17. A quadratic equation of the form 3x^2+bx+c=0 has roots of 6 plus or minus square root of 2. Determine the value of c.

The * values that make the solution a whole number are given as follows:

* = 4 or * = 9.

The divisibility rule for 5 states that a number is divisible by 5 if its ones digit (i.e., the digit in the units place) is either 0 or 5. In other words, if the number ends in 0 or 5, then it is divisible by 5.

The solution to the equation is obtained as follows:

5x = (516 + 49*)

x = (516 + 49*)/5

Hence it is needed one of these two following cases:

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The two smallest positive solutions for 4 sin(2x) = 2 are x = π/12 and x = 5π/12.

Starting with 4 sin (2x) = 2, we can simplify it by dividing both sides by 4 to get:

sin (2x) = 1/2

To solve for the two smallest positive solutions a and b, we need to find the values of 2x that satisfy sin (2x) = 1/2.

We know that sin (π/6) = 1/2, so one solution is 2x = π/6, which means x = π/12.

The next solution can be found by adding the period of sin (2x), which is π. Therefore, the next solution is 2x = π - π/6 = 5π/6, which means x = 5π/12.

Thus, the two smallest positive solutions for x are:

a = π/12 ≈ 0.26

b = 5π/12 ≈ 1.31

Therefore, the solution is a = 0.26 and b = 1.31.

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Answer: B. (x,y) → (2x - 4,2y-6)

Step-by-step explanation:

A transformation that preserves both distance and angle measure is called an isometry. An isometry preserves distance because the distance between any two points in the pre-image is the same as the distance between their corresponding points in the image. An isometry also preserves angle measure because the angle between any two intersecting lines in the pre-image is the same as the angle between their corresponding lines in the image.

Option (B) represents a transformation that preserves both distance and angle measure. This transformation is a combination of a horizontal and a vertical stretch (or compression) with a scale factor of 2 and a translation of 4 units to the right and 6 units down. Since a stretch (or compression) preserves angle measure, and a translation preserves distance and angle measure, this transformation preserves both distance and angle measure, and therefore, is an isometry.

Option (A) represents a horizontal stretch with a scale factor of 2 and a translation of 4 units to the left and 6 units down. This transformation does not preserve distance, since the horizontal distances are multiplied by a factor of 2, and it does not preserve angle measure, since the angles between intersecting lines are not necessarily preserved.

Option (C) represents a 90-degree rotation followed by a reflection across the x-axis, which preserves angle measure, but does not preserve distance, since the distances between corresponding points are not necessarily the same.

Option (D) represents a 90-degree counterclockwise rotation followed by a reflection across the y-axis, which preserves angle measure, but does not preserve distance, since the distances between corresponding points are not necessarily the same.

Therefore, the correct answer is option (B).

A two-tailed test is used in sample of 10 observations with an alpha level of 0.10. The critical value for this test is ± 1.697. These critical values are used to determine the rejection region of your hypothesis test.

In a two-tailed test with a sample of 10 observations and alpha equal to 0.10, the critical value would be ±1.697. This means that if the test statistic falls outside of this range, it would be considered statistically significant and we would reject the null hypothesis. The use of a two-tailed test means that we are interested in testing for the possibility of a difference in either direction, as opposed to a one-tailed test where we would only be interested in a difference in one specific direction.

Among the significance tests, single-ended and two-tailed tests are other ways of calculating the significance of the measurements determined from the data set under the test. A two-tailed test is appropriate if the predicted value is greater or less than the value on a test, i.e. whether the test taker will score some points higher or lower. This method is used to test the null hypothesis, if the predicted value is in the critical region, it accepts the alternative hypothesis instead of the null hypothesis. A one-tailed test is appropriate if the estimated value differs from the reference value in one direction (left or right) but not in two directions.

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The value of the car now is approximately $8,920.09.

The value of the car can now be determined using the compound interest formula:

A = P * (1 - r)^n

Where

The beginning sum is $16,490 in this instance, the yearly interest rate is 13%, or 0.13 as a decimal, and the number of years is 5.

We thus have:

A = 16490 * (1 - 0.13)^5 = 16490 * 0.541 = $8920.09

Therefore, the value of the car now is approximately $8,920.09.

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Answer: 400 miles

Step-by-step explanation: every hour, his distance decreases by 50 miles. After 4 hours, his distance is 200 miles, so 50 miles times 4 hours = 200 miles+the original 200 miles =400 miles.

The point estimate of the population mean is 18.4. Therefore, the correct answer is (d) is 18.4

When conducting a statistical study, it is important to have a good understanding of the population in question. In such cases, a sample of the population can be taken to infer information about the population.

One of the key parameters that is of interest is the population mean. The population mean represents the average value of a particular characteristic in the entire population. However, since it is usually not possible to collect data from the entire population, we use the sample mean as an estimate of the population mean.

In the given scenario, a simple random sample of a population was taken, and the following data was collected: 13, 17, 18, 21, 23. The point estimate of the population mean can be calculated by taking the mean of the sample.

The sample mean is calculated as follows:

(13 + 17 + 18 + 21 + 23) / 5 = 92 / 5 = 18.4

Therefore, the point estimate of the population mean is 18.4, option (d).

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The sampling distribution of the natural log of the odds ratio (lnOR) will be used for further statistical analyses to determine the relationship between smoking and breast cancer therefore lnOR  =0.86.

Based on the data provided from the case-control study of breast cancer and smoking, we can calculate the odds ratio (OR) to understand the association between smoking and breast cancer. Here's the data:

- Smoker: 25 cases, 15 controls
- Non-smoker: 75 cases, 85 controls

The odds ratio is calculated as (odds of exposure among cases) / (odds of exposure among controls), which is:

OR = (25/75) / (15/85) = 2.36

However, statistical inferences for odds ratios are based on the natural log of the odds ratio (lnOR) because the distribution for an odds ratio does not follow a normal distribution. To get the lnOR, we take the natural logarithm of the OR:

lnOR = ln(2.36) ≈ 0.86

The sampling distribution of the natural log of the odds ratio (lnOR) will be used for further statistical analyses to determine the relationship between smoking and breast cancer.

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The number of unique access keys that can be designed with the given numbers is 280 possible unique access keys.

This is a permutations problem which means it can be solved by the equation :

Number of permutations = n! / (n1! * n2! * ... * nk!)

Given the numbers, 1, 1, 1, 2, 3, 3, 3, 3, we can apply the formula to be :

Number of permutations = 8 ! / (3 ! x 1 ! x 4 !)

= 40, 320 / (6 x 1 x 24)

= 40, 320 / 144

= 280 possible access keys

In conclusion, a total of 280 possible unique access keys can be made.

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Answer:

2

Step-by-step explanation:

The region bounded above by the surface z=4 cos xsin y and below by the rectangle R:

We can use a double integral to find the volume of the region:

V = ∫∫R 4cos(x)sin(y) dA

where R is the rectangle defined by:

0 ≤ x ≤ π/2

0 ≤ y ≤ π/2

Then we can evaluate the integral as follows:

V = ∫∫R 4cos(x)sin(y) dA

= ∫0^(π/2) ∫0^(π/2) 4cos(x)sin(y) dxdy

= ∫0^(π/2) [4sin(x)](0 to π/2) dy

= ∫0^(π/2) 4sin(π/2) dy

= 4(sin(π/2))(π/2 - 0)

= 4(1)(π/2)

= 2π

Therefore, the volume of the region bounded above by the surface z=4 cos xsin y and below by the rectangle R is 2π.

x(0) = 0.25 in (initial displacement from equilibrium). x'(0) = 0 ft/s (initial velocity)

We can use Newton's second law to formulate the initial value problem that governs the motion of the mass. The equation is given by:

m*a = F_net

where m is the mass, a is the acceleration, and F_net is the net force acting on the mass.

The net force can be found as the sum of the spring force, the viscous resistance force, and the force due to gravity. Therefore, we have:

F_net = F_spring + F_viscous + F_gravity

where F_spring is the force exerted by the spring, F_viscous is the force due to the viscous resistance, and F_gravity is the force due to gravity.

The force exerted by the spring is given by Hooke's law:

F_spring = -k*x

where k is the spring constant and x is the displacement from the equilibrium position. Since the spring stretches 3 in under a weight of 4 lb, we can find k as:

k = F/x = 4/3 = 4/0.25 = 16 lb/in

Therefore, the force exerted by the spring is:

F_spring = -16*x

The force due to viscous resistance is proportional to the velocity of the mass and is given by:

F_viscous = -c*v

where c is the viscous damping coefficient and v is the velocity of the mass. Since the viscous resistance force is 6 lb when the velocity is 3 ft/s, we can find c as:

c = F_viscous/v = 6/3 = 2 lb·s/ft

Therefore, the force due to viscous resistance is:

F_viscous = -2*v

The force due to gravity is given by:

F_gravity = -m*g

where g is the acceleration due to gravity (32.2 ft/s^2).

Substituting these equations into the equation for net force, we get:

ma = -16x - 2v - mg

Since the displacement x and the velocity v are both functions of time t, we can rewrite this equation as a second-order ordinary differential equation in terms of x:

mx'' + 2cx' + kx = m*g

where x' and x'' denote the first and second derivatives of x with respect to t, respectively.

This is the initial value problem that governs the motion of the mass, subject to the initial conditions:

x(0) = 0.25 in (initial displacement from equilibrium)

x'(0) = 0 ft/s (initial velocity)

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Answer:

[tex]y \geqslant \frac{2}{3} x - 2[/tex]

[tex]y + 2 \geqslant \frac{2}{3} x[/tex]

[tex] \frac{2}{3}x - y \leqslant 2[/tex]

[tex]2x - 3y \leqslant 6[/tex]

To find the volume of the region in the first octant bounded by the coordinate planes, the plane y + z = 2, and the cylinder x = 4 − y^2, we need to first graph the given surfaces in 3D space.

The coordinate planes are the x-y, x-z, and y-z planes. In the first octant, these planes bound the region from below and on the sides.

The plane y + z = 2 is a plane that passes through the origin and intersects the y-z plane at y = 2 and z = 0, and the z-x plane at x = 2 and z = 0,The cylinder x = 4 − y^2 is a cylinder with radius 2 and centered at the origin, since it is a function of y^2 and only extends to y = 2 in the first octant.

To find the volume of the region bounded by these surfaces, we need to integrate over the region. We can do this by dividing the region into small rectangular prisms, and integrating over each prism.The limits of integration for x are 0 to 4 − y^2, the limits for y are 0 to 2, and the limits for z are 0 to 2 − y.
Therefore, the volume of the region is given by the triple integral: ∫∫∫ (dV) = ∫0^2 ∫0^(4-y^2) ∫0^(2-y) dz dxdy.

Evaluating the integral, we get: ∫∫∫ (dV) = ∫0^2 ∫0^(4-y^2) (2-y) dx dy
∫∫∫ (dV) = ∫0^2 (2-y)(4-y^2) dy
∫∫∫ (dV) = ∫0^2 8-4y^2-2y+ y^3 dy
∫∫∫ (dV) = [8y - 4y^3/3 - y^2 + y^4/4]0^2
∫∫∫ (dV) = 32/3, Therefore, the volume of the region in the first octant bounded by the coordinate planes, the plane y + z = 2, and the cylinder x = 4 − y^2 is 32/3 cubic units.

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The line through the point (2,0,0) and with direction vector (-1, a, 1) intersects the paraboloid z = x² + y² in only a single point for a = 0 or a = √(2).

To learn more about here:

To find the intersection points, we can substitute the equation of the line into the equation of the paraboloid:

z = x² + y²
z = (2-t)² + a*t²
x = 2-t
y = a*t
where t is the parameter for the line.

Substituting x and y into the equation of the paraboloid gives:
z = (2-t)² + a^2*t²

To find the values of a where the line intersects the paraboloid in only a single point, we need to find the values of a where this equation has exactly one solution. This occurs when the discriminant of the quadratic equation in t is zero:

a²*(2-a²) = 0
a = 0 or a = √(2)

Therefore, the line through the point (2,0,0) and with direction vector (-1, a, 1) intersects the paraboloid z = x² + y² in only a single point for a = 0 or a = √(2).

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The indefinite integral is ∫ x / √(x^2 - 6x + 39) dx = √(x^2 - 6x + 39)/2 + C.

To complete the square in the denominator of the integrand, we need to add and subtract a constant:

x^2 - 6x + 39 = (x^2 - 6x + 9) + 30 = (x - 3)^2 + 30

So we can rewrite the integrand as:

x / √[(x - 3)^2 + 30]

Next, we can use the substitution u = (x - 3)^2 + 30 to simplify the integral. Then, du/dx = 2(x - 3), which means dx = du/(2(x - 3)). Making this substitution gives:

∫ x / √[(x - 3)^2 + 30] dx = 1/2 ∫ du/√u

Now, we can use the substitution v = √u, which means dv/dx = 1/(2√u) du/dx = 1/(2√u)(2(x - 3)) = (x - 3)/√u, and dx = 2v dv/(x - 3). Making this substitution gives:

1/2 ∫ du/√u = 1/2 ∫ dv = 1/2 v + C

Substituting back for v and u, we get:

1/2 ∫ x / √[(x - 3)^2 + 30] dx = 1/2 ∫ dv = 1/2 √[(x - 3)^2 + 30] + C

Therefore, the indefinite integral of x / √(x^2 - 6x + 39) dx is:

∫ x / √(x^2 - 6x + 39) dx = √(x^2 - 6x + 39)/2 + C

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Answer:

7

Step-by-step explanation:

The magnitude of an earthquake that has the following intensity.

(a) 1,000 times that of I0 , R is 3.

(b) 100,000 times that of I0, R is 5.

What is ritcher scale?

The logarithm of the wave amplitude measured by seismographs is used to calculate the earthquake's Richter magnitude; adjustments are made to account for variations in the distances between different seismographs and the earthquake's epi-centre.

a) I = 1000. I₀

R = log(I / I₀)

= log(1000 I₀ / I₀)

= log(1000)

= log 10³

(i.e., log xⁿ = n log x)

= 3 log 10

R = 3

b) I = 100000 I₀

R = log(I / I₀)

=  log(100000 I₀ / I₀)

= log(100000)

= log 10⁵

(i.e., log xⁿ = n log x)

= 5 log 10

R = 5

The magnitude of an earthquake that has the following intensity.

(a) 1,000 times that of I0 , R is 3.

(b) 100,000 times that of I0, R is 5.

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Answer:

Yes, these two figures are similar because the ratios of corresponding sides are equal.

UR = ST = 2, RS = TU = 3

YV = WX = 4, VW = XY = 6

UR/RS = YV/VW

ST/TU = WX/XY

The sum of the series (x-y)^2 + x^2 + y^2 is 2x^2 - 2xy + 2y^2.

From the question, we have the following parameters that can be used in our computation:

(x-y)^2+x^2+y^2

The expression (x-y)^2 can be expanded as:

(x-y)^2 = x^2 - 2xy + y^2

Adding x^2 and y^2, we get:

(x-y)^2 + x^2 + y^2 = x^2 - 2xy + y^2 + x^2 + y^2

Combining like terms, we can simplify this expression to:

(x-y)^2 + x^2 + y^2 = 2x^2 - 2xy + 2y^2

Therefore, the sum of the series (x-y)^2 + x^2 + y^2 is 2x^2 - 2xy + 2y^2.

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Answer:

The system of equations is:

-5x + 9y = -9

5x - 8y = 4

The determinant of the coefficients is:

D = |-5 9|

| 5 -8|

D = (-5)(-8) - (9)(5) = 40 - 45 = -5

The determinant of x is found by replacing the x-coefficients with the constants:

Dx = |-9 9|

| 4 -8|

Dx = (-9)(-8) - (9)(4) = 72 - 36 = 36

The determinant of y is found by replacing the y-coefficients with the constants:

Dy = |-5 -9|

| 5 4|

Dy = (-5)(4) - (-9)(5) = -20 + 45 = 25

Using Cramer's Rule:

x = Dx/D = 36/-5 = -7.2

y = Dy/D = 25/-5 = -5

Therefore, the solution to the system is x = -7.2 and y = -5.

Hope this helps!

Answer:

   A1 = 160 degrees

   A2 = 20 degrees

   B1 = 70 degrees

   C1 = 110 degrees

   C2 = 70 degrees

   D1 = 20 degrees

   AD = 10 cm

   AE = 5√5 cm

   AB = 8 cm

Step-by-step explanation:

To solve this problem, we can start by using the fact that the diagonals of a rectangle are equal in length and bisect each other. Therefore, we know that:

- BD = AC = 10cm

- AE = EC = BD/2 = 5cm

- AB = CD = sqrt(AC^2 + BC^2) = sqrt(10^2 + 8^2) = sqrt(164) ≈ 12.81cm

- AD = BC = 8cm

To find the angles A1, A2, B1, C1, C2, and D1, we can use the following relationships:

- A1 = 180 - D2 = 180 - 20 = 160 degrees

- A2 = 180 - A1 = 180 - 160 = 20 degrees

- B1 = C2 = D2 = 20 degrees

- C1 = 180 - B1 = 180 - 20 = 160 degrees

- D1 = 180 - C2 = 180 - 20 = 160 degrees

Therefore:

- A1 = 160 degrees

- A2 = 20 degrees

- B1 = C2 = D2 = 20 degrees

- C1 = D1 = 160 degrees

Note that angles A1, C1, and D1 are all equal, as are angles A2, B1, and C2, because opposite angles in a rectangle are equal.

Finally, to find AD, we can use the Pythagorean theorem:

- AD = BC = 8cm

And to find AE, we can use the fact that diagonals bisect each other:

- AE = EC = BD/2 = 5cm

Therefore:

- AD = 8cm

- AE = 5cm

- AB ≈ 12.81cm

- A1 = 160 degrees

- A2 = 20 degrees

- B1 = C2 = D2 = 20 degrees

- C1 = D1 = 160 degrees

The calculated value of the surface area of the cube is 294 mm^2

From the question, we have the following parameters that can be used in our computation:

The cube

The side length of the cube is

Length = 7 mm

The surface area of the cube is calculated as

Surface area = 6 * Length^2

Substitute the known values in the above equation, so, we have the following representation

Surface area = 6 * 7^2

Evaluate

Surface area = 294 mm^2

Hence the surface area is 294 mm^2

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a. After 11 days, approximately 91.91% of the animals will be affected.

b. It will take around 20.83 days for 90% of the animals to become infected. The answer, rounded to two decimal places, is 20.83 days.

A logarithm is defined as the number of powers to which a number must be increased in order to obtain some other numbers. It is the simplest way to express enormous numbers. A logarithm has several key features that demonstrate that logarithm multiplication and division can also be represented in the form of logarithm addition and subtraction.

A) To find the percentage of animals infected after 11 days, we simply substitute t = 11 into the given equation for V(t):

V(11) = 100/ [tex](1 + 99e^{(-0.186*11)})[/tex]

Using a calculator, we get:

V(11) ≈ 91.91%

Therefore, about 91.91% of the animals will be infected after 11 days.

B) To find the time it takes until exactly 90% of the animals are infected, we need to solve the equation V(t) = 90 for t.

Substituting V(t) into the equation, we get:

90 = 100/ [tex](1 + 99e^{(-0.186t)})[/tex]

Multiplying both sides by [tex](1 + 99e^{(-0.186t)})[/tex], we get:

[tex]90 + 90e^{(-0.186t)} = 100[/tex]

Simplifying, we get:

[tex]e^{(-0.186t)} = 1/9[/tex]

Taking the natural logarithm of both sides, we get:

-0.186t = ln(1/9)

Solving for t, we get:

t ≈ 20.83 days

Therefore, about 20.83 days will elapse until exactly 90% of the animals are infected. Rounded to 2 decimal places, the answer is 20.83 days.

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Answer:

Add a dot (Point A) 5 units to the left of point C, a dot (Point B) 7 units below point C, and a final dot (Point D) that is 5 units across from Point B

Step-by-step explanation:

Using the definition of a rectangle (4 right angles)
- Length is usually viewed as left to right

- Width is usually viewed as top to bottom

Answer:

y = (-3/2)x + 4.

Step-by-step explanation:

y - 4 = (2/3)(x - 9)

y = (2/3)x - (2/3)(9) + 4

y = (2/3)x - 2

The slope of this line is 2/3.

A line that is perpendicular to this line will have a slope that is the negative reciprocal of 2/3. To find the negative reciprocal, we flip the fraction and change the sign:

slope of perpendicular line = -3/2

Now we can use the point-slope form of the equation of a line, which is:

y - y1 = m(x - x1)

where m is the slope of the line, and (x1, y1) is a point on the line.

Substituting the given point (6,-5) and the slope -3/2, we get:

y - (-5) = (-3/2)(x - 6)

y + 5 = (-3/2)x + 9

y = (-3/2)x + 4

Therefore, the equation of the line that is perpendicular to y - 4 = (2/3)(x - 9) and passes through the point (6,-5) is y = (-3/2)x + 4.

The probability that the sequence of rolls is 2, 6, 1, 4, 5 is 1/7776, or as a decimal rounded to four decimal places, it is approximately 0.0001.

To find the probability that the sequence of rolls is 2, 6, 1, 4, 5 when a fair 6-sided die is rolled five times, we can use the following steps:
1. Since there are 6 sides on a fair die, the probability of rolling any specific number is 1/6.
2. The probability of rolling a specific sequence of numbers is the product of the probabilities of rolling each number in that sequence.
3. In this case, the sequence is 2, 6, 1, 4, 5. The probability of rolling each number in this sequence is 1/6.
So, to calculate the probability of the sequence 2, 6, 1, 4, 5, we multiply the probabilities of each individual roll:

(1/6) * (1/6) * (1/6) * (1/6) * (1/6) = 1/7776

Rounded to four decimal places, the probability is approximately 0.0001. Therefore, the probability that the sequence of rolls is 2, 6, 1, 4, 5 is 0.0001, or 1/7776 as a fraction.

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The type of relationship between the sets M and N is intersection.

A set is a collection of well ordered items.

Given that there is in M prime number 2,3,5,7 and in N odd number 1,3,5,7,9 what is type of relation between set m and n?

So, the type of relationship between the sets M and N is intersection.

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Answer:

b) Dimensions of garden:
length = 10 feet
width = 3 feet

Step-by-step explanation:

I am only doing the part b. Part a is just a sketch of a rectangle with the dimensions computed

b) Find the dimensions of Amelia's garden

Let us use the variable L to represent the length and the variable W to represent the width of the garden

We are given that the length = 4 plus twice the width
In algebraic equation terms this would be
L = 4 + 2W

We are given that the area is 30 ft²
Area of a rectangle = LW
So
LW = 30

Substitute for L in terms of W:
(4 + 2W)W = 30
4W + 2W² = 30

Move 30 to the left and rearrange terms on the left
2W² + 4W - 30 = 0'

Divide by 2:
W² + 2W - 15 = 0

This is a quadratic equation that can be solved using factoring
Find the factors of -15 and see which of them when added will give a value of 2 which is the coefficient of W in the quadratic equation

Factors of -15 are
-15 1  => sum = 15 + 1 = - 14 X
-5 3  => sum = -5 + 3 = -2   X
5 -3  =>  sum = 5 + (-3) = 2  √

Given the correct factors we can rewrite the equation as
(W + 5)(W - 3) = 0

So either W + 5 = 0 or W - 3 =0

(If you multiply W + 5 by W - 3 you will get W² + 2W - 15)

Therefore the solutions to the quadratic equation are
W + 5 = 0   ==> W = -5  ; not possible, dimensions have to be positive
W - 3 = 0  ==> W = 3 ; this is the solution

So we have W, the width of the garden as 3 feet

Substitute W = 3 in the equation for length:
L = 4 + 2W

L = 4 + 2 x 3

L = 10

So length = 10 feet

The time period is given by

[tex]A(t) = -40e^{-5t-8.008}+80000[/tex]

a) Let's use the following variables:

t: time in minutes

A(t): amount of salt in the tank at time t in grams

V(t): volume of water in the tank at time t in gallons

Initially, the tank contains 300 grams of salt in 200 gallons of water, so the concentration of salt is:

[tex]C(0) = \frac{300g}{200gal} = 1.5g/gal[/tex]

As the brine solution is pumped into the tank at a rate of 5 gallons per minute and at a concentration of 0.4 kilograms of salt per gallon of water, the concentration of salt in the incoming solution is:

[tex]c_{in} = 0.4 kg/gal \times \frac{1000g}{1kg} \times \frac{1gal}{1L} = 400g/gal[/tex]

Let's assume that the tank is well-stirred, so the concentration of salt in the tank is uniform at any given time. Then, we can use the following differential equation to model the amount of salt in the tank:

[tex]\frac{dA}{dt} =c_(in) \times \frac{dV}{dt} - c(t) \times \frac{dV}{dt}[/tex]

where [tex]\frac{dV}{dt}\\[/tex] is the rate of change of the volume of water in the tank. We know that water is pumped into and out of the tank at the same rate of 5 gallons per minute, so [tex]\frac{dV}{dt} = 0[/tex], and the differential equation simplifies to:

[tex]\frac{dA}{dt} = c_(in) \times 5 -c(t) \times 5 = 2000 - 5c(t)[/tex]

This is a separable differential equation that we can solve by separating the variables and integrating:

[tex]\frac{dA}{2000-5c} = dt\\\\ \int \frac{dA}{2000-5c} = \int dt\\\\-\frac{1}{5} ln|2000 - 5c| = t+C\\\\c(t) = -\frac{1}{5} e^(-5t-5C) +400[/tex]

We can find the constant C by using the initial condition c(0) = 1.5, we get

[tex]C = ln3001.5 =8.008\\[/tex]

Therefore, the amount of salt in the tank at time t is,

[tex]A(t) = V(t) \times c(t)\\A(t) = 200 \times (-\frac{1}{5} e^{-5t-8.008}+400 )\\A(t) = -40e^{-5t-8.008}+80000[/tex]

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To find the probability p(1.6) for a random variable x with a gamma distribution where α=2 and β=1, you'll need to use the gamma probability density function. The gamma is given by:

f(x) = (β^α * x^(α-1) * e^(-βx)) / Γ(α)

where Γ(α) is the gamma function of α.

Now, plug in the values for α, β, and x:

f(1.6) = (1^2 * 1.6^(2-1) * e^(-1*1.6)) / Γ(2)

To calculate Γ(2), note that Γ(n) = (n-1)! for positive integers. In this case, Γ(2) = (2-1)! = 1! = 1.

f(1.6) = (1 * 1.6^1 * e^(-1.6)) / 1 = 1.6 * e^(-1.6)

Therefore, the probability density function value at x=1.6 for a random variable x with a gamma distribution where α=2 and β=1 is:

f(1.6) = 1.6 * e^(-1.6) ≈ 0.33013

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