.5

Step-by-step explanation:

Answer:

Hello There!!

Step-by-step explanation:

If 1 ton=2000 (1 ton to 4 is times by 4 so 2000

4 tons=8000 is times by 4)

hope this helps, have a great day!!

~Pinky~

The research title, "Seasonal distribution of endemic typhus in the Southern United States, 1922-1925" would most likely utilize a longitudinal ecologic study.

The correct option is: b. Longitudinal ecologic study

A longitudinal study is a research design that follows participants over an extended period to establish patterns of change or stability in development. The same group of people is monitored at various stages in their life. For example, a group of people may be followed from the age of 5 to the age of 50, and their health status, life choices, and other variables are monitored at each age interval.

The research title, "Seasonal distribution of endemic typhus in the Southern United States, 1922-1925," would most likely utilize a longitudinal ecologic study since the research is centered on an illness and covers four years. Therefore, researchers are likely to follow up with participants and monitor their health status over an extended period to obtain data on the seasonal distribution of endemic typhus.

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a) the value of the integral 1.664. b) the coefficient of x4 is (3/1280). c) the value of the integral  2.14 d) the percentage error in the approximation is -28.67%.

a) To evaluate the integral exactly, we make a substitution in the form of ax = sin 0 where a = (1/2).

Substitute x = (sin θ)/2, dx = (cos θ)/2 dθ, and 1 - 4x² = cos² θ in the integral to get it in terms of θ.

∫5√(1-4x²)dx = ∫(5/2) cos²θ dθApply the identity cos²θ = (1 + cos 2θ)/2 to simplify the integrand as shown.∫(5/2) cos²θ dθ = (5/4)∫(1 + cos2θ) dθ = (5/4)θ + (5/8)sin 2θ

Evaluate the above expression from 0 to π/2 to get the value of the integral.

(5/4)θ + (5/8)sin 2θ = (5/4) π/2 + (5/8) sin π = 1.664

b) The integrand is f(x) = 5√(1-4x²).We can write it as shown below, using the binomial series. f(x) = 5(1 - 4x²)^(1/2) = 5∑_(n=0)^∞〖(1/2)_n (2n)!/n! (1/16)^n x^(2n) 〗

The above expression is the Maclaurin Series expansion of f(x) as required.In the expansion, the coefficient of x4 is (1/2)_2 (2.4)/(2!) (1/16)^2 = (3/1280)

c) Integrating each term of the expansion, we obtain the following expression.(5/4)θ + (5/8)sin 2θ = (5/4) π/2 + (5/8) sin π = 1.664

We approximate the value of the integral using the first three terms of the series expansion, and then add the values of the integrated terms.

The terms up to x4 are included in this calculation. I=5[1+(1/2) (-4x²) +(1/2)_2 (-4x²)²]I=5[1-2x²+3x^4/4] = (5/4) (π/2 + (5/16)π²) ≈ 2.14

d) The percentage error in the approximation is given by:%Error = [(Exact value - Approximation value)/ Exact value] x 100Substitute the appropriate values to calculate.%Error = [(1.664 - 2.14)/1.664] x 100 = -28.67% (correct to 3 significant figures)Thus, the percentage error in the approximation is -28.67%.

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Without yield management, the theatre's revenues would be $176,100.

To calculate the theatre's revenues without yield management, we need to determine the number of tickets sold for each fare class and multiply it by the corresponding ticket price. The number of low fare class tickets sold is not provided in the information, so we cannot determine the exact revenue for that fare class. However, since there is ample demand for the low fare class, it is reasonable to assume that all 1761 seats are sold at the discounted price of $75. Therefore, the revenue from the low fare class would be 1761 seats × $75 = $132,075. For the high fare class, the demand is normally distributed with a mean of 1200 and a standard deviation of 150. To maximize revenues, the theatre should sell as many high fare class tickets as possible. Assuming all remaining seats (1761 - number of low fare class tickets) are sold at the high fare of $100, the revenue from the high fare class would be (1761 - number of low fare class tickets) × $100. To calculate the exact number of high fare class tickets sold, we need to know the cutoff point for the demand distribution that separates high fare demand from low fare demand. Since this information is not provided, we cannot determine the exact revenue from the high fare class. However, the only answer option that is close to the estimated revenue considering all high fare class tickets sold is $176,100.

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Answer:

13 + 1 1/2 - 1/4

Step-by-step explanation:

1 gallons = 4 quarts

3 1/4 = 13/4 x 4 = 13

Quarts left = original quart + quart added - quart drunk

13 + 1 1/2 - 1/4

Answer:

C and D

Step-by-step explanation:

Got it right on the test.

1 hexagon and 6 rectangles and 4 squares and two rectangles that are not squares are sets of shapes which can be used as the net of a three-dimensional figure. Option A and option D are correct.

Hexagon and 6 rectangles: This combination of shapes can be used to create the net of a hexagonal prism.

The hexagon will be used as the base of the prism, and the six rectangles can be attached to the sides of the hexagon to form the remaining faces of the prism.

When folded along the edges, the net will result in a hexagonal prism.

4 squares and two rectangles that are not squares: This combination can be used to create the net of a rectangular prism or a cuboid.

The four squares will be used as the top, bottom, and side faces of the prism, while the two rectangles (which are not squares) can be attached to the remaining sides to complete the net.

When the net is folded along the edges and assembled, it will form a rectangular prism or cuboid.

Hence, Option A and option D are correct. 1 hexagon and 6 rectangles and 4 squares and two rectangles that are not squares are sets of shapes which can be used as the net of a three-dimensional figure.

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Answer:

24 i believe :)

Answer:

16.39

Step-by-step explanation:

volume of sphere = (4/3) · 3.14(π) · r³

=> 2304 = 4/3 · 3.14 · r³

=> 2304 · 3/4 = 3.14 · r³

=> 1,728 = 3.14 · r³

=> r³ = 550.318

=> r = 8.19

diameter = 2 · r

=> diameter = 2 · 8.19 = 16.39

The probability of 2 cars crossing the bridge from X to Y in a 1-minute period is:

P(X = 2) = ((λ^x) / x!) * e^(-λ)

P(X = 2) = ((XX/100 * 10)^2 / 2!) * e^(-XX/100 * 10)

P(X = 2) = (XX/500) * e^(-XX/100 * 10)

P(X = 2) = (XX/500) * 0.000045

a) Let X be the number of cars that cross the bridge in a 1-minute period.

Since the Poisson process with a rate of µ = 600/hour, the rate of cars crossing the bridge is

λ = 600/60

= 10/min.

The probability that two cars will cross the bridge during the 1-minute period can be calculated by the following formula:

P(X = 2) = ((λ^x) / x!) * e^(-λ)

P(X = 2) = ((10^2) / 2!) * e^(-10)

P(X = 2) = 0.0045

b) Let Y be the number of cars that travel from X to Y in a 1-minute period.

Since the probability that a car travels from X to Y is p = 0.XX,

The probability that a car travels from Y to X is 1 - p.

As per the Poisson process, the probability of

λ = 10/min.

Let X be the number of cars that cross the bridge from X to Y in a 1-minute period.

Then X follows a Poisson distribution with a rate of

µ = 10*p

= XX/100 * 10.  

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Answer:

x - 4 x 2 = 3

Is your algebra expression.

Answer:

15°, 16°, 46°, 59°

Step-by-step explanation:

1.) so the entire angle is 89°, then you take off the 44° and the 30°

89° - 44° - 30° = 15°

So the answer to 1 is x = 15°

2.) 90° - 74° = 16°

3.) 180° - 134° = 46°

4.) the angles are vertical, so they're equal:

x = 59°

hope this helps:)

Answer:

You just have to combine like terms to get the most simplified v erosion of this equation.

The final simplifed answer would be

x+7

Step-by-step explanation:

Well just add everything up.

15x-14x=x

3+4=7

Combine

x+7

The independent groups t-test is appropriate for situations b and d. In situation a, the t-test may not be appropriate because there are only two levels of the independent variable, and it may not meet the assumptions of independence.

a. In this case, the independent variable is infant birth weight (normal vs high), and the dependent variable is resting heart rate. Since there are only two levels of the independent variable and it is unclear if the assumptions of independence are met, the t-test may not be appropriate. Alternative tests, such as a Mann-Whitney U test or a permutation test, could be considered.  b. Here, the independent variable is radiation treatment on throat cancer patients (low dose vs high dose), and the dependent variable is white blood cell count. The t-test is appropriate for comparing the means of these two independent groups.

c. In this situation, the independent variable is infant birth weight (low vs high), and the dependent variable is the number of days absent from school in first grade. Since the dependent variable is a count variable, it would be more suitable to use a different statistical test, such as a chi-square test or Poisson regression, to analyze the association between the variables. d. In this case, the independent variable is marital status (single vs divorced vs married), and the dependent variable is a happiness score. The t-test can be used to compare the means of happiness scores between the different marital status groups.

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Answer:

The area of the circle is 28.26 m

Step-by-step explanation:

A = π r^2

A = (3.14) × 3^2

A = 28.26

Area of the circle = 28.26 m

(I'm sorry but I think you made a mistake, PI is 3.14 not 3.15) :)