Galaxy B is moving away from us at a speed of 24,000 km/s.
The speed at which Galaxy B is moving away from us would be 24,000 km/s. Here's how we arrived at this conclusion:
Given: Galaxy A has a cosmological redshift in its spectrum of z = 0.01 indicating it is moving away from us at 3000 km/s.
Galaxy B has z = 0.08.We know that the redshift z is directly proportional to the speed at which the galaxy is moving away from us.
In other words, z ∝ v, where z is the redshift, and v is the speed.
Therefore, we can write:z₁/v₁ = z₂/v₂where z₁ and v₁ are the redshift and speed of Galaxy A, and z₂ and v₂ are the redshift and speed of Galaxy B.
Rearranging the formula, we get:v₂ = (z₂/z₁) x v₁
Substituting the values of z₁, v₁, and z₂ into the formula, we get:v₂ = (0.08/0.01) x 3000 km/sv₂ = 24,000 km/s
Therefore, Galaxy B is moving away from us at a speed of 24,000 km/s.
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One of Carla's friends suggests that she survey only eighth-graders because they are the
oldest and probably know more about the election than younger students. Do you think
this suggestion creates a random sample? Explain.
Answer:
This example is not a random example due to the fact that it is only questioning the eighth-graders. Random sample means that the sample is chosen simply randomly and everyone has an equal opportunity to be apart of the sampling.
Step-by-step explanation:
A parent interest group is looking at whether birth order affects scores on the ACT test. It was suggested that, on average, first-born children earn lower ACT scores than second-born children. After surveying a random sample of 100100 first-born children, the parents’ group found that they had a mean score of 23.823.8 on the ACT. A survey of 250250 second-born children resulted in a mean ACT score of 24.124.1. Assume that the population standard deviation for first-born children is known to be 1.21.2 points and the population standard deviation for second-born children is known to be 2.42.4 points. Is there sufficient evidence at the 1%1% level of significance to say that the mean ACT score of first-born children is lower than the mean ACT score of second-born children? Let first-born children be Population 1 and let second-born children be Population 2.
Step 1 of 3 :
State the null and alternative hypotheses for the test. Fill in the blank below.
Step 2: What is the Test Statistic
Step 3: Do we reject or fail to reject the null hypothesis? Do we have sufficient or insufficient data?
Step 1: To tell the null and alternative hypotheses for the test:
Null hypothesis (H0): The mean ACT score of first-born children is equal to or greater than the mean ACT score of second-born children. μ1 ≥ μ2Alternative hypothesis (Ha): The mean ACT score of first-born children is lower than the mean ACT score of second-born children. μ1 < μ2What are the other stepsStep 2: The test statistic for this scenario is the z-score, calculated as:
Step 3 involves comparing the chosen significance level (1% in this instance) with the test statistic to decide whether to reject or accept the null hypothesis. In case the test statistic falls within the critical region, the null hypothesis is rejected, but if it doesn't, the null hypothesis is not rejected.
We can also determine the p-value related to the test statistic and assess its compatibility with the selected level of significance. If the significance level exceeds the p-value, we do not reject the null hypothesis, but if the p-value is lower, then we reject the null hypothesis.
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Please answer and not give me links that make me download a whole bunch of stuff then just turn out to be an inappropriate picture (Yeah that's happened to me twice on here :/)
Answer:
3
Step-by-step explanation:
12/4=3
Graph y=2x+1
Can you please help.
Answer:
Here is the image you need to replicated
(っ◔◡◔)っ ♥ Hope It Helps ♥
please write all the steps... write clearly thanks
Determine the inverse Laplace transforms of: (b) 1 3s²+5s+1
The inverse Laplace transform of 1 / (3s² + 5s + 1) is f(t) = 1/2 × [tex]e^{(-t)[/tex]- 1/2 × [tex]e^{(-t/3)[/tex].
To find the inverse Laplace transform of the function F(s) = 1 / (3s² + 5s + 1), we can use partial fraction decomposition and reference tables for Laplace transforms.
Step 1: Factorize the denominator
Factorize the denominator of the function 3s² + 5s + 1 to find its roots:
3s² + 5s + 1 = (s + 1)(3s + 1)
Step 2: Write the partial fraction decomposition
Write the function F(s) as a sum of partial fractions:
F(s) = A / (s + 1) + B / (3s + 1)
Step 3: Determine the values of A and B
To find the values of A and B, we can multiply both sides of the equation by the common denominator and equate the numerators:
1 = A(3s + 1) + B(s + 1)
Expand the right side and collect like terms:
1 = (3A + B)s + (A + B)
By equating the coefficients of s and the constant terms on both sides, we get a system of equations:
3A + B = 0
A + B = 1
Solving this system of equations, we find A = 1/2 and B = -1/2.
Step 4: Write the inverse Laplace transform
Using the partial fraction decomposition, we can now write the inverse Laplace transform:
F(s) = 1/2 × (1 / (s + 1)) - 1/2 × (1 / (3s + 1))
Referring to Laplace transform tables, we find that the inverse Laplace transform of 1 / (s + a) is [tex]e^{(at)[/tex], and the inverse Laplace transform of 1 / (s - a) is [tex]e^{(at)[/tex]. Therefore, applying these results, we have:
f(t) = 1/2 × [tex]e^{(-t)[/tex] - 1/2 × [tex]e^{(-t/3)[/tex]
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Find the measure of the exterior 1.
A. 144°
B. 56°
C. 36°
D. 136°
= a) Prove that the given function u(x, y) = -8x3y + 8xy3 is harmonic b) Find v, the conjugate harmonic function and write f(z).
The function f(z) = u(x, y) + iv(x, y) simplifies to f(z) = -8x³y + 8xy³.
The function f(z) corresponding to the given function u(x, y) is f(z) = -8x³y + 8xy³.
To prove that the function u(x, y) = -8x³y + 8xy³ is harmonic, we need to show that it satisfies the Laplace's equation:
∇²u = ∂²u/∂x² + ∂²u/∂y² = 0
Let's calculate the second partial derivatives of u with respect to x and y:
∂u/∂x = -24x²y + 8y³
∂²u/∂x² = -48xy
∂u/∂y = -8x^3 + 24xy²
∂²u/∂y²= 48xy
Now, substitute these values into the Laplace's equation:
∂²u/∂x² + ∂²u/∂y² = -48xy + 48xy = 0
Since the Laplace's equation is satisfied, the function u(x, y) = -8x³y + 8xy³ is harmonic.
b) To find the conjugate harmonic function v, we need to find a function v(x, y) such that f(z) = u(x, y) + iv(x, y) is analytic, where z = x + iy.
Since the given function u(x, y) = -8x³y + 8xy³ is a real-valued function, the imaginary part of the conjugate harmonic function v(x, y) must be zero. Therefore, v(x, y) = 0, and the conjugate harmonic function is purely real.
Hence, the function f(z) = u(x, y) + iv(x, y) simplifies to f(z) = -8x³y + 8xy³.
Therefore, the function f(z) corresponding to the given function u(x, y) is f(z) = -8x³y + 8xy³.
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anthony can run at the rate (in meters per minute) shown in the graph below) which of the following best describes anthony’s rate of speed
Step-by-step explanation:
no this is a test I cant hlep i will tell ms.jules you are searching for the answers
The rank of the matrix of coefficients of a homogeneous system of m linear equations in n unknowns is never less than the rank of the augmented matrix. (a) Always true (b) Sometimes true (c) Never true, i.e., false (d) None of the above
The cοrrect answer is (c) Never true, i.e., false. "The rank οf the matrix οf cοefficients οf a hοmοgeneοus system οf m linear equatiοns in n unknοwns is never less than the rank οf the augmented matrix" is nοt always true.
What is a linear equatiοn?A linear equatiοn is an algebraic equatiοn that represents a straight line when graphed οn a Cartesian cοοrdinate plane. It is an equatiοn in which the highest pοwer οf the variable(s) is 1.
The rank οf the matrix οf cοefficients οf a hοmοgeneοus system οf m linear equatiοns in n unknοwns represents the maximum number οf linearly independent equatiοns in the system. It gives us infοrmatiοn abοut the dimensiοn οf the sοlutiοn space.
On the οther hand, the rank οf the augmented matrix οf the system includes bοth the matrix οf cοefficients and the cοlumn vectοr οf cοnstants. It represents the maximum number οf linearly independent rοws in the augmented matrix.
In general, the rank οf the matrix οf cοefficients can be less than, equal tο, οr greater than the rank οf the augmented matrix. It depends οn the specific system οf equatiοns.
Therefοre, the statement "The rank οf the matrix οf cοefficients οf a hοmοgeneοus system οf m linear equatiοns in n unknοwns is never less than the rank οf the augmented matrix" is nοt always true.
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During Year 3, Vernon Corporation reported after the net come of $3,595.000 During the year the number of shares of stock outstanding remained constant at 10.000 of $100 par 9 percent prefered stock and 399,000 shares of common stock. The company's kotel stockholders equity is $19,600,000 at December 31. Year 3. Vernas Corporation's common stock was selling at $54 per share at the end of is scal year. All clvidends for the year have been paid, including $4.20 per share so common stockholders.
Required
3. Compute the earings per share, (Round your answer to 2 decimal places.)
b. Compute the books te of common stock (Round your answer to 2 decimal places.) c. Compute the price warning Mic Round intermediate calculations and final answer to 2 decimal places) d. Compute the dividend yield (Round your percentage answer to 2 decimal places, fe, 0.2345 should be entered as 23.45).)
a The net income is $3505000
b The number of common shares outstanding is 399,000
c Earnings per share is $8.79
d Book value of common stock is $46.62
e. Price-earnings ratio (P/E ratio) is 6.14
How to calculate the valuea. Net income available to common stockholders:
Net income = $3,595,000
Dividends to preferred stockholders = (Number of preferred shares * Par value * Dividend rate) = (10,000 * $100 * 0.09) = $90,000
Net income available to common stockholders = Net income - Dividends to preferred stockholders
= $3,595,000 - $90,000
= $3,505,000
b. Weighted average number of common shares outstanding:
Number of common shares outstanding = 399,000
c. Earnings per share: EPS = Net income available to common stockholders / Weighted average number of common shares outstanding
= $3,505,000 / 399,000
≈ $8.79
d. Book value of common stock:
Total stockholders' equity = $19,600,000
Preferred stock equity = Number of preferred shares * Par value = 10,000 * $100 = $1,000,000
Common stock equity = Total stockholders' equity - Preferred stock equity
= $19,600,000 - $1,000,000
= $18,600,000
Book value of common stock = Common stock equity / Number of common shares outstanding
= $18,600,000 / 399,000
≈ $46.62
e. Price-earnings ratio (P/E ratio):
Price-earnings ratio = Price per share / Earnings per share
= $54 / $8.79
≈ 6.14
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Joe wants to rent a boat and spend less than $33. Boat cost $7 per hour, and Joe has a discount coupon for $9 off. What are the possible numbers of hours Joe could rent the boat? Use t for the number of hours. Write your answer as an inequality solved for t.
$7 per hour plus $9 off
$7x5=$35-$9 discount=$26
$26+$7=$33
6hours=$42-$9=$33
Joe could rent the boat for 6 hours
PLEASE HELP ME and the first person that answers correctly will get... BRAINLIEST I promise.
Step-by-step explanation:
V≈75.4
A≈100.53
How would u rewrite 4/5 and 1 2/3
Step-by-step explanation:
division fractions that's how u will rewrite them
What is the measure of the angle at the tail end of the kite? if the top area is where its pointed is 122?
Answer:
The angle is 58 degrees
Step-by-step explanation:
Given
See attachment for kite
Required
The angle at the tail end
Represent this angle with x.
From the attached kite, we have:
1 angle = 122
2 angles = right-angled
So, we have:
[tex]x + 122 +90+90 = 360[/tex] --- sum of angles in a kite
[tex]x + 302 = 360[/tex]
Solve for x
[tex]x =- 302 + 360[/tex]
[tex]x =58^\circ[/tex]
What is 7/4 as a mixed number
Find a formula for the exponential function passing through the points (-3,1/2) and (3,32).
The exponential function passing through the points (-3, 1/2) and (3, 32) can be represented by the equation [tex]f(x) = (2^{(5/6)}) * (2^{(x/6)})[/tex].
To find the exponential function passing through the given points, we can start by assuming the general form of an exponential function, [tex]f(x) = a * (b^x)[/tex], where a and b are constants to be determined. Plugging in the coordinates (-3, 1/2) into this equation gives us [tex]1/2 = a * (b^{(-3)})[/tex], and plugging in (3, 32) gives us [tex]32 = a * (b^3)[/tex].
Now, we can solve this system of equations to find the values of a and b. Taking the ratio of the two equations, we get [tex](1/2) / 32 = (a * (b^{(-3)})) / (a * (b^3))[/tex], which simplifies to [tex]1/64 = 1/b^6[/tex]. Solving for b, we find [tex]b = 2^{(1/6)[/tex].
Substituting this value back into either of the original equations, we can solve for a. Using the equation [tex]1/2 = a * (2^{(-3/6)})[/tex], we find [tex]a = 2^{(5/6)[/tex].
Therefore, the exponential function passing through the given points is [tex]f(x) = (2^{(5/6)}) * (2^{(x/6)})[/tex].
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SOMEONE PLEASE HELP I'VE BEEN ASKING FOR DAYS NOW WITH SCALE FACTOR!!!!! I will give brainliest if right!!!!!!
Answer:
It has been answered for you
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Help please I’ll mark brainiest
Answer:
V≈7853.98
Step-by-step explanation:
V=πr2h
r=10
h=25
Solution
V=πr2h=π·102·25≈7853.98163
Find the area of the shape shown below.
Answer:
42 units²
Step-by-step explanation:
Can someone please help me with this page
Answer:
2. there is no E it would be AB=CD
Moving to another question will save this response. uestion 2 15 point Pretty Lady Cosmetic Products has an average production process time of 40 days. Finished goods are kept on hand for an average of 15 days before they are sold. Accounts receivable are outstanding an avera and the firm receives 40 days of credit on its purchases from suppliers. Assume net sales of $1,200,000 and cost of goods sold of $900,000. Determine the average investment in accounts receivable, inventories, and accounts payable. What would be the net financing need conside three accounts? *Note: To solve this problem, you will need to first find the Inventory Period, the Receivables Period, and the Payment Period. $153.054.79 $154,054.79 $152.054.79 $152,154.80
The net financing need considering the three accounts is approximately $185,753.42.
To determine the average investment in accounts receivable, inventories, and accounts payable, we need to calculate the Inventory Period, Receivables Period, and Payment Period.
Inventory Period:
The Inventory Period is the average number of days it takes for finished goods to be sold. In this case, the average production process time is given as 40 days, and finished goods are kept on hand for an average of 15 days before they are sold. Therefore, the Inventory Period is the sum of these two periods:
Inventory Period = Production Process Time + Days Kept on Hand
Inventory Period = 40 days + 15 days
Inventory Period = 55 days
Receivables Period:
The Receivables Period is the average number of days it takes for accounts receivable to be collected. It is given that accounts receivable are outstanding for an average of 40 days. Therefore, the Receivables Period is 40 days.
Payment Period:
The Payment Period is the number of days the firm receives credit on its purchases from suppliers. It is given that the firm receives 40 days of credit. Therefore, the Payment Period is 40 days.
Now, we can calculate the average investment in accounts receivable, inventories, and accounts payable.
Average Investment in Accounts Receivable:
Average Investment in Accounts Receivable = (Net Sales / 365) * Receivables Period
Average Investment in Accounts Receivable = ($1,200,000 / 365) * 40
Average Investment in Accounts Receivable ≈ $131,506.85
Average Investment in Inventories:
Average Investment in Inventories = (Cost of Goods Sold / 365) * Inventory Period
Average Investment in Inventories = ($900,000 / 365) * 55
Average Investment in Inventories ≈ $142,191.78
Average Investment in Accounts Payable:
Average Investment in Accounts Payable = (Cost of Goods Sold / 365) * Payment Period
Average Investment in Accounts Payable = ($900,000 / 365) * 40
Average Investment in Accounts Payable ≈ $87,945.21
Finally, to calculate the net financing need, we subtract the average investment in accounts payable from the sum of the average investment in accounts receivable and inventories:
Net Financing Need = Average Investment in Accounts Receivable + Average Investment in Inventories - Average Investment in Accounts Payable
Net Financing Need = $131,506.85 + $142,191.78 - $87,945.21
Net Financing Need ≈ $185,753.42
Therefore, the net financing need considering the three accounts is approximately $185,753.42.
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A school is planning for an addition in some open space next to the current building. The existing building ends at the origin. The graph represents the system of equations that can be used to define the space for the addition. What is the system of equations that matches the graph?
y ≤ 3x
y > –2x – 1
y > 3x
y ≤ –2x – 1
y < –3x
y ≥ 2x – 1
y > –3x
y ≤ 2x – 1
Equation system that corresponds to the graph:
1. y ≤ 3x 2. y > –2x – 1
To find the system of equations that corresponds to the provided graph, we must first analyze it and locate the regions that fulfil the specified requirements.
1. Begin by locating the darkened region underneath the line y 3x. This line has a slope of 3 and intersects the origin (0,0). Shade the area beneath the line.
2. After that, locate the darkened region above the line y > -2x - 1. The slope of this line is -2, while the y-intercept is -1. The area above the line should be shaded.
3. The solution space that meets both requirements is represented by the overlapping shaded region between the two lines. The common area is located below y 3x and above y > -2x - 1.
4. The equation system that corresponds to this common area is: - y 3x - y > -2x - 1
The space for the addition in the open area next to the present building is defined by these two formulae.
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Solve the equation and determine the value of t.
t + 38 = 420
t =
Answer:
[tex]t=382[/tex]
Step-by-step explanation:
[tex]\mathrm{Subtract\:}38\mathrm{\:from\:both\:sides}[/tex]
[tex]t+38-38=420-38[/tex]
[tex]=382[/tex]
Answer:
[tex] \displaystyle t = 382[/tex]
Step-by-step explanation:
we are given a equation
we want to figure out t
to do so
recall that
addition equality propertySubstraction equality propertymultiplication equality propertydivision equality propertytherefore
if we substract 38 from both sides it won't affect the equality
so
[tex] \displaystyle t + 38 - 38 = 420 - 38 \\ \therefore \: t = 382[/tex]
let's verify it
[tex] \displaystyle382 + 38\stackrel{?}{=}420[/tex]
simplify addition:
[tex] \displaystyle420\stackrel{ \checkmark}{=}420[/tex]
hence, verified!
The blueprint for a fence shows a rectangle with the dimensions bin x
18in. The scale for the actual fence is 2in : 10ft. What is the length and
width of the actual fence?
Answer:
The width of the actual fence is 90ft
Step-by-step explanation:
If 2in = 10ft, you want to find the scale factor. To do this, you need to divide 10/2, so for every 5ft, it is 1in. Now, you would take how many inches you have and multiply it by 5. (in our case, 18in)
18 x 5 = 90.
So the actual fence length is 90ft.
Stonewall receives ¢250 per year in simple interest from an amount of money he invested in
ADB, Barclays and GCB. Suppose ADB pays an interest of 2%, Barclays pays an interest of
4% and GCB pays an interest of 5% per annum and an amount of ¢350 more was invested in
Barclays than the amount invested in ADB and GCB combined. Also, the amount invested in
Barclays is 2 times the amount invested in GCB.
a) Write down the three linear equations and represent them in the matrix form AX = B.
b) Find the amount of money Stonewall invested in ADB, Barclays and GCB using Matrix
Inversion
Answer:
The amount invested in ADB is ¢1363.[tex]\overline 3[/tex] =
The amount invested in Barclays is ¢3,427.[tex]\overline 3[/tex]
The amount invested in GCB is ¢1,713.[tex]\overline 3[/tex]
Step-by-step explanation:
The parameters of the investment Stonewall made are;
The amount in interest he receives from ADB, Barclays and GCB = ¢250
The amount of interest ADP pays = 2% per annum
The amount of interest Barclays pays = 4% per annum
The amount of interest GCB pays = 5% per annum
The amount invested in Barclays = The amount invested in ADB and GCB + ¢350
The amount invested in Barclays = 2 × The amount invested in GCB
a) Let 'x', represent the amount invested in ADB, 'y' represent the amount invested in Barclays, and 'z', represent the amount invested in GCB
We have;
y = x + z + 350
y = 2·z
0.02·x + 0.04·y + 0.05·z = 250
Therefore, we get the three linear equations as follows;
-x + y - z = 350...(1)
y - 2·z = 0...(2)
0.02·x + 0.04·y + 0.05·z = 250...(3)
Using Matrix inversion, we have;
[tex]\left[\begin{array}{ccc}-1&1&-1\\0&1&-2\\0.02&0.04&0.05\end{array}\right] \times \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}350&0&250\end{array}\right][/tex]
The transpose of the 3 by 3 matrix [tex]M^T[/tex] is given as follows;
[tex]M^T = \left[\begin{array}{ccc}-1&0&0.02\\1&1&0.04\\-1&-2&0.05\end{array}\right][/tex]
The Adjugate Matrix is given as follows;
[tex]Adj = \left[\begin{array}{ccc}0.13&-0.09&-1\\-0.04&-0.03&-2\\-0.02&0.06&-1\end{array}\right][/tex]
The inverse of the matrix = Adj/Det where, Det = -0.15, is therefore;
[tex]M^{-1} = \left[\begin{array}{ccc}\dfrac{-13}{15} &\dfrac{3}{5} &\dfrac{20}{3} \\\\\dfrac{4}{15} &\dfrac{1}{5} &\dfrac{40}{3} \\\\\dfrac{2}{15} &-\dfrac{2}{5} &\dfrac{20}{3} \end{array}\right][/tex]
We therefore, get the solution as follows;
[tex]\left[\begin{array}{ccc}\dfrac{-13}{15} &\dfrac{3}{5} &\dfrac{20}{3} \\\\\dfrac{4}{15} &\dfrac{1}{5} &\dfrac{40}{3} \\\\\dfrac{2}{15} &-\dfrac{2}{5} &\dfrac{20}{3} \end{array}\right]\times \left[\begin{array}{c}350&0&250\end{array}\right] = \left[\begin{array}{c}\dfrac{4,090}{3} \\&\dfrac{10,280}{3} \\ & \dfrac{5,140}{3} \end{array}\right][/tex]
[tex]\begin{array}{c}x = \dfrac{4,090}{3} \\&y = \dfrac{10,280}{3} \\ & z = \dfrac{5,140}{3} \end{array}[/tex]
The amount invested in ADB, x = ¢4,090/3 = ¢1363.[tex]\overline 3[/tex]
The amount invested in Barclays, y = ¢10,282/3 = ¢3,427.[tex]\overline 3[/tex]
The amount invested in GCB, z = ¢5,140/3 = ¢1,713.[tex]\overline 3[/tex]
Y’all NLE Choppa is in jail. Sending prayers and love
Answer:
NLE Choppa noooooooooooooooooo
Step-by-step explanation:
THE BEST RAPPPERRR!!!!!!!!!
Answer:
He wasnt in jail? I dont think cuz he was makin new music
Step-by-step explanation:
i need help i will give branliest please !!!
Can someone please help me please I really need help
Answer:
16 cans
Step-by-step explanation:
Hey,
I have to admit, this problem is pretty complicated. But I got you :).
To begin, we have to represent the small and large boxes with two separate variables. So...
y = large
x = small
(It really doesn't matter what variable you use, I just used these)
Now, we know that Estelle fills 3 large boxes and 5 small boxes and had a total of 170 cans in total. We can represent this by writing...
3y + 5x = 170
Then, we know that the boy worker filled 4 large boxes and 4 small boxes and had a total of 184 cans. We can represent this by writing...
4y + 4x = 184
As you can see, this is a system of equations.
3y + 5x = 170
4y + 4x = 184
We want to know how many cans each small box can hold, so we have to find a common number for x since x represents the small box.
To do this, we have to multiply the first equation by 4 and the second by 3. Here's what I mean...
4 (3y + 5x = 170)
3 (4y + 4x = 184)
When we do this you get...
12y + 20x = 680
12y + 12x = 552
Notice how the y's are now both 12. We had to do that in order to get rid of y, they had to equal the same number. Now, subtract...
8x = 128
Divide by 8...
x = 16
That means that...
YOUR ANSWER: Each small box can hold up to 16 cans.
I hope this helps :)
In how many ways can 6 adults and 3 children stand together in a line so that no two children are next to each other?
There are 168 ways for 6 adults and 3 children to stand together in a line with no adjacent children.
we can treat the 6 adults as distinct entities and place them in the line first. There are 6! (factorial) ways to arrange the adults. Now, we have 7 spaces between the adults (including the ends) where the children can be placed. We need to choose 3 out of these 7 spaces for the children, which can be done in 7C3 ways (7 choose 3). Therefore, the total number of arrangements is 6! × 7C3 = 720 × 35 = 16800. However, we need to exclude the cases where two or more children are adjacent. By considering the possible positions of two adjacent children, there are 48 such cases. Subtracting these cases from the total gives us 16800 - 48 = 16752. Therefore, there are 16752 ways for 6 adults and 3 children to stand together in a line with no adjacent children.
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how do i get all my points and brainlest back
.....um...i think u can't do this
Answer:
u should probably contact support or the creator of this site
Step-by-step explanation: