Find a linear inequality with the following solution set. Each grid line represents one unit. [asy] size(200); fill((-2,-5)--(5,-5)--(5,5)--(3,5)--cycle,yellow); real ticklen=3; real tickspace=2; real ticklength=0.1cm; real axisarrowsize=0.14cm; pen axispen=black+1.3bp; real vectorarrowsize=0.2cm; real tickdown=-0.5; real tickdownlength=-0.15inch; real tickdownbase=0.3; real wholetickdown=tickdown; void rr_cartesian_axes(real xleft, real xright, real ybottom, real ytop, real xstep=1, real ystep=1, bool useticks=false, bool complexplane=false, bool usegrid=true) { import graph; real i; if(complexplane) { label("$\textnormal{Re}$",(xright,0),SE); label("$\textnormal{Im}$",(0,ytop),NW); } else { label("$x$",(xright+0.4,-0.5)); label("$y$",(-0.5,ytop+0.2)); } ylimits(ybottom,ytop); xlimits( xleft, xright); real[] TicksArrx,TicksArry; for(i=xleft+xstep; i 0.1) { TicksArrx.push(i); } } for(i=ybottom+ystep; i 0.1) { TicksArry.push(i); } } if(usegrid) { xaxis(BottomTop(extend=false), Ticks("%", TicksArrx ,pTick=gray(0.1),extend=true),p=invisible);//,above=true); yaxis(LeftRight(extend=false),Ticks("%", TicksArry ,pTick=gray(0.1),extend=true), p=invisible);//,Arrows); } if(useticks) { xequals(0, ymin=ybottom, ymax=ytop, p=black, Ticks("%",TicksArry , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize)); yequals(0, xmin=xleft, xmax=xright, p=black, Ticks("%",TicksArrx , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize)); } else { xequals(0, ymin=ybottom, ymax=ytop, p=axispen, above=true, Arrows(size=axisarrowsize)); yequals(0, xmin=xleft, xmax=xright, p=axispen, above=true, Arrows(size=axisarrowsize)); } }; draw((-2,-5)--(3,5),dashed+red, Arrows(size=axisarrowsize)); rr_cartesian_axes(-5,5,-5,5); f

The equivalent form of the equation y=[tex]x^{2} -2x-15[/tex]given is y=(x+3)(x-5).

Given an equation y=[tex]x^{2}[/tex]-2x-15 and we are required to find the equivalent form of the equation.

Equation is like a relationship between two or more variables that are expressed in equal to form. Equation of two variables look like ax+by=c. It may be linear equation, quadratic equation, cubic equation or any other equation  depending on the powers of the variable.

To find the equivalent equations we are required to form factors of the equation. Equivalent equation are those equations which when solved gives the same solution as the equation when solved gives.

y=[tex]x^{2}[/tex]-2x-15

y=[tex]x^{2}[/tex]-5x+3x-15

y=x(x-5)+3(x-5)

y=(x+3)(x-5)

Hence the equivalent form of the equation y= [tex]x^{2}[/tex]-2x-15 given is

y=(x+3)(x-5).

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Question is incomplete as question should includes the equation

y= [tex]x^{2}[/tex]-2x-15.

Answer:

5a(a - 2)

2(a^2 - 4)

(x + 5)(x - 5)

(x - 4)(x^2 + 4x + 16)

(a - 3)(a - 2)

(x^2 + x + 1)(x^2 - x - 1)

Step-by-step explanation:

5a^2 - 10a     factor out GCF

5a(a - 2)

2a^2 - 8      factor out GCF

2(a^2 - 4)

x^2 - 25     difference of squares

(x + 5)(x - 5)

x^3 - 64      difference of cubes

(x - 4)(x^2 + 4x + 16)

a^2 - 5a + 6      factors of 6 that add to -5

a^2 - 2a - 3a + 6

a(a - 2) - 3(a - 2)

(a - 3)(a - 2)

x^4 - x^2 - 2x - 1      group last three terms

x^4 + (-x^2 - 2x - 1)

x^4 - (x^2 + 2x + 1)        factors of 1 that add to 2

x^4 - (x + 1)(x + 1)

x^4 - (x + 1)^2         difference of squares

(x^2 + (x + 1))(x^2 - (x + 1))        simplify

(x^2 + x + 1)(x^2 - x - 1)

The type of transformations that maps triangle ABC onto Triangle prime ABC is a  Reflection across the line y=x ; translation 10 units to the right and 4 units up.

Note that: Triangle ABC has vertices at points  which are:

A(-6,2), B(-2,6) and C(-4,2).

Therefore, the reflection across the line y=x has the rule of"

(x, y) ---(y, x).

Hence:

A(-6,2) -- A''(2,-6);

B(-2,6)--- B''(6,-2);

C(-4,2)----C''(2,-4).

2. The translation 10 units to the right and 4 units up is:

(x, y)----(x+10,y+4).

Hence

A''(2,-6)----A'(12,-2);

B''(6,-2)----B'(16,2);

C''(2,-4)----C'(12,0).

Therefore, Points A'B'C' are said to be exactly of the vertices of that of triangle A'B'C'.

Hence, The type of transformations that maps triangle ABC onto Triangle prime ABC is a  Reflection across the line y=x ; translation 10 units to the right and 4 units up.

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The recursive formula of h(n) is h(1) = 36 and h(n) = h(n -1) - 5

The function is given as:

h(n) = 41- 5n

Calculate h(1) and h(2)

h(1) = 41- 5(1)

h(1) = 36

h(2) = 41- 5(2)

h(2) = 31

Calculate the difference between h(1) and h(2)

d = 31 - 36

d = -5

This means that:

h(1) = 36 and h(n) = h(n -1) - 5

Hence, the recursive formula of h(n) is h(1) = 36 and h(n) = h(n -1) - 5

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a) The approximate area below the curve using five rectangles is 1280 square units.

b) The approximate area below the curve using ten rectangles is 1320 square units.

c) The approximate area below the curve using infinite number of rectangles is 1333.333 square units.

In this problem we must estimate the value of the area below the curve by finite number of rectangles using Riemann sums, whose expression is:

A ≈ [(b - a) / n] · ∑ f[a + i · (b - a) / n], for i = {0, 1, 2, 3, ..., n - 1}     (1)

Where:

The approximate area below the curve using five rectangles is: f(x) = 20 · x - x², a = 0, b = 20, n = 5

A ≈ [(20 - 0) / 5] · ∑ f[0 + i · (20 - 0) / 5]

A ≈ 4 · ∑ f(4 · i)

A ≈ 4 · [f(0) + f(4) + f(8) + f(12) + f(16)]

f(0) = 20 · 0 - 0² = 0

f(4) = 20 · 4 - 4² = 64

f(8) = 20 · 8 - 8² = 96

f(12) = 20 · 12 - 12² = 96

f(16) = 20 · 16 - 16² = 64

A ≈ 4 · (0 + 64 + 96 + 96 + 64)

A ≈ 1280

And using ten rectangles:

A ≈ [(20 - 0) / 10] · ∑ f[0 + i · (20 - 0) / 10]

A ≈ 2 · ∑ f(2 · i)

A ≈ 2 · [f(0) + f(2) + f(4) + f(6) + f(8) + f(10) + f(12) + f(14) + f(16) + f(18)]

f(0) = 20 · 0 - 0² = 0

f(2) = 20 · 2 - 2² = 36

f(4) = 20 · 4 - 4² = 64

f(6) = 20 · 6 - 6² = 84

f(8) = 20 · 8 - 8² = 96

f(10) = 20 · 10 - 10² = 100

f(12) = 20 · 12 - 12² = 96

f(14) = 20 · 14 - 14² = 84

f(16) = 20 · 16 - 16² = 64

f(18) = 20 · 18 - 18² = 36

A ≈ 2 · (0 + 36 + 64 + 84 + 96 + 100 + 96 + 84 + 64 + 36)

A ≈ 1320

And using infinite rectangles:

A ≈ [(b - a) / n] · ∑ f[a + i · (b - a) / n]

A ≈ (20 / n) · ∑ [20 · (20 / n) · i - (20 / n)² · i²]

A ≈ (20 / n) · ∑ [400 · i / n - 400 · i² / n²]

A ≈ ∑ (8000 · i / n² - 8000 · i² / n³)

A ≈ (8000 / n²) · ∑ i - (8000 / n³) · ∑ i²

A ≈ (8000 / n²) · [n · (n + 1) / 2] - (8000 / n³) · [n · (n + 1) · (2 · n + 1) / 6]

A ≈ (8000 / n²) · [(n² + n) / 2] - (8000 / n³) · [(n² + n) · (2 · n + 1) / 6]

A ≈ 4000 · (n² + 2) / n² - (8000 / n³) · [(2 · n³ + 3 · n² + n) / 6]

A ≈ 4000 · (1 + 2 / n²) - (4000 / 3) · [2 + 3 · (1 / n) + (1 / n²)]

As n → + ∞, then:

A ≈ 4000 - 8000 / 3

A ≈ 1333.333

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The correct answer is "all real numbers greater than or equal to 2".

Answer:

Step-by-step explanation:

The amplitude of a sine function is equal to one-half the distance between the maximum and minimum values of the function.

In this case, the maximum value is approximately 84 customers, and the minimum value is approximately 30 customers

Therefore, the value that is closest to the amplitude of the transformed function is 27 customers.

By taking the quotient between the total number of hours and the number of shifts each shift is 1.5 hours long.

We know that Paolo worked (2 + 3/4) hours this week, and that is equivalent to (1 + 5/6) shifts.

To find the number of hours that each shift takes, we need to take the quotient between the total number of hours and the number of shifts:

[tex]\frac{(2 + 3/4)h}{1 + 5/6} = 1.5 h[/tex]

This means that each shift is 1.5 hours long.

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Answer: 1/8

Step-by-step explanation:

Answer:

Domain: All real numbers (infinite) Range: [tex]y \leq -4[/tex]

Step-by-step explanation:

For the domain, the x-axis will continue to be used for the parabola, because the parabola can go on forever and can use an x- value on the x-axis. For the range, the largest number the parabola will go up to is -4. Therefore, if that is the highest point/ where the vertex is, the highest point on the y-axis is -4, resulting in the answer y can be equal to or less than -4 (y [tex]\leq[/tex] -4), since the parabola will continue to go down with the reflection over the x-axis.

ItItItIt takes Laura  15 minutes (0.25 hours) to cycle to the post office.

a. It takes Laura  15 minutes (0.25 hours) to cycle to the post office.

b. Speed:

Speed=Distance/Time

Speed=4/0.25

Speed=16 km/hr

c.  It takes Laura  20 minutes to cycle to the post office.

d. Time

Time=20 minutes (1/3 hour)

Speed:

Speed=Distance/Time

Speed=4÷1/3

Speed=12 km/hr

Therefore It takes Laura  15 minutes (0.25 hours) to cycle to the post office.

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The 15 foot ramp must make an angle 19.45° with the ground to reach the same window.

Since the height of the window above ground remains constant in both cases, it follows that by means of trigonometric identity sine; we have;

10sin30° = 15sinx

sinx = 5/15 = 0.333

x = sin-¹(0.333)

x = 19.45°.

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Answer: x = 2

Step-by-step explanation:

3x - 5 = 1

add 5 to both sides

3x - 5 (+5) = 1 (+5)

3x = 6

divide by 3 on both sides

3x/3 = 6/3

x = 2

The system of equations to the inverse of its coefficient matrix, A⁻¹, and the matrix of its solution, X is shown in the figure.

Given that the system of equations are shown in given figure.

The first system of equations are

[tex]\begin{aligned}4x+2y-z&=150\\x+y-z&=-100\\-3x-y+z&=600\\\end[/tex]

By writing in matrix AX=b, we get

Coefficient matrix [tex]A=\left[\begin{array}{lll}4&2&-1\\1&1&-1\\-3&-1&1\end{array}\right][/tex] and [tex]B=\left[\begin{array}{l}150&-100&600\end{array}\right][/tex]

Firstly, we will find the A⁻¹ by finding the determinant and adjoint of A and divide the adjoint with determinant, we get

[tex]\begin{aligned}|A|&=\left|\begin{array}{lll}4&2&-1\\1&1&-1\\-3&-1&1\end{array}\right|\\ &=4(1-1)-2(1-3)-1(-1+3)\\&=4(0)-2(-2)-1(2)\\ &=2\neq 0\end[/tex]

[tex]\begin{aligned}Adj A&=\left[\begin{array}{lll}0&2&2\\-1&1&-2\\-2&3&2\end{array}\right]^T\\&=\left[\begin{array}{lll}0&-1&-2\\2&1&3\\2&-2&2\end{array}\right]\end[/tex]

[tex]\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0&-0.5&-0.5\\1&0.5&1.5\\1&-1&1\end{array}\right]\end[/tex]

For a solution Consider [A B] and apply row operations, we get

[tex]\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{lll1}4&2&-1&150\\1&1&-1&-100\\-3&-1&1&600\end{array}\right]\\ R_{2}&\rightarrow 4R_{2}-R_{1},R_{3}\rightarrow 4R_{3}+3R_{1}\\ &\sim \left[\begin{array}{lll1}4&2&-1&150\\0&2&-3&-550\\0&2&1&2850\end{array}\right]\\ R_{3}&\rightarrow R_{3}-R_{2}\\ &\sim \left[\begin{array}{llll}4&2&-1&150\\0&2&-3&-550\\0&0&4&3400\end{array}\right]\end[/tex]

Thus, [tex]x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}-250\\1000\\850\end{array}\right][/tex]

The second system of equations are

[tex]\begin{aligned}x+y-z&=220\\5x-5y-z&=-640\\-x+y+z&=200\\\end[/tex]

Similarly, we will find for second system of equations

[tex]\begin{aligned}|A|&=\left|\begin{array}{lll}1&1&-1\\5&-5&-1\\-1&1&1\end{array}\right|\\ &=1(-5+1)-1(5-1)-1(5-5)\\&=1(-4)-1(4)-1(0)\\ &=-8\neq 0\end[/tex]

[tex]\begin{aligned}Adj A&=\left[\begin{array}{lll}-4&-4&0\\-2&0&-2\\-6&-4&-10\end{array}\right]^T\\&=\left[\begin{array}{lll}-4&-2&-6\\-4&0&-4\\0&-2&-10\end{array}\right]\end[/tex]

[tex]\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0.5&0.25&0.75\\0.5&0&0.5\\0&0.25&1.25\end{array}\right]\end[/tex]

[tex]\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{llll}1&1&-1&220\\5&-5&-1&-640\\-1&1&1&200\end{array}\right]\\ R_{2}&\rightarrow R_{2}-5R_{1},R_{3}\rightarrow R_{3}+R_{1}\\ &\sim \left[\begin{array}{llll}1&1&-1&220\\0&-10&4&-1740\\0&2&0&420\end{array}\right]\\ R_{3}&\rightarrow 5R_{3}+R_{2}\\ &\sim \left[\begin{array}{llll}1&1&-1&220\\0&-10&4&-1740\\0&0&4&360\end{array}\right]\end[/tex]

Thus, [tex]x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}100\\210\\90\end{array}\right][/tex]

The third system of equations are

[tex]\begin{aligned}2x+2y-z&=290\\x+y-3z&=500\\x-y+2z&=600\\\end[/tex]

Similarly, we will find for third system of equations

[tex]\begin{aligned}|A|&=\left|\begin{array}{lll}2&2&-1\\1&1&-3\\1&-1&2\end{array}\right|\\ &=2(2-3)-2(2+3)-1(-1-1)\\&=2(-1)-2(5)-1(-2)\\ &=-10\neq 0\end[/tex]

[tex]\begin{aligned}Adj A&=\left[\begin{array}{lll}-1&-5&-2\\-3&5&4\\-5&5&0\end{array}\right]^T\\&=\left[\begin{array}{lll}-1&-3&-5\\-5&5&5\\-2&4&0\end{array}\right]\end[/tex]

[tex]\begin{aligned}A^{-1}&=\frac{Adj A}{|A|}\\ &=\left[\begin{array}{lll}0.1&0.3&0.5\\0.5&-0.5&-0.5\\0.2&-0.4&0\end{array}\right]\end[/tex]

get

[tex]\begin{aligned}\left[A\right.\text{ }\left.B\right]&=\left[\begin{array}{llll}2&2&-1&290\\1&1&-3&500\\1&-1&2&600\end{array}\right]\\ R_{2}&\rightarrow 2R_{2}-R_{1},R_{3}\rightarrow 2R_{3}-R_{1}\\ &\sim \left[\begin{array}{llll}2&2&-1&290\\&0&-5&710\\0&-4&5&910\end{array}\right]\end[/tex]

Thus, [tex]x=\left[\begin{array}{l}x\\y\\z\end{array}\right]=\left[\begin{array}{l}479\\-405\\-142\end{array}\right][/tex]

Hence, each system of equations to the inverse of its coefficient matrix, A⁻¹, and the matrix of its solution, X.

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The steps in evaluating each of the following expressions is shown below.

An expression is a mathematical equation which shows the relationship that exist between two or more numerical quantities or variables.

15 - 35/7 - 2 + 3 - 4

15 - (35/7) - 2 + 3 - 4 (bracket and division)

15 - 5 - 2 + 3 - 4 (regroup)

15 + 3 - 5 - 2 - 4 (subtract and add)

18 - 11 = 7.

10 + 2(9 - 5) - 16/18

10 + (2 × 4) - 8/9 (bracket and division)

10 + 8 - 8/9 (add)

18 - 8/9 (subtract)

162/9 - 8/9 = 17 1/9 or 154/9.

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Complete Question:

Show your steps in evaluating each of the following expressions. The steps count 4 points each, the answer is 1 point.

A. 15 - 35/7 -2 + 3 -4

B. 10 + 2(9 - 5) - 16/18

The radius of the cylinder exists 1.93mm.

Let V be the volume of the cylinder exists 200 mm³

r be the radius

h be the height exists 17

Volume of cylinder, V = π r²h

200 = π r² (17)

200 = 53.40707 r²

200 =  53.41 r²

simplifying the above equation, we get

r² = 200/53.41

r² = 3.74461 = 3.74

r² = 3.74

r = 1.933907961 = 1.93

Therefore, r = 19.3

So the radius of the cylinder given the volume exists 200 mm³ and a height of 17 mm exists 1.93 mm.

Therefore, the radius of the cylinder exists 1.93mm.

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Using a system of equations, it is found that Oscar bought 6 markers, 6 notebooks and 15 packs of sticky notes.

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

He bought the same number of markers as notebooks, hence:

x = y.

He bought three more packs of sticky notes than markers and notebooks combined, hence:

z = 3 + x + y

z = 3 + 2x.

He paid $1.05 for each marker, $2.25 for each notebook, and the packs of sticky notes were $1.95. He spent $49.05 in total, hence:

1.05x + 2.25y + 1.95z = 49.05.

Then:

1.05x + 2.25x + 1.95(3 + 2x) = 49.05.

7.2x = 43.2

x = 43.2/7.2

x = 6.

The amounts are as follows:

He bought 6 markers, 6 notebooks and 15 packs of sticky notes.

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The rule which describe the composition of transformations that

maps ΔBCD to ΔB"C"D" is:

Reflection across the y-axis composition translation of 6 units x,

negative 5 units y ⇒ last answer

Step-by-step explanation:

Let us revise the reflection across the y-axis , horizontal translation

and vertical translation

1. If point (x , y) is reflected across the y-axis, then its image is (-x , y)

2. If point (x , y) is translated h units to the right, then its image is

  (x + h , y), if translated h units to the left, then its image is (x - h , y)

3. If point (x , y) is translated k units up, then its image is (x , y + k),

  if translated k units down, then its image is (x , y - k)

∵ The vertices of triangle BCD are (1 , 4) , (1 , 2) , (5 , 3)

∵ The vertices of triangle B'C'D' are (-1 , 4) , (-1 , 2) , (-5 , 3)

∵ The x-coordinates of the vertices of Δ B'C'D' have the same

  magnitude of x-coordinates of Δ BCD and opposite signs

∴ Δ B'C'D' is the image of Δ ABC after reflection across the y-axis

∵ The vertices of triangle B'C'D' are (-1 , 4) , (-1 , 2) , (-5 , 3)

∵ The vertices of triangle B''C''D'' are (5 , -1) , (5 , -3) , (1 , -2)

∵ The image of -1 is 5 and the image of -5 is 1

∴ The x-coordinates of the vertices of triangle B'C'D' are added by 6

∵ The image of 4 is -1 , image of 2 is -3 and the image of 3 is -2

∴ The y-coordinates of the vertices of triangle B'C'D' are subtracted

  by 5

∴ Δ B"C"D" is the image of Δ B'C'D' by translate 6 units to the right

 and 5 units down ⇒ (x + 6 , y - 5)

The rule which describe the composition of transformations that

maps ΔBCD to ΔB"C"D" is:

Reflection across the y-axis composition translation of 6 units x,

negative 5 units y

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The composition of transformations that maps BCD to B"C"D" is described by the following rule:

Composition translation of 6 units x, negative 5 units y across the y-axis last response

Let us rewrite the y-axis reflection, horizontal translation, and vertical translation.

1. If point (x, y) is mirrored across the y-axis, the image of that point is (-x , y)

2. If point (x, y) is translated h units to the right, its image is (x + h, y), and if it is translated h units to the left, its image is (x + h, y) (x - h , y)

3. If point (x, y) is translated k units up, its image is (x, y + k), but if it is translated k units down, its image is (x , y - k)

∵ Triangle BCD has (1, 4) vertices, (1, 2) vertices, and (5, 3) vertices. Triangle B'C'D' has (-1, 4) vertices, (1, 2) vertices, and (5, 3) vertices. The x-coordinates of the B'C'D' vertices have the same magnitude as the x-coordinates of BCD and opposite signs. B'C'D' is the picture of ABC after it has been reflected across the y-axis.

The vertices of triangle B'C'D' are (-1, 4), (-1, 2), (-5, 3), (1, -2)

The vertices of triangle B"C"D" are (5, -1), (5, -3), (1, -2)

The x-coordinates of the triangle B'C'D' vertices are added by 6.

The image of 4 is -1, the image of 2 is -3, and the image of 3 is -2.

The y-coordinates of triangle B'C'D' vertices are subtracted by 5.

∴ Δ B"C"D" is the image of Δ B'C'D' by translating 6 units to the right

and 5 units down ⇒ (x + 6 , y - 5)

The rule which describe the composition of transformations that

maps ΔBCD to ΔB"C"D" is:

Reflection across the y-axis composition translation of 6 units x,

negative 5 units y

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Answer:

Yes, multiplying every length of a Pythagorean triple by the same whole number results in a Pythagorean triple.

Step-by-step explanation:

Another example of  multiplying a Pythagorean triple is

Take the known Pythagorean triple 5, 12 and 13:

5 12 13   * 2  = 10 24 26

and 26^2 = 10^2 + 24^2

     676 = 100 + 576 = 676

The width of the deck is 6.5 feet.

The perimeter of a two-dimensional shape is the total length of the outline. To find the perimeter of a rectangle, we add the lengths of all four sides. Since opposite sides of a rectangle are always equal, we need to find the dimensions of length and width to find the perimeter of a rectangle. We can write the perimeter of the rectangle as twice the sum of its length and width. The perimeter is a linear measure and has units as meters, centimeters, inches, feet, etc.

Assuming the deck is rectangle

Perimeter = 2(length + width)

33ft = 2 (10 + width)

16.5 = 10 + width

width = 6.5 ft

Thus the width of the deck is 6.5 feet.

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Answer:

14

Step-by-step explanation:

An integers are whole numbers and  their opposites.  ...-3,-2,-2,0,1,2,3...

I just played with the numbers and found that 14 + 14^2 (14 squared is 14 x 14).

14 + (14)(14) = 210

Answer:

refer to the attachment

Health club members or were smokers is [tex]334[/tex] from a A survey of [tex]400[/tex] nurses yielded.

How can we find the health club members or were smokers ?

Events of health club members is [tex]P(M)=279[/tex]

Events of smokers is [tex]P(S)=215[/tex]

Probability of health club members and smokers is [tex]p(M and S)=160[/tex]

Probability of health club members or were smokers is =?

So we use the formula

[tex]p(M or S)=p(M)+p(S)-p(M and S)\\ = 279+215-160\\ = 334[/tex]

Probability of health club members or were smokers is [tex]334[/tex]

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Answer:

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Let the number of boys be b and girls be g.

The ratio is:

And the number of girls is:

Substitute the number into ratio and find the value of b:

The volume of the container is 5969 cm³. The correct option is D. 5,969 cm³

From the question, we are to determine the volume of the container

Volume of the container = Volume of the cone - Volume of the small cone

The volume of a cone is given by the formula,

V = 1/3πr²h

Where V is the volume

r is the radius

and h is the height

From the given information,

For the small cone,

Diameter = 10 cm

∴ Radius, r = 10cm/ 2 = 5 cm

h = 6 cm

For the big cone

diameter = diameter of the container = 30 cm

∴ Radius = 30cm/ 2

Radius = 15cm

h = height of small cone + height of container

h = 6 cm + 20 cm

h = 26 cm

Putting the parameters into

Volume of the container = Volume of the cone - Volume of the small cone

We get,

Volume of the container = 1/3π × 15² ×26 - 1/3π × 5² × 6

Volume of the container = 1/3π (15² ×26 -  5² × 6)

Volume of the container = 1/3π (5850 -  150)

Volume of the container = 1/3π (5700)

Volume of the container = 1/3 × π × 5700

Volume of the container = 5969.026 cm³

Volume of the container ≈ 5969 cm³

Hence, the volume of the container is 5969 cm³. The correct option is D. 5,969 cm³

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Answer: C

Step-by-step explanation:

[tex]\frac{5\pi}{4}[/tex] is in the third quadrant, so both x and y are negative.

Therefore, the only possible answer is C.