The equation of the cone should be [tex]z=\sqrt{x^2+y^2}[/tex]. Parameterize [tex]S[/tex] by
[tex]\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u\,\vec k[/tex]
with [tex]1\le u\le2[/tex] and [tex]0\le v\le2\pi[/tex]. Take the normal vector to [tex]S[/tex] to be
[tex]\vec s_v\times\vec s_u=u\cos v\,\vec\imath+u\sin v\,\vec\jmath-u\,\vec k[/tex]
Then the integral of [tex]\vec F[/tex] across [tex]S[/tex] is
[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_1^2(-u\cos v\,\vec\imath-u\sin v\,\vec\jmath+u^3\,\vec k)\cdot(u\cos v\,\vec\imath+u\sin v\,\vec\jmath-u\,\vec k)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_0^{2\pi}\int_1^2(-u^2-u^4)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle-2\pi\int_1^2(u^2+u^4)\,\mathrm du\,\mathrm dv=\boxed{-\frac{256\pi}{15}}[/tex]
To evaluate the surface integral, we need to find the flux of the vector field F across the oriented surface S. Given that F(x, y, z) = −xi − yj + z3k, and S is the part of the cone z = x2 + y2 between the planes z = 1 and z = 2 with downward orientation, we can proceed as follows: First, find the unit normal vector to the surface. Next, calculate the dot product between the vector field F and the unit normal vector. Finally, integrate the dot product over the surface S using the downward orientation.
Explanation:To evaluate the surface integral, we need to find the flux of the vector field F across the oriented surface S. Given that F(x, y, z) = −xi − yj + z3k, and S is the part of the cone z = x2 + y2 between the planes z = 1 and z = 2 with downward orientation, we can proceed as follows:
First, we need to find the unit normal vector to the surface. In this case, the unit normal vector is -∇(z - x^2 - y^2)/|∇(z - x^2 - y^2)|. By calculating the gradient and normalizing it, we get the unit normal vector as (2x, 2y, -1)/√(1 + 4x^2 + 4y^2).Next, we calculate the dot product between the vector field F and the unit normal vector. The dot product is -2x - 2y + z^3.Finally, we integrate the dot product over the surface S using the downward orientation. The integral is given by ∫∫S (-2x - 2y + z^3)dS.Learn more about Surface Integral here:https://brainly.com/question/32088117
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Write y = 2x^2 + 8x + 3 in vertex form.
y = 2(x – 2)^2 – 5
y = (x – 4)^2 + 3
y = 2(x + 2)^2 – 5
y = (x + 4)^2 + 3
Answer:
y = 2(x + 2)^2 – 5
Step-by-step explanation:
When y = ax^2 +bx +c is written in vertex form, it becomes ...
y = a(x +b/(2a))^2 +(c -b^2/(4a))
The constant term in the squared binomial is b/(2a) = 8/(2(2)) = +2. Only one answer choice matches:
y = 2(x +2)^2 -5
rain gutter is 36 feet long, 8 inches in height,3 inches across base, 12 inches across top..how many gallons of water will it hold when full?
Answer:
112.77 gallons of water
Step-by-step explanation:
We will calculate the area by Trapezoid formula :
[tex]A=\frac{a+b}{2}\times h\times l[/tex]
Given Base a = 12 inches = 1 feet
Base b = 3 inches = 0.25 feet
height = 8 inches = 0.67 feet
length = 36 feet = 36 feet
[tex]Area=\frac{1+2.5}{2}\times 0.67\times 36[/tex]
= 0.625 × 0.67 × 36
= 15.075 cubic feet.
As we know 1 cubic feet = 7.48052 per liquid gallon
Therefore, 15.075 cubic feet = 15.075 × 7.48052
= 112.768839 ≈ 112.77 liquid gallon
When full it will hold 112.77 gallons of water
Suppose that on the average, 7 students enrolled in a small liberal arts college have their automobiles stolen during the semester. What is the probability that more than 1 student will have his automobile stolen during the current semester? Round your answer to four decimal places.
To find the probability that more than 1 student will have their automobile stolen during the current semester, we can use a Poisson distribution. The average number of automobile thefts per semester is 7. By calculating 1 minus the probability of 0 or 1 thefts, we find that the probability of more than 1 theft is approximately 0.9991.
Explanation:To find the probability that more than 1 student will have their automobile stolen during the current semester, we can use a Poisson distribution. The Poisson distribution models the number of events that occur in a fixed interval of time or space. In this case, the average number of automobile thefts per semester is given as 7. We want to find the probability that more than 1 student will have their automobile stolen, so we need to calculate 1 minus the probability that 0 or 1 student will have their automobile stolen.
The probability mass function of a Poisson distribution is given by P(X=k) = e^{-λ} * (λ^k) / k! where λ is the average number of events. In this case, λ = 7.
So, the probability that more than 1 student will have their automobile stolen is P(X>1) = 1 - (P(X=0) + P(X=1)).
To calculate P(X=0) and P(X=1), we can substitute k=0 and k=1 into the probability mass function and sum them up.
P(X=0) = e^{-7} * (7^0) / 0! = e^{-7}
P(X=1) = e^{-7} * (7^1) / 1! = 7e^{-7}
Therefore, P(X>1) = 1 - (e^{-7} + 7e^{-7}).
Rounding this answer to four decimal places, we get P(X>1) ≈ 0.9991.
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PLEASE HELP I ONLY NEED THIS ONE TO FINISH THE SECTION
The functions r and s are defined as follows.
r(x) = -2x + 1
s(x) = -x^2 + 2
Find the value of .
r(s(3))
Answer:
The value of r(s(3)) = -21
Step-by-step explanation:
It is given that,
r(x) = -2x + 1
s(x) = -x^2 + 2
To find the value of r(s(3))
s(x) = -x^2 + 2
s(3) = (-3)^2 + 2 [Substitute 3 instead of x]
= 9 + 2
= 11
Therefore s(3) = 11
r(x) = -2x + 1
r(s(3)) = r(11) [Substitute 11 instead of x]
= -2(11) + 1
= -22 + 1
= -21
Therefore the value of r(s(3)) = -21
The answer is:
[tex]r(s(3))=15[/tex]
Why?To solve the problem, first, we need to compose the functions, and then evaluate the obtained function. Composing function means evaluating a function into another function.
We have that:
[tex]f(g(x))=f(x)\circ g(x)[/tex]
From the statement we know the functions:
[tex]r(x)=-2x+1\\s(x)=-x^{2}+2[/tex]
We need to evaluate the function "s" into the function "r", so:
[tex]r(s(x))=-2(-x^2+2)+1\\\\r(s(x))=2x^{2}-4+1=2x^{2}-3[/tex]
Now, evaluating the function, we have:
[tex]r(s(3))=2(3)^{2}-3=2*9-2=18-3=15[/tex]
Have a nice day!
You are in charge of purchases at the student-run used-book supply program at your college, and you must decide how many introductory calculus, history, and marketing texts should be purchased from students for resale. Due to budget limitations, you cannot purchase more than 700 of these textbooks each semester. There are also shelf-space limitations: Calculus texts occupy 2 units of shelf space each, history books 1 unit each, and marketing texts 4 units each, and you can spare at most 1,200 units of shelf space for the texts. If the used book program makes a profit of $10 on each calculus text, $4 on each history text, and $8 on each marketing text, how many of each type of text should you purchase to maximize profit? HINT [See Example 3.]
calculus text(s) =
history text(s) =
marketing text(s) =
What is the maximum profit the program can make in a semester?
Answer:
Calculus texts: 600History texts: 0Marketing texts: 0Step-by-step explanation:
Each Calculus text returns $10/2 = $5 per unit of shelf space. For History and Marketing texts, the respective numbers are $4/1 = $4 per unit, and $8/4 = $2 per unit. Using 1200 units of shelf space for 600 Calculus texts returns ...
$5/unit × 1200 units = $6000 . . . profit
Any other use of units of shelf space will reduce profit.
The College should purchase only 600 Calculus textbooks to maximize College profits.
Data and Calculations:
Budget limit on the number of textbooks per semester = 700
Shelf space for the purchased textbooks = 1,200 units
Calculus History Marketing
Shelf-space occupied
by each textbook 2 1 4
Profit per textbook $10 $4 $8
Profit per shelf-space $5 ($10/2) $4 ($4/1) $2 ($8/4)
The highest profit per shelf space is $5 generated by Calculus.
The highest profit that the College can make over the purchase of used Calculus textbooks = $6,000 ($5 x 1,200) or ($10 x 600).
Thus, for the College to maximize its profits, it should purchase 600 Calculus textbooks, which will not exceed the textbook limit for the semester, and at the same time maximally utilize the shelf space available.
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Use Gauss's approach to find the following sum
4+10+16+22+...+70
The sum of the sequence is
Each consecutive term in the sum is separated by a difference of 6, so the [tex]n[/tex]-th term is [tex]4+6(n-1)=6n-2[/tex] for [tex]n\ge1[/tex]. The last term is 70, so there are [tex]6n-2=70\implies n=12[/tex] terms in the sum.
Now,
[tex]S=4+10+\cdots+64+70[/tex]
but also
[tex]S=70+64+\cdots+10+4[/tex]
Doubling the sum and grouping terms in the same position gives
[tex]2S=(4+70)+(10+64)+\cdots+(64+10)+(70+4)=12\cdot74[/tex]
[tex]\implies\boxed{S=444}[/tex]
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 3. f(t) = 10 + 20 / (t + 1).
Answer:
Step-by-step explanation:Find the slope of the line that passes through the points shown in the table.
The slope of the line that passes through the points in the table is
.
If $9,400 is invested at an interest rate of 8% per year, find the value of the investment at the end of 5 years if interested is compounded annually (once a year), semiannually (twice a year), monthly (12 times a year), daily (assume 365 days a year), or continuously. Round to the nearest cent. For each, use the correct compound interest formula from the following. A = P ( 1 + r n ) n t or A = P e r t
(a) Annual:
(b) Semiannual:
(c) Monthly:
(d) Daily:
(e) Continuously:
Answer:
(a) $13811.68
(b) $13914.30
(c) $14004.55
(d) $14022.54
(e) $14023.15
Step-by-step explanation:
Since, the amount formula in compound interest,
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
Where,
P = Principal amount,
r = annual rate,
n = number of periods,
t = number of years,
Here, P = $ 9,400, r = 8% = 0.08, t = 5 years,
If the amount is compounded annually,
n = 1,
Hence, the amount of investment would be,
[tex]A=9400(1+\frac{0.08}{1})^5=9400(1.08)^5=\$ 13811.6839219\approx \$ 13811.68[/tex]
(a) If the amount is compounded annually,
n = 1,
The amount of investment would be,
[tex]A=9400(1+\frac{0.08}{1})^5=9400(1.08)^5=\$ 13811.6839219\approx \$ 13811.68[/tex]
(b) If the amount is compounded semiannually,
n = 2,
The amount of investment would be,
[tex]A=9400(1+\frac{0.08}{2})^{10}=9400(1.04)^{10}=\$13914.2962782\approx \$ 13914.30[/tex]
(c) If the amount is compounded Monthly,
n = 12,
The amount of investment would be,
[tex]A=9400(1+\frac{0.08}{12})^{60}=9400(1+\frac{1}{150})^{60}=\$ 14004.549658\approx \$ 14004.55[/tex]
(d) If the amount is compounded Daily,
n = 365,
The amount of investment would be,
[tex]A=9400(1+\frac{0.08}{365})^{365\times 5}=9400(1+\frac{2}{9125})^{1825}=\$ 14022.5375476\approx \$ 14022.54[/tex]
(e) Now, the amount in compound continuously,
[tex]A=Pe^{rt}[/tex]
Where, P = principal amount,
r = annual rate,
t = number of years,
So, the investment would be,
[tex]A=9400 e^{0.08\times 5}=9400 e^{0.4}=\$14023.1521578\approx \$14023.15[/tex]
A vehicle factory manufactures cars. The unit cost C (the cost in dollars to make each car) depends on the number of cars made. If x cars are made, then the unit cost is given by the function C(x) = x^2 - 400x + 45,377 . What is the minimum unit cost?
Do not round your answer.
Answer:
5377
Step-by-step explanation:
C(x) = x^2 - 400x + 45,377
To find the location of the minimum, we take the derivative of the function
We know that is a minimum since the parabola opens upward
dC/dx = 2x - 400
We set that equal to zero
2x-400 =0
Solving for x
2x-400+400=400
2x=400
Dividing by2
2x/2=400/2
x=200
The location of the minimum is at x=200
The value is found by substituting x back into the equation
C(200) = (200)^2 - 400(200) + 45,377
=40000 - 80000+45377
=5377
Answer:
The minimum unit cost is 5377
Step-by-step explanation:
Note that we have a cudratic function of negative principal coefficient.
The minimum value reached by this function is found in its vertex.
For a quadratic function of the form
[tex]ax ^ 2 + bx + c[/tex]
the x coordinate of the vertex is given by the following expression
[tex]x=-\frac{b}{2a}[/tex]
In this case the function is:
[tex]C(x) = x^2 - 400x + 45,377[/tex]
So:
[tex]a=1\\b=-400\\c=45,377[/tex]
Then the x coordinate of the vertex is:
[tex]x=-\frac{-400}{2(1)}[/tex]
[tex]x=200\ cars[/tex]
So the minimum unit cost is:
[tex]C(200) = (200)^2 - 400(200) + 45,377[/tex]
[tex]C(200) = 5377[/tex]
The centers for disease control and prevention reported that 25% of baby boys 6-8 months old in the united states weigh more than 20 pounds. A sample of 16 babies is studied. What is the probability that fewer than 3 weigh more than 20 pounds?
Answer:
0.1971 ( approx )
Step-by-step explanation:
Let X represents the event of weighing more than 20 pounds,
Since, the binomial distribution formula is,
[tex]P(x)=^nC_r p^r q^{n-r}[/tex]
Where, [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
Given,
The probability of weighing more than 20 pounds, p = 25% = 0.25,
⇒ The probability of not weighing more than 20 pounds, q = 1-p = 0.75
Total number of samples, n = 16,
Hence, the probability that fewer than 3 weigh more than 20 pounds,
[tex]P(X<3) = P(X=0)+P(X=1)+P(X=2)[/tex]
[tex]=^{16}C_0 (0.25)^0 (0.75)^{16-0}+^{16}C_1 (0.25)^1 (0.75)^{16-1}+^{16}C_2 (0.25)^2 (0.75)^{16-2}[/tex]
[tex]=(0.75)^{16}+16(0.25)(0.75)^{15}+120(0.25)^2(0.75)^{14}[/tex]
[tex]=0.1971110499[/tex]
[tex]\approx 0.1971[/tex]
With a short time remaining in the day, a delivery driver has time to make deliveries at 7 locations among the 9 locations remaining. How many different routes are possible?
There are 9,072 different routes possible for the delivery driver.
Explanation:To find the number of different routes possible, we can use the concept of permutations. Since the driver has to deliver to 7 out of the 9 remaining locations, we can calculate the number of ways to choose 7 out of 9 and then multiply it by the number of ways to arrange those 7 locations. The formula for permutations is P(n, r) = n! / (n - r)!. In this case, n = 9 and r = 7.
P(9, 7) = 9! / (9 - 7)! = 9! / 2! = 9 × 8 × 7 × 6 × 5 × 4 × 3 = 9,072
Therefore, there are 9,072 different routes possible for the delivery driver.
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To find the number of different routes possible, we can use the combination formula: C(n, k) = n! / (k!(n-k)!). Plugging in the values, we get C(9, 7) = 36. Therefore, there are 36 different routes possible.
Explanation:To find the number of different routes possible, we can use the combination formula:
C(n, k) = n! / (k!(n-k)!)
Where n is the total number of locations remaining (9) and k is the number of locations the driver can make deliveries to (7).
Plugging in the values, we get:
C(9, 7) = 9! / (7!(9-7)!)
= 9! / (7!2!)
= (9 * 8 * 7!)/(7! * 2!)
= (9 * 8)/(2 * 1)
= 36
Therefore, there are 36 different routes possible.
Scores on the Critical Reading part of the SAT exam in a recent year were roughly Normal with mean 495495 and standard deviation 118118 . You choose an SRS of 100100 students and average their SAT Critical Reading scores. If you do this many times, the mean of the average scores you get will be close to 495118√=45.57495118=45.57 . 495495 . 495100√=49.5495100=49.5 . 495100=4.95495100=4.95 .
Answer: [tex] 495[/tex]
Step-by-step explanation:
We know that if the population is normally distributed with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], then the sampling distribution of the sample mean , [tex]\overline{x}[/tex] is also normally distribution with :-
Mean : [tex]\mu_{\overlien{x}}=\mu[/tex]
Given : The scores on the Critical Reading part of the SAT exam in a recent year were roughly Normal with mean [tex]\mu= 495[/tex]
Then , the mean of the average scores you get will be close to [tex] 495[/tex]
Use spherical coordinates to find the volume of the region that lies outside the cone z = p x 2 + y 2 but inside the sphere x 2 + y 2 + z 2 = 2. Write the answer as an exact answer, which should involve π and √ 2. Do not round or use a calculator.
I assume the cone has equation [tex]z=\sqrt{x^2+y^2}[/tex] (i.e. the upper half of the infinite cone given by [tex]z^2=x^2+y^2[/tex]). Take
[tex]\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]
The volume of the described region (call it [tex]R[/tex]) is
[tex]\displaystyle\iiint_R\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^{2\pi}\int_0^{\sqrt2}\int_{\pi/4}^\pi\rho^2\sin\varphi\,\mathrm d\varphi\,\mathrm d\rho\,\mathrm d\theta[/tex]
The limits on [tex]\theta[/tex] and [tex]\rho[/tex] should be obvious. The lower limit on [tex]\varphi[/tex] is obtained by first determining the intersection of the cone and sphere lies in the cylinder [tex]x^2+y^2=1[/tex]. The distance between the central axis of the cone and this intersection is 1. The sphere has radius [tex]\sqrt2[/tex]. Then [tex]\varphi[/tex] satisfies
[tex]\sin\varphi=\dfrac1{\sqrt2}\implies\varphi=\dfrac\pi4[/tex]
(I've added a picture to better demonstrate this)
Computing the integral is trivial. We have
[tex]\displaystyle2\pi\left(\int_0^{\sqrt2}\rho^2\,\mathrm d\rho\right)\left(\int_{\pi/4}^\pi\sin\varphi\,\mathrm d\varphi\right)=\boxed{\frac43(1+\sqrt2)\pi}[/tex]
To win at LOTTO in one state, one must correctly select 7 numbers from a collection of 48 numbers (1 through 48). The order in which the selection matter. How many different selections are possible? made does not There are different LOTTO selections.
The number of different possible 7-number selections from a pool of 48, where order does not matter, can be calculated using the mathematical concept of combinations. The formula to calculate the total number of combinations is C(n, k) = n! / [k!(n-k)!], where n is the total number of options and k is the number of options selected.
Explanation:In the LOTTO game you described, you must select 7 numbers from a pool of 48, and the order of the numbers does not matter. This is a problem of combinations in mathematics. The formula for combinations is C(n, k) = n! / [k!(n-k)!], where n is the total number of options, k is the number of options selected, and '!' denotes a factorial. In this case, n=48 (the numbers 1 through 48) and k=7 (the seven numbers you select).
By plugging these values into the formula, we can calculate the total number of different selections possible: C(48, 7) = 48! / [7!(48-7)!]. This calculation would give us the total number of combinations of 7 numbers that can be selected from a pool of 48, which represents all the different possible LOTTO selections. It should be remembered that factorials such as 48! or 7! represent a product of an integer and all the integers below it, down to 1.
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There are 85,900,584 different selections possible in the LOTTO game.
Explanation:To calculate the number of different selections possible in the LOTTO game, we need to focus on the concept of permutations.
With 48 numbers to choose from and 7 numbers to be selected, the number of permutations can be calculated using the formula P(n, r) = n! / (n-r)!, where n is the total number of items and r is the number of items to be selected.
In this case, n = 48 and r = 7.
Using the formula, we have P(48, 7) = 48! / (48-7)! = (48 * 47 * 46 * 45 * 44 * 43 * 42) / (7 * 6 * 5 * 4 * 3 * 2 * 1)
= 85,900,584 different selections possible.
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A car is travelling at exactly 6 miles per hour and accelerates at a constant rate to exactly 65 miles per hour.
This acceleration takes exactly 14.8 seconds.
What is the rate of acceleration, in miles per hour2?
Answer:
[tex]14878.04878miles/hours^2[/tex]
Step-by-step explanation:
Let's find a solution by understanding the following:
The acceleration rate is defined as the change of velocity within a time interval, which can be written as:
[tex]A=(Vf-Vi)/T[/tex] where:
A=acceleration rate
Vf=final velocity
Vi=initial velocity
T=time required for passing from Vi to Vf.
Using the problem's data we have:
Vf=65miles/hour
Vi=6miles/hour
T=14.8seconds
Using the acceleration rate equation we have:
[tex]A=(65miles/hour - 6miles/hour)/14.8seconds[/tex], but look that velocities use 'hours' unit while 'T' uses 'seconds'.
So we need to transform 14.8seconds into Xhours, as follows:
[tex]X=(14.8seconds)*(1hours/60minutes)*(1minute/60seconds)[/tex]
[tex]X=0.0041hours[/tex]
Using X=0.0041hours in the previous equation instead of 14.8seconds we have:
[tex]A=(65miles/hour - 6miles/hour)/0.0041hours[/tex]
[tex]A=(61miles/hour)/0.0041hours[/tex]
[tex]A=(61miles)/(hour*0.0041hours)[/tex]
[tex]A=61miles/0.0041hours^2[/tex]
[tex]A=14878.04878miles/hours^2[/tex]
In conclusion, the acceleration rate is [tex]14878.04878miles/hours^2[/tex]
A bookmark has a perimeter of 46 centimeters and an area of 102 square centimeters. What are the dimensions of the bookmark?
Answer:
6 cm by 17 cm
Step-by-step explanation:
The area is the product of the dimensions; the perimeter is double the sum of the dimensions.
So, we want to find two numbers whose product is 102 and whose sum is 23.
102 = 1·102 = 2·51 = 3·34 = 6·17
The last of these factor pairs has a sum of 23.
The dimensions are 6 cm by 17 cm.
calculate simple interest and maturity value
Principal is. 6,000. interest rate 5%. the time is 15. what is the solution
Answer:
Simple interest=4500
maturity value= 10500
Step-by-step explanation:
Given: Principal P = 6000
Rate % R = 5%
time T = 15 years
we know that simple interest [tex]SI=\frac{PRT}{100}[/tex]
⇒[tex]SI=\frac{6000\times5\times15}{100}[/tex]
on calculating we get SI = 4500
and maturity value = Principal amount + Simple interest
maturity value = 6000+4500 = 10500
hence the final answers are
Simple interest= 4500
and maturity value= 10500
hope this helps!!
PLEASE HELP
Identify the radius and center
x^2 +y^2 -6x -2y + 1 = 0
Answer:
The center is the point (3,1) and the radius is 3 units
Step-by-step explanation:
we know that
The equation of a circle in standard form is equal to
[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]
we have
[tex]x^{2}+y^{2}-6x-2y+1=0[/tex]
Convert to standard form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}-6x)+(y^{2}-2y)=-1[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
[tex](x^{2}-6x+9)+(y^{2}-2y+1)=-1+9+1[/tex]
[tex](x^{2}-6x+9)+(y^{2}-2y+1)=9[/tex]
Rewrite as perfect squares
[tex](x-3)^{2}+(y-1)^{2}=9[/tex]
The center is the point (3,1) and the radius is 3 units
At Information and Communications University: 31 female seniors were on the dean’s list 62 women were on the dean’s list who were not seniors 45 male seniors were on the dean’s list 87 female seniors were not on the dean’s list 96 male seniors were not on the dean’s list 275 women were not senior and were not on the dean’s list 88 men were on the dean’s list who were not seniors 227 men were not seniors and were not on the dean’s list (a) How many were seniors? (b) How many were women? (c) How many were on the dean’s list? (d) How many were seniors on the dean’s list? (e) How many were female seniors? (f) How many were women on the dean’s lists? (g) How many were students at the college?
Step-by-step explanation:
There's a lot of information here, so first things first, let's get organized.
Let's start by assigning variable names to each group. We want the variables to be short but easy to understand.
For example, let's say the number of female seniors on the dean's list is FSD (F for female, S for senior, and D for dean's list).
FSD = 31
Sticking to this naming scheme:
FD = 62
MSD = 45
FS = 87
MS = 96
F = 275
MD = 88
M = 227
Now we can begin.
a) We want to know how many seniors there are. So all we have to do is add up all the variables with an S in them.
FSD + MSD + FS + MS
31 + 45 + 87 + 96
259
b) We want to know how many women there are. So add up all the variables with an F in them.
FSD + FD + FS + F
31 + 62 + 87 + 275
455
c) We want to know how many are on the dean's list. So add up all the variables with a D in them.
FSD + FD + MSD + MD
31 + 62 + 45 + 88
226
d) Now we want to know how many are seniors AND on the dean's list. So add up all the variables that have both an S and a D.
FSD + MSD
31 + 45
76
e) We want to know how many female seniors there are, so add up all the variables with both an F and an S.
FSD + FS
31 + 87
118
f) We want to know how many women were on the dean's list, so add up all the variables with both an F and a D.
FSD + FD
31 + 62
93
g) Finally, we want to know how many students there are total. So add up all the variables.
FSD + FD + MSD + FS + MS + F + MD + M
62 + 45 + 87 + 96 + 275 + 88 + 227
880
Using arithmetic operations based on the provided data, we deduced the total number of seniors, women, students on the dean's list, and overall students at the college, among other specific categorizations.
Explanation:To solve these problems, we will first interpret the given data, then perform simple arithmetic operations based on the information provided to answer each part of the question.
(a) To find how many were seniors, we sum the number of female and male seniors on the dean's list and those not on the dean's list: 31 (female seniors on the dean's list) + 87 (female seniors not on the dean's list) + 45 (male seniors on the dean's list) + 96 (male seniors not on the dean's list) = 259.(b) To calculate how many were women, we add the women on the dean's list and those not, including seniors and non-seniors: 31 (female seniors on the dean's list) + 62 (women on the dean's list who were not seniors) + 87 (female seniors not on the dean's list) + 275 (women not senior and not on the dean's list) = 455.(c) How many were on the dean's list? Summing the numbers for all students on the dean's list gives us: 31 (female seniors) + 62 (women not seniors) + 45 (male seniors) + 88 (men not seniors) = 226.(d) How many were seniors on the dean's list? Adding together the number of female and male seniors on the dean's list: 31 (female seniors) + 45 (male seniors) = 76.(e) How many were female seniors? Adding the number of female seniors on the dean's list and those not gives: 31 (on the dean's list) + 87 (not on the dean's list) = 118.(f) How many were women on the dean's list? This is the total number of women, both seniors and not, on the dean's list: 31 (female seniors) + 62 (women not seniors) = 93.(g) Summing students from all categories to find how many were students at the college: (259 seniors, both genders) + (275 women not senior/not on dean's list) + (88 men on the dean's list who were not seniors) + (227 men who were not seniors and not on the dean's list) = 849.A pig farmer owns a 20-arce farm and started business with 16 pigs. After one year, he has 421 pigs. Assuming a constant growth rate, how many pigs would he have after a total of 5.1 years?
Answer:
The correct answer is 206550 pigs.
Step-by-step explanation:
First of all you need to calculate the constant growth rate, which is given by the formula [tex]p=((\frac{f}{s} )^{\frac{1}{y} } -1)*100[/tex] where f is the value at the end of the year, s is the start value of that year, and y is the number of years.
From the excercise facts, we know that for 1 year (y=1), the final value is 421 (f=421), and the start value is 16 (s=16). Replacing them in the formula we get : [tex]((\frac{421}{16}) ^{\frac{1}{1} } -1)*100 = 2531.25[/tex] So, the constant growth rate equals 2531.25.
Next, we have to multiply the starting amount of pigs times the constant growth rate times the amount of time that passed to get the final quantity of pigs. This would be [tex]16*2531.25*5.1[/tex] and this gives us a total amount of 206550 pigs.
Have a nice day.
2. Which of the following are terms of the series with nth term T-3n +17? a) 80 b) 170 c)217 d) 312 e) 278 f) 3566
Answer:
The correct options are a,b,e and f.
Step-by-step explanation:
It is given that the nth terms of the series is defined as
[tex]T_n=3n+17[/tex]
Subtract 17 from both the sides.
[tex]T_n-17=3n[/tex]
Divide both sides by 3.
[tex]\frac{T_n-17}{3}=n[/tex]
The term Tₙ is a term of given series if n is a positive integer.
(a) The given term is 80.
[tex]n=\frac{80-17}{3}=21[/tex]
Since n is a positive integer, therefore 80 is a term of given series.
(b) The given term is 170.
[tex]n=\frac{170-17}{3}=51[/tex]
Since n is a positive integer, therefore 170 is a term of given series.
(c) The given term is 217.
[tex]n=\frac{217-17}{3}=66.67[/tex]
Since n is not a positive integer, therefore 217 is a term of given series.
(d) The given term is 312.
[tex]n=\frac{312-17}{3}=98.33[/tex]
Since n is not a positive integer, therefore 312 is a term of given series.
(e) The given term is 278.
[tex]n=\frac{278-17}{3}=87[/tex]
Since n is a positive integer, therefore 278 is a term of given series.
(f) The given term is 3566.
[tex]n=\frac{3566-17}{3}=1183[/tex]
Since n is a positive integer, therefore 3566 is a term of given series.
Thus, the correct options are a, b, e and f.
What is the mode for the set of data?
Ages
Stem Leaves
1 0, 3, 6
2 0, 1, 3, 7, 7, 8, 9
3 0, 2, 3, 3, 3, 3, 8, 9
4 6, 6, 6, 8
1|0 = 10 years old
3
33
38
46
Answer:
33
Step-by-step explanation:
The mode is the number that sets up the most in an answer.
Remember, the stem is the number that is in the "tens" place value, while the leaves is the number in the "ones" place value.
Expand the stem-leaves:
10, 13, 16, 20, 21, 23, 27, 27, 28, 29, 30, 32, 33, 33, 33, 33, 38, 39, 46, 46, 46, 48
The number that shows up the most is 33, and is your answer.
~
what is the solution if the inequality shown below? a+2<-10
Answer:
a < -12
Step-by-step explanation:
Isolate the variable, a. Treat the < as a equal sign, what you do to one side, you do to the other. Subtract 2 from both sides:
a + 2 < -10
a + 2 (-2) < -10 (-2)
Simplify:
a < -10 - 2
a < -12
a < -12 is your answer.
~
Answer:
[tex]\huge \boxed{a<-12}\checkmark[/tex]
Step-by-step explanation:
Subtract by 2 both sides of equation.
[tex]\displaystyle a+2-2<-10-2[/tex]
Simplify, to find the answer.
[tex]-10-2=-12[/tex]
[tex]\displaystyle a<-12[/tex], which is our answer.
The concept of determining which reactant is limiting and which is in excess is akin to determining the number of sandwiches that can be made from a set number of ingredients. Assuming that a cheese sandwich consists of 2 slices of bread and 3 slices of cheese, determine the number of whole cheese sandwiches that can be prepared from 44 slices of bread and 69 slices of cheese.
_____ sandwiches
Which of the two ingredients limits the number of sandwiches that can be made?
slices of cheese
slices of bread
What quantity of the ingredient in excess remains?
With 22 packed cheese sandwiches made, bread is the 'limiting' ingredient, and we have three slices of cheese in excess.
Explanation:The number of sandwiches that can be made depends on the availability of the critical ingredients needed to complete a sandwich: 2 slices of bread and three slices of cheese. To calculate the number of sandwiches, we can determine how many sandwiches each ingredient batch could provide, then find the minimum of those two, as the 'limiting' factor will be the ingredient that finishes first.
For bread, we have 44 pieces/ 2 pieces per sandwich = 22 sandwiches. With cheese, we have 69 pieces/ 3 pieces per sandwich = 23 sandwiches. Therefore, we can fully make a maximum of 22 sandwiches because we would run out of bread first. This means the bread is the 'limiting' ingredient, while cheese is in excess. With 22 sandwiches made, we would use three slices of cheese per sandwich for 66 slices used, leaving three slices of cheese remaining. The quantity in excess is three slices of cheese.
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There are 0 slices of bread in excess, and there are 3 slices of cheese in excess.
Let's start by calculating the maximum number of sandwiches that can be made using 44 slices of bread:
Each sandwich requires 2 slices of bread.
So, the maximum number of sandwiches that can be made using 44 slices of bread is 44/2 = 22 sandwiches.
Now, let's calculate the maximum number of sandwiches that can be made using 69 slices of cheese:
Each sandwich requires 3 slices of cheese.
So, the maximum number of sandwiches that can be made using 69 slices of cheese is 69/3 = 23 sandwiches.
Now, we compare the results:
With 44 slices of bread, we can make 22 sandwiches.
With 69 slices of cheese, we can make 23 sandwiches.
Since we can only make 22 sandwiches due to the limitation of the bread, the bread is the limiting ingredient.
The quantity of the ingredient in excess can be found by subtracting the number of sandwiches that can be made using the limiting ingredient from the total number of slices of that ingredient.
For the bread:
Quantity of bread in excess = Total slices of bread - (Number of sandwiches × Slices of bread per sandwich)
Quantity of bread in excess = 44 - (22 *2)
= 44 - 44
= 0
For the cheese:
Quantity of cheese in excess = Total slices of cheese - (Number of sandwiches × Slices of cheese per sandwich)
Quantity of cheese in excess = 69 - (22 *s 3)
= 69 - 66
= 3
So, there are 0 slices of bread in excess, and there are 3 slices of cheese in excess.
a. Find dy/dx if y^2 + x^2 = 16 b. Find the equation of the tangent line that contains the point (2, 2 squareroot 3).
Answer:
[tex]x+\sqrt{3}y=8[/tex]
Step-by-step explanation:
Given equation of curve,
[tex]y^2+x^2=16[/tex]
[tex]\implies y^2=16-x^2[/tex]
Differentiating with respect to x,
[tex]2y\frac{dy}{dx}=-2x[/tex]
[tex]\implies \frac{dy}{dx}=-\frac{x}{y}[/tex]
Since, the tangent line of the curve contains the point (2, 2√3),
Thus, the slope of the tangent line,
[tex]m=\left [ \frac{dy}{dx} \right ]_{(2, 2\sqrt{3})}=-\frac{1}{\sqrt{3}}[/tex]
Hence, the equation of tangent line would be,
[tex]y-2\sqrt{3}=-\frac{1}{\sqrt{3}}(x-2)[/tex]
[tex]\sqrt{3}y-6=-x+2[/tex]
[tex]\implies x+\sqrt{3}y=8[/tex]
What is a regular tessellation? How many regular tessellations are possible? Why aren’t there infinitely many regular tessellations?
Answer:
Step-by-step explanation:
What is a regular tessellation?
A regular tessellation is a pattern made by repeating a regular polygon. In simpler words regular tessellations are made up entirely of congruent regular polygons all meeting vertex to vertex.
How many regular tessellation are possible?
There are only 3 regular tessellation.
1. Triangle
2. Square
3. Hexagon
Why aren't there infinitely many regular tessellations?
Not more than 3 regular tessellations are possible because the sums of the interior angles are either greater than or less than 360 degrees....
Final answer:
A regular tessellation is a repeating pattern of a regular polygon that fills a plane without gaps or overlaps, with only three possible: equilateral triangles, squares, and regular hexagons. There aren't infinitely many because a tessellation requires the angles at a vertex to sum to 360 degrees, a condition only satisfied by these three shapes.
Explanation:
A regular tessellation is a pattern made by repeating a regular polygon to fill a plane without any gaps or overlaps. There are exactly three regular tessellations possible, which are constituted by:
Equilateral trianglesSquaresRegular hexagonsThere aren't infinitely many regular tessellations because for a regular polygon to tessellate, the sum of the angles at a vertex where polygons meet must be exactly 360 degrees.
This criterion is satisfied only by triangles (each angle is 60 degrees), squares (each angle is 90 degrees), and hexagons (each angle is 120 degrees).
Polygons with more sides have larger interior angles, such that the sum exceeds 360 degrees, preventing them from tessellating regularly.
A town recently dismissed 9 employees in order to meet their new budget reductions. The town had 7 employees over 50 years of age and 15 under 50. If the dismissed employees were selected at random, what is the probability that exactly 1 employee was over 50? Express your answer as a fraction or a decimal number rounded to four decimal places.
Using the hypergeometric distribution, the required probability is 0.0129.
Given that, the town had 7 employees over 50 years of age and 15 under 50.
Let, n₁=7 and n₂=15
Total employees in the town = 7+15 = 22
We have to find the probability that exactly 1 dismissed employee was over 50.
That means the remaining 8 dismissed employees are under 50.
Let the number of dismissed employees under 50 is denoted by x and above 50 denoted by y.
Then, using the hypergeometric distribution, the required probability can be calculated as:
[tex]P(x=1, y=8) = \dfrac{\binom{7}{1}+\binom{15}{8}}{\binom{22}{9}}[/tex]
[tex]P(x=1, y=8) = \dfrac{7+6435}{ 497420}[/tex]
[tex]P(x=1, y=8) = \dfrac{6442}{ 497420}[/tex]
[tex]P(x=1, y=8) = 0.0129[/tex]
Hence, the probability that exactly 1 employee was over 50 is 0.0129.
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In the following probability distribution, the random variable x represents the number of activities a parent of a 6 th dash to 8 th -grade student is involved in. Complete parts (a) through (f) below.
x 0 1 2 3 4
P(x) 0.216 0.072 0.322 0.289 0.101
Compute and interpret the mean of the random variable x.
The mean is ___ activities.
(Type an integer or a decimal.)
(d) Compute the standard deviation of the random variable x.
The standard deviation is _____ activities.
(Round to one decimal place as needed.)
(e) What is the probability that a randomly selected student has a parent involved in three activities?
The probability is _____ (Type an integer or a decimal.)
(f) What is the probability that a randomly selected student has a parent involved in three or four activities?
The probability is_____ (Type an integer or a decimal.)
The mean/expected value is
[tex]E[X]=\displaystyle\sum_x x\,P(X=x)=0P(X=0)+1P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)[/tex]
[tex]\implies E[X]=\boxed{1.987}[/tex]
d. The standard deviation is the square root of the variance, which itself is
[tex]V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2[/tex]
We have
[tex]E[X^2]=\displaystyle\sum_xx^2\,P(X=x)=0^2P(X=0)+1^2P(X=1)+2^2P(X=2)+3^2P(X=3)+4^2P(X=4)[/tex]
[tex]\implies E[X^2]=5.577[/tex]
Then the variance is
[tex]V[X]=5.577-1.987^2\approx1.629[/tex]
and so the standard deviation is
[tex]\sqrt{V[X]}\approx\boxed{1.276}[/tex]
e. We know this immediately from the table:
[tex]P(X=3)=\boxed{0.289}[/tex]
f. A parent can be involved in either 3 or 4 activities, but not simultaneously 3 and 4 activities (i.e. these events are disjoint), so
[tex]P(X=3\text{ or }X=4)=P(X=3)+P(X=4)=\boxed{0.39}[/tex]
Applying statistical concepts, we have that:
c) The mean is of 1.987 activities.
d) The standard deviation is of 1.67 activities.
e) The probability is 0.289.
f) The probability is 0.39.
The probability distribution is given by:
[tex]P(X = 0) = 0.216[/tex]
[tex]P(X = 1) = 0.072[/tex]
[tex]P(X = 2) = 0.322[/tex]
[tex]P(X = 3) = 0.289[/tex]
[tex]P(X = 4) = 0.101[/tex]
Item c:
The mean is given by the sum of the multiplications of each outcome by it's probability, thus:
[tex]E(X) = 0.216(0) + 0.072(1) + 0.322(2) + 0.289(3) + 0.101(4) = 1.987[/tex]
The mean is of 1.987 activities.
Item d:
The standard deviation is given by the square root of the sum of the squares of each outcome subtracted by the mean, multiplied by it's probability, thus:
[tex]\sqrt{V(X)} = \sqrt{0.216(0-1.987)^2 + 0.072(1-1.987)^2) + 0.322(2-1.987)^2) + ...}[/tex]
[tex]\sqrt{V(X)} = 1.67[/tex]
The standard deviation is of 1.67 activities.
Item e:
This probability is P(X = 3), thus 0.289.
Item f:
This probability is:
[tex]p = P(X = 3) + P(X = 4) = 0.289 + 0.101 = 0.39[/tex]
A similar problem is given at https://brainly.com/question/14395358
Marla Opper currently earns $95,000 a year and is offered a job in another city for $118,000. The city she would move to has 11 percent higher living expenses than her current city.
What amount must Marla earn in the new city to maintain her current buying power?
Answer:
Since living expenses are 11% higher in the new city. Marla would need an 11% increase in her current income to maintain her current buying power when moving. We can solve this by simply multiplying her current income by 11%.
Step-by-step explanation:
[tex]95,000*1.11 = 105,450[/tex]
If we multiply her current income of $95,000 by 11% we see that Marla would need an income of $105,450 in order to maintain her buying power in the new city. Since her new job offer is paying $118,000 she will have more buying power in the new city.
Answer: Marla must earn an Income of $105,450 to maintain her buying power.
Marla would need to earn $105,450 in the new city to maintain her current buying power after taking into account an 11 percent increase in living expenses.
Explanation:The student has asked how much Marla Opper must earn at the new job in another city to maintain her current buying power, considering that the new city has living expenses that are 11 percent higher. To calculate this, we first need to figure out what amount of income in the new city would be equivalent to Marla's current income of $95,000, given the increase in living expenses.
Firstly, we find 11 percent of Marla's current income:
11% of $95,000 = 0.11 × $95,000 = $10,450This means Marla needs an additional $10,450 on top of her current income to maintain the same buying power in the new city. So, the amount Marla needs to earn in the new city is:
$95,000 + $10,450 = $105,450Thus, Marla would need to earn $105,450 in the new city to keep her current buying power.
Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2
[tex]y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}[/tex]
Divide both sides by [tex]\dfrac13y^{-2}(x)[/tex]:
[tex]3y^2y'-y^3=xe^x\ln x[/tex]
Substitute [tex]v(x)=y(x)^3[/tex], so that [tex]v'(x)=3y(x)^2y'(x)[/tex].
[tex]v'-v=xe^x\ln x[/tex]
Multiply both sides by [tex]e^{-x}[/tex]:
[tex]e^{-x}v'-e^{-x}v=x\ln x[/tex]
The left side can be condensed into the derivative of a product.
[tex](e^{-x}v)'=x\ln x[/tex]
Integrate both sides to get
[tex]e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C[/tex]
Solve for [tex]v(x)[/tex]:
[tex]v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x[/tex]
Solve for [tex]y(x)[/tex]:
[tex]y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x[/tex]
[tex]\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}[/tex]