Find the equation of the line passing through the point P
(2,1,-1) and orthogonal to the plane 2x-y+3z=10? [use X=P+TD
vector, X=x,y,z]

Y(y) = dF_Y(y)/dy = -y^2

a. Let X be the number of picked points that is smaller than 1/4. Determine the distribution of X.Random variables and distributions Let A be the set of all possible 3-tuples (a1, a2, a3) picked from (0, 1).The number of the picked points that are smaller than 1/4 is random with X: A → {0, 1, 2, 3}X((a1, a2, a3)) = |{i ∈ {1, 2, 3} : ai < 1/4}|This is, X is the number of i's such that ai < 1/4. We would like to compute the distribution of X.Let us count the number of 3-tuples (a1, a2, a3) for which X = 0, 1, 2, or 3. These counts are as follows:X = 0: There is only one tuple for which all ai ≥ 1/4.X = 1: Each of the ai can be either ≥ 1/4 or < 1/4, and there are three ways to choose which one is less than 1/4. Therefore, there are 2^3 · 3 = 24 3-tuples (a1, a2, a3) such that exactly one ai < 1/4.X = 2: Either all ai < 1/4, or two ai's < 1/4 and the other one ≥ 1/4. If all ai < 1/4, then there is one tuple for this. If two ai's < 1/4 and the other one ≥ 1/4, then there are 3 ways to choose which ai is ≥ 1/4, and for each such choice there are 3 · 2 = 6 ways to pick the other two ai's. Thus, there are 1 + 3 · 6 = 19 tuples for which X = 2.X = 3: All ai's < 1/4. There is only one such tuple.Therefore, the probability mass function of X is as follows:P(X = 0) = 1/8P(X = 1) = 3/8P(X = 2) = 19/32P(X = 3) = 1/32b. Let Y be the middle one of the three points. Determine the cdf of Y. (It is a function with domain R!)The range of Y is the interval (0, 1). We can assume that a1 ≤ a2 ≤ a3, since any 3-tuple can be sorted in this way. If Y < y, then the largest of the three points must be less than y. Therefore, we need to find the probability that the largest point is less than y, given that the three points are picked independently at random from (0, 1) and are ordered as a1 ≤ a2 ≤ a3.Let A be the set of all possible 3-tuples (a1, a2, a3) picked from (0, 1), where a1 ≤ a2 ≤ a3. Let B(y) be the set of all 3-tuples (a1, a2, a3) ∈ A such that a3 < y. Then P(Y < y) = P(B(y)).To compute P(B(y)), let us count the number of 3-tuples (a1, a2, a3) ∈ A such that a3 < y. Fix a1 and a2. Then a3 < y if and only if a3 ∈ (0, y), so there are exactly y(1 - y) choices for a3. Therefore,|B(y)| = ∫∫A y(1 - y) da1 da2 = ∫0^1 ∫a1^1 y(1 - y) da2 da1 = ∫0^1 y(1 - y) (1 - a1) da1 = 1/6 - y^3/3Thus,P(Y < y) = P(B(y)) = |B(y)|/|A| = [1/6 - y^3/3]/1 = 1/6 - y^3/3This is the cdf of Y for y ∈ (0, 1). We can extend this function to all of R by setting F_Y(y) = 0 if y ≤ 0 and F_Y(y) = 1 if y ≥ 1.c. Determine the pdf of Y.The cdf of Y isF_Y(y) = 1/6 - y^3/3for y ∈ (0, 1). Therefore, the pdf of Y isf_Y(y) = dF_Y(y)/dy = -y^2for y ∈ (0, 1). Thus, the density is constant, and Y has a uniform distribution on (0, 1).

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According to the information, the values that complete the table are: 4, 6, and 150.

To calculate the missing numbers in the table we must analyze the existing data. Once we have analyzed this data we can establish that it is a constant relationship, so we can find the missing numbers by means of a rule of three as shown below:

30 = 2

60 =?

60 * 2/30 = 4

30 = 2

90 =?

90 * 2/30 = 6

2 = 30

10 =?

10 * 30/2 = 150

According to the above, the missing values in the box from left to right are: 4, 6, and 150.

Note: This question is incomplete. Here is the complete information:

Attached image

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The annual interest rate is 6.5%

The first step is to write out the parameters given in the question

Joseph places $5500 in a savings account for 30 months

He rans $893.75 in interest

The annual interest rate can be calculated by multiplying the amount in the savings by the number of months which is 30

= 5500 × 30y
= 165,000y

= 165,000y/12

= 13,750y

Equate 13,750y  with the amount of interest

13,750y= 898.75

y= 898.75/ 13,750

y= 0.065 × 100

= 6.5%

Hence the annual interest rate is 6.5%

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Answer:

AC is about 33 inches.

Step-by-step explanation:

Set your calculator to degree measure.

[tex] \tan(35) = \frac{x}{47} [/tex]

[tex]x = 47 \tan(35) = 32.91[/tex]

Based on the calculation we know that

The value of x as 7− is 35.

The value of x as 7+ is 35.

Suppose that f( x ) = 7 x − 7. We can complete the following statements by plugging in the values of x and evaluating the function.

As x → 7 − , f ( x ) → 42 − 7 = 35
As x approaches 7 from the left, the function f(x) approaches 35.

As x → 7 + , f ( x ) → 42 − 7 = 35
As x approaches 7 from the right, the function f(x) also approaches 35.

Therefore, the function f(x) = 7x - 7 has a horizontal asymptote at y = 35 as x approaches 7 from both the left and the right.

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The possible rational zeros of h(x) are 1, -1, 3, -3, 1/2, -1/2, 3/2, and -3/2.

According to the Rational Zeros Theorem, if a polynomial has any rational zeros, they must be in the form of p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

For the given polynomial h(x)=-2x^(4)+7x^(3)-8x^(2)+9x+3, the constant term is 3 and the leading coefficient is -2.

The factors of 3 are 1 and 3, and the factors of -2 are 1, 2, -1, and -2.

Therefore, the possible rational zeros of h(x) are: p/q = ±1/1, ±3/1, ±1/2, ±3/2

Simplifying these values, we get the following list of possible rational zeros:

±1, ±3, ±1/2, ±3/2

So, the possible rational zeros of h(x) are 1, -1, 3, -3, 1/2, -1/2, 3/2, and -3/2.

Be sure to check each of these values by plugging them back into the original polynomial to see if they are actually zeros.

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By answering the above question, we may infer that So Kendra's water equation bottle holds 64 fluid ounces.

A mathematical equation links two statements and utilises the equals sign (=) to indicate equality. In algebra, an equation is a mathematical assertion that proves the equality of two mathematical expressions. For instance, in the equation 3x + 5 = 14, the equal sign separates the numbers by a gap. A mathematical formula may be used to determine how the two sentences on either side of a letter relate to one another. The logo and the particular piece of software are usually identical. like, for instance, 2x - 4 = 2.

The volume of a quart is 32 fluid ounces. Kendra's water bottle, which holds 2 quarts of water, therefore has:

64 fluid ounces are equal to 2 quarts when each is 32 fluid ounces.

So Kendra's water bottle holds 64 fluid ounces.

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By answering the above question, we may state that We are unable to solve this equation for both l and w since it has only one solution and two unknowns.

Mathematicians research numbers, their variants, equations, forms, and related structures, as well as possible locations for these things. The relationship between a group of inputs, each of which has a corresponding output, is referred to as a function. Every input contributes to a single, distinct output in a connection between inputs and outputs known as a function. A domain, codomain, or scope is assigned to each function. Often, functions are denoted with the letter f. (x). The key is an x. There are four main categories of accessible functions: on functions, one-to-one capabilities, so many capabilities, in capabilities, and on functions.

We are aware of the frame's 42-inch circumference but are unaware of the picture's or frame's shape, as well as if the frame's breadth fluctuates or is constant.

With the assumption that the frame is rectangular and has a constant width, we might create the equation shown below:

2l + 2w + 4x = 42

where L stands for the picture's length, W for its width, and X for the frame's width (assuming there are four sides with the same width).

We are unable to solve this equation for both l and w since it has only one solution and two unknowns.

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The point in which the line through the origin perpendicular to the plane 2x - y - z = 4 meets the plane 3x - 5y + 2z = 6 is (4/3, -2/3, -2/3).

To find the point in which the line through the origin perpendicular to the plane 2x - y - z = 4 meets the plane 3x - 5y + 2z = 6, we can use the following steps:

Find the normal vector of the plane 2x - y - z = 4. This is given by the coefficients of the x, y, and z terms, which are 2, -1, and -1, respectively. So the normal vector is (2, -1, -1).

Since the line through the origin is perpendicular to the plane, it must be parallel to the normal vector of the plane. Therefore, the direction vector of the line is (2, -1, -1).

The equation of the line through the origin with direction vector (2, -1, -1) is given by x = 2t, y = -t, and z = -t, where t is a parameter.

Substitute the equations of the line into the equation of the plane 3x - 5y + 2z = 6 to find the value of t:

3(2t) - 5(-t) + 2(-t) = 6

6t + 5t - 2t = 6

9t = 6

t = 2/3

Substitute the value of t back into the equations of the line to find the point of intersection:

x = 2(2/3) = 4/3

y = -(2/3) = -2/3

z = -(2/3) = -2/3

Therefore, the point in which the line through the origin perpendicular to the plane 2x - y - z = 4 meets the plane 3x - 5y + 2z = 6 is (4/3, -2/3, -2/3).

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Yes, it is better to use percentages to compare groups when making a graph. This is because percentages allow for a more accurate comparison of the data, as they take into account the relative size of each group. If we only use frequencies, we may get a skewed view of the data, as one group may be much larger than the other.

Here is the side by side bar graph showing the men's and women's responses to the question "Do you play sports?":

Gender Yes No

Male 9 7

Female 7 14

The graph indicates that a higher percentage of males play sports compared to females. However, it also shows that there are more females who do not play sports compared to males. This suggests that there may be a gender difference in the participation of sports among students.

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The simplified expression is x^(454)y^(584) * 2^(109).

To simplify the given expression, we need to use the laws of exponents. The laws of exponents are:

- (a^m)^n = a^(mn)
- a^m * a^n = a^(m+n)
- a^m / a^n = a^(m-n)

Using these laws, we can simplify the given expression as follows:

(x^3y^3)^31 * (32x^13y^18)^31 / (2xy^3)^2 * (2xy^3)^2 * (2x)^2 * (2x^3y^5)^3 * (4xy)^5 * (2x^3y^5)^3 * (4x)^5

= x^(3*31)y^(3*31) * 32^31x^(13*31)y^(18*31) / 2^2x^2y^(3*2) * 2^2x^2y^(3*2) * 2^2x^2 * 2^(3*2)x^(3*3)y^(5*3) * 4^5x^5y^(5*5) * 2^(3*2)x^(3*3)y^(5*3) * 4^5x^5

= x^(93)y^(93) * 32^31x^(403)y^(558) / 2^6x^6y^6 * 2^6x^6y^6 * 2^2x^2 * 2^6x^9y^15 * 4^5x^5y^25 * 2^6x^9y^15 * 4^5x^5

= x^(93+403)y^(93+558) * 32^31 / 2^6 * 2^6 * 2^2 * 2^6 * 4^5 * 2^6 * 4^5 * x^(6+6+2+9+5+9+5) * y^(6+6+15+25+15)

= x^(496)y^(651) * 32^31 / 2^(6+6+2+6+6) * 4^(5+5) * x^42 * y^67

= x^(496-42)y^(651-67) * 32^31 / 2^26 * 4^10

= x^(454)y^(584) * 32^31 / 2^26 * 4^10

= x^(454)y^(584) * 2^(5*31) / 2^26 * 2^(2*10)

= x^(454)y^(584) * 2^(155) / 2^(26+20)

= x^(454)y^(584) * 2^(155-46)

= x^(454)y^(584) * 2^(109)

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Answer:A

Step-by-step explanation:

because

Answer:

Step-by-step explanation:

The method to find the midpoint of two points is (a + b)/2 a and b both being x+x and y+y for the differentiating coordinates respectively.

here, it would be x = (-2 + 4)/2 and y = (5-5)/2

giving you the coordinate [ 1 , 0 ]

Answer:

Ans = (1.50 x 29 movies) + 16

Step-by-step explanation:

y represents total costs

x represents how many movie he rents

$1.50 is the rent of each movie

$16 is the fixed fee

thus

y = 1.50 x + 16

Ans = (1.50 x 29 movies) + 16

For Michael to pay cash for the tractor priced at R160,000 with the exchange rate as US$1 = R17.96, the amount he must change is US$8,908. 69.

An exchange rate is the unit rate at which one country's currency is exchanged for another.

Exchange rates are based on a country's economic performance or indices in comparison to other countries.

The price of the tractor = R160,000

Exchange Rate: US$ = R17.96

R160,000 = US$8,908. 69 (R160,000/R17.96)

Thus, Michael needs to exchange US$8,908. 69 to purchase the tractor costing R160,000.

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For 15:
Sine: √2/2
Cosine: √2/2
Tangent: 1

For 17:
Sine: -1/2
Cosine: √3/2
Tangent: -1/√3

For 19:
Sine: (1+√3)/2
Cosine: (1-√3)/2
Tangent: √3

For 21:
Sine: -1/2
Cosine: -√3/2
Tangent: 1/√3

For 23:
Sine: √3/2
Cosine: 1/2
Tangent: √3

For 25:
Sine: -√2/2
Cosine: -√2/2
Tangent: -1

For 27:
Sine: (1-√3)/2
Cosine: (1+√3)/2
Tangent: -√3

For 29:
Sine: -(√3-1)/2
Cosine: (1+√3)/2
Tangent: -√3

For 16:
Sine: √3/2
Cosine: -1/2
Tangent: -√3

For 18:
Sine: -√2/2
Cosine: √2/2
Tangent: -1

For 20:
Sine: (√3-1)/2
Cosine: (1+√3)/2
Tangent: √3

For 22:
Sine: -(1+√3)/2
Cosine: (√3-1)/2
Tangent: -1/√3

For 24:
Sine: 1/2
Cosine: √3/2
Tangent: 1/√3

For 26:
Sine: √2/2
Cosine: -√2/2
Tangent: 1

For 28:
Sine: (√3+1)/2
Cosine: (1-√3)/2
Tangent: -1/√3

For 30:
Sine: -(1-√3)/2
Cosine: (√3+1)/2
Tangent: √3

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The p-value for our test is 0.0120, accurate to 4 decimal places.

We are conducting a study to see if the proportion of men over 50 who regularly have their prostate examined is significantly different from 0.49. We are performing a two-tailed test and our sample data produce the test statistic z=2.517.

To find the p-value, we need to use a standard normal distribution table or calculator. We will use the table for this answer.

The first step is to find the area to the left of the test statistic z=2.517 in the standard normal distribution table. This gives us the probability of obtaining a z-score less than or equal to 2.517. The closest value in the table is 2.51, which gives us an area of 0.9940.

Since we are conducting a two-tailed test, we need to find the area in both tails of the distribution. To do this, we subtract the area to the left of the test statistic from 1: 1-0.9940 = 0.0060. This gives us the area in the right tail of the distribution.

Finally, we multiply this value by 2 to get the p-value for our two-tailed test: 0.0060 * 2 = 0.0120.

Therefore, the p-value for our test is 0.0120, accurate to 4 decimal places.

p-value = 0.0120

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The members of the second set are 0, 1, 4, 9, 16, 25, 36, 49, 64, and 81.

The members of the first set, y|y=(x/x+1), x ϵ Z+, x<12, are the values of y that satisfy the given equation and conditions. The members of the second set, {x | x is the square of a whole number and x < 100}, are the values of x that satisfy the given conditions.

For the first set, we need to find the values of y that satisfy the equation y=(x/x+1) for positive integers x less than 12. The values of x that satisfy this condition are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11. We can plug these values into the equation to find the corresponding values of y:

y = (1/1+1) = 0.5
y = (2/2+1) = 0.6667
y = (3/3+1) = 0.75
y = (4/4+1) = 0.8
y = (5/5+1) = 0.8333
y = (6/6+1) = 0.8571
y = (7/7+1) = 0.875
y = (8/8+1) = 0.8889
y = (9/9+1) = 0.9
y = (10/10+1) = 0.9091
y = (11/11+1) = 0.9167

Therefore, the members of the first set are 0.5, 0.6667, 0.75, 0.8, 0.8333, 0.8571, 0.875, 0.8889, 0.9, 0.9091, and 0.9167.

For the second set, we need to find the values of x that are the square of a whole number and less than 100. The whole numbers that satisfy this condition are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. We can square these values to find the corresponding values of x:

x = 0^2 = 0
x = 1^2 = 1
x = 2^2 = 4
x = 3^2 = 9
x = 4^2 = 16
x = 5^2 = 25
x = 6^2 = 36
x = 7^2 = 49
x = 8^2 = 64
x = 9^2 = 81

Therefore, the members of the second set are 0, 1, 4, 9, 16, 25, 36, 49, 64, and 81.

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The equation of the polynomial is f(x) = (1/36)(x - 3)2(x + 4)2(x + 1)3.

The equation of the polynomial can be found using the given information about the degree, zeroes, and y-intercept. The general form of a polynomial is:

f(x) = a(x - x1)n1(x - x2)n2...(x - xk)nk

Where x1, x2, ..., xk are the zeroes of the polynomial, n1, n2, ..., nk are the multiplicities of the zeroes, and a is the leading coefficient.

Using the given information, we can plug in the values for the zeroes and their multiplicities:

f(x) = a(x - 3)2(x + 4)2(x + 1)3

The degree of the polynomial is 7, which is the sum of the multiplicities of the zeroes. So, the equation is correct in terms of the degree.

Now, we need to find the value of a using the y-intercept. The y-intercept is the point where the graph of the polynomial crosses the y-axis, which means that x = 0. So, we can plug in x = 0 and the given y-intercept (0,4) into the equation and solve for a:

4 = a(0 - 3)2(0 + 4)2(0 + 1)3

4 = a(9)(16)(1)

4 = 144a

a = 4/144

a = 1/36

Now, we can plug in the value of a back into the equation to get the final answer:

f(x) = (1/36)(x - 3)2(x + 4)2(x + 1)3

Therefore, the equation of the polynomial is f(x) = (1/36)(x - 3)2(x + 4)2(x + 1)3.

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Answer: 7

Step-by-step explanation: The average rate of change of a function f(x) over an interval [a,b] is defined as the ratio of “change in the function values” to the "change in the endpoints of the interval"1. In other words, it is the slope of the line that passes through two points on the graph of f(x)2.

To find the average rate of change of f(x) = x² + x + 4 from x = 2 to x = 4, we can use this formula:

Average rate of change = [f(4) - f(2)] / (4 - 2)

First, we need to plug in x = 4 and x = 2 into f(x) and simplify:

f(4) = (4)² + (4) + 4 f(4) = 16 + 8 f(4) = 24

f(2) = (2)² + (2) + 4 f(2) = 4 + 6 f(2) = 10

Next, we need to subtract f(2) from f(4), and divide by (4 - 2):

Average rate of change = [24 - 10] / (4 - 2) Average rate of change = 14 / 2 Average rate of change = 7

Answer:

x=5, y=-15

Step-by-step explanation:

7x+2y=5

3x+2y=-15

4x=20

 x=5

7(5)+2y=5

       2y=-30

         y=-15

To create a graph and write an equation to represent the list of ordered pairs given, we need to plot each point on a coordinate system.

x | y

--|--

2 |-3

1 |-1

-1| 3

3 |-5

Plotting these points on a coordinate plane, we get:

![Graph of ordered pairs]

To write an equation to represent these ordered pairs in the form y = mx + b, we need to find the slope of the line and the y-intercept.

To find the slope:

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

Let's use the points (1, -1) and (3, -5) to find the slope:

m = (-5 - (-1)) / (3 - 1)

m = -4 / 2

m = -2

Now that we know the slope, we can use any point and the slope to find the y-intercept (b).

Using the point (1, -1):

y = mx + b

-1 = (-2)(1) + b

b = 1

So the equation that represents these ordered pairs in the form y = mx + b is:

y = -2x + 1

And the graph of this equation looks like:

![Graph of equation y = -2x + 1]

The cumulative distribution function for N when a coin is flipped 3 times with  P[H]=0. 3 is given by ,

F(N≤ 0) = 0.343 , F(N≤1) = 0.784 , F(N≤2) = 0.973 , F(N≤3) = 1.000.

Cumulative distribution function (CDF) of N,

Probability of getting n or fewer heads in three coin flips,

(All values of n from 0 to 3)

Probability of getting exactly n heads in three flips.

Apply binomial distribution,

P(N = n) = ³Cₙ ×(0.3)ⁿ × (0.7)³⁻ⁿ

For cumulative distribution function, add up the probabilities of getting 0, 1, 2, or 3 heads in three flips,

F(N≤ 0)

=P(N=0)

=³C₀ × (0.3)⁰ × (0.7)³

= 1 × 1 × 0.343

= 0.343

For F(N≤1) is equal to,

⇒F(N≤1)

= P(N=0) + P(N=1)

 =  ³C₀ × (0.3)⁰ × (0.7)³ + ³C₁ × (0.3)¹ × (0.7)²

= 0.343 + 0.441

= 0.784

F(N ≤ 2)

= P(N=0) + P(N=1) + P(N=2)

= ³C₀ × (0.3)⁰ × (0.7)³

+ ³C₁ × (0.3)¹ × (0.7)²

+³C₂ × (0.3)² × (0.7)¹

= 0.343 + 0.441 + 0.189

= 0.973

This implies CDF for N is equal to,

F(N ≤n) = [tex]\sum_{0}^{x}[/tex]³Cₓ × (0.3)ˣ × (0.7)³⁻ˣ

And F(N≤3) is equal to,

⇒F(N≤3)

= P(N=0) + P(N=1) + P(N=2) + P(N=3)

=  ³C₀ × (0.3)⁰ × (0.7)³

+ ³C₁ × (0.3)¹ × (0.7)²

+ ³C₂ × (0.3)² × (0.7)¹

+ ³C₃× (0.3)³ × (0.7)⁰

= 0.343 + 0.441 + 0.189 + 0.027

= 1.000

Therefore, the cumulative distribution function for N is equal to,

F(N≤ 0) = 0.343

F(N≤1) = 0.784

F(N≤2) = 0.973

F(N≤3) = 1.000

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The Initial value problem is defined

as [tex] \frac{dy}{dt} =\alpha y(1 -y), y(0) = y_0[/tex]

a) Equilibrium points or critical values are y = 0 and y = 1. Also, y = 0, is unstable and y = 1, is asymptomatic stable.

b) The solution of above initial value problem is y = 1 , which means at the end the disease will spread through the entire population.

We have a population data which can be divided into two parts. Let consider x be the proportion of susceptible individuals and y the proportion of infectious individuals; then x + y = 1. Now, Initial value problem ( that a differential equation) is, [tex] \frac{dy}{dt} = \alpha y(1 -y), y(0) = y_0[/tex]

a) we have to determine equilibrium points and nature of asymptote for above equation. To determine the equilibrium solution of equation we must put, dy/dt = 0, for all t values. At equilibrium, dy/dt = 0

=> αy(1 - y ) = 0

=> y( 1 - y) = 0

=> either y = 0 or 1 - y = 0

=> y = 0 or y = 1

so, y = 0 is unstable and y = 1 , asymptomatic stable.

b) Now, we have to solve initial value problem, [tex]\frac{dy}{y(1 - y)} = \alpha dt [/tex]

Using partial fraction decomposition,

[tex] \frac{1}{y(1 - y) }= \frac{1}{y} - \frac{1}{1 - y}[/tex]

integrating both sides,

[tex]\int {( \frac{1}{y} - \frac{1}{(1 - y)})dy } = \int{ \alpha dt }[/tex]

[tex]ln (y) - ln (1 - y) = \alpha t + c[/tex]

[tex] ln( \frac{ y}{1- y}) \: = \alpha t + c [/tex]

[tex] \frac{y}{1 - y}= e^{\alpha t} c_1 [/tex]

using initial condition, [tex]y(0) = y_0 [/tex]

[tex] \frac{y_0}{1 - y_0}= 1 ×c_1 [/tex]

[tex]c_1 = \frac{ y_0}{1 - y_0}[/tex]

[tex]so, \frac{y}{1 - y} = \frac{ y_0}{1 - y_0}e^{\alpha t} [/tex]

cross multiplication

[tex]y(1 - y_0) = ((1 - y) y_0 )e^{\alpha t} [/tex]

[tex]y - yy_0 = y_0e^{\alpha t} - yy_0 e^{\alpha t} [/tex]

[tex]y = \frac{ y_0}{y_0 + ( 1 - y_0) e^{-\alpha t} }[/tex]

as [tex]t→ ∞ , e^{- \alpha t } → 0[/tex]

[tex]y = \frac{ y_0}{y_0 }= 1 [/tex]

So, y = 1, means that ultimately the disease spreads through the entire population.

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Kim should use 8/3 (or 2.67) ounces of oil in the recipe.

what is a linear equation?

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a variable raised to the first power (i.e., the exponent of the variable is 1).

Based on the table, we can see that the recipe calls for a total of 24 ounces of dressing, with a ratio of 3 parts oil to 1 part vinegar.

To calculate how much oil Kim should use, we can set up a proportion:

3 parts oil : 1 part vinegar = x ounces of oil : 8 ounces of vinegar

Cross-multiplying, we get:

3x = 8

Dividing both sides by 3, we get:

x = 8/3

To complete the table, we can fill in the remaining values based on the given ratio:

Total Ounces of Dressing Ounces of Oil       Ounces of Vinegar

               12                                    9                         3

               16                                   12                      4

               20                                 15                         5

               24                                 18                         6

Therefore, Kim should use 8/3 (or 2.67) ounces of oil in the recipe.

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The table can be filled up accordingly:

1.  5 years compounded semi-annually:

b = (1 + 0.06/2)

y= 1800 (1 + 0.03)^10

x = 2

2. For the 5 years compounded quarterly:

b =  (1 + 0.06/4)

y =  1800 (1 + 0.015)^20

x =4

3. For the 5 years compounded monthly:

b = (1 + 0.06/12)

y = 1800 (1.005)^60

x = 12

4. For 10 years compounded annually:

b = (1 + 0.06/1)

x =  1800 (1.06)^10

y = 1

5. For 10 years compounded quarterly:

b =  (1 + 0.06/4)

y =1800 (1 + 0.005)^60

x = 4

6. For 10 years compounded monthly:

b = (1 + 0.06/12)

y = 1800 (1 + 0.005)^120

x = 12

To solve the compound interest we use the formula:

P(1+r/n)^(n*t).

For the 5 years compounded semi-annually:

A = 1800 (1 + 0.06/2)^2*5

A = 1800 (1 + 0.03)^10

A = 1800 (1.03)^10

A = 1800 (1.344)

A = 2419

For the 5 years compounded quarterly:

A = 1800 (1 + 0.06/4)^4*5

A = 1800 (1 + 0.015)^20

A = 1800 (1.015)^20

A= 1800 (1.3468)

A= 2424.24

For the 5 years compounded monthly:

A = 1800 (1 + 0.06/12)^12*5

A = 1800 (1 + 0.005)^60

A = 1800 (1.005)^60

A = 1800 (1.34885)

A = 2427.93

For 10 years compounded annually:

A= 1800 (1 + 0.06/1)^10

A = 1800 (1.06)^10

A = 1800 (1.7908)

A = 3223.53

For 10 years compounded semi-annually:

A = 1800 (1 + 0.06/2)^20

A = 1800 (1 + 0.03)^20

A = 1800 (1.03)^20

A =1800 (1.806)

A = 3251

For 10 years compounded quarterly:

A= 1800 (1 + 0.06/4)^4*10

A = 1800 (1 + 0.015)^40

A= 1800 (1.015)^40

A = 1800 (1.814)

A = 3265.23

For 10 years compounded monthly:

A= 1800 (1 + 0.06/12)^120

A = 1800 (1 + 0.005)^120

A = 1800 (1.005)^20

A= 1800 (1.8194)

A = 3274.9

The best pay period is that with the highest returns which is 10 years compounded monthly.

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Answer:

Step-by-step explanation:

I don't know if this is right but i'm just trying to help you. So I did 27 times 9 because the blue part is 9cm and the pink part H is 27cm, I got 243. Then the whole thing is 243cm but you see how it says in blue area=30cm well that means that that small peace if 30cm out of the whole 243cm, so what I did was i subtracted 243cm by 30cm and I got a answer of 213 cm. If that was what you had needed help on. I hope i helped you even a little bit, please let me know if i did help you. And also um i don't know how to friend someone so if you can tell me how to friend someone i was thinking about friending you. And ill also just you some points if you want, say mabey like 10 or 20 points. Thanks and hoped that helped and good luck buddy. :)

The 42 grams of aluminum implies that the chemist has 0.06429ml of aluminum

The density of material shows the denseness of that material in a specific given area. A material’s density is defined as its mass per unit volume. Density is essentially a measurement of how tightly matter is packed together. It is a unique physical property of a particular object. The principle of density was discovered by the Greek scientist Archimedes. It is easy to calculate density if you know the formula and understand the related units The symbol ρ represents density or it can also be represented by the letter D.

How to determine the amount of aluminum?

The mass is given as:  Mass = 42 grams

The density of aluminum is: Density = 2.7 g/cm³

So, we have: Volume = Density/Mass

This gives, Volume = 2.7/42

Evaluate : Volume = 0.06429cm³

Convert to ml

Volume = 0.06429ml

Hence, the chemist has 0.06429ml of aluminum

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The rate of change of f(x) = 2(2) is equal to the rate of change of the function in the graph.

A function is a relation between two sets of values in mathematics. It is a mathematical process that takes an input and produces an output. It is represented as an equation which describes the relationship between the input and the output. A function can also be used to represent a mathematical rule or procedure that takes a set of inputs and produces a set of outputs.

The rate of change, or slope, is the measure of how quickly one variable changes when the other changes. In the graph, the rate of change is a constant, meaning that for every one unit the x-value increases the y-value increases by two. This is the same as the rate of change of f(x) = 2(2), as for every two units the x-value increases the y-value increases by four. Both the graph and the equation of the function have a constant rate of change which is equal to two, so the rate of change of f(x) = 2(2) is equal to the rate of change of the function in the graph.

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