Answer:
50% chance
Step-by-step explanation:
Answer:
1/3
Step-by-step explanation:
There are 3 coins, and each coin has 2 possible outcomes. So there are a total of 6 possible outcomes.
Of these 6 outcomes, 3 are heads. Of these 3, only 1 has tails on the opposite face.
So the probability is 1/3.
We can also show this using conditional probability:
P(A|B) = P(A∩B) / P(B)
Probability that A occurs, given that B has occurred = Probability that both A and B occur / Probability that B occurs
Here, A = tails on opposite face and B = heads.
P(A|B) = (1/6) / (3/6)
P(A|B) = 1/3
A mass weighing 1 slug is attached to the end of a spring with spring constant 26 lb/ft. Damping force is numerically equal to 2 times the velocity. The mass is pulled down 6 ft and then released from rest. Write down and solve the initial value problem for the position x(t)
Answer:
x = e^(-t) [1.2 sin(5t) + 6 cos(5t)]
Step-by-step explanation:
A damped oscillator has the equation of motion:
m d²x/dt² + β dx/dt + k x = 0
This is an example of a second order linear ordinary differential equation with constant coefficients.
d²x/dt² + b dx/dt + c x = 0
Notice the leading coefficient is 1.
If b²−4c > 0, then the solution is:x = C₁ e^(-½ t (b + √(b²−4c) )) + C₂ e^(-½ t (b − √(b²−4c) ))
If b²−4c = 0, then the solution is:x = (C₁ t + C₂) e^(-bt/2)
If b²−4c < 0, then the solution is:x = e^(-bt/2) [C₁ sin(½ t √(4c−b²)) + C₂ cos(½ t √(4c−b²))]
Given that m = 1, β = 2, and k = 26:
d²x/dt² + 2 dx/dt + 26 x = 0
Here, b = 2 and c = 26, so:
b²−4c = (2)²−4(26) = -100 < 0
The general solution is:
x = e^(-2t/2) [C₁ sin(½ t √100) + C₂ cos(½ t √100)]
x = e^(-t) [C₁ sin(5t) + C₂ cos(5t)]
To find the values of C₁ and C₂, first find dx/dt, then plug in the initial conditions.
dx/dt = e^(-t) [5C₁ cos(5t) − 5C₂ sin(5t)] − e^(-t) [C₁ sin(5t) + C₂ cos(5t)]
dx/dt = e^(-t) [5C₁ cos(5t) − 5C₂ sin(5t) − C₁ sin(5t) − C₂ cos(5t)]
dx/dt = e^(-t) [(5C₁ − C₂) cos(5t) − (5C₂ + C₁) sin(5t)]
Given x(0) = 6:
6 = e^(0) [C₁ sin(0) + C₂ cos(0)]
6 = C₂
Given x'(0) = 0:
0 = e^(0) [(5C₁ − C₂) cos(0) − (5C₂ + C₁) sin(0)]
0 = 5C₁ − C₂
0 = 5C₁ − 6
C₁ = 1.2
So the solution is:
x = e^(-t) [1.2 sin(5t) + 6 cos(5t)]
Here's the graph:
desmos.com/calculator/bavfsoju5c
What is the possible solution?
[tex]\sin(3x+13)=\cos(4x)\\\sin(3x+13)=\cos(90-4x)\\3x+13=90-4x\\7x=77\\x=11[/tex]
Oscar has lost his dog in either forest A (with a priori probability 0.4) or in forest B (with a priori probability 0.6). On any given day, if the dog is in A and Oscar spends a day searching for it in A, the conditional probability that he will find the dog that day is 0.25. Similarly, if the dog is in B and Oscar spends a day looking for it there, the conditional probability that he will find the dog that day is 0.15. The dog cannot go from one forest to the other. Oscar can search only in the daytime, and he can travel from one forest to the other only at night. In which forest should Oscar look to maximize the probability he finds his dog on the first day of the search? Given that Oscar looked in A on the first day but didn't find his dog, what is the probability that the dog is in A? If Oscar flips a fair coin to determine where to look on the first day and finds the dog on the first day, what is the probability that he looked in A?
Answer:
a
Step-by-step explanation:
still a 10% chance for oscar
a) Oscar should look in forest A to maximize the probability of finding his dog on the first day. b) So, the probability that the dog is in forest A, given that Oscar looked in A on the first day but didn't find his dog, is [tex]\( \frac{1}{3} \)[/tex]. c) The probability that Oscar looked in forest A, given that he finds the dog on the first day, is approximately [tex]\(0.6579\)[/tex].
a) To maximize the probability of finding his dog on the first day, Oscar should choose the forest where the conditional probability of finding the dog on that day is higher.
For forest A:
Probability of finding the dog on the first day = Probability of the dog being in forest A × Conditional probability of finding the dog in forest A on the first day
[tex]\[ = 0.4 \times 0.25 = 0.1 \][/tex]
For forest B:
Probability of finding the dog on the first day = Probability of the dog being in forest B × Conditional probability of finding the dog in forest B on the first day
[tex]\[ = 0.6 \times 0.15 = 0.09 \][/tex]
Comparing these probabilities, Oscar should look in forest A to maximize the probability of finding his dog on the first day.
b) Given that Oscar looked in forest A on the first day but didn't find his dog, we need to find the conditional probability that the dog is in forest A.
Using Bayes' theorem:
[tex]\[ P(\text{Dog in A} | \text{Search in A, not found}) = \frac{P(\text{Search in A, not found} | \text{Dog in A}) \times P(\text{Dog in A})}{P(\text{Search in A, not found})} \][/tex]
[tex]\[ = \frac{(1 - 0.25) \times 0.4}{1 - (0.4 \times 0.25)} \][/tex]
[tex]\[ = \frac{0.75 \times 0.4}{1 - 0.1} \][/tex]
[tex]\[ = \frac{0.3}{0.9} \][/tex]
[tex]\[ = \frac{1}{3} \][/tex]
c) If Oscar flips a fair coin to determine where to look on the first day and finds the dog on the first day, we need to find the probability that he looked in forest A.
Let [tex]\( H \)[/tex] be the event that the coin flip resulted in heads (i.e., he looked in forest A), and [tex]\( D \)[/tex] be the event that he finds the dog on the first day.
Using Bayes' theorem again:
[tex]\[ P(H | D) = \frac{P(D | H) \times P(H)}{P(D)} \][/tex]
[tex]\[ = \frac{P(D | H) \times P(H)}{P(D | A) \times P(A) + P(D | B) \times P(B)} \][/tex]
Since he finds the dog on the first day, [tex]\( P(D) = P(D | A) \times P(A) + P(D | B) \times P(B) \)[/tex].
[tex]\[ P(D) = (0.25 \times 0.4) + (0.15 \times 0.6) \][/tex]
[tex]\[ P(D) = 0.1 + 0.09 \][/tex]
[tex]\[ P(D) = 0.19 \][/tex]
Now, substitute the values:
[tex]\[ P(H | D) = \frac{0.25 \times 0.5}{0.19} \][/tex]
[tex]\[ P(H | D) = \frac{0.125}{0.19} \][/tex]
[tex]\[ P(H | D) \approx 0.6579 \][/tex]
The complete question is
Oscar has lost his dog in either forest A (with a priori probability 0.4) or in forest B (with a priori probability 0.6). On any given day, if the dog is in A and Oscar spends a day searching for it in A, the conditional probability that he will find the dog that day is 0.25. Similarly, if the dog is in B and Oscar spends a day looking for it there, the conditional probability that he will find the dog that day is 0.15. The dog cannot go from one forest to the other. Oscar can search only in the daytime, and he can travel from one forest to the other only at night.
a) In which forest should Oscar look to maximize the probability he finds his dog on the first day of the search?
b) Given that Oscar looked in A on the first day but didn't find his dog, what is the probability that the dog is in A?
c) If Oscar flips a fair coin to determine where to look on the first day and finds the dog on the first day, what is the probability that he looked in A?
p 3 • p 2 • p
Multiply or divide as indicated.
heeelllpppp
For this case we have the following expression:
[tex]p ^ 3 * p ^ 2 * p =[/tex]
By definition of multiplication of powers of the same base, we have to put the same base and add the exponents, that is:
[tex]a ^ n * a ^ m = a ^ {n + m}[/tex]
So:
[tex]p ^ 3 * p ^ 2 * p = p^{ 3 + 2 + 1} = p^6[/tex]
Answer:
[tex]p^6[/tex]
What is the present value of $3,000 per year for 9 years discounted back to the present at 10 percent?
Answer:
$17,277.07
Step-by-step explanation:
Present value of annuity is the present worth of cash flow that is to be received in the future, if future value is known, rate of interest is r and time is n then PV of annuity is
PV of annuity = [tex]\frac{P[1-(1+r)^{-n}]}{r}[/tex]
= [tex]\frac{3000[1-(1+0.10)^{-9}]}{0.10}[/tex]
= [tex]\frac{3000[1-(1.10)^{-9}]}{0.10}[/tex]
= [tex]\frac{3000[1-0.4240976184]}{0.10}[/tex]
= [tex]\frac{3000(0.5759023816)}{0.10}[/tex]
= [tex]\frac{1,727.7071448}{0.10}[/tex]
= 17,277.071448 ≈ $17,277.07
Consider the following equation. f(x, y) = e−(x − a)2 − (y − b)2 (a) Find the critical points. (x, y) = a,b (b) Find a and b such that the critical point is at (−3, 8). a = b = (c) For the values of a and b in part (b), is (−3, 8) a local maximum, local minimum, or a saddle point?
a.
[tex]f(x,y)=e^{-(x-a)^2-(y-b)^2}\implies\begin{cases}f_x=-2(x-a)e^{-(x-a)^2-(y-b)^2}\\f_y=-2(y-b)e^{-(x-a)^2-(y-b)^2}\end{cases}[/tex]
Critical points occur where [tex]f_x=f_y=0[/tex]. The exponential factor is always positive, so we have
[tex]\begin{cases}-2(x-a)=0\\-2(y-b)=0\end{cases}\implies(x,y)=\boxed{(a,b)}[/tex]
b. As the previous answer established, the critical point occurs at (-3, 8) if [tex]\boxed{a=-3}[/tex] and [tex]\boxed{b=8}[/tex].
c. Check the determinant of the Hessian matrix of [tex]f(x,y)[/tex]:
[tex]\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}[/tex]
The second-order derivatives are
[tex]f_{xx}=(-2+4(x-a)^2)e^{-(x-a)^2-(y-b)^2}[/tex]
[tex]f_{xy}=4(x-a)(y-b)e^{-(x-a)^2-(y-b)^2}[/tex]
[tex]f_{yx}=4(x-a)(y-b)e^{-(x-a)^2-(y-b)^2}[/tex]
[tex]f_{yy}=(-2+4(y-b)^2)e^{-(x-a)^2-(y-b)^2}[/tex]
so that the determinant of the Hessian is
[tex]\det\mathbf H(x,y)=f_{xx}f_{yy}-{f_{xy}}^2=\left((4(x-a)^2-2)(4(y-b)^2-2)-16(x-a)^2(y-b)^2\right)e^{-2(x-a)^2-2(y-b)^2}[/tex]
[tex]\det\mathbf H(x,y)=(16(x-a)^2(y-b)^2-8(x-a)^2-8(y-b)^2)+4)e^{-2(x-a)^2-2(y-b)^2}[/tex]
The sign of the determinant is unchanged by the exponential term so we can ignore it. For [tex]a=x=-3[/tex] and [tex]b=y=8[/tex], the remaining factor in the determinant has a value of 4, which is positive. At this point we also have
[tex]f_{xx}(-3,8;a=-3,b=8)=-2[/tex]
which is negative, and this indicates that (-3, 8) is a local maximum.
The critical point of the given function is (a, b). For it to be at (-3, 8), a and b should be -3 and 8 respectively. This point (-3, 8) is a local maximum for the function.
Explanation:The critical point of the given equation f(x, y) = e−(x − a)² − (y − b)² are the coordinates (a, b).
To find the values of a and b such that the critical point is at (−3, 8), we simply set (a, b) = (−3, 8). This implies that a = -3 and b = 8.In the context of the function f(x, y), the given point (-3,8) represents a local maximum. This is because the function f(x, y) achieves its maximum value when its exponent is at a minimum, which occurs at x=a and y=b. Hence we can say that (-3, 8) is a local maximum for the function f(x, y) with a = -3 and b = 8.Learn more about Critical points here:https://brainly.com/question/32077588
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The city of Raleigh has 10500 registered voters. There are two candidates for city council in an upcoming election: Brown and Feliz. The day before the election, a telephone poll of 450 randomly selected registered voters was conducted. 237 said they'd vote for Brown, 190 said they'd vote for Feliz, and 23 were undecided.a. What is the population of this survey? b. What is the size of the population?c. What is the size of the sample?e. Give the sample statistic for the proportion of voters surveyed who said they'd vote for Brown. f. Based on this sample, we might expect how many of the 9500 voters to vote for Browng. Is this data qualitative or quantitative?
Answer:
quantitative
Step-by-step explanation:
Question 7 4 pts To borrow money, you pawn your mountain bike. Based on the value of the bike, the pawnbroker loans you $552. One month later, you get the bike back by paying the pawnbroker $851. What annual interest rate did you pay? Write as a percentage.
Answer:
The annual rate of interest is 650 %.
Step-by-step explanation:
Given,
The amount of loan = $ 552,
Total amount paid after one month = $ 851,
So, the interest for one month = $ 851 -$ 552 = $ 299,
Thus, the monthly interest = [tex]\frac{\text{Interest for a month}}{\text{Total amount of loan}}\times 100[/tex]
[tex]=\frac{299}{552}\times 100[/tex]
[tex]=\frac{29900}{552}[/tex]
Since, 1 year = 12 month ⇒ 1 month = 1/12 year,
Hence, the annual rate of interest = [tex]\frac{29900}{552}\times 12=\frac{358800}{552}=650\%[/tex]
Answer:
650.2%
Step-by-step explanation:
We have to calculate annual interest rate by this formula :
A = P( 1 + rt )
A = Future value of loan ( $851 )
P = Principal amount ( $552 )
r = Rate of interest
t = Time in years
As we know, 1 year = 12 months . By converting 1 month to year we get
1 month = [tex]\frac{1}{12}[/tex] year = 0.0833 year
Now we put the values in the formula
$851 = $552( 1 + r × 0.0833 )
= [tex]\frac{851}{552} =\frac{552(1+(0.0833r))}{552}[/tex]
= 1.5417 = 1 + 0.0833r
= 1.5417 - 1 = 0.0833r
= 0.5417 = 0.0833r
r = 6.502
r is in decimal form so we have to multiply with 100 to convert the value in percentage.
6.502 × 100 = 650.2%
The annual interest rate that you pay is 650.2%
Which has the greater energy, light of wavelength 519 nm or light with a frequency of 5.42 x 10^8 sec^-1?
Try this solution:
the rule: if f1>f2, then E1>E2, where f1;f2 - frequency, E1;E2 - energy of light.
The formula is L=c/f, where L - the wavelength, c - 3*10⁸ m/s, f - frequency.
Frequency for the wavelength 519 nm. is:
[tex]f=\frac{c}{L}=\frac{3*10^8}{519*10^{-9}}=\frac{3*10^17}{519}=578034682080924=5.78*10^{14}( \frac{1}{sec})[/tex]
Answer: the energy of light of wavelength 519 nm.
Find the given higher-order derivative.
f''(x) = 9- 3/x
f'''(x)=
Answer:
[tex]f'''(x)=\frac{3}{x^{2}}[/tex]
Step-by-step explanation:
We are given with Second-order derivative of function f(x).
[tex]f''(x)=9-\frac{3}{x}[/tex]
We need to find Third-order derivative of the function f(x).
[tex]f''(x)=9-\frac{3}{x}=9-3x^{-1}[/tex]
We know that,
f'''(x) = (f''(x))'
So,
[tex]f'''(x)=\frac{\mathrm{d}\,f''(x)}{\mathrm{d} x}[/tex]
[tex]f'''(x)=\frac{\mathrm{d}\,(9-3x^{-1})}{\mathrm{d} x}[/tex]
[tex]f'''(x)=\frac{\mathrm{d}\,9}{\mathrm{d} x}-\frac{\mathrm{d}\,3x^{-1}}{\mathrm{d} x}[/tex]
[tex]f'''(x)=0-3\frac{\mathrm{d}\,x^{-1}}{\mathrm{d} x}[/tex]
[tex]f'''(x)=-3(-1)x^{-1-1}[/tex]
[tex]f'''(x)=3x^{-2}[/tex]
[tex]f'''(x)=\frac{3}{x^{2}}[/tex]
Therefore, [tex]f'''(x)=\frac{3}{x^{2}}[/tex]
What are the solutions of the following system?
Answer:(-6,312), (6,312)
Step by Step explanation:
Solve the first equation for y.10x^2-y=48
y=-48+10x^2
Substitute the given value of y into the equation 2y=16x^2+482(-48+10x^2)=16x^2+48
Solve the equation for x.
x=-6
x=6
Substitute the given value of x into the equation y=-48+10(-6)^y=-48+10(-6)^2
y=-48+10×6^2
Solve the equation for yy=312
y=312
(08.03 LC) Factor completely: x2 + 10x + 24 (5 points) Prime (x + 12)(x + 2) (x + 3)(x + 8) (x + 6)(x + 4)
Answer:
Option C is correct
Step-by-step explanation:
We need to factorize the expression:
[tex]x^2+10x+24[/tex]
For factorization we need to break the middle term such that the product is equal to 24x^2 and the sum is equal to 10x
We know that 6*4 = 24 and 6+4 =10
So, solving
[tex]=x^2+6x+4x+24[/tex]
Taking common
[tex]=x(x+6)+4(x+6)[/tex]
[tex]=(x+4)(x+6)[/tex]
So, the factors of
[tex]x^2+10x+24[/tex]
are
[tex](x+4)(x+6)[/tex]
Hence Option C is correct
The correct answer is actually D. The person who had answered before me had accurate math, but they were confused. The answer
(x + 6)(x + 4) is on D, not C.
The volume of Saturn is about 8.27 x 10 cubic kilometers. The volume of Earth is about 1.09 x 102 cubic kilometers. The number of Earths that can fit
inside Saturn can be found by dividing Saturn's volume by Earth's volume. Find this quotient and express the answer in scientific notation.
A: 9.01 x 10^26
B: 75.9 x 10^1
C:759
D:7.59x10^2
Answer:
Option D: [tex]7.59*10^{2}[/tex]
Step-by-step explanation:
we know that
The volume of Saturn is about [tex]8.27*10^{14}\ km^{3}[/tex]
The volume of Earth is about [tex]1.09*10^{12}\ km^{3}[/tex]
Dividing Saturn's volume by Earth's volume
[tex]\frac{8.27*10^{14}}{1.09*10^{12}} =(\frac{8.27}{1.09})*10^{14-12} =7.59*10^{2}[/tex]
The correct answer is option D. The quotient in scientific notation is [tex]75.9 \times 10^2[/tex]
To find the number of Earths that can fit inside Saturn, we divide the volume of Saturn by the volume of Earth. Given the volumes:
Saturn's volume [tex]= 8.27 \times 10^{26}\ km^3[/tex]
Earth's volume [tex]= 1.09 \times 10^{24}\ km^3[/tex]
The calculation is as follows:
Number of Earths in Saturn = Saturn's volume / Earth's volume
[tex]=\frac{ (8.27 \times 10^{26})} {(1.09 \times 10^{24})}[/tex]
To simplify this, we divide the coefficients and subtract the exponents of 10:
[tex]= \frac{8.27}{ 1.09} \times \frac{10^{26}} { 10^{24}}[/tex]
[tex]= 7.59 \times 10^2[/tex]
(1 point) The players on a soccer team wear shirts, with each player having one of the numbers 1, 2, ..., 11 on their backs. The set A contains players with even numbers on their shirts. The set B comprises players wearing an odd number less than 7. The set C contains the defenders, which are those wearing numbers less than 6. Select the correct set that corresponds to each of the following. Part a) A∩(B∪C) A. {1,2,3,4,5} B. ∅ C. {1,3,5} D. {2,4} E. {2} Part b) (A∩Bc)∪(B∩C)c A. {6,7,8,9,11} B. {2,4,6,7,8,9,10,11} C. {2,3,4,5,6,8,10} D. {1,2,3,4,5,6,8,10} E. {6,7,8,10,11}
This question involves operations on sets to identify specific members based on conditions. Part a) resolves to D. {2,4}, while part b) finds the solution to be B. {2,4,6,7,8,9,10,11}, highlighting the application of intersection, union, and complement operations in set theory.
Explanation:To solve these problems, we need to understand the operations on sets such as intersection (A∩B), union (A∪B), and the complement of a set (Bc). For part a), we identify set A as {2,4,6,8,10}, B as {1,3,5}, and C as {1,2,3,4,5}. A∩(B∪C) means we're looking for the intersection of A with the union of B and C. Since B∪C = {1,2,3,4,5}, intersecting this with A gives us D. {2,4} as the answer.
For part b), (A∩Bc)∪(B∩C)c means we're looking at elements in A but not in B, combined with elements not in both B and C. Since Bc = {6,7,8,9,10,11} and (B∩C)c = {6,7,8,9,10,11}, union these two gives us answer B. {2,4,6,7,8,9,10,11}, by including A∩Bc = {2,4,6,8,10} and excluding duplicates when union with (B∩C)c.
Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based at 0. a.Give your answer using summation notation. b.Give the interval on which the series converges.
In this question (https://brainly.com/question/12792658) I derived the Taylor series for [tex]\mathrm{sinc}\,x[/tex] about [tex]x=0[/tex]:
[tex]\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}[/tex]
Then the Taylor series for
[tex]f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt[/tex]
is obtained by integrating the series above:
[tex]f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}[/tex]
We have [tex]f(0)=0[/tex], so [tex]C=0[/tex] and so
[tex]f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}[/tex]
which converges by the ratio test if the following limit is less than 1:
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}[/tex]
Like in the linked problem, the limit is 0 so the series for [tex]f(x)[/tex] converges everywhere.
The Taylor series for the function f(x) = ∫ sinc(t)dt based at 0 is derived from the Taylor series of sinc(x) by integrating it term by term, given in summation notation as ∑ (-1)ⁿ * xⁿ⁺¹ / (n+1)! for n=0 to n=∞. The series converges for all real numbers (-∞, ∞).
Explanation:In order to find the Taylor series for the function f(x) = ∫ sinc(t)dt based at 0, one can use the Taylor series for sinc(x) and integrate term by term. We know the Taylor series for sinc(x) is x - x³/3! + x⁵/5! - ..., so the Taylor series for f(x) can be written as x²/2 - x⁴/4*3! + x⁶/6*5! - ... . In summation notation, this is ∑ (-1)ⁿ * xⁿ⁺¹ / (n+1)! for n=0 to n=∞.
The Taylor series for any function converges to the function itself within a certain interval called the radius of convergence. For the Taylor series of sinc(x), due to the nature of sine being bounded between -1 and 1, the series will converge for all real numbers (-∞, ∞).
Learn more about Taylor series here:https://brainly.com/question/32235538
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Calculate the annual effective interest rate of a 12 % nominal annual interest rate compound monthly
Answer:
12.683%
Step-by-step explanation:
The effective annual rate is given by ...
(1 +r/n)^n -1
where r is the nominal annual rate, and n is the number of compoundings per year. Filling in the given numbers, we have ...
effective rate = (1 +0.12/12)^12 -1 ≈ 0.12683 = 12.683%
Assume that 1400 births are randomly selected and 1378 of the births are girls. Use subjective judgment to describe the number of girls as significantly high, significantly low, or neither significantly low nor significantly high. Choose the correct answer below. A. The number of girls is neither significantly low nor significantly high. B. The number of girls is significantly high. C. The number of girls is significantly low. D. It is impossible to make a judgment with the given information.
Answer: Hence, Option 'B' is correct.
Step-by-step explanation:
Since we have given that
Number of births = 1400
Number of birth of girls = 1378
Number of birth of boys is given by
[tex]1400-1378\\\\=22[/tex]
so, the number of girls is significantly higher than the number of boys.
So, the number of births of girls is significantly high.
Hence, Option 'B' is correct.
B. The number of girls is significantly high.
When evaluating whether the number of girl births in a sample is significantly high or low, we can reference the expected natural ratio of girls to boys, which is typically 100:105. Given that in the scenario 1378 out of 1400 births were girls, this significantly deviates from the expected natural ratio. For comparison, an article in Newsweek states that the natural ratio is 100:105, and in China, it is 100:114, which equals 46.7 percent girls. If we consider a sample where out of 150 births, there are 60 girls (or 40%), this is lower than the expected percentage based on China's statistics but not implausible. However, in the case of the scenario with 1378 girls out of 1400 births, the proportion of girls is approximately 98.43%, which seems very unlikely given the natural ratio, suggesting an unusual or non-random process may be involved.
Therefore, based on subjective judgment and without applying more precise statistical tests, the number of girls being 1378 out of 1400 births is significantly high compared to natural birth rates or the stated birth rate in China. This leads us to select the correct answer: B. The number of girls is significantly high.
The GMAC Insurance company reported that the mean score on the National Drivers Test was 69.9 with a standard deviation of 3.7 points. The test scores are approximately bell-shaped. Approximately 68% of all test scores were between two values A and B. What is the value of A? Write only a number as your answer. Round to one decimal place.
Answer:
66.2
Step-by-step explanation:
We know that the middle 68% of the distribution is between -1 and +1 standard deviations from the mean. -1 standard deviation is a score of ...
69.9 - 3.7 = 66.2
_____
Comment on the question
The way the question is worded, "A" can be any value less than 68.1, down to -∞. The question does does not require the A-B range to be symmetrical.
Explain how each of the following rates satisfies the definition of ratio. Given an example of how each is used.
A. 1580 people/square mile
B. 360 kilowatt-hour/4 months
C. 450 people/year
D. 355 calories/6 ounces
Answer:
Step-by-step explanation:
A ratio is defined as a comparison of two amounts by division.It is of the form of [tex]\frac{p}{q}[/tex] where p and q are the quantities.
A. 1580 people/square mile in division form
1580[tex]\frac{people}{\text{square mile}}[/tex]
This satisfies ratio definition as this compares two quantities People and Square mile and is of the form [tex]\frac{p}{q}[/tex]
Where,
p: People, q: Square mile
Example,
[tex]\frac{1580\times people}{1\times squaremile}[/tex]
expresses that per 1 square mile there are 1580 people. Thus rate satisfies the ratio definition.
B. 360 kilowatt-hour/4 months
in division form [tex]\frac{\text{360 kilowatt-hour}}{\text{4months}}[/tex]
This satisfies ratio definition as this compares two quantities kilowatt-hour and months and is of the form [tex]\frac{p}{q}[/tex]
where, p: kilowatt-hour and q: people
Example,
[tex]\frac{\360\times kilowatt-hour}{4\times months}[/tex]
expresses that per 4 months there are 360 kilowatt-hour. Thus, rate satisfies the ratio definition.
C. 450 people/year
In division form 450[tex]\frac{people}{year}[/tex]
This satisfies ratio definition as this compares two quantities people and year and is of the form [tex]\frac{p}{q}[/tex]
where, p: people and q: year
Example,
[tex]\frac{450\times people}{1\times\text{year}}[/tex]
expresses that per year there are 450 people. Thus, rate satisfies the ratio definition.
D. 355 calories/6 ounces
In division form [tex]\frac{355clories}{6ounces}[/tex]
This satisfies ratio definition as this compares two quantities calories and ounces and is of the form [tex]\frac{p}{q}[/tex]
where p: calories and q: ounces
Example,
[tex]\frac{355\times calories}{6\times\text{ounces}}[/tex]
expresses that per 6 ounces there are 355 calories. Thus, rate satisfies the ratio definition.
Find the volume of the wedge cut from the first octant by the cylinder z=12-3y^2 and the plane x+y=2.
Answer:
The wedge cut from the first octant ⟹ z ≥ 0 and y ≥ 0 ⟹ 12−3y^2 ≥ 0 ⟹ 0 ≤ y ≤ 2
0 ≤ y ≤ 2 and x = 2-y ⟹ 0 ≤ x ≤ 2
V = ∫∫∫ dzdydx
dz has changed from zero to 12−3y^2
dy has changed from zero to 2-x
dx has changed from zero to 2
V = ∫∫∫ dzdydx = ∫∫ (12−3y^2) dydx = ∫ 12(2-x)-(2-x)^3 dx =
24(2)-6(2)^2+(2-2)^4/4 -(2-0)^4/4 = 20
Step-by-step explanation:
It can be deduced that the volume of the wedge cut from the first octant will be 20.
How to calculate the volumeFrom the information, the wedge cut from the first octant will be z ≥ 0 and y ≥ 0 = 12−3y² ≥ 0 = 0 ≤ y ≤ 2
Also, it can be deduced that 0 ≤ y ≤ 2 and x = 2-y ⟹ 0 ≤ x ≤ 2. Therefore, V = ∫dzdydx. In this case,
dz = 12−3y
dy = 2-x
dx = 2
V = ∫dzdydx = ∫(12−3y²) dydx = ∫12(2-x)-(2-x)³dx
= [24(2) - 6(2)² + (2-2)⁴/4 -(2-0)⁴]/4
= 20
In conclusion, the volume of the wedge cut from the first octant will be 20.
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(a + 3)(a - 2)
hurry helpppp asapppp
(a+3)(a-2)
Multiply the two brackets together
a^2-2a+3a+3*-2
a^2+a-6
Answer is a^2+a-6
ANSWER
[tex]{a}^{2} + a - 6[/tex]
EXPLANATION
The given expression is
[tex](a + 3)(a - 2)[/tex]
We expand using the distributive property to obtain:
[tex]a(a - 2) + 3(a - 2)[/tex]
We multiply out the parenthesis to get:
[tex] {a}^{2} - 2a + 3a - 6[/tex]
Let us now simplify by combining the middle terms to obtain;
[tex]{a}^{2} + a - 6[/tex]
A scientist has two solutions, which she has labeled Solution A and Solution B. Each contains salt. She knows that Solution A is 70% salt and Solution B is 95% salt. She wants to obtain 150 ounces of a mixture that is 75% salt. How many ounces of each solution should she use?
Answer:
Solution A: 120 ounces
Solution B: 30 ounces
Step-by-step explanation:
Let's call A the amount of Solution A. Solution A is 70% salt
Let's call B the amount of Solution B. Solution A is 95% salt
The resulting mixture should have 75% salt and 150 ounces .
Then we know that the total amount of mixture will be:
[tex]A + B = 150[/tex]
Then the total amount of salt in the mixture will be:
[tex]0.7A + 0.95B = 0.75 * 150[/tex]
[tex]0.7A + 0.95B = 112.5[/tex]
Then we have two equations and two unknowns so we solve the system of equations. Multiply the first equation by -0.95 and add it to the second equation:
[tex]-0.95A -0.95B = 150 * (-0.95)[/tex]
[tex]-0.95A -0.95B =-142.5[/tex]
[tex]-0.95A -0.95B =-142.5[/tex]
+
[tex]0.7A + 0.95B = 112.5[/tex]
--------------------------------------
[tex]-0.25A = -30[/tex]
[tex]A = \frac{-30}{-0.25}[/tex]
[tex]A = 120\ ounces[/tex]
We substitute the value of A into one of the two equations and solve for B.
[tex]120 + B = 150[/tex]
[tex]B = 30\ ounces[/tex]
3.17 Scores on stats final. Below are final exam scores of 20 Introductory Statistics students. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 57, 66, 69, 71, 72, 73, 74, 77, 78, 78, 79, 79, 81, 81, 82, 83, 83, 88, 89, 94 (a) The mean score is 77.7 points. with a standard deviation of 8.44 points. Use this information to determine if the scores approximately follow the 68-95-99.7% Rule. (b) Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below.
Answer:
Yes they do.
And yes they do follow a normal distribution.
Percentages are close to 68-95-99.7%, we can declare that yes, the 68-95-99.7% rule is roughly followed and yes data appear to follow a normal distribution.
What is a normal distribution?It's the probability curve of a continuous distribution that's most likely symmetric around the mean. On the Z curve, at Z=0, the chance is 50-50. A bell-shaped curve is another name for it.
We have a data of final exam scores of 20 Introductory.
a) Range of 1 standard deviation:
(77.7 – 8.44, 77.7 + 8.44) [69.3, 86.1]
Range of 2 standard deviation:
(77.7 – 2(8.44), 77.7 + 2(8.44)) [60.8, 94.6]
Range of 3 standard deviation:
(77.7 – 3(8.44), 77.7 + 3(8.44)) [52.4, 103.0]
Number of data points lie within 1 standard deviation = 14
Percent of data points lie within 1 SD = (14/20)×100 = 70%
Number of data points lie within 2 SD = 19
Percent of data points lie within 1 SD = (19/20)×100 = 95%
Number of data points lie within 3 SD = 20
Percent of data points lie within 1 SD = (20/20)×100 = 100%
Because these percentages are close to 68-95-99.7%, we can declare that yes, the 68-95-99.7% rule is roughly followed.
b)
Because the histogram in the graph is symmetric, and the normal probability plot reveals that the points are very close to a straight line, the data appears to follow a normal distribution.
Thus, percentages are close to 68-95-99.7%, we can declare that yes, the 68-95-99.7% rule is roughly followed and yes data appear to follow a normal distribution.
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Angle measures and segment lengths someone please explain!!!
Intersecting chords theorem:
[tex]7\cdot7=10x[/tex]
i.e. when two chords intersect, the products of the resulting line segments' lengths are equal. Then
[tex]10x=49\implies x=\dfrac{49}{10}=\boxed{4.9}[/tex]
Answer:
4.9
Step-by-step explanation:
got it right
Piney Woods Conservation is a company that attempts to help offset the effects of deforestation. A local forest contains approximately 500,000 trees. Lumber companies are continuously clearing the forest at a rate of 4.7% per year. Piney Woods Conservation is about to begin planting trees in the region throughout each year at an average rate of 15,000 trees per year. They are curious to know how long it will be before the number of trees they have planted will be equal to the number of trees still remaining in the forest.
Answer:
15.7 years
Step-by-step explanation:
we know that
The deforestation is a exponential function of the form
[tex]y=a(b)^{x}[/tex]
where
y ----> the number of trees still remaining in the forest
x ----> the number of years
a is the initial value (a=500,000 threes)
b is the base
b=100%-4.7%=95.3%=95.3/100=0.953
substitute
[tex]y=500,000(0.953)^{x}[/tex]
The linear equation of planting threes in the region is equal to
[tex]y=15,000x[/tex]
using a graphing tool
Solve the system of equations
The intersection point is (15.7,235,110)
see the attached figure
therefore
For x=15.7 years
The number of trees they have planted will be equal to the number of trees still remaining in the forest
Answer: There is only one solution, and it is viable.
Step-by-step explanation:
A particle moves along line segments from the origin to the points (2, 0, 0), (2, 4, 1), (0, 4, 1), and back to the origin under the influence of the force field. F(x, y, z) = z^2i + 3xyj + 2y^2k. Find the work done.
The work is equal to the line integral of [tex]\vec F[/tex] over each line segment.
Parameterize the paths
from (0, 0, 0) to (2, 0, 0) by [tex]\vec r_1(t)=t\,\vec\imath[/tex] with [tex]0\le t\le2[/tex],from (2, 0, 0) to (2, 4, 1) by [tex]\vec r_2(t)=2\,\vec\imath+4t\,\vec\jmath+t\,\vec k[/tex] with [tex]0\le t\le1[/tex],from (2, 4, 1) to (0, 4, 1) by [tex]\vec r_3(t)=(2-t)\,\vec\imath+4\,\vec\jmath+\vec k[/tex] with [tex]0\le t\le2[/tex], andfrom (0, 4, 1) to (0, 0, 0) by [tex]\vec r_4(t)=(4-4t)\,\vec\jmath+(1-t)\,\vec k[/tex] with [tex]0\le t\le1[/tex]The work done by [tex]\vec F[/tex] over each segment (call them [tex]C_1,\ldots,C_4[/tex]) is
[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0[/tex]
[tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3[/tex]
[tex]\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2[/tex]
[tex]\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3[/tex]
Then the total work done by [tex]\vec F[/tex] over the particle's path is 46.
The total work done by the force field as the particle follows this path is 4/3 units of work.
Explanation:To find the total work done by the force field as the particle moves along the path, we'll calculate the line integrals for each segment and then sum them up:
For the first segment (from the origin to (2, 0, 0)):
Parameterization: r1(t) = (2t, 0, 0), t from 0 to 1.
Work on this segment: ∫[0,1] F(r1(t)) · r1'(t) dt
F(r1(t)) = (0, 0, 0), as z = 0.
r1'(t) = (2, 0, 0).
∫[0,1] F(r1(t)) · r1'(t) dt = ∫[0,1] (0) dt = 0.
For the second segment (from (2, 0, 0) to (2, 4, 1)):
Parameterization: r2(t) = (2, 4t, t), t from 0 to 1.
Work on this segment: ∫[0,1] F(r2(t)) · r2'(t) dt
F(r2(t)) = (t^2, 0, 8t^2).
r2'(t) = (0, 4, 1).
∫[0,1] F(r2(t)) · r2'(t) dt = ∫[0,1] (0 + 0 + 8t^2) dt = ∫[0,1] 8t^2 dt = [8t^3/3] from 0 to 1 = (8/3).
For the third segment (from (2, 4, 1) to (0, 4, 1)):
Parameterization: r3(t) = (2 - 2t, 4, 1), t from 0 to 1.
Work on this segment: ∫[0,1] F(r3(t)) · r3'(t) dt
F(r3(t)) = (1, 0, 8).
r3'(t) = (-2, 0, 0).
∫[0,1] F(r3(t)) · r3'(t) dt = ∫[0,1] (-2) dt = [-2t] from 0 to 1 = -2.
For the fourth segment (from (0, 4, 1) back to the origin):
Parameterization: r4(t) = (0, 4 - 4t, 1), t from 0 to 1.
Work on this segment: ∫[0,1] F(r4(t)) · r4'(t) dt
F(r4(t)) = (1, 0, 32 - 8t).
r4'(t) = (0, -4, 0).
∫[0,1] F(r4(t)) · r4'(t) dt = ∫[0,1] (0 + 0 + 0) dt = 0.
Now, sum up the work done on each segment:
Total work = 0 + (8/3) - 2 + 0 = 4/3.
So, the total work done by the force field as the particle follows this path is 4/3 units of work.
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Which is an equation for the nth terms of the sequence 12,15,18,21
[tex]\bf 12~~,~~\stackrel{12+3}{15}~~,~~\stackrel{15+3}{18}~~,~~\stackrel{18+3}{21}~\hspace{10em}\stackrel{\textit{common difference}}{d=3} \\\\[-0.35em] ~\dotfill\\\\ n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} a_1=12\\ d=3 \end{cases} \\\\\\ a_n=12+(n-1)3\implies a_n=12+3n-3\implies a_n=3n+9[/tex]
Answer:
tₙ = 3(3 + n)
Step-by-step explanation:
Points to remember
nth term of an AP
tₙ = a + (n - 1)d
Where a - first term of AP
d - Common difference of AP
To find the nth term
The given series is,
12,15,18,21 .....
Here a = 12 and d = 15 - 12 = 3
tₙ = a + (n - 1)d
= 12 + (n - 1)3
=12 + 3n - 3
= 9 + 3n
= 3(3 + n)
Therefore tₙ = 3(3 + n)
The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2, y2 = y1(x) e−∫P(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). y'' + 2y' + y = 0; y1 = xe−x
First confirm that [tex]y_1=xe^{-x}[/tex] is a solution to the ODE,
[tex]y''+2y'+y=0[/tex]
We have
[tex]{y_1}'=e^{-x}-xe^{-x}=(1-x)e^{-x}[/tex]
[tex]{y_1}''=-e^{-x}-(1-x)e^{-x}=(-2+x)e^{-x}[/tex]
Substituting into the ODE gives
[tex](-2+x)e^{-x}+2(1-x)e^{-x}+xe^{-x}=0[/tex]
Suppose [tex]y_2(x)=v(x)y_1(x)[/tex] is another solution to this ODE. Then
[tex]{y_2}'=v'y_1+v{y_1}'[/tex]
[tex]{y_2}''=v''y_1+2v'{y_1}'+v{y_1}''[/tex]
and substituting these into the ODE yields
[tex](v''y_1+2v'{y_1}'+v{y_1}'')+2(v'y_1+v{y_1}')+vy_1=0[/tex]
[tex]xe^{-x}v''+2e^{-x}v'=0[/tex]
[tex]xv''+2v'=0[/tex]
Let [tex]w(x)=v'(x)[/tex]. Then the remaining ODE is linear in [tex]w[/tex]:
[tex]xw'+2w=0[/tex]
Multiply both sides by the integrating factor, [tex]x[/tex], and condense the left hand side as a derivative of a product:
[tex]x^2w'+2xw=(x^2w)'=0[/tex]
Integrate both sides with respect to [tex]x[/tex] and solve for [tex]w[/tex]:
[tex]x^2w=C_1\implies w=C_1x^{-2}[/tex]
Back-substitute and integrate both sides with respect to [tex]x[/tex] to solve for [tex]v[/tex]:
[tex]v'=C_1x^{-2}\implies v=-C_1x^{-1}+C_2[/tex]
Back-substitute again to solve for [tex]y_2[/tex]:
[tex]\dfrac{y_2}{y_1}=C_2-\dfrac{C_1}x[/tex]
[tex]\implies y_2=C_2xe^{-x}-C_1e^{-x}[/tex]
[tex]y_1[/tex] already captures the solution [tex]xe^{-x}[/tex], so the remaining one is
[tex]\boxed{y_2=e^{-x}}[/tex]
A differential equation shows the relationship between functions and their derivatives.
The equation of [tex]y_2(x)[/tex] is: [tex]y_2 = -e^{-x}[/tex]
The given parameters are:
[tex]y_2 = y_1(x) \int\frac{ e^{(\int -P(x)\ dx)} }{ y_1^2(x) }dx[/tex]
[tex]y" + 2y' + y = 0[/tex]
[tex]y_1 = xe^{-x}[/tex]
The general equation is:
[tex]y" + P(x) y' + Q(x)y = 0[/tex]
Compare the above equation to [tex]y" + 2y' + y = 0[/tex]
[tex]P(x) = 2[/tex]
Integrate:
[tex]\int\limits^x_0 P(x') dx'= \int\limits^x_0 2 dx'[/tex]
[tex]\int\limits^x_0 P(x') dx'= 2x|\limits^x_0[/tex]
[tex]\int\limits^x_0 P(x') dx'= 2[x - 0][/tex]
[tex]\int\limits^x_0 P(x') dx'= 2[x ][/tex]
[tex]\int\limits^x_0 P(x') dx'= 2x[/tex]
We have:
[tex]y_2 = y_1(x) \int\frac{ e^{(\int -P(x)\ dx)} }{ y_1^2(x) }dx[/tex]
The above equation becomes:
[tex]y_2 = y_1(x) \int\frac{e^{(\int -2 dx)} }{ y_1^2(x) }dx[/tex]
Substitute [tex]y_1 = xe^{-x}[/tex]
[tex]y_2 = xe^{-x} \int\frac{ e^{(\int -2 dx)} }{ (xe^{-x})^2 }dx[/tex]
Integrate
[tex]y_2 = xe^{-x} \int\frac{ e^{-2x}}{ (xe^{-x})^2 }dx[/tex]
Evaluate the exponents
[tex]y_2 = xe^{-x} \int\frac{ e^{-2x}}{ x^2e^{-2x} }dx[/tex]
Cancel out common factors
[tex]y_2 = xe^{-x} \int\frac{1}{ x^2 }dx[/tex]
Rewrite as:
[tex]y_2 = xe^{-x} \int x^{-2}dx[/tex]
Integrate
[tex]y_2 = xe^{-x} \times -\frac{1}{x} + c[/tex]
[tex]y_2 = -e^{-x} + c[/tex]
Set c to 0.
[tex]y_2 = -e^{-x} + 0[/tex]
[tex]y_2 = -e^{-x}[/tex]
Hence, the equation of [tex]y_2(x)[/tex] is:
[tex]y_2 = -e^{-x}[/tex]
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1 -For what value of x is line a parallel to line b
2-For what value of x is line a parallel to line b
Answer:
1) x = 17, line a parallel to line b
2) x = 18, line a parallel to line b
Step-by-step explanation:
If line a parallel to line b then (10x - 40) + 50 = 180
Solve for x
10x - 40 + 50 = 180
Combine like terms
10x + 10 = 180
10x = 170
x = 17
x = 17, line a parallel to line b
-------------------------------------------------
If line a parallel to line b then 5x - 16 = 74
Solve for x
5x - 16 = 74
5x = 90
x = 18
x = 18, line a parallel to line b
The value of x which makes line a parallel to line b can be found by equating the slopes of the two lines and solving for x.
Explanation:In mathematics, two lines a and b are parallel if and only if their slopes are equal. When we are given the equations of the lines and are asked to find the value of x that make the lines parallel, we start by setting the slopes of the two lines equal to each other. Let's assume now that line a is represented by y = mx + b1 and line b by y = nx + b2. In order for line a to be parallel to line b, m must be equal to n. Therefore, you solve for x from the equation m=n.
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Please help someone
Answer:
1. Y
2. N
3. N
4. N
Step-by-step explanation:
Let's use the second equation, since it seems to be easier to use.
To check if an ordered pair is a solution, plug it in to the equation.
1. [tex]-10+18=8[/tex] --> Y
2. [tex]25-12=13[/tex] --> N
3. [tex]0-9=-9[/tex] --> N
4. [tex]35-27=8[/tex] --> N
Edit : The 4th equation doesn't work for the first equation, whereas the first one still does.
Answer:
(-2,-6)
Step-by-step explanation:
-9x +2y = 6
5x - 3y = 8
1) Make one of the coefficients the same - y.
-9x +2y = 6 * 3
5x - 3y = 8 * 3
-27x +6y = 18
10x - 6y = 16
2) Add the new equations.
(-27x +6y = 18) + (10x - 6y = 16)
(-27x +6y) + (10x - 6y) = 18 + 16
-17x = 34
3) Divide to find the value of x
-17x = 34
x = 34/-17
x = -2
4) Substitute x into either equation to find the value of y.
-9(-2) +2y = 6
18 +2y = 6
2y = -12
y = -12/2
y = -6
5(-2) - 3y = 8
-10 - 3y = 8
-3y = 18
y = 18/-3
y = -6
Your answer is (-2,-6).