Let A and B be events. The symmetric difference A?B is defined to be the set of all elements that are in A or B but not both.
In logic and engineering, this event is also called the XOR (exclusive or) of A and B.
Show that P(AUB) = P(A) + P(B)-2P(AnB), directly using the axioms of probability.

Answer:

12/27

Step-by-step explanation:

Count all letters and all vowels then divide vowels by letters

The probability that a vowel is randomly selected in the experiment of selecting a letter from the phrase "Sean wants to eat at Olive Garden", is 4/9.

The probability of any event suppose A, in an experiment is given as:

P(A) = n/S,

where P(A) is the probability of event A, n is the number of favorable outcomes to event A in the experiment, and S is the total number of outcomes in the experiment.

In the question, we are given an experiment of selecting a letter from the phrase "Sean wants to eat at Olive Garden".

We are asked to find the probability that the selected letter is a vowel.

Let the event of selecting a vowel from the experiment of selecting a letter from the phrase "Sean wants to eat at Olive Garden" be A.

We can calculate the probability of event A by the formula:

P(A) = n/S,

where P(A) is the probability of event A, n is the number of favorable outcomes to event A in the experiment, and S is the total number of outcomes in the experiment.

The number of outcomes favorable to event A (n) = 12 (Number of vowels in the phrase)

The total number of outcomes in the experiment (S) = 27 (Number of letters in the phrase).

Now, we can find the probability of event A as:

P(A) = 12/27 = 4/9

∴ The probability that a vowel is randomly selected in the experiment of selecting a letter from the phrase "Sean wants to eat at Olive Garden", is 4/9.

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Answer:

[tex]\approx \bold{6544\ in^3/sec}[/tex]

Step-by-step explanation:

Given:

Rate of change of radius of cylinder:

[tex]\dfrac{dr}{dt} = +7\ in/sec[/tex]

(This is increasing rate so positive)

Rate of change of height of cylinder:

[tex]\dfrac{dh}{dt} = -6\ in/sec[/tex]

(This is decreasing rate so negative)

To find:

Rate of change of volume when r = 20 inches and h = 16 inches.

Solution:

First of all, let us have a look at the formula for Volume:

[tex]V = \pi r^2h[/tex]

Differentiating it w.r.to 't':

[tex]\dfrac{dV}{dt} = \dfrac{d}{dt}(\pi r^2h)[/tex]

Let us have a look at the formula:

[tex]1.\ \dfrac{d}{dx} (C.f(x)) = C\dfrac{d(f(x))}{dx} \ \ \ (\text{C is a constant})\\2.\ \dfrac{d}{dx} (f(x).g(x)) = f(x)\dfrac{d}{dx} (g(x))+g(x)\dfrac{d}{dx} (f(x))[/tex]

[tex]3.\ \dfrac{dx^n}{dx} = nx^{n-1}[/tex]

Applying the two formula for the above differentiation:

[tex]\Rightarrow \dfrac{dV}{dt} = \pi\dfrac{d}{dt}( r^2h)\\\Rightarrow \dfrac{dV}{dt} = \pi h\dfrac{d }{dt}( r^2)+\pi r^2\dfrac{dh }{dt}\\\Rightarrow \dfrac{dV}{dt} = \pi h\times 2r \dfrac{dr }{dt}+\pi r^2\dfrac{dh }{dt}[/tex]

Now, putting the values:

[tex]\Rightarrow \dfrac{dV}{dt} = \pi \times 16\times 2\times 20 \times 7+\pi\times 20^2\times (-6)\\\Rightarrow \dfrac{dV}{dt} = 22 \times 16\times 2\times 20 +3.14\times 400\times (-6)\\\Rightarrow \dfrac{dV}{dt} = 14080 -7536\\\Rightarrow \dfrac{dV}{dt} \approx \bold{6544\ in^3/sec}[/tex]

So, the answer is: [tex]\approx \bold{6544\ in^3/sec}[/tex]

Answer:

a) dV(s)  =  15,386 cm³

b) dS(s) = 4,396 cm²

c) dV(s)/V(s) = 1,07 %    and   dS(s)/ S(s)  =  0,71 %

Step-by-step explanation:

a) The volume of the sphere is

V(s) = (4/3)*π*x³        where x is the radius

Taking derivatives on both sides of the equation we get:

dV(s)/ dr  =  4*π*x²    or

dV(s)  =  4*π*x² *dr

the possible propagated error in cm³ in computing the volume of the sphere is:

dV(s)  = 4*3,14*(7)²*(0,025)

dV(s)  =  15,386 cm³

b) Surface area of the sphere is:

V(s) = (4/3)*π*x³  

dV(s) /dx  =  S(s) = 4*π*x³

And

dS(s) /dx  = 8*π*x

dS(s) = 8*π*x*dx

dS(s) = 8*3,14*7*(0,025)

dS(s) = 4,396 cm²

c) The approximates errors in a and b are:

V(s) =  (4/3)*π*x³     then

V(s) = (4/3)*3,14*(7)³

V(s) = 1436,03 cm³

And  the possible propagated error in volume is from a)  is

dV(s)  =  15,386 cm³

dV(s)/V(s)  = [15,386 cm³/1436,03 cm³]* 100

dV(s)/V(s) = 1,07 %

And for case b)

dS(s) = 4,396 cm²

And the surface area of the sphere is:

S(s) =  4*π*x³        ⇒   S(s) =  4*3,14*(7)²    ⇒ S(s) = 615,44 cm²

dS(s) = 4,396 cm²

dS(s)/ S(s)  =  [ 4,396 cm²/615,44 cm² ] * 100

dS(s)/ S(s)  =  0,71

Answer:

Find answer below

Step-by-step explanation:

f(x)=2x-6

Domain of 2x-6: {solution:-∞<x<∞, interval notation: -∞, ∞}

Range of 2x-6: {solution:-∞<f(x)<∞, interval notation: -∞, ∞}

Parity of 2x-6: Neither even nor odd

Axis interception points of 2x-6: x intercepts : (3, 0) y intercepts (0, -6)

inverse of 2x-6: x/2+6/2

slope of 2x-6: m=2

Plotting : y=2x-6

Answer:

5.215 units (rounded up to three decimal places)

Step-by-step explanation:

To find the distance between points (3.1 , 0.3) and (2.7, -4.9)

We use the Pythagoras Theorem which states that for a right triangle of sides a,b and c then;

a² + b²  = c² ,  Where c is the hypotenuse.

In our case, the distance between the two points is the hypotenuse of triangle formed by change in y-axis and change in x-axis.

The distance (hypotenuse) squared = (-4.9 - 0.3)² + (2.7 - 3.1)² = 27.04 + 0.16 = 27.2

Hypotenuse (the distance between) = [tex]\sqrt{27.2}[/tex] = 5.215 units (rounded up to three decimal places)

[tex]\dfrac{x}{5}=-2\\\\x=-10[/tex]

Answer:

[tex]\huge \boxed{{x=-10}}[/tex]

Step-by-step explanation:

[tex]\displaystyle \frac{x}{5} =-2[/tex]

We need the x variable to be isolated on one side of the equation, so we can find the value of x.

Multiply both sides of the equation by 5.

[tex]\displaystyle \frac{x}{5}(5) =-2(5)[/tex]

Simplify the equation.

[tex]x=-10[/tex]

The value of x that makes the equation true is -10.

Answer: 3.2 feet.

Step-by-step explanation:

Given: The end of a hose was resting on the ground, pointing up an angle. Sal measured the path of the water coming out of the hose and found that it could be modeled using the equation[tex]f(x) = -0.3x^2 + 2x[/tex], where [tex]f(x)[/tex] is the height of the path of the water above the ground, in feet, and [tex]x[/tex] is the horizontal distance of the path of the water from the end of the hose, in feet.

At x= 4 , we get

[tex]f(x) = -0.3(4)^2 + 2(4)=-0.3(16)+8 =-4.8+8=3.2[/tex]

Hence, when the water was 4 feet from the end of the hose,  its height above the ground is 3.2 feet.

Answer:

3.2 feet.

Step-by-step explanation:

Hi there! :)

Answer:

[tex]\huge\boxed{2(14x - 41y)}[/tex]

(75x - 67y) - (47x + 15y)

Distribute the '-' sign with the terms inside of the parenthesis:

75x - 67y - (47x - (15y))

75x - 67y - 47x - 15y

Combine like terms:

28x - 82y

Distribute out the greatest common factor:

2(14x - 41y)

Answer:

2.7 in²

Step-by-step explanation:

Since ∆BAC and ∆EDF are similar, therefore, the ratio of their area = square of the ratio of their corresponding side lengths.

Thus, if area of ∆EDF = x, area of ∆BAC = 6 in², EF = 2 in, BC = 3 in, therefore:

[tex] \frac{6}{x} = (\frac{3}{2})^2 [/tex]

[tex] \frac{6}{x} = (1.5)^2 [/tex]

[tex] \frac{6}{x} = 2.25 [/tex]

[tex] \frac{6}{x}*x = 2.25*x [/tex]

[tex] 6 = 2.25x [/tex]

[tex] \frac{6}{2.25} = \frac{2.25x}{2.25} [/tex]

[tex] 2.67 = x [/tex]

[tex] x = 2.7 in^2 [/tex] (nearest tenth)

Answer:

Total hours that Jenny ran = 3.63 hours.

Step-by-step explanation:

Jenny ran on Tuesday for = 2/5 hours or 0.4 hours.

Time consumed to run on Thursday = 11/6 hours or 1.83 hours.

Time consumed to run on Saturday = 21/ 15 hours or 1.4 hours.

Here, the total hours can be calculated by just adding all the running hours. So the running hours of Tuesday, Thursday, and Saturday will be added to find the total hours.

Total hours that Jenny ran = 0.4 + 1.83 + 1.4 = 3.63 hours.

Answer:

-3

Step-by-step explanation:

[tex]8+3y = -1\\3y = -9\\y = -3[/tex]

Answer:

y = -3

Step-by-step explanation:

-1=3y+8

3y+8=-1

3y=-9

y=-3

Answer and Step-by-step explanation:

The computation of points on the ellipse is shown below:-

Distance between any point on the ellipse

[tex](3 cos t, sin t) and (\frac{4}{3},0) is\\\\ d = \sqrt{(3 cos\ t - \frac{4}{3}^2) } + (sin\ t - 0)^2\\\\ d^2 = (3 cos\ t - \frac{4}{3})^2 + sin^2 t[/tex]

To minimize

[tex]d^2, set\ f' (t) = 0\\\\2(3cos\ t - \frac{x=4}{3} ).3(-sin\ t) + 2sin\ t\ cos\ t = 0\\\\ 8 sin\ t - 16 sin\ t\ cos\ t = 0\\\\ 8 sin\ t (1 - 2 cos\ t) = 0\\\\ sin\ t = 0, cos\ t = \frac{1}{2} \\\\ t= 0, \ 0, \pi,2\pi,\frac{\pi}{3} , \frac{5\pi}{3}[/tex]

Now we create a table by applying the critical points which are shown below:

t            [tex]d^{2} = (3\ cos t - \frac{4}{3})^{2} + sin^{2}t[/tex]

0           [tex]\frac{25}{9}[/tex]

[tex]\pi[/tex]           [tex]\frac{169}{9}[/tex]

[tex]2\pi[/tex]         [tex]\frac{25}{9}[/tex]

[tex]\frac{\pi}{3}[/tex]          [tex]\frac{7}{9}[/tex]

[tex]\frac{5\pi}{3}[/tex]         [tex]\frac{7}{9}[/tex]

When t = [tex]\frac{\pi}{3}[/tex], x is [tex]\frac{3}{2}[/tex] and y is [tex]\frac{\sqrt{3} }{2}[/tex]. So, the required points are [tex](\frac{3}{2},\frac{\sqrt{3} }{2})[/tex]

When t = [tex]\frac{5\pi}{3}[/tex], x is [tex]\frac{3}{2}[/tex] and y is [tex]\frac{-\sqrt{3} }{2}[/tex]. So, the required points are [tex](\frac{3}{2},\frac{-\sqrt{3} }{2})[/tex]

Answer:

[tex]x = 7[/tex]

Step-by-step explanation:

We can solve the equation [tex]5x+3+8x-4=90[/tex] by isolating the variable x on one side. To do this, we must simplify the equation.

[tex]5x+3+8x-4=90[/tex]

Combine like terms:

[tex]13x - 1 = 90[/tex]

Add 1 to both sides:

[tex]13x = 91[/tex]

Divide both sides by 13:

[tex]x = 7[/tex]

Hope this helped!

Answer:

x = 7

Step-by-step exxplanation:

5x + 3 + 8x - 4 = 90

5x + 8x = 90 - 3 + 4

13x = 91

x = 91/13

x = 7

probe:

5*7 + 3 + 8*7 - 4 = 90

35 + 3 + 56 - 4 = 90

Answer:

0.05m^2

Step-by-step explanation:

5 divided by 100

Answer: a. 0.4 × 0.15 = 0.060

Step-by-step explanation: Probability of the complement of an event is the one that is not part of the event.

For P(A):

P(A') = 1 - 0.6

P(A') = 0.4

For P(B):

P(B') = 1 - 0.85

P(B') = 0.15

To determine probability of A' and B':

P(A' and B') = P(A')*P(B')

P(A' and B') = 0.4*0.15

P(A' and B') = 0.06

Probability of the complement of the event is 0.060

Answer:

b

Step-by-step explanation:

Answer:

Ok, as i understand it:

for a point P = (x, y)

The values of x and y can be randomly chosen from the set {1, 2, ..., 10}

We want to find the probability that the point P lies on the second quadrant:

First, what type of points are located in the second quadrant?

We should have a value negative for x, and positive for y.

But in our set;  {1, 2, ..., 10}, we have only positive values.

So x can not be negative, this means that the point can never be on the second quadrant.

So the probability is 0.

Answer:

[tex]Probability = \frac{3}{7}[/tex]

Step-by-step explanation:

Given

Electrician = 6

Mechanic = 8

Required

Determine the probability of selecting an electrician

First, we need the total number of employees;

[tex]Total = n(Electrician) + n(Mechanic)[/tex]

[tex]Total = 6 + 8[/tex]

[tex]Total = 14[/tex]

Next, is to determine the required probability using the following formula;

[tex]Probability = \frac{n(Electrician)}{Total}[/tex]

[tex]Probability = \frac{6}{14}[/tex]

Divide numerator and denominator by 2

[tex]Probability = \frac{3}{7}[/tex]

Hence, the probability of selecting an electrician is 3/7

Answer:

Step-by-step explanation:

-6x²+26x+20

=-2(3x²-13x-10)

=-2(3x²-15x+2x-10)

=-2[3x(x-5)+2(x-5)]

=-2(x-5)(3x+2)

Complete Question

A population has a mean mu μ equals = 77 and a standard deviation σ = 14. Find the mean and standard deviation of a sampling distribution of sample means with sample size n equals = 26

Answer:

The mean of sampling distribution of the sample mean ( [tex]\= x[/tex]) is [tex]\mu_{\= x } = 77[/tex]

The standard deviation of sampling distribution of the sample mean ( [tex]\= x[/tex]) is  

    [tex]\sigma _{\= x} = 2.746[/tex]

Step-by-step explanation:

From the question we are told that

    The population mean is  [tex]\mu = 77[/tex]

     The  standard deviation is  [tex]\sigma = 14[/tex]

     The sample size is  [tex]n = 26[/tex]

Generally the standard deviation of sampling distribution of the sample mean ( [tex]\= x[/tex]) is  mathematically represented as

           [tex]\sigma _{\= x} = \frac{ \sigma }{ \sqrt{n} }[/tex]

substituting values  

          [tex]\sigma _{\= x} = \frac{ 14}{ \sqrt{26} }[/tex]

          [tex]\sigma _{\= x} = 2.746[/tex]

Generally the mean of sampling distribution of the sample mean ( [tex]\= x[/tex]) is  equivalent to the population mean i.e  

      [tex]\mu_{\= x } = \mu[/tex]

      [tex]\mu_{\= x } = 77[/tex]

Answer: 0%

Step-by-step explanation:

There's 40 students, and 40 students read physics. That means that every student reads physics. So, no student could read only chemistry.

Answer:

x= -3     x = 1/2     x=-2

Step-by-step explanation:

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

Set equal to zero

0 =(x+3) (2x-1)(x+2)

Using the zero product property

x+3 =0   2x-1 =0    x+2 =0

x= -3    2x =1       x = -2

x= -3     x = 1/2     x=-2

The equation is represented 3 units to the left of the complex plane and 2 units up.

A complex equation is an equation that involves complex numbers when solving it. A complex number is a number that has both a real part and an imaginary part.

Well to see how this is represented, we first need to multiply it out so we can see how it looks when it is simplified!

[tex]=(3-2i)(i^2)\\\\\\i^2=-1\\\\\\=(3-2i)(-1)\\\\\\=(-3+2i)[/tex]

We know that on a complex plane, our imaginary numbers are represented on the vertical axis.

So the original expression, (3-2i) would have been 3 units to the right on a complex graph and 2 units downward!

The equation I input above should be pretty straightforward, but one thing I didn't mention was that i^2 should = -1 when dealing with complex numbers!

Therefore, the equation 3-2i * i^2 is equal to -3 + 2i, this is graphed 3 units to the left and to units upward!

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Answer:

hkkr

need school the long said

Answer:

That would indicate 20.0 ml

id appreciate a rating thanks XP

Answer:

[tex]Slanted\ Roof = 20.77\ ft[/tex]

Step-by-step explanation:

The question has missing attachment (See attachment 1 for complete figure)

Given

Width, W = 20ft

Let the taller height be represented with H and the shorter height with h

H = 10ft

h = 8ft

Overhang = 8 inch

Required

Determine the length of the slanted roof

FIrst, we have to determine the distance between the tip of the roof and the shorter height;

Represent this with

This is calculated by

[tex]D = H - h[/tex]

Substitute 10 for H and 8 for h

[tex]D = 10 - 8[/tex]

[tex]D = 2ft[/tex]

Next, is to calculate the length of the slant height before the overhang;

See Attachment 2

Distance L can be calculated using Pythagoras theorem

[tex]L^2 = 2^2 + 20^2[/tex]

[tex]L^2 = 4 + 400[/tex]

[tex]L^2 = 404[/tex]

Take Square root of both sides

[tex]\sqrt{L^2} = \sqrt{404}[/tex]

[tex]L = \sqrt{404}[/tex]

[tex]L = 20.0997512422[/tex]

[tex]L = 20.10\ ft[/tex] -------Approximated

The full length of the slanted roof is the sum of L (calculated above) and the overhang

[tex]Slanted\ Roof = L + 8\ inch[/tex]

Substitute 20.10 ft for L

[tex]Slanted\ Roof = 20.10\ ft + 8\ inch[/tex]

Convert inch to feet to get the slanted roof in feet

[tex]Slanted\ Roof = 20.1\ ft + 8/12\ ft[/tex]

[tex]Slanted\ Roof = 20.10\ ft + 0.67\ ft[/tex]

[tex]Slanted\ Roof = 20.77\ ft[/tex]

Hence, the total length of the slanted roof in feet is approximately 20.77 feet

Answer:

[tex]Mean = 53.25[/tex]

Step-by-step explanation:

Given

Low Temperature : 40−44 || 45−49 ||  50−54 || 55−59 || 60−64

Frequency: --------------- 3 -----------6----------- 1-----------3--- -----7

Required

Determine the mean

The first step is to determine the midpoints of the given temperatures

40 - 44:

[tex]Midpoint = \frac{40+44}{2}[/tex]

[tex]Midpoint = \frac{84}{2}[/tex]

[tex]Midpoint = 42[/tex]

45 - 49

[tex]Midpoint = \frac{45+49}{2}[/tex]

[tex]Midpoint = \frac{94}{2}[/tex]

[tex]Midpoint = 47[/tex]

50 - 54:

[tex]Midpoint = \frac{50+54}{2}[/tex]

[tex]Midpoint = \frac{104}{2}[/tex]

[tex]Midpoint = 52[/tex]

55- 59

[tex]Midpoint = \frac{55+59}{2}[/tex]

[tex]Midpoint = \frac{114}{2}[/tex]

[tex]Midpoint = 57[/tex]

60 - 64:

[tex]Midpoint = \frac{60+64}{2}[/tex]

[tex]Midpoint = \frac{124}{2}[/tex]

[tex]Midpoint = 62[/tex]

So, the new frequency table is as thus:

Low Temperature : 42 || 47 ||  52 || 57 || 62

Frequency: ----------- 3 --||- -6-||- 1-||- --3- ||--7

Next, is to calculate mean by

[tex]Mean = \frac{\sum fx}{\sum x}[/tex]

[tex]Mean = \frac{42 * 3 + 47 * 6 + 52 * 1 + 57 * 3 + 62 * 7}{3+6+1+3+7}[/tex]

[tex]Mean = \frac{1065}{20}[/tex]

[tex]Mean = 53.25[/tex]

The computed mean is greater than the actual mean

Answer:

Step-by-step explanation:

perimeter of triangle=sum of lengths of sides=5x-7+3x-4+2x-6=10x-17

Answer:

10x - 17

Step-by-step explanation:

To find the perimeter of a triangle, add up all three sides

( 5x-7) + ( 3x-4) + ( 2x-6)

Combine like terms

10x - 17

Answer:

Step-by-step explanation:

60°=2×30°

one angle is double the angle of the same right angled triangle.

so hypotenuse is double the smallest side.

Hypotenuse=10

smallest side=10/2=5

third side =√(10²-5²)=5√(2²-1)=5√3

(a) Let [tex]A(t)[/tex] denote the amount of sugar in the tank at time [tex]t[/tex]. The tank starts with only pure water, so [tex]\boxed{A(0)=0}[/tex].

(b) Sugar flows in at a rate of

(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min

and flows out at a rate of

(A(t)/1080 kg/L) * (7 L/min) = 7A(t)/1080 kg/min

so that the net rate of change of [tex]A(t)[/tex] is governed by the ODE,

[tex]\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}[/tex]

or

[tex]A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}[/tex]

Multiply both sides by the integrating factor [tex]e^{7t/1080}[/tex] to condense the left side into the derivative of a product:

[tex]e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}[/tex]

[tex]\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}[/tex]

Integrate both sides:

[tex]e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt[/tex]

[tex]e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C[/tex]

Solve for [tex]A(t)[/tex]:

[tex]A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}[/tex]

Given that [tex]A(0)=0[/tex], we find

[tex]0=\dfrac{378}5+C\implies C=-\dfrac{378}5[/tex]

so that the amount of sugar at any time [tex]t[/tex] is

[tex]\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}[/tex]

(c) As [tex]t\to\infty[/tex], the exponential term converges to 0 and we're left with

[tex]\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5[/tex]

or 75.6 kg of sugar.

Answer:

B. FALSE

Step-by-step explanation:

Surface area of cone = πr(r + l)

Where,

r = r

l = 3r

S.A of cone = πr(r + 3r)

= πr² + 3πr²

S.A of cone = 4πr²

Surface area of cylinder = 2πrh + 2πr² = 2πr(h + r)

Where,

r = r

h = 2r

S.A of cylinder = 2πr(2r + r)

= 4πr² + 2πr²

S.A of cylinder = 6πr²

The surface are of the cone and that of the cylinder are not the same. The answer is false.

Answer:false

Step-by-step explanation:

False