Answer:
7/36
Step-by-step explanation:
7/9* 1/4
Multiply the numerator
7*1 = 7
9*4 = 36
Put the numerator over the denominator
7/36
Answer:
Brunliest plzStep-by-step explanation:
7/9 x 1/4 = 7/36
if a square field is 35m find its perimeter
Answer:
i think its 140m
Step-by-step explanation:
If researchers report that the results from their study were significant, p< .05, this means that:
A.) If they conducted the study over they would get the same results less than 5 times in 100.
B.) The results are untrustworthy
C.) We would expect the results to occur by chance less than 95 times out of 100
D.) We would expect the results to occur by chance less than 5 times out of 100.
If researchers report that the results from their study were significant, p < .05, this means that we would expect the results to occur by chance less than 5 times out of 100.
What does it indicate when researchers report results with p < .05?When researchers report that the results from their study were significant, p < .05, it means that we would expect the results to occur by chance less than 5 times out of 100. This threshold is commonly used in statistical hypothesis testing.
In statistical hypothesis testing, the p-value represents the probability of obtaining results as extreme as the observed results, assuming that the null hypothesis is true. If the p-value is less than the predetermined significance level of .05 (or 5%), it is considered statistically significant. This means that the observed results are unlikely to have occurred by chance alone, and there is evidence to support the alternative hypothesis.
Therefore, option D, "We would expect the results to occur by chance less than 5 times out of 100," is the correct interpretation of reporting significant results with p < .05.
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Sunshine had $29 and used $10,how much does she have left?
Answer:
Sunshine has $19 left after spending $10.
Step-by-step explanation:
29 - 10 = x
Subract 10 from 29 to solve for x.
x = 19
I really need help on this ASAP
A hostel of 500 boys has provisions for 4 days. If the provisions were continued for
20 days, how many students have gone out of the hostel ?
plzzzzzzzzzzzzzz tell me answer now I need help
Given :
A hostel of 500 boys has provisions for 4 days.
To Find :
If the provisions were continued for 20 days, how many students have gone out of the hostel.
Solution :
Let, number of students left after 20 days are x.
Now, provisions are given by :
\(\text{Provisions = Number of students}\times \text{Number of days}\)
Also, provisions remains same.
So, initial provision = final provision
\(500\times 4 = x \times 20\\\\x=100\)
Therefore, 100 students have gone out of the hostel.
Hence, this is the required solution.
One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction y of the population who have heard the rumor and the fraction who have not heard the rumor. (a) Write a differential equation that is satisfied by y. (Use k for the constant of proportionality.)
dy/dt = ____
(b) Solve the differential equation. Assume y(0) = C. y = _____
(c) A small town has 1300 inhabitants. At 8 AM, 100 people have heard a rumor. By noon half the town has heard it. At what time will 90% of the population have heard the rumor? (Do not round k in your calculation. Round the final answer to one decimal place.) ______hours after the beginning
(a) The differential equation that is satisfied by y is:
\(\frac{dy}{dt} = ky(1-y)\)
(b) To solve the differential equation, we separate the variables and integrate both sides:
\(\frac{dy}{y*(1-y)} = k*dt\)
Integrating both sides, we get:
\(\frac{lnly}{1-y} = k*t +c1\)
where C1 is an arbitrary constant of integration.
We can rewrite the equation in terms of y:
\(\frac{y}{1-y} = e^{(k*t+c1)}\)
Multiplying both sides by (1-y), we get:
\({y} = e^{(k*t+c1)} *(1-y)\)
\(y= \frac{C}{(1+(c-1)e^{-kt} }\)
where C = y(0) is the initial fraction of the population who have heard the rumor.
(c) In this case, the initial fraction of the population who have heard the rumor is y(0) = \(\frac{100}{1300}\) = 0.077. At noon, half the town has heard the rumor, so y(4) = 0.5.
Substituting these values into the equation from part (b), we get:
\(0.5= \frac{0.077}{1+(0.777-1) e^{-k4} }\)
Solving for k, we get:
\(k= ln(\frac{12.857}{4} )\)
Substituting this value of k into the equation from part (b), and setting y = 0.9 (since we want to find the time at which 90% of the population has heard the rumor), we get:
\(0.9= \frac{0.077}{1+(0.777-1) e^{-ln(12.857}*\frac{t}{4} }\))
Solving for t, we get:
t = 8.7 hours after the beginning (rounded to one decimal place)
A differential equation is a mathematical equation that relates a function to its derivatives. It is a powerful tool used in many fields of science and engineering to describe how physical systems change over time. The equation typically includes the independent variable (such as time) and one or more derivatives of the dependent variable (such as position, velocity, or temperature).
Differential equations can be classified based on their order, which refers to the highest derivative present in the equation, and their linearity, which determines whether the equation is a linear combination of the dependent variable and its derivatives. Solving a differential equation involves finding a function that satisfies the equation. This can be done analytically or numerically, depending on the complexity of the equation and the available tools.
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Find the amount of time.
I=$10, P=$100, r=2%
Solve -14x+7y=28 for y
Answer: y= 4+2x
Step-by-step explanation: Move all terms that don't contain y to the right side and solve.
Carlisle Transport had $4,520 cash at the beginning of the period. During the period, the firm collected $1,654 in receivables, paid $1,961 to supplier, had credit sales of $6,916, and incurred cash expenses of $500. What was the cash balance at the end of the period?
To calculate the cash balance at the end of the period, we need to consider the cash inflows and outflows.
Starting cash balance: $4,520
Cash inflows: $1,654 (receivables collected)
Cash outflows: $1,961 (payments to suppliers) + $500 (cash expenses)
Total cash inflows: $1,654
Total cash outflows: $1,961 + $500 = $2,461
To calculate the cash balance at the end of the period, we subtract the total cash outflows from the starting cash balance and add the total cash inflows:
Cash balance at the end of the period = Starting cash balance + Total cash inflows - Total cash outflows
Cash balance at the end of the period = $4,520 + $1,654 - $2,461
Cash balance at the end of the period = $4,520 - $807
Cash balance at the end of the period = $3,713
Therefore, the cash balance at the end of the period is $3,713.
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Suppose p(A|B) = 0.7, P(A|BC) = 0.3, and p(A) = 0.5- What is p(B) (write it up to first decimal place)?_____
p(B) = 0.5 up to the first decimal place.
We can use Bayes' theorem to find p(B) given the conditional probabilities provided:
\(p(A|B) = 0.7, p(A|BC) = 0.3, and p(A) = 0.5\)
First, we can use the law of total probability to find p(A|B) and p(A|BC):
\(p(A) = p(A|B)p(B) + p(A|BC)p(B^c)\)
where \(B^c\) denotes the complement of B, i.e., not B. Substituting the given values, we get:
\(0.5 = 0.7p(B) + 0.3p(B^c)\)
We can rearrange this equation to solve for p(B):
\(0.2 = 0.4p(B) - 0.3p(B^c)\\0.5 = p(B) + p(B^c)\)
Substituting \(p(B^c) = 1 - p(B)\), we get:
\(0.5 = p(B) + (1 - p(B))\)
Solving for p(B), we get:
\(p(B) = 0.5 - p(B^c) = 0.5 - (1 - p(B)) = 2p(B) - 0.5\\0.5 = p(B)\)
Therefore, p(B) = 0.5 up to the first decimal place.
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Find the arcumierence of the circle
iready area and circumference of a circle
Answer:
72.22
Step-by-step explanation:
C = \(\pi \: d\)
3.14×23=72.22
The circumference of a circle whose radius is 11.5 inches.
Thus, option (a) is correct.
Given:
Diameter = 23 inch
So, Radius of circle = diameter/2
=23/2
= 11.5 inch
Now, the formula for circumference of a circle is
= 2πr
Substituting the value of r= 11.5 into the formula as
= 2 x 3.14 x 11.5
= 72.22 inches
Thus, the circumference of circle is 72.22 inches.
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E is between D and F. Solve for each segmant length
Answer:
DE = 32
EF = 34
Step-by-step explanation:
3x + 5 + 4x - 2 = 66
7x + 3 = 66
7x = 63
x = 9
DE = 3(9)+5 = 32
EF = 4(9)-2 = 34
check: 32 + 34 = 66
Answer:
DE = 32
EF = 34
DF = 66
Step-by-step explanation:
DE + EF = DF from the segment addition postulate
3x+5 + 4x-2 = 66
Combine like terms
7x+3 = 66
Subtract 3 from each side
7x+3-3 = 66-3
7x = 63
Divide each side by 7
7x/7 = 63/7
x = 9
DE = 3x+5 = 3(9)+5 = 27+5 = 32
EF = 4x-2 = 4(9) -2 = 36-2 = 34
DF = 66
Find the sum.
eleven and eight hundredths added to two and twenty-five hundredths
The sum is .
Answer:
Step-by-step explanation:
11 8/100 +2 25/100 is 13 33/100
Marie can find the volume of a cube by using the formula V = 53, where s represents the side length of the cube. If Marie's
cube has a side length of 2 centimeters, what is the volume of her cube? Show your work. Write your answer as a fraction.
Answer:
The volume of her cube is 15.625cm^3
Step-by-step explanation:
The formula V = s³, where s represents the side length of the cube.
Each morning Bill leaves home between 6:30 and 8:00 to drive to work at University of Texas. The time it takes Bill to drive to work (TIME) depends on the departure time when he leaves after 6:30 (DEPART), the number of red lights on the way (REDS) and the number of trains that he has to wait for at the crossing (TRAINS). Observations for these variables are for 231 working days in 2006. TIME is measured in minutes after 6:30 that Bill departs. The estimated regression model is as follows; TIME -19.9166+0.3692DEPART+1.3353REDS +2.7548TRAINS R¹ -0.634 s.e (1.2548) (0.3038) (0.01553) (0.1390) a) What is the average estimated time in minutes to drive to work for Bill when he leaves on time at 6:30 and there are no red lights and no trains at the crossroad to wait?
( b) Interpret the estimated coefficients of REDS and TRAINS. c) Using a 5% significance level, test the hypothesis that each train delays Bill by 3 minutes. State your conclusion.
a) The average estimated time for Bill to drive to work when he leaves on time at 6:30 with no red lights and no trains to wait for is approximately -19.9166 minutes. b) The estimated coefficients of REDS and TRAINS in the regression model are 1.3353 (REDS). c) The absolute value of the calculated t-value (-1.7733) is less than the critical t-value (1.9719), we fail to reject the null hypothesis.
a) To find the average estimated time in minutes for Bill to drive to work when he leaves on time at 6:30 and there are no red lights and no trains at the crossroad to wait, we substitute the values into the regression model:
TIME = -19.9166 + 0.3692(DEPART) + 1.3353(REDS) + 2.7548(TRAINS)
Given:
DEPART = 0 (as he leaves on time at 6:30)
REDS = 0 (no red lights)
TRAINS = 0 (no trains to wait for)
Substituting these values:
TIME = -19.9166 + 0.3692(0) + 1.3353(0) + 2.7548(0)
= -19.9166
Therefore, the average estimated time for Bill to drive to work when he leaves on time at 6:30 with no red lights and no trains to wait for is approximately -19.9166 minutes. However, it's important to note that negative values in this context may not make practical sense, so we should interpret this as Bill arriving approximately 19.92 minutes early to work.
b) The estimated coefficients of REDS and TRAINS in the regression model are:
1.3353 (REDS)
2.7548 (TRAINS)
Interpreting the coefficients:
- The coefficient of REDS (1.3353) suggests that for each additional red light, the estimated time to drive to work increases by approximately 1.3353 minutes, holding all other factors constant.
- The coefficient of TRAINS (2.7548) suggests that for each additional train Bill has to wait for at the crossing, the estimated time to drive to work increases by approximately 2.7548 minutes, holding all other factors constant.
c) To test the hypothesis that each train delays Bill by 3 minutes, we can conduct a hypothesis test.
Null hypothesis (H0): The coefficient of TRAINS is equal to 3 minutes.
Alternative hypothesis (Ha): The coefficient of TRAINS is not equal to 3 minutes.
We can use the t-test to test this hypothesis. The t-value is calculated as:
t-value = (coefficient of TRAINS - hypothesized value) / standard error of coefficient of TRAINS
Given:
Coefficient of TRAINS = 2.7548
Hypothesized value = 3
Standard error of coefficient of TRAINS = 0.1390
t-value = (2.7548 - 3) / 0.1390
= -0.2465 / 0.1390
≈ -1.7733
Using a significance level of 5% (or alpha = 0.05) and looking up the critical value for a two-tailed test, the critical t-value for 230 degrees of freedom is approximately ±1.9719.
Since the absolute value of the calculated t-value (-1.7733) is less than the critical t-value (1.9719), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that each train delays Bill by 3 minutes.
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please help would appreciate so much
Help pleasee giving brainliest!!
Answer:
x=21
<1=120
Step-by-step explanation:
So to find x set up an equation like this:
(5x+15)+(3x-3)=180
since the bottom and top lines are congruent they are 'interior angles on the same side of a transversal' which are always supplementary, so that's where 180 comes from.
calculate the equation and you'll get x=21 :D
next to find 1 just put 21 in place of x: 3(21)-3=60
since it's supplementary to angle 1 you take 180-60=120
angle <1=120!
i need help 9th grade math
Answer:
I have reason to believe its a
Step-by-step explanation:
Easy 9nth grade math :)
evaluate 9c3 and 12p3
We need the formulas for combinations and permutations
\(\begin{gathered} \text{nCk}=\frac{n!}{k!(n-k)!} \\ \text{nPk}=\frac{n!}{(n-k)!} \end{gathered}\)Replacing the information we have
\(\begin{gathered} 9C3=\frac{9!}{3!(9-3)!}=\frac{9!}{3!6!}=84 \\ 12P3=\frac{12!}{(12-3)!}=\frac{12!}{9!}=1320 \end{gathered}\)Then 9C3=84 and 12P3=1320.
Several students decide to start a T-shirt company. After initial expenses of $280, they purchase t-shirts wholesale for $3.99. They sell each T-shirt for $10.99. How many must they sell to break even?
Answer:
must sell atleast 26 T shirts
Step-by-step explanation:
First, let's create an expression.
x = # of t-shirts.
$10.99x - ($280 + $3.99x)
Now, let's set our expression equal to 0.
$10.99x - ($280 + $3.99x) = 0
Distribute the negative sign.
$10.99x - $280 - $3.99x = 0
Combine like terms.
$7x - $280 = 0
Add $280 to both sides.
$7x = $280
Divide both sides by $7
x = 40
The students would have to sell 40 t-shirts to break even.
put these fractions from least to greatest please!! 2/3 7/12 17/24 3/4
Answer:
7/12 < 2/3 < 17/24 < 3/4
Step-by-step explanation:
least to greatest
The perimeter of a rectangle is 72, and the width is 4 less
than 3 times the length. What is the length?
O 12
O 10
O 16
O 14
Answer is 10
Explaination:
10x3=30
30-4=26
Perimeter = length times 2 + width times 2
26 x 2= 52
10 x 2= 20
52+20=72
Please mark brainliest
section 0101 of a course contains 120 students, while section 0201 contains 60 students. if 12 students show up at a review session, how likely is it that at least 10 of those are from section 0101?
The probability that at least 10 of those students are from section 0101 is given as follows:
P(X >= 10) = 0.172 = 17.2%.
What is the hypergeometric distribution formula?The mass probability formula, giving the probability of x successes out of n options, is presented as follows:
\(P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}\)
\(C_{n,x} = \frac{n!}{x!(n-x)!}\)
The parameters are listed and explained as follows:
N is the size of the population from which the sample of the problem is taken.n is the size of the sample of the problem.k is the total number of desired outcomes in the population of the problem.The parameter values for this problem are given as follows:
N = 180, k = 120, n = 12.
The probability of at least 10 is given as follows:
P(X >= 10) = P(X = 10) + P(X = 11) + P(X = 12)
Using a hypergeometric distribution calculator with the above parameters, the probability is given as follows:
P(X >= 10) = 0.172 = 17.2%.
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help with my geometry please
Answer:
x = 11
z = 86
Step-by-step explanation:
8x + 6 and 10x-16 are vertical angles
Vertical angles are pairs of angles that are opposite each other and have the same vertex, or point of intersection. They are formed when two lines intersect at a point, and are always congruent, or of equal measure.
To solve this equation, we need to isolate the variable x on one side of the equation. To do this, we can start by subtracting 6 from both sides of the equation:
8x + 6 - 6 = 10x - 16 - 6
8x = 10x - 22
Now we can subtract 8x from both sides of the equation:
8x - 8x = 10x - 22 - 8x
0 = 2x - 22
To solve for x, we can add 22 to both sides of the equation:
0 + 22 = 2x - 22 + 22
22 = 2x
Finally, we can divide both sides of the equation by 2 to find the value of x: 22 / 2 = 2x / 2
x = 11
Therefore, the solution to the equation is x = 11.
Now that we have x, z is a supplementary angle to 8x + 6 (or you could do 10x - 16)
Supplementary angles are pairs of angles that add up to 180 degrees. They are formed when two lines intersect at a point, and the angles formed at the intersection are supplementary.
First plug in x, 8x + 6 = 8(11) + 6 = 88 + 6 = 94
180 - 94 = z
z = 86
Circle theorem. Find the following angles.
The measure of the angle ∠CDE in the cyclic quadrilateral AEDC and the angle ∠BFC indicates;
(a) \(G\widehat{A}E\) = 38°
(b) \(A \widehat{E}B\) = 63°
What is a cyclic quadrilateral?A cyclic quadrilateral is an inscribed quadrilateral of a circle.
(a) The quadrilateral ACDE is a cyclic quadrilateral, therefore;
∠CDE and ∠AEF are supplementary angles
m∠CDE + m∠AEF = 180° (Definition of supplementary angles
m∠CDE = 128°
Therefore; m∠EAF = 180° - 128° = 52°
AC is a diameter of the circle, therefore FA is a radius of the circle and ∠GAF is a right angle (90°) (Properties of a tangent to a circle)
∠GAF = ∠GAE + ∠EAF (Angle addition postulate)
m∠GAE = m∠GAF - m∠EAF
m∠GAE = 90° - 52° = 38°
Angle \(G\widehat{A}E\) = 38°
(b) Angle AEB, angle AFE and angle EAF are the interior angles of the triangle AEF, therefore;
m∠AEB + m∠AFE + m∠EAF = 180° (Angle sum property of a triangle)
m∠AEB = 180° - (m∠EAF + m∠AFE)
m∠AEB = 180° - (52° + m∠AFE)
∠AFE and ∠BFC are vertical angles, therefore;
m∠AFE = m∠BFC = 65°
m∠AEB = 180° - (52° + 65°) = 63°
\(A\widehat{E}B\) = 63°
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In the figure M and N are mid point at QP and PR prove that: respectively (1) MN = 1/2 QR →>> (ii) MN // QR M Q P N R
By the property of the mid point theorem of triangle:
(1) MN = 1/2 QR and (ii) MN || QR
Explain about the mid point theorem:A point that divides a line segment into two equal halves is the midway of the segment. The line segment's midpoint is located precisely in the middle.
According to the midpoint theorem, a segment is half the length of a triangle's third side if it is created by joining the midpoints of two of its sides.According to the midpoint theorem, a segment must be parallel toward the third side if it is created by joining the midpoints of two triangle's sides. Furthermore, it mentions that the length of that section is equal to that of the third side.(1) MN = 1/2 QR
By the property of the mid point theorem of triangle:
MN = 1/2 QR , As M and Na are the mid points sides QP and PR of triangle .
(ii) MN || QR
And thus, both the lines MN and QR will be parallel.
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complete question:
In the figure M and N are mid point at QP and PR prove that: respectively
(1) MN = 1/2 QR
(ii) MN || QR
The figure is attached.
Show that if U is open in X, and A is closed in X, then UA is open in X, and A\U is closed in X.
The intersection of N(x) and N'(x), denoted by N(x)∩N'(x), is an open neighborhood of x. Since N(x)∩N'(x) ⊆ N(x) ⊆ U and N(x)∩N'(x) ⊆ N'(x) ⊆ X\U, we can conclude that N(x)∩N'(x) ⊆ UA.
Since N(x)∩A is a non-empty set contained in A\U, we have shown that every point in (A\U)' has a neighborhood contained in A\U. Therefore, (A\U)' is open in X, which implies that A\U is closed in X.
To show that if U is open in X and A is closed in X, then UA is open in X and A\U is closed in X, we need to prove two statements:
UA is open in X.A\U is closed in X.Let's prove these statements one by one:
To show that UA is open in X, we need to prove that for every point x in UA, there exists an open neighborhood around x that is completely contained within UA.Let x be an arbitrary point in UA. Since x is in UA, it must belong to U as well as A. Since U is open in X, there exists an open neighborhood N(x) of x that is completely contained within U. Now, since x is in A, it is also in X\U (complement of U in X). As A is closed in X, X\U is closed in X, which means its complement, U, is open in X. Therefore, there exists an open neighborhood N'(x) of x that is completely contained within X\U.
Now, consider the intersection of N(x) and N'(x), denoted by N(x)∩N'(x). This intersection is an open neighborhood of x. Since N(x)∩N'(x) ⊆ N(x) ⊆ U and N(x)∩N'(x) ⊆ N'(x) ⊆ X\U, we can conclude that N(x)∩N'(x) ⊆ UA.
Since N(x)∩N'(x) is an open neighborhood of x completely contained within UA, we have shown that UA is open in X.
To show that A\U is closed in X, we need to prove that its complement, (A\U)', is open in X.Let x be an arbitrary point in (A\U)'. Since x is not in A\U, it means that x must either be in A or in U (or both). If x is in A, then x is not in A\U. Therefore, x is in U.
Since x is in U and U is open in X, there exists an open neighborhood N(x) of x that is completely contained within U. Now, consider the intersection of N(x) and A. Since x is in A, N(x)∩A is a non-empty set. Let y be any point in N(x)∩A.
We know that N(x)∩A ⊆ U∩A ⊆ A\U, because if y was in U, it would contradict the assumption that y is in A. Therefore, N(x)∩A is a subset of A\U.
Since N(x)∩A is a non-empty set contained in A\U, we have shown that every point in (A\U)' has a neighborhood contained in A\U. Therefore, (A\U)' is open in X, which implies that A\U is closed in X.
Hence, we have shown that if U is open in X and A is closed in X, then UA is open in X, and A\U is closed in X.
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A standard pair of six-sided dice is rolled. what is the probability of rolling a sum less than 8? express your answer as a fraction or a decimal number rounded to four
The answer is 0.5833 using probability.
What is probability?
By simply dividing the favorable number of possibilities by the entire number of possible outcomes, the probability of an occurrence can be determined using the probability formula. Because the favorable number of outcomes can never exceed the entire number of outcomes, the chance of an event occurring might range from 0 to 1.
The probability of an event can be calculated using the following formula.
Probability(Event) = Positive Results/Total Results = x/n
When two dice are rolled sample space is of 36 elements.
The elements which give sum less than 8 are given as follow
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5)
(3,1),(3,2),(3,3),(3,4)
(4,1),(4,2),(4,3)
(5,1),(5,2)
(6,1)
Here the total number of elements that satisfy conditions is 21
Hence probability is given by (21/36)
Probability=0.5833
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Consider a sagmal x[n] having the corresponding Fourier transform X(c j
). What world be the fourier transfotm of the signal y(n)=3x[n]n n
) Select one X(e −(v−3n)
) 3X(e 3(∗+∗ 2
)
) π(x −λ(e−3)
) 3πX 2
(e s
) 3X(e −1(ein)
)
The correct Fourier transform of the signal y(n) = 3x[n]n (n) is 3X(e−j(v−3n)).
Explanation:
Given information: a signal x[n] having the corresponding Fourier transform X(c j), and another signal y(n) = 3x[n]n (n) .
We know that, the Fourier transform of y(n) is given by:
Y(e^jv)=sum from - infinity to infinity y(n)e^jvn.
where y(n) = 3x[n]n (n)
Substituting y(n) in the above equation, we get:
Y(e^jv) = 3 * sum from - infinity to infinity x[n]n (n) * e^jvn.
We know that, the Fourier transform of x[n]n (n) is X(e^j(v-2pi*k)/3).
Therefore, substituting the value of y(n) in the above equation, we get:
Y(e^jv) = 3 * sum from - infinity to infinity x[n]n (n) * e^jvn
= 3X(e^j(v-3n)).
Hence, the Fourier transform of the signal y(n) = 3x[n]n (n) is 3X(e−j(v−3n)).
Conclusion: The correct Fourier transform of the signal y(n) = 3x[n]n (n) is 3X(e−j(v−3n)).
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I need help solving for x please
Answer:
x is 55 degrees because the whole angle is a total of 180 degrees. All you have to do is do 180 minus the given angle, 125