2. A recent SAT test was distributed N(518, 116). a) What percentage should
have grades <511?; b) What percentage should have grades between 410
and 597? (MUST use the normal table.)
HROW JA WOH2 TRUM.230AU JAMD30AU03 HTW 21J0238 JA
Answers
a) Approximately 25.78% of students should have grades less than 511. b) Approximately 99.99% of students should have grades between 410 and 597.
What is a normal table?A normal table, also known as a standard normal table or a z-table, is a mathematical table that provides values of the cumulative distribution function (CDF) of the standard normal distribution. The standard normal distribution has a mean of 0 and a standard deviation of 1, and is a common distribution used in statistics and probability theory. A normal table is used to calculate probabilities for a normal distribution by looking up values in the table based on the standard deviation and mean of the distribution. The values in the table represent the area under the curve of the normal distribution to the left of a given z-score, and can be used to find probabilities and percentiles for specific z-scores or ranges of z-scores.
a) To find the percentage of students who scored less than 511, we need to standardize the score using the formula:
z = (x - μ) / σ
where x is the score obtained, μ is the mean, and σ is the standard deviation.
Substituting the given values, we get:
z = (511 - 518) / 10.77 = -0.65 (\(\sqrt{116} = 10.77\))
Using the normal table, we can find the area to the left of z = -0.65, which is 0.2578 or 25.78%. Therefore, approximately 25.78% of students should have grades less than 511.
b) To find the percentage of students who scored between 410 and 597, we need to standardize both scores and find the area between the two z-scores.
For x = 410:
z₁ = (410 - 518) / 10.77 = -10.04
For x = 597:
z₂ = (597 - 518) / 10.77 = 7.34
Using the normal table, we can find the area to the left of z1 and z2, which are both very close to 0. We can then subtract the smaller area from the larger area to find the area between the two z-scores.
Area to the left of the z₁ = 0.0001
Area to the left of the z₂ = 0.9999
Area between z₁ and z₂ = 0.9999 - 0.0001 = 0.9998
Therefore, approximately 99.99% of students should have grades between 410 and 597.
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Related Questions
I’m triangle PQR, SP = SQ. What is SQR? Help please.
Answers
Answer:
25
Step-by-step explanation:
we have to subtract SPQ from SRQ=SQR
mark me as brainliest
Please help me as soon as possible!
Thank you in advance!
Answers
Answer:
A) 15
Step-by-step explanation:
a = 3 b = 5 c = 7
a + b + c
3 + 5 + 7 = 15
Answer:
I believe it is A 15
Step-by-step explanation:
A grocery store sells a bag of 6 oranges for $3.90. What is the unit cost?
Pls Answer and I shall give u a- Thank You - Vielen Dank
Answers
Answer:
it would be 0.64
Step-by-step explanation:
please help asap!!!!
Answers
Answer:
Step-by-step explanation:
Given functions are,
f(x) = \(\sqrt{x} +3\)
g(x) = 4 - \(\sqrt{x}\)
22). (f - g)(x) = f(x) - g(x)
= \(\sqrt{x}+3-(4 - \sqrt{x} )\)
= \(\sqrt{x} +3-4+\sqrt{x}\)
= \(2\sqrt{x}-1\)
Domain of the function will be [0, ∞).
23). (f . g)(x) = f(x) × g(x)
= \((\sqrt{x}+3)(4-\sqrt{x} )\)
= \(4(\sqrt{x}+3)-\sqrt{x}(\sqrt{x}+3)\)
= \(4\sqrt{x} +12-x-3\sqrt{x}\)
= \(-x+\sqrt{x}+12\)
Domain of the function will be [0, ∞).
Which of the following is basically a promissory note, or a promise to repay a certain amount of money at some point in the future?
-Bond
-CD
-Mutual fund
-Stock
Answers
Answer:
Bond
Step-by-step explanation:
A promissory note or a promise to repay a certain amount of money at some point in the future is basically a bond.
A bond is a debt security that represents a loan made by an investor to a borrower, which is usually a corporation or government agency. It is a fixed-income investment, meaning that the borrower promises to pay a specific amount of interest over a set period of time and return the principal amount of the loan on the date of maturity. Bonds are issued for various purposes, such as raising capital, funding new projects, or refinancing debt.
CD (Certificate of Deposit) is a savings instrument issued by a bank or credit union that generally pays a fixed rate of interest over a set term. Mutual fund is an investment vehicle that pools money from multiple investors to purchase a portfolio of securities, such as stocks, bonds, or both. Stock is an ownership share in a company that represents a claim on part of the company's assets and earnings.
You want to put a logo on each cell phone holder you make. Sew-on logos cost 2.30 for 10. Stick-on logos cost 9.40 for 100. Divide the cost by the number of logos for both types
Answers
Answer:
\(Rate = \$0.23\) and \(Rate = \$0.094\)
Step-by-step explanation:
Given
Sew-on logos
Cost = $2.30
Units = 10
Stick-on logos
Cost = $9.40
Units = 100
Required
Divide cost by units, in both cases
When we divide cost by units, it gives the unit rate.
i.e.
\(Rate = \frac{Cost}{Units}\)
FOR SEW-ON LOGOS
Substitute $2.30 for Cost and 10 for Units
\(Rate = \frac{\$2.30}{10}\)
\(Rate = \$0.23\)
FOR STICK-ON LOGOS
Substitute $9.40 for Cost and 100 for Units
\(Rate = \frac{\$9.40}{100}\)
\(Rate = \$0.094\)
Hence, the solution is
\(Rate = \$0.23\) and \(Rate = \$0.094\)
100 points & brainliest if you get it correct!
Answers
\(\\ \sf\longmapsto \dfrac{12.19}{5.7}\)
\(\\ \sf\longmapsto 2.14/lb\)
Brand 22.15/lb+2=2.17/lb
Brand 2 is a better deal and 2.17lb is the unit price
PLEASE HELP MEEE!!
5, 18, 6, 18, 13 find the mean absolute deviation (MAD)
(SHOW WORK)
Answers
60/5 =12
The quadrilateral below is a kite. Round your answer to the nearest tenth. If QR=13 and PT=8, find QT
Answers
Answer:
The answer is "10.25"
Step-by-step explanation:
Please find the graph in the attachment:
In the given kite diagram the PT= TR so,
in \(\Delta QTR\) we applying the Pythagoras theory:
\(\to QR^2=QT^2+TR^2\\\\\to 13^2=QT^2+8^2\\\\\to QT^2= 13^2-8^2\\\\\to QT^2= 169-64\\\\\to QT^2=105\\\\\to QT=\sqrt{105}\\\\\to QT= 10.2469508\\\\ \to QT= 10.25\)
Let V and W be vector spaces, and let T: V W be a linear transformation. Given a subspace U of V, let T(U) denote the set of all images of the form T(x), where x is in U. Show that T(U) is a subspace of W. To show that T(U) is a subspace of W, first show that the zero vector of wis n TU. Choose the correct answer below. a. Since V s a subspace of U the zero vector of u ou is in V. Since T s inear T Ou .w, where is the zero vector of w.?oms T(U) b. Since U is a subspace of W, the zero vector of w, ow, is in U. Since T is linear, T(0w) = 0v, where 0v is the zero vector of V So 0w is in T(U). c. Since V is a subspace of U, the zero vector of V, 0v is in U. Since T is linear, T(0v-0w where o,. is the zero vector of W Som s in T(U). d. Since U is a subspace of V, the zero vector of V, 0y, is in U.Since T is linear, Toy)-ow where Ow is the zero vector of W. So Ow is in T(U)
Answers
The T(U) is a subspace of W.
Since U is a subspace of V, the zero vector of V, 0v, is in U. Since T is linear, T(0v) = 0w, where 0w is the zero vector of W. So 0w is in T(U). Therefore, T(U) is a subspace of W.
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How far can light travel in 7 minutes?
Answers
Answer: 78120000 miles
Step-by-step explanation:
PLS HELP!
45-[38-{60÷3-(6-9÷3)÷3}]
Answers
Answer:
26
Step-by-step explanation:
If you need step by step, reply to this comment.
a square of area 36 cm squared is cut to make two rectangles a and b the ratio of area a to area b is 2:1 what is the diemensions of rectangles a and b
Answers
Rectangle A has dimensions of 6 cm by 4 cm, and Rectangle B has dimensions of 6 cm by 2 cm.
Let's assume the side length of the square is x. Since the area of the square is 36 cm², we have x² = 36. Solving this equation, we find x = 6.
Now, we need to cut the square into two rectangles, A and B. Let's assume the dimensions of Rectangle A are length L and width W, and the dimensions of Rectangle B are length L' and width W'. According to the given information, the ratio of the area of A to the area of B is 2:1.
Since the area of a rectangle is given by length multiplied by width, we have the equation LW = 2(L'W').
Substituting the values, we have 6L = 2(6W'), simplifying to 3L = 6W', and further simplifying to L = 2W'.
Since the dimensions of Rectangle A and Rectangle B satisfy this condition, we can choose any values that satisfy the equation.
One possible solution is L = 4 cm and W = 6 cm for Rectangle A, and L' = 2 cm and W' = 6 cm for Rectangle B.
Therefore, the dimensions of Rectangle A are 6 cm by 4 cm, and the dimensions of Rectangle B are 6 cm by 2 cm.
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PLS HELP WILL GUVE 100 POINTS FOR BOTH
Answers
Answer:
44. 7in
45. 35in
Step-by-step explanation:
44.
\(24^{2} + x^{2} = 25^{2} \\576 + x^{2} = 625\\576 - 625 = -x^{2} \\\sqrt{49} = x\\x = 7\)
45.
\(15^{2} + x^{2} = 25^{2} \\225 + x^{2} = 625\\225 - 625 = -x^{2} \\-400 = -x^{2} \\\sqrt{400} = x\\x = 20\)
\(25in*15in = 35in^{2}\)
if you draw two cards from a deck, what is the probability that you will get the ace of diamonds and a black card?
Answers
The probability that the ace of diamonds and a black card will come is 1 / 51 .
Case A :
probability of first getting an ace of diamonds = 1 / 52
probability of getting a black card given first card is ace of diamond = 26 / 51
Hence , probability that the ace of diamonds and a black card will come is ( 1 / 52 ) * ( 26 / 51 ) = ( 1 / 102 )
Case B :
probability of first getting a black card = ( 26 / 52 )
probability of getting an ace , given first card is a black card = ( 1 / 51 )
Hence , the probability that the ace of diamonds and a black card will come is ( 26 / 52 ) * ( 1 / 51 ) = ( 1 / 102 )
P ( A ∪ B ) = P ( A ) + P (B ) + P ( A ∩ B )
= ( 1 / 102 ) + ( 1 / 102 )
= 1 / 51 .
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The soft goods department of a large department store sells 175 units per month of a certain large bath towel. The unit cost of a towel to the store is \( \mathbf{S 2 . 5 0} \) and the cost of placing
Answers
The optimal order quantity is 1256 units and the minimum total cost is S3150.04. The soft goods department of a large department store sells 175 units per month of a certain large bath towel. The unit cost of a towel to the store is S2.50 and the cost of placing an order is S375.
In this problem, the order quantity (Q) will be calculated using the economic order quantity (EOQ) formula as follows:
EOQ = √[(2DS)/H] Where: D = Annual Demand, S = Cost of placing an order, H = Carrying cost per unit per year, Carrying cost per unit per year can be computed using the following formula: H = iC
Where: i = Annual carrying charge rate, C = Unit cost of a towel to the store
Hence, H = 0.12 x S2.50H
= S0.30D is already given as 175 units per month, so the annual demand (D) will be:
D = 175 x 12D
= 2100 units per year
Substitute all values into the EOQ formula:
EOQ = √[(2 x 2100 x 375)/0.30]EOQ
= √[1,575,000]EOQ
= 1255.13 units
Rounding up, the optimal order quantity is 1256 units.
The minimum total cost will be calculated using the following formula:
TC = DH + (Q/2)S + (D/Q) x HC
Where: TC = Total cost H = Carrying cost per unit per year, S = Cost of placing an order, Q = Order quantity, D = Annual demand.
HC = Holding cost per unit per year
TC = (2100 x S2.50 x 0.3) + (1256/2 x S2.50) + (2100/1256 x 0.12 x S2.50)TC
= S 1575 + S1570 + S5.04TC
= S3150.04
Therefore, the optimal order quantity is 1256 units and the minimum total cost is S3150.04.
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A hospital purchases a new magnetic resonance imaging (MRI) machine for $500,000. The depreciated value (reduced value) y after t years is y = 500,000 - 47,000t, for 0 ≤ t ≤ 9.
(a) Use the constraints of the model and a graphing utility to graph the equation using an appropriate viewing window.
(b) Use the value feature of the graphing utility to determine the value of y when t = 5.8. Verify your answer algebraically.
(c) Use the zoom and trace features of the graphing utility to determine the value of t when y = 156,900. Verify your answer algebraically.
Answers
The depreciated value (reduced value) y after t years is y = 500,000 - 47,000t, for 0 ≤ t ≤ 9 is about 6.6.
(a) The graph of the equation y = 500,000 - 47,000t, for 0 ≤ t ≤ 9, can be plotted as follows using a graphing calculator: Now, we need to find the value of y when t = 5.8 using the value feature of the graphing utility. We will do that in part (b).
(b) Using the value feature of the graphing utility, we obtain: y = 500,000 - 47,000t = 500,000 - 47,000(5.8) ≈ $229,400We can verify our answer algebraically by substituting t = 5.8 into the equation: y = 500,000 - 47,000t = 500,000 - 47,000(5.8) = 500,000 - 272,600 = $227,400(rounded to the nearest hundred).
(c) Using the zoom and trace features of the graphing utility, we can determine the value of t when y = 156,900. We will do that in part (c).Now, we will find the value of t using the trace feature of the graphing utility. The value of t is about 6.6.We can verify our answer algebraically by substituting y = 156,900 into the equation and solving for t:156,900 = 500,000 - 47,000tt = (500,000 - 156,900)/47,000t ≈ 6.66... ≈ 6.6
Therefore, the value of t when y = 156,900 is about 6.6.
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HElp pLS i LAVA YOUUU!!!!!!!!
Answers
Answer:
The annual rate of interest on the musician's loan for the trumpet is approximately 12%.
Step-by-step explanation:
To find the annual rate of interest, we can rearrange the formula for simple interest, I = Prt, to solve for the interest rate (r).
Given that the principal (P) is $2,200, the time (t) is 3 years, and the total interest (I) is $792, we can substitute these values into the formula:
792 = 2200 * r * 3
To solve for r, divide both sides of the equation by (2200 * 3):
r = 792 / (2200 * 3)
r ≈ 0.12
To express the interest rate as a percentage, we multiply r by 100:
r * 100 ≈ 0.12 * 100 ≈ 12
Therefore, the annual rate of interest on the musician's loan for the trumpet is approximately 12%.
I don't get my question for my homework. Here is the questions, "You have 46 gold coins, 115 diamonds, and 184 rubies. You need to put them in treasure chests, and each chest must contain the same number of each individual item. What is the greatest number of treasure chests you can fill? How many gold coins in each chest? How many diamonds in each chest? How many rubies in each chest? " This is the type of question. There is also one more problem I'm stuck on. "You and some friends took a metal detector to the beach every day for a week to search for coins. You managed to find 246 nickels, 312 dimes, and 204 quarters. When you divided them up, you realized that each person got the same exact amount of each coin with no coins remaining. How many are in your group? How many nickels did each person receive? How many dimes did each person receive? How many quarters did each person receive?" Please help me out! Thanks
Answers
Answer:
Question 1
The greatest number of treasure chest you can fill = 23
Step-by-step explanation:
Gold coins = 46
Diamonds = 115
Rubies = 184
To determine the greatest number of treasure chests you can fill, find the highest common factors of 46, 115 and 184
46 = 1, 2, 23, 46.
125 = 1, 5, 23, 115
184 = 1,2,4,8,23,46,92,184
The highest common factors of 46, 115 and 184 is 23
The greatest number of treasure chest you can fill = 23
Number of gold coins in each chest = 46/23
= 2
Number of diamonds in each chest = 115 / 23
= 5
Number of rubies in each chest = 184 /23
= 8
Question 2
Nickels = 246
Dimes = 312
Quarters = 204
Find the highest common factor of 246, 312, 204
246 = 1, 2, 3, 6, 41, 82, 123, 246
312 = 1, 2, 3, 4, 6, 8, 12, 13, 24, 26, 39, 52, 78, 104, 156, 312.
204 = 1, 2, 3, 4, 6, 12, 17, 34, 51, 68, 102, 204
The highest common factor of 246, 312, 204 is 6
How many are in your group?
6 members
There are 6 members in your group that each person got the same exact amount of each coin with no coins remaining.
Number of nickels each person receive = 246 / 6
= 41
Number of dimes each person receive = 312 / 6
= 52
Number of quarters each person receive = 204 / 6
= 34
Plz help ASAP!!! WILL MARK YOU BRAINLIEST IF YOU ANSWER IS CORRECT!!! Question#6
Answers
Answer:
B
Step-by-step explanation:
Recall that the domain is the span of x-values covered by the graph, while the range is the span of y-values covered by the graph.
From the graph, we can see that the span of x-values covered goes from x=-2 to x=1.
However, note that while at x=-2, the dot is closed, while at x=1, the dot is open. This means that we do not include the point at x=1. Therefore, the domain is all values of x between x=-2 and x=1 not including x=1.
In set notation, this is:
\(\{x|-2\leq x<1\}\)
Thus, the domain is all values greater than or equal to -2 but less than 1.
For the range, we can see that the graph spans from y=-4 to y=2. Again, at y=2, the dot is open. Thus, we do not include y=2 in our range.
In set notation, this is:
\(\{y|-4\leq y<2\}\)
This is interpreted as all y-values greater than or equal to -4 but less than 2.
The answer is B
Help, its just two questions...
Answers
Answer:
In Q39, the value 3x should be positive and in Q40, the final answer is missing 8x and the fact that 4x^2 is negative (-4x^2)
Step-by-step explanation:
39:
1. \(-(-3x)\) is presented as \(-3x\) when it should be \(+3x\), as 2 negatives make a positive.
2. Again, \(-3x\) is shown instead of \(+3x\)
3. The answer \(-x^2-2x\) should instead be \(-x^2+4x\), as \(-3x\) is used to reach this incorrect answer instead of the correct \(+3x\)
40:
1. \((x^3-4x^2+3)+(-3x^3+8x-2)=(x^3-3x^3)-4x^2+8x+(3-2)=-2^3-4x^2+8x+1\),
The final answer is wrong as it is missing the 8x and the fact that it is \(-4x^2\)and not \(+4x^2\)
The width of a rectangle is 5 units less than the length. If the area is 150 square units the find the dimensions of the rectangle
Answers
Answer:
Step-by-step explanation:
width is 10 length is 15.
If the graph of the linear equation ax-8y=15 has an x-intercept at (3,0), then what is the value of a?
Answers
====================================
Work Shown:
Plug in (x,y) = (3,0)
ax - 8y = 15
a(3) - 8y = 15 .... replace x with 3
3a - 8(0) = 15 ... replace y with 0
3a - 0 = 15
3a = 15
a = 15/3
a = 5
The equation updates to 3x - 8y = 15.
(MATH!!!) HELP!!! One endpoint of a segment is (-9,-1). The midpoint of the segment is (-2,4). What is the second endpoint of this segment?
Answers
Answer:
C; (5, 7)
Step-by-step explanation:
A solution of the initial value problem Dy(t)/dt + 8y(t) = 1 + e-6t is a. x(t) = 1/8 + + 1/2 e6t - 5/8 e8t
b. x(t) = 1/8 + 1/2 e-6t - 5/8 e-8t
c. x(t) = 1/8 - 1/2 e6t + 5/8 e8t
d. x(t) = 1/4 + 1/2 e6t - 5/8 e8t
Answers
The solution of the initial value problem Dy(t)/dt + 8y(t) = 1 + e-6t is option (c) y(t) = (1/8) - (1/8) * e^(-8t).
To solve the given initial value problem, we can use the method of integrating factors.
The given differential equation is:
\(dy(t)/dt + 8y(t) = 1 + e^(-6t)\)
First, we write the equation in the standard form:
\(dy(t)/dt + 8y(t) = 1 + e^(-6t)\)
The integrating factor (IF) is given by the exponential of the integral of the coefficient of y(t), which is 8 in this case:
IF = \(e^(∫8 dt)\)
=\(e^(8t)\)
Now, we multiply both sides of the differential equation by the integrating factor:
\(e^(8t) * dy(t)/dt + 8e^(8t) * y(t) = e^(8t) * (1 + e^(-6t))\)
Next, we can simplify the left side by applying the product rule of differentiation:
\((d/dt)(e^(8t) * y(t)) = e^(8t) * (1 + e^(-6t))\)
Integrating both sides with respect to t gives:
\(∫(d/dt)(e^(8t) * y(t)) dt = ∫e^(8t) * (1 + e^(-6t)) dt\)
Integrating the left side gives:
\(e^(8t) * y(t) = ∫e^(8t) dt\)
\(= (1/8) * e^(8t) + C1\)
For the right side, we can split the integral and solve each term separately:
\(∫e^(8t) * (1 + e^(-6t)) dt = ∫e^(8t) dt + ∫e^(2t) dt\)
\(= (1/8) * e^(8t) + (1/2) * e^(2t) + C2\)
Combining the results, we have:
\(e^(8t) * y(t) = (1/8) * e^(8t) + C1\)
\(y(t) = (1/8) + C1 * e^(-8t)\)
Now, we can apply the initial condition y(0) = 0 to find the value of C1:
0 = (1/8) + C1 * e^(-8 * 0)
0 = (1/8) + C1
Solving for C1, we get C1 = -1/8.
Substituting the value of C1 back into the equation, we have:
\(y(t) = (1/8) - (1/8) * e^(-8t)\)
Therefore, the solution to the initial value problem is:
\(y(t) = (1/8) - (1/8) * e^(-8t)\)
The correct answer is option (c) \(y(t) = (1/8) - (1/8) * e^(-8t).\)
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Multiply:3/10 * 7/2
Answers
In order to multiply two fractions, we can follow the steps below:
0. find the product between the two numerators; ,(3 * 7 = 21)
,1. find the product between the two denominators; ,(10 * 2 = 20)
,2. the product of the two fractions will be the result of step 1 (the numerator of the final result), divided by the result of step 2 (the denominator of the final result):
\(\frac{3}{10}\cdot\frac{7}{2}=\frac{3\cdot7}{10\cdot2}=\frac{21}{20}\)Therefore, the answer is:
\(\frac{21}{20}\)What number represents the same as 1 hundred plus 5 tens plus 12 ones?
Answers
Answer:
162
Step-by-step explanation:
1 hundred (100) + 5 tens (50) + 12 ones (10+2)
100+50+12 = 162
Which statement is true? Question 5 options: A) |–13| = 13 B) – |13| = 13 C) |13| = – 13 D) – |–13| = 13
Answers
Step-by-step explanation:
|-13| = 13
13 = 13 Option A is true
-|13| = 13
-13 = 13
-13 ≠ 13 Option B isn't true
|13| = -13
13 = -13
13 ≠ -13 Option C isn't true
-|-13| = 13
-13 = 13
-13 ≠ 13 Option D isn't true
What are not exponential functions?.
Answers
It is not of exponential order for h(t) = et2. F is of exponential order and has order. F is piecewise continuous. No function's Laplace transform is possible. g(t) = eat, where t [0,.
Non-exponential form:
To represent a scientifically notated number as a non-exponential quantity: • Remove the exponential component of the number by moving the decimal point the same number of places as the exponent's value. The decimal is moved to the right by the same number if the exponent is positive.
Here are a few instances of functions other than exponential ones. y = 3 1 x as a result. n = 0 3 p as a result. Because y = ( 4) x. Since b = 1, y = 6 0 x.
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In a coordinate plane, if points C(2, 5), D(- 1, 2) and E(x, y) lie on line 4, which of the following would be the coordinate of point E?A. (0, 1)B. (1,1)C. (0, 2)D, (1, 4)
Answers
To answer this, we will find the equation of the line that passes through point C and D and check which option is in the same line, this will be the answer.
First, let's find the slope using th coordinates of the points C and D:
\(s=\frac{y_D-y_C}{x_D-x_C}=\frac{2-5}{-1-2}=\frac{-3}{-3}=1\)Now, we can use point C, for example, to right in the slope-point form:
\(\begin{gathered} (y-y_C)=s(x-x_C) \\ y-5=1(x-2) \\ y=x-2+5 \\ y=x+3 \end{gathered}\)With this equation, we can input values of x and check is we get the same y.
Alternative A and C have x = 0, so let's check it:
\(\begin{gathered} y=x+3 \\ y=0+3 \\ y=3 \end{gathered}\)I got y = 3, neither alternative have y = 3, so it isn't A nor C.
B and D have x = 1, let's check it:
\(\begin{gathered} y=x+3 \\ y=1+3 \\ y=4 \end{gathered}\)We got y = 4, which matches alternative D.
So, the correct alternative is D.