Answer:
A transitive property
Step-by-step explanation:
There isn't much to this.
This is the the transitive property.
I guess I can go through each choice and tell you what the property looks like or postulate.
A) If x=y and y=z, then x=z.
This is the exact form of your conditional.
x is AB here
y is BC here
z is DE here
B) Segment Addition Postulate
If A,B, and C are collinear with A and B as endpoints, then AB=AC+CB.
Your conditional said nothing about segment addition (no plus sign).
C) Distributive property is a(b+c)=ab+ac.
This can't be applied to any part of this. There is not even any parenthesis.
D) The symmetric property says if a=b then b=a.
There is two parts to our hypothesis where this is only part to the symmetric property for the hypothesis .
The statement 'If AB = BC and BC = DE, then AB = DE' is justified by the Transitive Property of Equality, stating that, if two quantities both equal a third, they are equal to each other.
Explanation:The justification for the statement 'If AB = BC and BC = DE, then AB = DE' is the Transitive Property of Equality. This property states that if two quantities are both equal to a third quantity, then they are equal to each other. In this case, AB and DE are both equal to BC, therefore, according to the transitive property, AB must be equal to DE.
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The random variables X and Y have the joint PMF pX,Y(x,y)={c⋅(x+y)2,0,if x∈{1,2,4} and y∈{1,3},otherwise. All answers in this problem should be numerical. Find the value of the constant c .
In order for [tex]p_{X,Y}(x,y)[/tex] to be a valid PMF, its integral over the distribution's support must be equal to 1.
[tex]p_{X,Y}(x,y)=\begin{cases}\dfrac{c(x+y)}2&\text{for }x\in\{1,2,4\}\text{ and }y\in\{1,3\}\\\\0&\text{otherwise}\end{cases}[/tex]
There are 3*2 = 6 possible outcomes for this distribution, so that
[tex]\displaystyle\sum_{x,y}p_{X,Y}(x,y)=\sum_{x\in\{1,2,4\}}\sum_{y\in\{1,3\}}\frac{c(x+y)}2=1[/tex]
[tex]1=\displaystyle\frac c2\sum_{x\in\{1,2,4\}}((x+1)+(x+3))=\frac c2\sum_{x\in\{1,2,4\}(2x+4)[/tex]
[tex]1=\displaystyle\frac c2((2+4)+(4+4)+(8+4))[/tex]
[tex]1=13c\implies\boxed{c=\dfrac1{13}}[/tex]
1) For the following problem: Let f={(-2,3),(-1,1),(0,0),(1,-1), (2,-3)} and let g-{-3,1),(-1,-2), (0, 2),(2, 2),(3,1)}.Find the following a) f(1) and g-1) b) (gof (1) c) (gofof)(-1) d) (fog)(3)
Answer:
1. f(1)=-1 and g(-1)=-2.
2. (gof)(1)=-2.
3. (gofof)(-1)=-2
4. (fog)(3)=-1
Step-by-step explanation:
The given functions are defined as
f={(-2,3),(-1,1),(0,0),(1,-1), (2,-3)}
g={(-3,1),(-1,-2), (0, 2),(2, 2),(3,1)}
1.
The value of function f at x=1 is -1, So, f(1)=-1.
The value of function g at x=-1 is -2, So, g(-1)=-2.
Therefore the value of f(1) is -1 and the value of g(-1) is -2.
2.
[tex](g\circ f)(1)=g(f(1))[/tex] [tex][\because (g\circ f)(x)=g(f(x))][/tex]
[tex](g\circ f)(1)=g(-1)[/tex] [tex][\because f(1)=-1][/tex]
[tex](g\circ f)(1)=-2[/tex] [tex][\because g(-1)=-2][/tex]
Therefore the value of (gof)(1) is -2.
3.
[tex](g\circ f\circ f)(-1)=(g\circ f)(f(-1))[/tex] [tex][\because (g\circ f)(x)=g(f(x))][/tex]
[tex](g\circ f\circ f)(-1)=(g\circ f)(1)[/tex] [tex][\because f(-1)=1][/tex]
[tex](g\circ f\circ f)(-1)=-2[/tex] [tex][\because \text{From part 2}, (g\circ f)(1)=-2][/tex]
Therefore the value of (gofof)(1) is -2.
4.
[tex](f\circ g)(3)=f(g(3))[/tex] [tex][\because (f\circ g)(x)=f(g(x))][/tex]
[tex](f\circ g)(3)=f(1)[/tex] [tex][\because g(3)=1][/tex]
[tex](f\circ g)(3)=-1[/tex] [tex][\because f(1)=-1][/tex]
Therefore the value of (fog)(3) is -1.
22 Which of the following polynomials has a factor of x-1? A) p(x)=x^3 +x^2 -2x+1 B) q(x)=2x^3-x^2 +x-1 (Crx)= 3x^3-x-2 D) S(x)=-3x^3+ 3x +1
Answer:
The correct option is C.
Step-by-step explanation:
If (x-c) is a factor of a polynomial f(x), then f(c)=0.
It is given that (x-1) is a factor of the polynomial. It means the value of the function at x=1 is 0.
In option A,
The given function is
[tex]p(x)=x^3+x^2-2x+1[/tex]
Substitute x=1 in the given function.
[tex]p(1)=(1)^3+(1)^2-2(1)+1=1+1-2+1=1[/tex]
Since p(1)≠0, therefore option A is incorrect.
In option B,
The given function is
[tex]q(x)=2x^3-x^2+x-1[/tex]
Substitute x=1 in the given function.
[tex]q(1)=2(1)^3-(1)^2+(1)-1=2-1+1-1=1[/tex]
Since q(1)≠0, therefore option B is incorrect.
In option C,
The given function is
[tex]r(x)=3x^3-x-2[/tex]
Substitute x=1 in the given function.
[tex]r(1)=3(1)^3-(1)-2=3-1-2=0[/tex]
Since r(1)=0, therefore option C is correct.
In option D,
The given function is
[tex]s(x)=-3x^3+3x+1[/tex]
Substitute x=1 in the given function.
[tex]s(1)=-3(1)^3+3(1)+1=-3+3+1=1[/tex]
Since s(1)≠0, therefore option D is incorrect.
Express the confidence interval left parenthesis 0.023 comma 0.085 right parenthesis(0.023,0.085) in the form of ModifyingAbove p with caret minus Upper E less than p less than ModifyingAbove p with caret plus Upper Ep−E
Final answer:
To express the confidence interval (0.023, 0.085) in the requested form, calculate the midpoint (p') as 0.054 and the margin of error (EBP) as 0.031. The confidence interval will thus be expressed as 0.054 - 0.031 < p < 0.054 + 0.031.
Explanation:
The question asks to express the confidence interval (0.023, 0.085) in the form of ModifyingAbove p with caret minus Upper E less than p less than ModifyingAbove p with caret plus Upper E. To do this, we first need to calculate the midpoint (p') and the margin of error (EBP) of the given interval.
The midpoint (p') is the average of the lower and upper bounds of the interval, calculated as (0.023 + 0.085) / 2 = 0.054. The margin of error (EBP) is the difference between p' and either boundary of the interval, calculated as 0.085 - 0.054 = 0.031 or 0.054 - 0.023 = 0.031. Therefore, the interval in the requested format is 0.054 - 0.031 < p < 0.054 + 0.031.
The confidence interval (0.023, 0.085) can be expressed in the form [tex]\( 0.054 - 0.031 < p < 0.054 + 0.031 \)[/tex].
To express the confidence interval (0.023, 0.085) in the form [tex]\( \hat{p} - E < p < \hat{p} + E \),[/tex] where [tex]\( \hat{p} \)[/tex] is the point estimate and ( E ) is the margin of error, we need to find the point estimate and the margin of error.
Step 1: Find the point estimate [tex]\( \hat{p} \)[/tex].
The point estimate is the midpoint of the confidence interval. We can calculate it using the formula:
[tex]\[ \hat{p} = \frac{\text{Lower bound} + \text{Upper bound}}{2} \][/tex]
Substitute the values:
[tex]\[ \hat{p} = \frac{0.023 + 0.085}{2} \][/tex]
[tex]\[ \hat{p} = \frac{0.108}{2} \][/tex]
[tex]\[ \hat{p} = 0.054 \][/tex]
So, the point estimate [tex]\( \hat{p} \)[/tex] is 0.054.
Step 2: Find the margin of error \( E \).
The margin of error is half the width of the confidence interval. We can calculate it using the formula:
[tex]\[ E = \frac{\text{Upper bound} - \text{Lower bound}}{2} \][/tex]
Substitute the values:
[tex]\[ E = \frac{0.085 - 0.023}{2} \][/tex]
[tex]\[ E = \frac{0.062}{2} \][/tex]
[tex]\[ E = 0.031 \][/tex]
So, the margin of error ( E ) is 0.031.
Step 3: Express the confidence interval in the required form.
Now, we can write the confidence interval in the form [tex]\( \hat{p} - E < p < \hat{p} + E \):[/tex]
[tex]\[ 0.054 - 0.031 < p < 0.054 + 0.031 \][/tex]
[tex]\[ 0.023 < p < 0.085 \][/tex]
Therefore, the confidence interval (0.023, 0.085) can be expressed in the form [tex]\( 0.054 - 0.031 < p < 0.054 + 0.031 \)[/tex].
A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.
Answer:
Step-by-step explanation:
With the information you have provided, the volume can only be stated as a third degree polynomial. We do not know the measurements of the square cut out of each corner, we can only call it "x".
If each side of the square measures 3 feet wide and we cut out 2 squares from a side, the length of that side is 3 - 2x. Volume is length times width times height. The length and the width will both be 3 - 2x, and the height is the measurement of x. But since we don't know it, the height is just x. Multiplying the length times the width times the height looks like this:
(3 - 2x)(3 - 2x)(x)
When you FOIL all this together you get a third degree polynomial
[tex]4x^3-12x^2+9x[/tex]
If you know the measurement of the squares cut out, plug that value in for x and get the volume in a number.
Let A, B, and C be arbitrary sets within a universal set, U. For each of the following statements, either prove that the statement is always true or show a counterexample to prove it is not always true. When giving a counterexample, you should define the three sets explicitly and say what the left-hand and right-hand sides of the equation are for those sets, to make it clear that they are not equal.
(A \ B) × C = (A × C) \ (B × C)
Answer with Step-by-step explanation:
Let A, B and C are arbitrary sets within a universal set U.
We have to prove that [tex]( A/B)\times C=(A\times C)/(B\times C)[/tex] is always true.
Let [tex](x,y)\in (A/B)\times C[/tex]
Then [tex] x\in(A/B) [/tex] and [tex] y\in C[/tex]
Therefore, [tex] x\in A[/tex] and [tex] x\notin B[/tex]
Then, (x,y) belongs to [tex] A\times C[/tex]
and (x,y) does not belongs to [tex] B\times C[/tex]
Hence,[tex] (x,y)\in(A\times C)/(B\times C)[/tex]
Conversely ,Let (x ,y)belongs to [tex] (A\times C)/(B\times C)[/tex]
Then [tex] (x,y)\in (A\times C)[/tex] and [tex] (x,y)\notin (B\times C)[/tex]
Therefore,[tex] x\in A,y\in C[/tex] and [tex] x\notin B,y\in C[/tex]
[tex] x\in(A/B)[/tex] and [tex]y\in C[/tex]
Hence, [tex] (x,y)\in(A/B)\times C[/tex]
Therefore,[tex] (A/B)\times C=(A\times C)/(B\times C)[/tex] is always true.
Hence, proved.
Consider the experiment of rolling a dice. Find the probability of getting an even number 3) and a number that is multiple of 3.
The required probability of rolling a dice and getting an even number and a number that is a multiple of 3 is 1/6.
To find the probability of rolling a dice and getting an even number (3, 4, or 6) and a number that is a multiple of 3 (3 or 6), we need to consider the outcomes that satisfy both conditions.
There are three even numbers on a standard six-sided dice: 2, 4, and 6. Out of these three even numbers, two of them are multiples of 3 (3 and 6).
The probability of rolling an even number is 3/6 (3 even numbers out of 6 possible outcomes) or simply 1/2.
The probability of rolling a number that is a multiple of 3 is 2/6 (2 multiples of 3 out of 6 possible outcomes) or 1/3.
Now, to find the probability of both events occurring together (rolling an even number and a number that is a multiple of 3), we multiply the individual probabilities:
Probability of getting an even number AND a number that is a multiple of 3 = (1/2) * (1/3) = 1/6.
So, the probability of rolling a dice and getting an even number and a number that is a multiple of 3 is 1/6.
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The probability of rolling an even number is 1/2, while the probability of rolling a multiple of 3 is 1/3. However, the probability of rolling a number that is both an even number and a multiple of 3 (which is only the number 6) is 1/6 or approximately 16.67%.
Explanation:The experiment in question involves rolling a single die and finding the probability of getting an even number as well as a number that is a multiple of 3. A standard die has six faces, numbered from 1 to 6. An even number on a die could be 2, 4, or 6, while a multiple of 3 could be 3 or 6.
To find the probability of getting an even number, we count the favorable outcomes for an even number, which are three (2, 4, 6), and divide by the total number of outcomes possible on a die, which is six. This gives us a probability of ½ or 50% for rolling an even number.
To find the number that is a multiple of 3, we have two outcomes (3 and 6). Again, dividing by the total number of outcomes we get a probability of ⅓ or approximately 33.33%.
But in this case, we are interested in the occurrence of both events together, which means we are looking for the probability of getting a number that is both an even number and a multiple of 3. The only number on a die that is both even and a multiple of 3 is the number 6.
Since there is only one favorable outcome (6) and six possible outcomes in total, the probability of rolling a number that is both even and a multiple of 3 is ⅖ or approximately 16.67%.
The Earth is 4.6 billion years old, but human civilization (the building of cities) began only about 10,000 years ago. If you represent the history of Earth with a line one mile long (63,360 inches), how long must the line be to represent the history of human civilization?
Answer:
0.13774 inches ( approx ).
Step-by-step explanation:
Given,
The age of earth = 4.6 billion years = 4600000000 years,
The represented age of earth = 63,360 inches,
That is, scale factor in representation of age
[tex]=\frac{\text{Represented age}}{\text{Actual age}}[/tex]
[tex]=\frac{ 63,360}{4600000000}[/tex]
Now, the age of human civilization = 10000 years,
Thus, the represented age of human civilization = actual age × scale factor
[tex]=10000\times \frac{ 63,360}{4600000000}[/tex]
[tex]=\frac{633600000}{4600000000}[/tex]
[tex]=0.137739130435[/tex]
[tex]\approx 0.13774\text{ inches}[/tex]
Which established actor stared in The Godfather as Don Vito? a. Marlon Brando b. John Wayne c. Henry Fonda d. None of the above
Answer:
a) Marlon Brando
Step-by-step explanation:
Don Vito Corleone was played by Marlon Brando in the film Godfather released in the year 1972 directed by Francis Ford Coppola based on the 1969 book by Mario Puzo. Paramount studios was against the casting of Marlon Brando but the director insisted on giving him a screen test. The screen test convinced the studio to cast him in the role.
Marlon Brando starred as Don Vito Corleone in The Godfather, earning an Academy Award for his performance.
Explanation:The established actor who starred in The Godfather as Don Vito Corleone is Marlon Brando. Brando's performance in this iconic film is widely recognized as one of his greatest roles. Released in 1972, The Godfather is a crime film directed by Francis Ford Coppola and based on the novel of the same name by Mario Puzo. Marlon Brando's portrayal of the mafia patriarch earned him an Academy Award for Best Actor, which further solidified his legacy as one of Hollywood's legendary actors. The other actors listed, John Wayne and Henry Fonda, were not part of the cast of The Godfather.
Comparing yourself to a triangle with 3 possible occupation/career choices, could this be compared to "congruent" rather than "equal"? Explain your thinking...
Answer:
Yes
Step-by-step explanation:
If I am understanding the question correctly, then I would have to say it is compared to "congruent".
Congruent: two triangles are compared and have the same sides and angles
Equilateral: all 3 sides of the triangle are the same.
Basing my opinion on the definitions above. I believe this situation could be compared to "congruent" because as mentioned in the question we are comparing ourselves to a triangle that has 3 DIFFERENT career choices.
If this was compared to "Equal" there would be 1 career choice repeated 3 times.
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). SOLUTION (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) We start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. If we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .
Answer:
See explanation.
Step-by-step explanation:
The hyperbolic sine and cosine functions are defined as follows:
[tex] \sinh(x) = \frac{ {e}^{x} - {e}^{ - x} }{2} [/tex]
[tex]\cosh(x) = \frac{ {e}^{x} + {e}^{ - x} }{2} [/tex]
We want to show that:
[tex]\cosh^{2} (x) - \sinh^{2} (x) = 1[/tex]
We use the definition by substituting the expressions into the left hand side and simplify to obtain the RHS.
[tex] \cosh^{2} (x) - \sinh^{2} (x) = {( \frac{ {e}^{x} + {e}^{ - x} }{2} )}^{2} + {( \frac{ {e}^{x} - {e}^{ - x} }{2} )}^{2} [/tex]
[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} +2 {e}^{x} {e}^{ - x} + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 {e}^{x} {e}^{ - x} + {e}^{ - 2x} }{4} [/tex]
[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} +2 + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 + {e}^{ - 2x} }{4} [/tex]
[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} + {e}^{ - 2x} - {e}^{2x} + {e}^{ - 2x }+ 2 +2 }{4} [/tex]
[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ 4 }{4} [/tex]
[tex]\cosh^{2} (x) - \sinh^{2} (x)=1[/tex]
b)
If we start with the identity in a) and we divide both sides by cosh²x we get:
[tex] \frac{\cosh^{2} (x) }{\cosh^{2} (x) } -\frac{\sinh^{2} (x) }{\cosh^{2} (x) } =\frac{1}{\cosh^{2} (x) } [/tex]
This simplifies to:
[tex]1 - \tanh ^{2} (x) = \sec \: h ^{2} (x) [/tex]
Insert parentheses to make the following sentence true. 4 • 5 + 16 - 2 = 82
Answer:
[tex]4 \cdot(5 + 16) - 2 = 82[/tex]
Step-by-step explanation:
The given equation is
[tex]4 \cdot5 + 16 - 2 = 82[/tex]
We want to insert parenthesis, so that this equation will be true.
We insert parenthesis to obtain:
[tex]4 \cdot(5 + 16) - 2 [/tex]
We now use PEDMAS to simplify.
[tex]4 \cdot(21) - 2 =82[/tex]
[tex]84 - 2 = 82[/tex]
[tex]82 = 82[/tex]
Find the unpaid balance on the debt. (Round your answer to the nearest cent.)
After 6 years of monthly payments on $190,000 at 3% for 25 years.
Answer:
unpaid balance is $156439.86
Step-by-step explanation:
present value= $190000
rate = 3% monthly = 3/12 = 0.25 % = 0.0025
time = 25 year = 25 × 12 300
to find out
the unpaid balance
solution
first we calculate monthly payment
payment = principal ( 1- (1+r)^(-t) /r )
and we put all value here
190000 = principal ( 1- (1+0.0025)^(-300) / 0.0025 )
principal = 190000 / ( 1- (1+0.0025)^(-300) / 0.0025 )
principal = 901
so after 6 year unpaid is
present value = principal ( 1- (1+r)^(-t) /r )
put here principal 901 and rate 0.0025 and time 300 - (year )
time = 300 - (6*12) = 300 - 96 = 228
payment = 901 ( 1- (1+0.0025)^(-228) / 0.0025 )
payment = 901 ( 713.63)
so unpaid balance is $156439.86
When Bill makes a sandwich, he may choose from among 3 kinds of rolls, 4 varieties of meat, and 2 types of sliced cheese. If he chooses one roll, one meat, and one type of cheese, how many different kinds of sandwiches can he make?
Answer: Bill can make 24 different kinds of sandwiches
Step-by-step explanation:
Given : The number of kinds of rolls = 3
The number of varieties of meat = 4
The number of types of sliced cheese = 2
If he chooses one roll, one meat, and one type of cheese, then the number of different kinds of sandwiches he can make is given by :-
[tex]3\times2\times4=24[/tex]
Hence, Bill can make 24 different kinds of sandwiches.
The Vertex of a parabola is at (8-1), and it's why intercept is negative 17, which function represents the parabola
Answer:
The function is equal to [tex]y=-(1/4)(x-8)^{2}-1[/tex]
Step-by-step explanation:
we know that
The equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
a is a coefficient
(h,k) is the vertex
In this problem we have
(h,k)=(8,-1)
substitute
[tex]y=a(x-8)^{2}-1[/tex]
Find the value of a
Remember that we have the y-intercept
The y-intercept is the point (0,-17)
substitute
x=0,y=-17
[tex]-17=a(0-8)^{2}-1[/tex]
[tex]-17=64a-1[/tex]
[tex]64a=-17+1[/tex]
[tex]64a=-16[/tex]
[tex]a=-16/64[/tex]
[tex]a=-1/4[/tex]
therefore
The function is equal to
[tex]y=-(1/4)(x-8)^{2}-1[/tex]
see the attached figure to better understand the problem
The measurement of the height of 600 students of a college is normally distributed with a mean of
175 centimeters and a standard deviation of 5 centimeters.
What percent of students are between 180 centimeters and 185 centimeters in height?
12.5
13.5
34
68
Answer: Second Option
[tex]P(180<X <185)=13.5\%[/tex]
We know that the mean is:
[tex]\mu=175[/tex]
and the standard deviation is:
[tex]\sigma=5[/tex]
We are looking at the percentage of students between 180 centimeters and 185 centimeters in height.
This is:
[tex]P(180<X <185)[/tex]
We calculate the Z-score using the formula:
[tex]Z=\frac{X-\mu}{\sigma}[/tex]
For [tex]X=180[/tex]
[tex]Z_{180}=\frac{180-175}{5}[/tex]
[tex]Z_{180}=1[/tex]
For [tex]X=185[/tex]
[tex]Z_{185}=\frac{185-175}{5}[/tex]
[tex]Z_{185}=2[/tex]
Then we look at the normal table
[tex]P(1<Z<2)[/tex]
[tex]P(1<Z<2)=P(Z<2)-P(Z<1)[/tex]
[tex]P(1<Z<2)=0.9772-0.8413[/tex]
[tex]P(1<Z<2)=0.135[/tex]
[tex]P(180<X <185)=13.5\%[/tex]
Note: You can get the same conclusion using the empirical rule
Look at the attached image for [tex]\mu+ 1\sigma <\mu <\mu + 2\sigma[/tex]
Find an equation for the tangent to the curve at P and the horizontal tangent to the curve at Q. y = 5 + cot x - 2 csc x 0 1 2 3 0 2 4 x y Upper QUpper P left parenthesis StartFraction pi Over 2 EndFraction comma 3 right parenthesis
Answer with explanation:
The given function in x and y is,
y= 5 +cot x-2 Cosec x
To find the equation of tangent, we will differentiate the function with respect to x
[tex]y'= -\csc^2 x+2 \csc x\times \cot x[/tex]
Slope of tangent at (π/2,3)
[tex]y'_{(\frac{\pi}{2},3)}= -\csc^2\frac{\pi}{2} +2 \csc \frac{\pi}{2}\times \cot \frac{\pi}{2}\\\\=-1+2\times 1 \times 0\\\\= -1[/tex]
Equation of tangent passing through (π/2,3) can be obtained by
[tex]\rightarrow \frac{y-y_{1}}{x-x_{1}}=m(\text{Slope})\\\\ \rightarrow \frac{y-3}{x-\frac{\pi}{2}}=-1\\\\\rightarrow 3-y=x-\frac{\pi}{2}\\\\\rightarrow x+y-3-\frac{\pi}{2}=0[/tex]
⇒There will be no Horizontal tangent from the point (π/2,3).
Final answer:
The equation of the tangent to the curve y = 5 + cot x - 2 csc x at a given point can be found using the derivative, but for point P (π/2, 3), the slope is undefined because cot x and csc x are undefined at π/2. A horizontal tangent occurs where the derivative is zero, and point Q must be determined by solving y' = 0.
Explanation:
To find the equation of the tangent at point P, we first need to compute the derivative of the given function y = 5 + cot x - 2 csc x to find the slope of the tangent at any point on the curve. For the given curve y = 5 + cot x - 2 csc x, the derivative y' will give us the slope of the tangent at any point x. The slope of the tangent at point P, where P is (π/2, 3), can be found by evaluating the derivative at x = π/2.
However, since cot x and csc x are undefined at x = π/2, this curve will not have a well-defined slope and therefore not a conventional tangent line at x= π/2. On the other hand, a horizontal tangent will occur at points where the derivative y' is equal to zero. To find point Q, we must solve y' = 0 for x, and then use the x-value to find the corresponding y-coordinate on the curve.
All but 1/13 of the students enrolled at a particular elementary school participated in "Family Fun Night" activities. If a total of 396 students were involved in the evening's activities, how many students attend the school?
Answer: 429
Step-by-step explanation:
If all but 1/13 of the students were involved in the evening's activities, that means that 12/13 of the students were involved.
To calculate the total number of students, first you do a simple rule of three to know the number of students that weren't involved in the activities:
If 12/13 of the students represent 396 students, then how many students are 1/13 of the students?
[tex]\frac{\frac{1}{13} *396}{\frac{12}{13} } =\frac{396}{12} =33[/tex]
Then, you add this number to the number of students that were involved in the activities:
[tex]396+33=429[/tex]
So, 429 is the number of students that attend the school.
You can also calculate it directly by doing the following rule of three:
If 12/13 of students represent 396 students, then how many students are 13/13 of the students?
[tex]\frac{\frac{13}{13} *396}{\frac{12}{13} } =\frac{396*13}{12} =\frac{5148}{12} =429[/tex]
Answer:429
Step-by-step explanation: let the total number of students in the elementary school be X
X - X/13 = 396
12X/13=396
X=396/*13/12
X=5148/12
X=429
Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. (If an answer does not exist, enter DNE.) x = t2 − t + 9, y = t3 − 3t
Answer:
The horizontal tangents occur at: (9,-2) and (11,2)
The vertical tangent occurs at (8.75,-1.375)
See attachment
Step-by-step explanation:
The given parametric equations are:
[tex]x=t^2-t+9[/tex] and [tex]y=t^3-3t[/tex]
The slope function is given by:
[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]
[tex]\frac{dy}{dx}=\frac{3t^2-3}{2t-1}[/tex]
The tangent is vertical when [tex]\frac{dx}{dt}=0[/tex]
[tex]\implies 2t-1=0[/tex]
[tex]t=\frac{1}{2}[/tex]
When [tex]t=\frac{1}{2}[/tex], [tex]x=(\frac{1}{2})^2-\frac{1}{2}+9=8.75[/tex], [tex]y=0.5^3-3(0.5)=-1.375[/tex]
The vertical tangent occurs at (8.75,-1.375)
The tangent is horizontal when [tex]\frac{dy}{dt}=0[/tex]
[tex]3t^2-3=0[/tex]
[tex]\implies t=\pm1[/tex]
When t=1, [tex]x=(1)^2-1+9=9[/tex], [tex]y=1^3-3(1)=-2[/tex]
When t=-1, [tex]x=(-1)^2+1+9=11[/tex], [tex]y=(-1)^3-3(-1)=2[/tex]
The horizontal tangents occur at: (9,-2) and (11,2)
(1.)Find the slope of the line that passes through the given pair of points. (If an answer is undefined, enter UNDEFINED.) (?a + 3, b ? 3) and (a + 3, ?b) *******(2.)If the line passing through the points (a, 1) and (6, 5) is parallel to the line passing through the points (2, 7) and (a + 2, 1), what is the value of a?
Answer:
1. The slope of the line is [tex]m=\frac{-2b+3}{2a}[/tex].
2. The value of a is 18.
Step-by-step explanation:
If a line passes through two points, then the slope of the line is
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
(1)
It is given that the line passes through the points (-a + 3, b - 3) and (a + 3, -b). So, the slope of the line is
[tex]m=\frac{-b-(b-3)}{a+3-(-a+3)}[/tex]
[tex]m=\frac{-b-b+3}{a+3+a-3)}[/tex]
[tex]m=\frac{-2b+3}{2a}[/tex]
The slope of the line is [tex]m=\frac{-2b+3}{2a}[/tex].
(2)
If the line passing through the points (a, 1) and (6, 5), then the slope of the line is
[tex]m_1=\frac{5-1}{6-a}=\frac{4}{6-a}[/tex]
If the line passing through the points (2, 7) and (a + 2, 1), then the slope of the line is
[tex]m_2=\frac{1-7}{a+2-2}=\frac{-6}{a}[/tex]
The slopes of two parallel lines are same.
[tex]m_1=m_2[/tex]
[tex]\frac{4}{6-a}=\frac{-6}{a}[/tex]
On cross multiplication we get
[tex]4a=-6(6-a)[/tex]
[tex]4a=-36+6a[/tex]
[tex]4a-6a=-36[/tex]
[tex]-2a=-36[/tex]
Divide both sides by -2.
[tex]a=18[/tex]
Therefore the value of a is 18.
Pulam, Inc. prepared the following master budget items for July:
Production and sales 36,000 units
Variable manufacturing costs:
Direct materials $ 36,000
Direct labor $ 72,000
Variable manufacturing overhead $ 72,000
Fixed manufacturing costs $ 180,000
Total manufacturing costs $ 360,000
During July, Pulam actually sold 42,000 units. Prepare a flexible budget for Pulam based on actual sales.(Do not round your intermediate calculations).
Production and sales units
Variable manufacturing costs:
Direct materials $
Direct labor $
Variable manufacturing overhead $
Fixed manufacturing costs $
Total manufacturing costs $
Answer:
During July, Pulam actually sold 42,000 units.
Variable Manufacturing Costs for 42,000 units are-
1. Direct Material : [tex](36000/36000)\times42000 =42000[/tex] dollars
2. Direct Labor: [tex](72000/36000)\times42000 =84000[/tex] dollars
3. Variable Manufacturing Overhead : [tex](72000/36000)\times42000 =84000[/tex] dollars
4. Fixed manufacturing costs : $ 180000
5. Total manufacturing costs : [tex]42000+84000+84000+180000=390000[/tex] dollars
Students are collecting canned goods for a local food pantry. Last year 12 students were able to distribute 1000 flyers in nine hours. This year there are 15 students handing out the same number of flyers. How long should it take them?
Answer:
7.2 h
Step-by-step explanation:
The time required is inversely proportional to the number of students.
t = k/n or
tn = k
Let t1 and n1 represent last year
and t2 and n2 represent this year. Then
t1n1 = t2n2
Data:
t1 = 9 h; n1 = 12 students
t2 = ?; n2 = 15 students
Calculation:
9 × 12 = t2 × 15
108 = 15t2
t2 = 108/15 = 7.2 h
It will take the students 7.2 h to distribute the flyers.
An investor can invest money with a particular bank and earn a stated interest rate of 15.40%; however, interest will be compounded quarterly. What are the nominal, periodic, and effective interest rates for this investment opportunity?
Answer:
Nominal rate = 15.40%, Periodic rate = 3.85% and effective interest rate = 16.31%
Step-by-step explanation:
An investor can invest money and earn a stated interest rate = 15.40%
We have to find the nominal, periodic and effective interest rates for this investment.
Nominal rate = Stated interest rate is also called annual percentage rate or nominal interest rate. Therefore nominal interest rate = 15.40%.
Periodic interest rate = Periodic interest rate is the annual interest rate divided by number of compounding periods.
Periodic rate = ( [tex]\frac{15.40}{4}[/tex] ) = 3.85%
Effective interest rate = We have to calculate Effective interest rate (EAR) by the formula = [tex]EAR=[(\frac{1+i}{n})^{n}-1][/tex]
where i = rate of interest and n = number of compounding periods
[tex]EAR=[(1+\frac{0.154}{4})^4]-1[/tex]
[tex]EAR=[(1+.0385)^{4}]-1[/tex]
[tex]EAR=[(1.0385)^{4}]-1[/tex]
EAR = 1.1631 - 1
EAR = 0.1631 = 16.31%
Therefore, Nominal rate = 15.40%, Periodic rate = 3.85% and effective interest rate = 16.31%
A company that makes thing-a-ma-bobs has a start up cost of $18263. It costs the company $1.14 to make each thing-a-ma-bob and the company charges $4.06 for each thing-a-ma-bob. Let x represent the number of thing-a-ma-bobs made. Write the cost function for this company. Write the revenue function for this company. Write the profit function for this company.
Answer:
c(x) = 18263 +1.14xr(x) = 4.06xp(x) = 2.92x -18263Step-by-step explanation:
cost function
Cost for x tambs is the sum of start-up cost and per-unit cost multiplied by the number of units.
c(x) = 18263 +1.14x
revenue function
Revenue from the sale of x tambs is the product of their price and the number sold.
r(x) = 4.06x
profit function
Profit from the sale of x tambs is the difference between the revenue and cost:
p(x) = r(x) -c(x)
p(x) = 4.06x -(18263 +1.14x)
p(x) = 2.92x -18263
The cost function, C(x), for the company is [tex]\( C(x) = 1.14x + 18263 \)[/tex] .The revenue function, R(x), for the company is [tex]\( R(x) = 4.06x[/tex] .The profit function, P(x), for the company is [tex]( P(x) = R(x) - C(x) = 4.06x - (1.14x + 18263) \)[/tex] .
To determine the profit function for the company, we need to calculate the total cost and total revenue for making and selling 'x' thing-a-ma-bobs, respectively, and then subtract the total cost from the total revenue.1. The cost function, C(x), is the sum of the variable cost (cost per unit times the number of units) and the fixed cost (start-up cost). Since it costs $1.14 to make each thing-a-ma-bob and the start-up cost is $18263, the cost function is:[tex]\[ C(x) = (\text{Cost per unit}) \times x + \text{Fixed cost} \]\[ C(x) = 1.14x + 182632[/tex]
. The revenue function, R(x), is the amount of money the company earns from selling 'x' thing-a-ma-bobs at $4.06 each. Therefore, the revenue function is[tex]:\[ R(x) = (\text{Selling price per unit}) \times x \][ R(x) = 4.06x3.[/tex]The profit function, P(x), is the revenue minus the cost. To find the profit for 'x' thing-a-ma-bobs, we subtract the cost function from the revenue function:[tex]\[ P(x) = R(x) - C(x) \]\[ P(x) = 4.06x - (1.14x + 18263) \]\[ P(x) = 4.06x - 1.14x - 18263 \]\[ P(x) = 2.92x - 18263 \]So, the profit function for the company is \( P(x) = 2.92x - 18263 \),[/tex]
where 'x' represents the number of thing-a-ma-bobs made and sold.
There are 40 students in a course: 15 freshmen, 8 sophomores, 12 juniors, and 5 seniors. One student is selected at random to attend a workshop and report back to the class. What is the probability that the chosen student is a junior
Answer:
The probability that the chosen student is a junior is [tex]\frac{3}{10}[/tex]
Step-by-step explanation:
Given,
The total number of students = 40,
In which, freshmen = 15,
Sophomores = 8,
Juniors = 12,
And, seniors = 5,
Thus, the probability that junior student when a student is chosen
[tex]=\frac{\text{Total possible arrangement of 1 junior student }}{\text{Total possible arrangement of 1 student}}[/tex]
[tex]=\frac{^{12}C_1}{^{40}C_1}[/tex]
[tex]=\frac{12}{40}[/tex]
[tex]=\frac{3}{10}[/tex]
The probability of selecting a junior student is 0.3 or 30%.
Explanation:To find the probability of selecting a junior student, we need to divide the number of junior students by the total number of students. There are 12 junior students out of a total of 40 students, so the probability can be calculated as:
Probability = Number of junior students / Total number of students
Probability = 12 / 40
Probability = 0.3
Therefore, the probability of selecting a junior student is 0.3 or 30%.
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A rumor spreads among a group of 400 people. The number of people, N(t), who have heard the rumor by the time t in hours since the rumor started to spread can be approximated by a function of the form
N(t) = 400 / (1+399e^(-0.4t))
Approximately how long will it take until half the people have heard the rumor?
Answer:
about 15 hours
Step-by-step explanation:
You want to find t such that N(t)=200. Fill in the equation with that information and solve for t.
200 = 400/(1 +399e^(-0.4t))
1 +399e^(-0.4t) = 400/200 = 2 . . . . . multiply by (1+399e^(-0.4t))/200
399e^(-0.4t) = 1 . . . . . . . . . . . . . . . . . . subtract 1
e^(-0.4t) = 1/399 . . . . . . . . . . . . . . . . . .divide by 399
-0.4t = ln(1/399) . . . . . . . take the natural log
t = ln(399)/0.4 ≈ 14.972 . . . . . . . divide by -0.4, simplify
Rounded to tenths, it will take 15.0 hours for half the people to have heard the rumor.
The time it will take for half the group, or 200 people, to have heard the rumor based on the given exponential decay function is approximately 1.72 hours.
Explanation:The problem presented is a mathematical question involving a model of exponential decay. In this context, we want to know when half of the people, or 200 people, will have heard the rumor. To find out, we substitute N(t) with 200 in the equation given:
200 = 400 / (1+399e^(-0.4t))
After simplifying and solving for t, t comes out to be approximately 1.72 hours. So, it takes approximately 1.72 hours for half the people to have heard the rumor.
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Solve this differential Equation by using power series
y''-x^2y=o
We're looking for a solution
[tex]y=\displaystyle\sum_{n=0}^\infty a_nx^n[/tex]
which has second derivative
[tex]y''=\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}=\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n[/tex]
Substituting these into the ODE gives
[tex]\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^\infty a_nx^{n+2}=0[/tex]
[tex]\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=2}^\infty a_{n-2}x^n=0[/tex]
[tex]\displaystyle2a_2+6a_3x+\sum_{n=2}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=2}^\infty a_{n-2}x^n=0[/tex]
[tex]\displaystyle2a_2+6a_3x+\sum_{n=2}^\infty\bigg((n+2)(n+1)a_{n+2}-a_{n-2}\bigg)x^n=0[/tex]
Right away we see [tex]a_2=a_3=0[/tex], and the coefficients are given according to the recurrence
[tex]\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_2=0\\a_3=0\\n(n-1)a_n=a_{n-4}&\text{for }n\ge4\end{cases}[/tex]
There's a dependency between terms in the sequence that are 4 indices apart, so we consider 4 different cases.
If [tex]n=4k[/tex], where [tex]k\ge0[/tex] is an integer, then[tex]k=0\implies n=0\implies a_0=a_0[/tex]
[tex]k=1\implies n=4\implies a_4=\dfrac{a_0}{4\cdot3}=\dfrac2{4!}a_0[/tex]
[tex]k=2\implies n=8\implies a_8=\dfrac{a_4}{8\cdot7}=\dfrac{6\cdot5\cdot2}{8!}a_0[/tex]
[tex]k=3\implies n=12\implies a_{12}=\dfrac{a_8}{12\cdot11}=\dfrac{10\cdot9\cdot6\cdot5\cdot2}{12!}a_0[/tex]
and so on, with the general pattern
[tex]a_{4k}=\dfrac{a_0}{(4k)!}\displaystyle\prod_{i=1}^k(4i-2)(4i-3)[/tex]
If [tex]n=4k+1[/tex], then[tex]k=0\implies n=1\implies a_1=a_1[/tex]
[tex]k=1\implies n=5\implies a_5=\dfrac{a_1}{5\cdot4}=\dfrac{3\cdot2}{5!}a_1[/tex]
[tex]k=2\implies n=9\implies a_9=\dfrac{a_5}{9\cdot8}=\dfrac{7\cdot6\cdot3\cdot2}{9!}a_1[/tex]
[tex]k=3\implies n=13\implies a_{13}=\dfrac{a_9}{13\cdot12}=\dfrac{11\cdot10\cdot7\cdot6\cdot3\cdot2}{13!}a_1[/tex]
and so on, with
[tex]a_{4k+1}=\dfrac{a_1}{(4k+1)!}\displaystyle\prod_{i=1}^k(4i-1)(4i-2)[/tex]
If [tex]n=4k+2[/tex] or [tex]n=4k+3[/tex], then[tex]a_2=0\implies a_6=a_{10}=\cdots=a_{4k+2}=0[/tex]
[tex]a_3=0\implies a_7=a_{11}=\cdots=a_{4k+3}=0[/tex]
Then the solution to this ODE is
[tex]\boxed{y(x)=\displaystyle\sum_{k=0}^\infty a_{4k}x^{4k}+\sum_{k=0}^\infty a_{4k+1}x^{4k+1}}[/tex]
Find the equation of the line that passes through the points (12,21) and (46,33). V.35r-162
Answer:
the answer is Y=(6/17)X + (285/17)
Step-by-step explanation:
1. Identify the X and Y coordinate of each point.
for example, the first point could be (12,21), therefore X1 = 12 and Y1 = 21 and the second point is then (46,33), therefore X2 = 46 and Y2 = 33
2. to find the equation of the line it is necessary to find the slope.
knowing that the equiation of the slope is M = Y2-Y1/X2-X1 we replace
[tex]m=\frac{33-21}{46-12}=\frac{12}{34}[/tex] and then we simplify the result to [tex]\frac{12}{34}[/tex]
3. Using the Point-slope equation which is Y-Y1=M(X-X1) we replace
[tex]Y-33=\frac{6}{17}(X-46)[/tex]
[tex]Y=\frac{6}{17}X-\frac{276}{17}+33[/tex]
[tex]Y=\frac{6}{17}X+\frac{285}{17}[/tex]
find integers x and y such that 115x + 30y = gcd(431, 29)
Answer:
There are none.
Step-by-step explanation:
gcd(431,29)=1
The reasons I came to this conclusion is because 29 is prime.
29 is not a factor of 431 so we are done.
So now we want to find (x,y) such that 115x+30y=1.
I'm going to use Euclidean's Algorithm.
115=30(3)+25
30=25(1)+5
25=5(5)
So we know we are done when we get the remainder is 0 and I like to look at the line before the remainder 0 line to see the greatest common divisor or 115 and 30 is 5.
So 115x+30y=5 has integer solutions (x,y) where d=5 is the smallest possible positive such that 115x+30y=d will have integer solutions (x,y).
So since 1 is smaller than 5 and we are trying to solve 115x+30y=1 for integer solutions (x,y), there there is none.
Furthermore, 115x+30y=1 can be written as 5(23x+6y)=1 and we know that 5 is not a factor of 1.
Real estate ads suggest that 75 % of homes for sale have garages, 29 % have swimming pools, and 13 % have both features. What is the probability that a home for sale has a) a pool or a garage? b) neither a pool nor a garage? c) a pool but no garage?
Answer: a) 91% b) 9% c) 16%
Step-by-step explanation:
Let A be the event that homes for sale have garages and B be the event that homes for sale have swimming pools.
Now, given :[tex]P(A)=75\%=0.75[/tex]
[tex]P(B)=29\%=0.29[/tex]
[tex]P(A\cap B)=13\%=0.13[/tex]
a) [tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\\Rightarrow\ P(A\cup B)=0.75+0.29-0.13=0.91[/tex]
Hence, the probability that a home for sale has a pool or a garage is 91%.
b) The probability that a home for sale has neither a pool nor a garage is given by :-
[tex]1-P(A\cup B)=1-0.91=0.09[/tex]
Hence, the probability that a home for sale has neither a pool nor a garage is 9%.
c) The probability that a home for sale has a pool but no garage is given by :-
[tex]P(B)-P(A\cap B)=0.29-0.13=0.16[/tex]
Hence, the probability that a home for sale has a pool but no garage is 16%.
Final answer:
The probability a home for sale has a pool or garage is 91%, the probability of having neither feature is 9%, and the probability of having a pool but no garage is 16%.
Explanation:
The question involves solving problems related to probability and set theory. To find out the probability of events involving homes for sale with certain features, we can use the principle of inclusion and exclusion for probabilities.
Answering Part a)
The probability that a home has a pool or a garage is given by the formula P( A or B ) = P( A ) + P( B ) - P( A and B ). Thus, the probability is:
P( Pool or Garage ) = P( Pool ) + P( Garage ) - P( Pool and Garage )
= 0.29 + 0.75 - 0.13
= 0.91 or 91%
Answering Part b)
To find a home with neither a pool nor a garage, we subtract the probability of having either from 1. Therefore:
P( Neither ) = 1 - P( Pool or Garage )
= 1 - 0.91
= 0.09 or 9%
Answering Part c)
The probability of a home having a pool but no garage is calculated by taking the probability of having a pool and subtracting the probability of having both a pool and a garage:
P( Pool but no Garage ) = P( Pool ) - P( Pool and Garage )
= 0.29 - 0.13
= 0.16 or 16%