there are 15 students in a class and 6 of them will be chosen to go on a field trip. how many ways can these students be chosen?

Answer:

[tex] \frac{4}{3} \pi {r}^{3} = 2881[/tex]

[tex]r = \sqrt[3]{ \frac{2881}{ \frac{4}{3} \pi} } = 8.83[/tex]

The radius of the beach ball is about 8.83 cm.

Answer:

[tex] \sqrt{72 {m}^{5} {n}^{2} } [/tex]

[tex] \sqrt{2 \times 36 \times {m}^{4} \times m \times {n}^{2} } [/tex]

[tex]6 {m}^{2} n \sqrt{2m} [/tex]

C is the correct answer.

The slope (m) of the line y = x + 3 is 1, while the y-intercept of the line is 3.

If an equation of a line is expressed in slope-intercept form as y = mx + b, we can easily determine its slope and the y-intercept which are:

m is the slope

b is the y-intercept.

Given the equation y = x + 3, therefore:

the coefficient of x is the slope (m), which is 1.

the y-intercept (b) of the line is 3.

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The Maclaurin series through degree 4 of f(x)=(1+x)^(-4/3) is approximately 1 - (4/3)x + (14/9)x^2 - (56/27)x^3 + (208/81)x^4.

To find the Maclaurin series of f(x)=(1+x)^(-4/3), we start by using the formula for the Maclaurin series of a function f(x) centered at x=0:

f(x) = ∑[n=0 to infinity] f^(n)(0) * x^n / n!

where f^(n)(0) represents the nth derivative of f(x) evaluated at x=0.

We begin by finding the first few derivatives of f(x):

f(x) = (1+x)^(-4/3)

f'(x) = (-4/3)(1+x)^(-7/3)

f''(x) = (28/9)(1+x)^(-10/3)

f'''(x) = (-224/27)(1+x)^(-13/3)

f''''(x) = (2912/81)(1+x)^(-16/3)

Evaluating these derivatives at x=0, we get:

f(0) = 1

f'(0) = -4/3

f''(0) = 14/9

f'''(0) = -56/27

f''''(0) = 208/81

Substituting these values into the Maclaurin series formula, we get:

f(x) ≈ 1 - (4/3)x + (14/9)x^2 - (56/27)x^3 + (208/81)x^4

This gives us the Maclaurin series through degree 4 of f(x)=(1+x)^(-4/3).

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The derivative is undefined when the denominator is 0, which occurs when x = ±7. So the critical values are x = -7, 0, and 7.

(A) To find the critical values, we need to find where the derivative of f(x) equals zero or is undefined. Taking the derivative of f(x), we get:

f'(x) = 7(x² - 49) - 7x(2x) / (x² - 49)²
f'(x) = 0 when x = 0 (undefined at x = ±7)

So the critical values of f(x) are x = 0.

(B) f(x) is increasing on the intervals (-∞, -7) and (7, ∞).

(C) f(x) is decreasing on the intervals (-7, 0) and (0, 7).

(D) There are no local maxima.

(E) There is one local minimum at x = -7.

(F) f(x) is concave up on the intervals (-∞, -7/√2) and (7/√2, ∞).

(G) f(x) is concave down on the intervals (-7/√2, 7/√2).

(H) The inflection points of f are x = ±7.

(I) There are two horizontal asymptotes: y = 0 and y = 7.

(J) There are two vertical asymptotes: x = -7 and x = 7.

(K) Graph complete.

Critical values of f(x) are the values of x where the derivative f'(x) is either 0 or undefined. f'(x) = (-49x) / (x^2 - 49)^2.

Setting the numerator equal to 0, we get x = 0. The derivative is undefined when the denominator is 0, which occurs when x = ±7. So the critical values are x = -7, 0, and 7.

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The derivative is a. y' = [7 * cos(7x) * tan(3x) - 3 * sin(7x) * sec²(3x)] / [tan²(3x)] and the derivative of second funtction is b. (ln(π) * [tex]\pi^(3x^2-7)[/tex]) * (6x) + (9 - x²) / (x²+9)² - 32 / sqrt(1 - (32x+1)²).

a. y = sin(7x)/tan(3x)

To differentiate this function, we can use the quotient rule, which states that if we have a function in the form f(x) = g(x)/h(x), where g(x) and h(x) are differentiable functions, the derivative of f(x) is given by:

f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))²

In this case, g(x) = sin(7x) and h(x) = tan(3x). Let's differentiate both g(x) and h(x) first:

g'(x) = d/dx [sin(7x)] = 7 * cos(7x)

h'(x) = d/dx [tan(3x)] = 3 * sec²(3x)

Now we can substitute these derivatives into the quotient rule formula:

y' = [(7 * cos(7x) * tan(3x)) - (sin(7x) * 3 * sec²(3x))] / (tan(3x))²

Simplifying further, we get:

y' = [7 * cos(7x) * tan(3x) - 3 * sin(7x) * sec²3x)] / [tan²(3x)]

b. y = [tex]\pi^{(3x^2-7)[/tex] + x/(x²+9) + cos⁻¹(32x+1)

To differentiate this function, we can use the sum and chain rules. Let's differentiate each term separately:

For the first term, y₁ = [tex]\pi^{(3x^2-7)[/tex]:

y₁' = d/dx [[tex]\pi^{(3x^2-7)[/tex]]

Using the chain rule, the derivative is:

y₁' = (ln(π) * [tex]\pi^{(3x^2-7)[/tex]) * (6x)

For the second term, y₂ = x/(x²+9):

y₂' = d/dx [x/(x²+9)]

Using the quotient rule, the derivative is:

y₂' = [(1 * (x²+9)) - (x * 2x)] / (x²+9)²

Simplifying further, we get:

y₂' = (9 - x²) / (x²+9)²

For the third term, y₃ = cos⁻¹(32x+1):

y₃' = d/dx [cos⁻¹(32x+1)]

Using the chain rule, the derivative is:

y₃' = -32 / sqrt(1 - (32x+1)²)

Now, we can add all the derivatives together to find the derivative of the function:

y' = y₁' + y₂' + y₃'

y' = (ln(π) * [tex]\pi^{(3x^2-7)[/tex])) * (6x) + (9 - x²) / (x²+9)² - 32 / sqrt(1 - (32x+1)²)

The complete question is:
a. Differentiate the function: [tex]y=\frac{sin(7x)}{tan(3x)}[/tex].

b. Differentiate the function: [tex]\pi^{(3x^2-7)[/tex] + x/x²+9+cos⁻¹(32x+1)

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The volume of the reservoir with an isosceles triangle surface and trapezoid cross-sections is 375,000 cubic meters.

To find the volume of the reservoir with an isosceles triangle surface and trapezoid cross-sections, we'll follow these steps:

1. Identify the dimensions of the trapezoid: Given that the top side (base) of the trapezoid is 100 m, its bottom side (smaller base) will be half of that, so 50 m. Similarly, the height of the trapezoid is half the length of the top side, so 50 m.

2. Calculate the area of the trapezoid cross-section: To find the area of a trapezoid, we use the formula A = (1/2)(b1 + b2)h, where A is the area, b1 and b2 are the lengths of the two bases, and h is the height. In our case, A = (1/2)(100 + 50)(50) = (1/2)(150)(50) = 3750 square meters.

3. Determine the length of the reservoir: The length of the reservoir is given as 100 m.

4. Calculate the volume of the reservoir: Finally, to find the volume, we multiply the area of the trapezoid cross-section by the length of the reservoir. V = A * L = 3750 * 100 = 375,000 cubic meters.

So, the volume of the reservoir with an isosceles triangle surface and trapezoid cross-sections is 375,000 cubic meters.

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This equation holds true, which confirms y2 = x^4 as a solution. Therefore, both y1 = x^(-1) and y2 = x^4 are valid solutions to the homogeneous Euler-Cauchy equation provided.

You are given that two solutions of the homogeneous Euler-Cauchy equation are y1 = x^(-1) and y2 = x^4. The general form of the Euler-Cauchy equation is: x^2 * y''(x) + p * x * y'(x) + q * y(x) = 0

To confirm the given solutions are correct, we need to substitute y1 and y2 into the equation and check if the equation holds true (i.e., equals zero). For y1 = x^(-1), we first find its derivatives: y1'(x) = -x^(-2) y1''(x) = 2x^(-3)

Now, substitute y1 and its derivatives into the Euler-Cauchy equation: x^2 * (2x^(-3)) - 2 * x * (-x^(-2)) - 4 * (x^(-1)) = 0 Simplifying the equation: 2 - 2 + 4 = 0

This equation holds true, which confirms y1 = x^(-1) as a solution. For y2 = x^4, we find its derivatives: y2'(x) = 4x^3 y2''(x) = 12x^2 Now, substitute y2 and its derivatives into the Euler-Cauchy equation: x^2 * (12x^2) - 2 * x * (4x^3) - 4 * (x^4) = 0

Simplifying the equation: 12x^4 - 8x^4 - 4x^4 = 0

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There are initially 750 bacteria. There are approximately 2.11 x 10^49 bacteria after 15 minutes. It takes approximately 0.111 minutes for the number of bacteria to double.

a. The initial number of bacteria (when t=0) can be found by plugging t=0 into the equation A(t) = 750e^(6.25t). So, A(0) = 750e^(6.25*0) = 750e^0 = 750*1 = 750. Thus, there are initially 750 bacteria.

b. To find the number of bacteria after 15 minutes, plug t=15 into the equation: A(15) = 750e^(6.25*15). A(15) ≈ 2.11 x 10^49. So, there are approximately 2.11 x 10^49 bacteria after 15 minutes.

c. To find the time it takes for the number of bacteria to double, set A(t) equal to twice the initial amount, 2 * 750 = 1500: 1500 = 750e^(6.25t). Solve for t by dividing both sides by 750, then taking the natural logarithm: ln(2) = 6.25t. Finally, divide by 6.25: t ≈ 0.111. Thus, it takes approximately 0.111 minutes for the number of bacteria to double.

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To find the mass of the lamina, we need to integrate the density function over the region d. the center of mass of the lamina is at the point (4/9 l, 8/13).

The density function is given as:

ρ(x,y) = 13y

Integrating this over the region d, we get:

m = ∫∫d ρ(x,y) dA

where dA is the differential area element in the region d.

To perform this integration, we need to split the region d into small rectangles and integrate over each rectangle. Since the region is defined by the inequality y ≤ sin x, we can split it into rectangles with base dx and height sin x - 0 = sin x. Therefore, we have:

m = ∫0l ∫0sinx ρ(x,y) dy dx
 = ∫0l ∫0sinx 13y dy dx
 = 13 ∫0l [y^2/2]0sinx dx
 = 13 ∫0l (sin^2x)/2 dx
 = 13/4 [x - (1/2)sin(2x)]0l
 = 13/4 l

Therefore, the mass of the lamina is (13/4)l.

To find the center of mass, we need to find the moments of the lamina about the x- and y-axes, and then divide them by the total mass.

The moment of the lamina about the x-axis is given by:

Mx = ∫∫d y ρ(x,y) dA

Integrating this over the region d, we get:

Mx = ∫0l ∫0sinx yρ(x,y) dy dx
  = ∫0l ∫0sinx 13y^2 dy dx
  = 13/3 ∫0l [y^3/3]0sinx dx
  = 13/3 ∫0l (sin^3x)/3 dx
  = 13/9 [3x - 4sin(x) + sin(3x)]0l
  = 13/9 l

Therefore, the x-coordinate of the center of mass is given by:

x = Mx/m = (13/9)l / (13/4)l = 4/9 l

Similarly, the moment of the lamina about the y-axis is given by:

My = ∫∫d x ρ(x,y) dA

Integrating this over the region d, we get:

My = ∫0l ∫0sinx xρ(x,y) dy dx
  = ∫0l ∫0sinx 13xy dy dx
  = 13/2 ∫0l [y^2x/2]0sinx dx
  = 13/2 ∫0l (sin^3x)/3 dx
  = 13/6 [cos(x) - cos^3(x)]0l
  = 13/6

Therefore, the y-coordinate of the center of mass is given by:

y = My/m = (13/6) / (13/4) = 8/13

Hence, the center of mass of the lamina is at the point (4/9 l, 8/13).

To find the mass and center of mass of the lamina that occupies the region D with the given density function (x, y) = 13y, we need to compute the mass (M) and the coordinates of the center of mass (x bar, y bar).

First, let's find the mass (M):
M = ∬D (x, y) dA = ∫(0 to l) ∫(0 to sin(x)) 13y dy dx

To find the center of mass, we need to compute x bar and y bar:

x bar = (1/M) * ∬D x * (x, y) dA = (1/M) * ∫(0 to l) ∫(0 to sin(x)) x * 13y dy dx

y bar = (1/M) * ∬D y * (x, y) dA = (1/M) * ∫(0 to l) ∫(0 to sin(x)) y * 13y dy dx

Compute the integrals above to obtain the mass M and the coordinates of the center of mass (x bar, y bar).

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The calculated value of the mean of a and b is 10

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

20 - b = a

a = 16

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

20 - b = 16

Evaluate the like terms

b = 4

The mean of a and b is calculated as

Mean = (a + b)/2

So, we have

Mean = (16 + 4)/2

Evaluate

Mean = 10

Hence, the mean value of a and b is 10

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The number of outcomes possible when a spinner has three equally sized sections labeled from 1 to 3 and the spinner is spun three times is 27

The total number of outcomes refers to the possible events that can occur if an event takes place. These are helpful in calculating probability. For example, when a coin is tossed, the outcomes possible are heads and tails.

In the given question, the possible outcomes when a spinner has three equally sized sections labeled from 1 to 3 and the spinner is spun three times are calculated by the number of outcomes raised to the power the number of times the event occurs.

Possible outcome possible in  event = 3

Number of times the event occurs =  3

Thus the number of outcomes = [tex]3^3[/tex] = 27

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To find the first five terms of the sequence of partial sums for the given expression (-5)n+1/n!, we'll calculate each term and add them cumulatively.

1. S1: When n=1, term T1 = (-5)(1+1)/1! = -5/1 = -5
  So, S1 = T1 = -5

2. S2: When n=2, term T2 = (-5)(2+1)/2! = 15/2 = 7.5
  So, S2 = S1 + T2 = -5 + 7.5 = 2.5

3. S3: When n=3, term T3 = (-5)(3+1)/3! = -20/6 = -3.3333
  So, S3 = S2 + T3 = 2.5 - 3.3333 = -0.8333

4. S4: When n=4, term T4 = (-5)(4+1)/4! = 25/24 = 1.0417
  So, S4 = S3 + T4 = -0.8333 + 1.0417 = 0.2084

5. S5: When n=5, term T5 = (-5)(5+1)/5! = -30/120 = -0.25
  So, S5 = S4 + T5 = 0.2084 - 0.25 = -0.0416

The first five terms of the sequence of partial sums are: S1 = -5, S2 = 2.5, S3 = -0.8333, S4 = 0.2084, and S5 = -0.0416.

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Q1. The value of lim f(x) as x approaches 17 is 3.

Q2. The value of lim f(x) as x approaches infinity does not exist.

Q1. To find the value of lim f(x) as x approaches 17, we substitute 17 for x in the expression f(x) = x(2-x)/(sqr(11x)+1). This gives us:

lim f(x) = lim [x(2-x)/(sqr(11x)+1)] as x approaches 17

= 17(2-17)/(sqr(11*17)+1)

= -15/2(187)+1

= 3

Q2. To find the value of lim f(x) as x approaches infinity, we can use L'Hopital's rule. Taking the derivative of the numerator and denominator with respect to x, we get:

lim f(x) = lim [(2-x)/(2sqr(11x)+x)] as x approaches infinity

= lim [-(1)/(22sqr(11x)+1)] as x approaches infinity (by applying L'Hopital's rule again)

As x approaches infinity, the denominator approaches infinity, so the limit of the expression is 0. Therefore, the limit of f(x) as x approaches infinity does not exist.

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b. False

The t-test for dependent means is not the appropriate statistical analysis in this case because it is used to compare the means of two related groups. Here, parents and children are two independent groups. Instead, a t-test for independent means would be more suitable for comparing the values of parents and children.

The t-test for dependent means, also known as paired-samples t-test, is used to compare the means of two related groups. The relatedness of the two groups implies that the observations in one group are matched or paired with the observations in the other group, and the differences between the paired observations are analyzed.

This test is appropriate when the same subjects are measured twice, before and after an intervention, or when two measurements are taken from each subject under different conditions.

In contrast, the t-test for independent means, also known as unpaired or two-sample t-test, is used to compare the means of two independent groups. The independence of the two groups means that the observations in one group are not related to the observations in the other group.

This test is appropriate when the two groups are formed by different subjects, or when the same subjects are assigned to different conditions or treatments.

In the given case, parents and children are two independent groups, as they are not related in any way. Thus, the t-test for dependent means is not appropriate for comparing the values of parents and children. Instead, the t-test for independent means should be used, which would provide a statistical test of whether the means of the two groups are different from each other.

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The company can fence off a maximum square/rectangular area of 48,400 square meters. To find the maximum square/rectangular area that the company can fence off, they need to use all 880m of fencing available.

Let's call the length and width of the fenced area "L" and "W", respectively.
For a square, L = W, so we can write:
4L = 880
L = 220m
The maximum square area would be:
A = L x W = 220m x 220m = 48,400m²

For a rectangle, we need to use the fact that the perimeter (2L + 2W) equals 880m. We can solve for one variable (let's say L) in terms of the other (W), and then substitute it into the area equation:
2L + 2W = 880
L = 440 - W
A = L x W = (440 - W) x W = 440W - W²

To find the maximum area, we need to find the vertex of the quadratic equation. We can do this by finding the value of W that makes the derivative of the equation equal to 0:
dA/dW = 440 - 2W = 0
W = 220m
L = 440 - 220 = 220m
The maximum rectangular area would be:
A = L x W = 220m x 220m = 48,400m²
Therefore, the company can fence off a maximum square/rectangular area of 48,400 square meters.

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The surface area of the given figure is about 743.18 square feet.

The bottom most plane is a circle with radius 6 ft.

So the area of bottom most surface = 2π(6)² = 72π = 226.19 square ft. (Rounding to nearest hundredth)

The area of lateral surface of the bottom circular shape = 2π*6*4 = 150.80 square ft. (Rounding to nearest hundredth)

The surface area of top most pentagonal shape = (1/4)*√(5(5 + 2√5))*(4)² = 27.53 square ft. (Rounding to nearest hundredth)

The surface area of the contact surface of pentagonal and circular cylinder is = 226.19 - 27.53 = 198.66 square ft.

The surface area of lateral surface of pentagonal cylinder = 5*7*4 = 140 square ft.

So total surface area is about = 226.19 + 150.80 + 27.53 + 198.66 + 140 = 743.18 square ft.

Hence, surface area is about 743.18 square ft. (Rounding off to nearest hundredth).

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The slope of the tangent line to the given polar curve at θ = π/3 is √3/5.

Given the polar curve r = 5 + 4cosθ and the point θ = π/3, we'll follow these steps:

1. Convert polar coordinates to rectangular coordinates: x = rcosθ and y = rsinθ
2. Differentiate x and y with respect to θ
3. Find the slope dy/dx

Step 1: Convert polar coordinates to rectangular coordinates:
x = rcosθ = (5 + 4cosθ)cosθ
y = rsinθ = (5 + 4cosθ)sinθ

Step 2: Differentiate x and y with respect to θ:
dx/dθ = -4cosθ(sinθ + cosθ)
dy/dθ = 4cos^2θ - 4sin^2θ - 5sinθ

Step 3: Find the slope dy/dx at θ = π/3:
dy/dx = (dy/dθ) / (dx/dθ) at θ = π/3
= (-4cos(π/3)(sin(π/3) + cos(π/3))) / (4cos^2(π/3) - 4sin^2(π/3) - 5sin(π/3))
= -√3 / -5

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To determine the correct two-column table for the given data of Tamara's reading pages and time taken, we need to compare the given data with the values in each row of tables. The table with "Pages Time: 25 36, 63 48, 74.5 52" is the correct one. So, the correct answer is C).

Identify the data given, Tamara read 25 pages in 36 minutes, 48 pages in 63 minutes, and 52 pages in 74.5 minutes.

Based on the given data, create a two-column table that has one column for the number of pages Tamara read and another column for the time it took her to read them.

Compare the values in each row of the table to the given data to make sure they match.

The first table, "Pages Time: 25 36, 63 48, 74.5 52" matches the given data and has two columns for the number of pages and the time taken to read them, so it is the correct answer.

The other tables do not match the given data or do not have two columns for the number of pages and the time taken to read them.

Therefore, the table "Pages Time: 25 36, 63 48, 74.5 52" is the correct two-column table for the given data. So, the correct option is C).

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The volume of the volleyball is about 120 times greater than that of the golf ball rounding to the nearest number we get the exact value of 129 times greater. Thus, option B is correct.

Diameter of Volleyball = 8.5 in

Diameter of Golfball = 1.68 in

The volume of a sphere is calculated by using the formula,

V = (4/3) * π * [tex]r^{3}[/tex]

Volume of volleyball = (4/3) * π * [tex](4.25)^3[/tex]

The volume of the volleyball =  635.5 cubic inches

Volume of golf ball = (4/3) * π * [tex](0.84)^3[/tex]

The volume of the golf ball = 0.61 cubic inches

The ratio of the volume of the volleyball to that of the golf ball is:

volleyball / golf ball = 635.5 / 0.61 =  120

Therefore, we can conclude that the volume of the volleyball is about 120 times greater than that of the golf ball.

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The complete question is:

There is a volleyball with a diameter of 8. 5 in. And a golf ball with a diameter of 1. 68 in. Find how many times greater the volume of the volleyball is than that of the golf ball.

a. It is about 85. 2 times greater.

b. It is about 129 times greater.

c. It is about 25. 6 times greater.

d. It is about 13. 1 times greater.

To find the p-value for the regression equation that fits the given data on house prices, perform a multiple linear regression analysis using statistical software or calculator, and check the p-values associated with each variable in the model.

The question asks to find the p-value for the regression equation that fits the given data about house prices, which includes square footage, number of bedrooms, age, and selling price. The data provided contains information about different houses with various square footage, bedrooms, age, and selling prices. To find the p-value for the regression equation, you need to perform a multiple linear regression analysis using a statistical software or calculator. However, as an AI, I cannot directly perform these calculations. I would recommend using statistical software like R, Python, or Excel to input your data and perform the regression analysis. Once you have the results, you can find the p-value associated with each variable in the model. The p-value indicates the statistical significance of each variable in predicting the selling price of the houses. A lower p-value (typically less than 0.05) suggests that the variable significantly impacts the selling price, while a higher p-value (greater than 0.05) indicates that the variable is not significant in predicting the selling price.

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For the price of a gallon of milk which is rising about 1. 36% per year since 2000,

a) If cost of milk is $4.70 at present then the cost of milk to next year is 4.76.

b) The time taken for the price to top $5 is equals to 4.6 years.

The increasing rate of prices of a gallon of milk since 2000 = 1.36% per year

Now, we see price is compounding annually like simple interest does, so let's consider a function, F = P(1 + \frac{I}{k})ⁿ

where I = rate of change per year, k = the compounding periods per year = 1, n= the number of compounding time period beyond year 2000, P = price in the year 2000, and F = the price in the future 2000 as the present.

a) If milk cost is equals to $4.70, then n = 1, k = 1, P = $47.0, I = 1.36%, Future cost of milk in next year, F = 4.70( 1 + 0.0136)

= 4.70 × 1.0136

= 4.76392

b) Now, future value, F = $5, P = $4.70, I = 0.0136, k = 1, we have to determine the value of n. So, 5 = 4.70( 1 + 0.0136)ⁿ

=> 5/4.7= 1.0136ⁿ

=> 1.064 = 1.0136ⁿ

Taking logarithm both sides

=> ln( 1.064) = n ln( 1.0136)

=> 0.0620 = 0.01351 × n

=> n = 4.6

Hence, required value is 4.6 years.

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The given expression is unclear and contains symbols that are difficult to interpret. It is not possible to provide a brief solution without a clear understanding of the equation and the meaning of the symbols.

The provided equation is not well-defined and contains several symbols that are not clearly defined. In order to provide an explanation.

It is necessary to have a clear and properly formatted equation, along with the definitions and relationships of the symbols involved.

Without this information, it is not possible to analyze the equation or provide a meaningful explanation. Please provide a clear and well-defined equation for further analysis.

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The answers are:

a.   5 ft

b.   No

c.   [0, ∞)

d.   [5, 55]

From the graph:

x-axis:  t = time (in seconds)

y-axis:  H(t) = height (in feet)

a) When H(0), t = 0  ⇒  y-intercept.

Therefore, from inspection of the graph, H(0) = 5 ft

b) H(t) = 0 does not have a solution as the curve never touches the y-axis

(when y = 0).

c) Domain: set of all possible input values (x-values)

The domain is time (t) (x-axis), therefore the domain is [0, ∞)

d) Range: set of all possible output values (y-values)

From inspection of the graph, the lowest value of H(t) is 5 and the highest value of H(t) is 55.  

Therefore, the range is [5, 55]

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The margins of error are smaller for a larger sample size

a) The margin of error for a 95% self belief interval can be calculated the use of the formula:

ME = z√((p(1-p))/n)

Where,

z is the z-score corresponding to the preferred degree of self assurance (95% in this case),

p is the estimated percentage of college students in prefer of the proposed make bigger and n is the pattern measurement (250 in this case).

Using this formula, we can assemble the following desk of margins of error for p values of 0.1, 0.3, 0.5, 0.7, and 0.9:

p          ME

0.1      0.052

0.3     0.044

0.5     0.040

0.7     0.044

0.9     0.052

b) With a pattern measurement of 500, the margin of error calculations can be repeated the usage of the equal method as in section (a), however with n equal to five hundred rather of 250.

p       ME

0.1    0.036

0.3    0.030

0.5    0.027

0.7    0.030

0.9    0.036

As expected, the margins of error are smaller for a larger sample size.

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The margins of error are smaller for a large pattern size

a) The margin of error for a 95% self faith interval can be calculated the use of the formula:

ME = z√((p(1-p))/n)

Where,

z is the z-score corresponding to the favored diploma of self assurance (95% in this case),

p is the estimated share of university college students in select of the proposed make higher and n is the sample size (250 in this case).

Using this formula, we can collect the following desk of margins of error for p values of 0.1, 0.3, 0.5, 0.7, and 0.9:

p ME

0.1 0.052

0.3 0.044

0.5 0.040

0.7 0.044

0.9 0.052

b) With a sample dimension of 500, the margin of error calculations can be repeated the utilization of the equal technique as in area (a), then again with n equal to 5 hundred as a substitute of 250.

p ME

0.1 0.036

0.3 0.030

0.5 0.027

0.7 0.030

0.9 0.036

As expected, the margins of error are smaller for a large pattern size.

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Answer:

x=4

Step-by-step explanation:

Based on the given conditions, formulate: 28(6-x)+30x = 176

Apply the Distributive Property: 168 - 28x+30x = 176

Combine like terms: 168+2x=176

Rearrange variables to the left side of the equation: 2x=176-168

Calculate the sum or difference: 2x=8

Divide both sides of the equation by the coefficient of variable:x=8/2

Cross out the common factor: x=4

Answer:90 miles

Step-by-step explanation: multiply 15 by 6

The total calories in a meal is 755 Calories.

We have,

110 g of carbohydrates, 25 g of protein, 20 g of fat, and 5 g of alcohol.

Now,  110 g carbohydrates

= 110 x 4

= 440 calories

and, 25 g protein

= 25 x 4

= 100 calories

and,20 g fat

= 20 x 9

= 180 calories

and, 5 g alcohol  

= 5 x 7

=  35 calories.

So, the total calories in a meal  

= 440+100+180+35

= 755.

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The coordinates of k so that the ratio of JK to KL is 7 to 1 is k(18,142)

Simultaneous Equations are sets of algebraic equations that share common variables and are solved at the same time (that is, simultaneously). They can be used to calculate what each unknown actually represents and there is one solution that satisfies both equations

The given coordinates are

J(-2, 2),  K(x, y) and L(30, -22)

This implies that

Using slope formula, we have

(y-2)/ (x+2) = 7/1

Cross and multiply to get

1(y-2) = 7(x+2)

y-2 = 7x +14

y-7x = 14+2

y-7x = 16 ..................1

Also

(-22-y) / (30-x) = 7/1

-22-y = 210 -7x

-y+7x=210+22

-y+7x=232......................2

From equation 1

y = 16+7x

Therefore in equation 2

-16+7x+7x=232

14x = 232+16

14x=248

x = 248/14

x= 18

Then y = 16+7x

y = 16+7(18)

y = 142

Therefore k(18,142)

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