Nonlinear Differential equation The discriminating equation said to be non direct if any product exists between the dependent variable and its derivations or between the derivations themselves.
The given differential equation,
(3)dydx = y(9+x2)/x(3-7y)
dy/dx 9+x2)/y (3-7y)
1 () dy 9+x2) dx (3-7y) V = 0
The given differential equation contains term product between the dependent variable (y) themselves.
So It is nonlinear differential equation.
Alternative ( With all questions answer ) :
1. Given differential equation is -
--> Ordinary
--> Of order 2
--> with x as dependent and t as independent variable
--> Linear
2. Given differential equation is -
--> Ordinary
--> Of order 2
--> with y as dependent and x as independent variable
--> Linear
3. Given differential equation is -
--> Ordinary
--> Of order 1
--> with y as dependent and x as independent variable
--> NOT Linear
1) The secondary dy dx is called ordinary.
The secondary ду хе is called partial. This isn't set up in any of the equations in question. Hence all are ordinary discriminational equations.
2) The secondary dy dx is of order 1 and the secondary day dx2 is of order 2. This way we can have third,
fourth. order derivations.
Our first 2 equations have maximum outgrowth of order 2. Hence similar ODE is of order 2. The last question equation has loftiest order of outgrowth = 1 hence it's first order ODE.
3) The variables whose derivations do in the equation are dependent and remaining variables are independent. Then in first question outgrowth of x is taken. Hence x is dependent. Same description applies for the remaing ODEs.
4) The linearity depends on 2 facctors. The power( indicator or exponent) of dy dx or day dx2( whichever occers in the equation) is called degree of ODE.( elect the loftiest order outgrowth to find degree). In any of the question we do not have degree further than 1. So this first criteria of linearity is satisfied by all given.
The alternate criteria is in any of terms containing dy dx or day dx2 the measure( constant or variable) mustn't contain the dependent variable. In our last question, if simplified we get the dependent variable( 1/ y) in addition with dy dx. Hence it's NOT direct. The terms without derivations can contain y but there also power of y must be 1. Look at question 2. The last term is 2y. If it were y2 or y3 or 1 y that equation couldn't be direct.
Simplification of (3)
dy dx y(2-3x x(1-3 y )
1-3y dy (2-3x) dx у
(2-3x) :- -3) dy dx
This can be further integrated to solve, but the equation is NOT liner.
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show me neat sketch of two graphs on the same set of axes that shows how the total amount repaid (T) depends on the number of months (N) for both of the instalment factors.
Answer:
Step-by-step explanation:
I can explain how the two graphs would look like.
If we are comparing two installment factors with respect to the total amount repaid (T) over time (N), we can plot two separate graphs on the same set of axes. The x-axis would represent the number of months (N) and the y-axis would represent the total amount repaid (T).
For each installment factor, we would plot a line that represents the total amount repaid over time. The shape and slope of each line would depend on the interest rate, the amount borrowed, and the number of payments per year.
The two lines would likely have a similar shape, but different slopes and intercepts. The slope of each line would depend on the interest rate and the amount borrowed, with steeper slopes indicating higher interest rates or larger amounts borrowed. The intercept of each line would depend on the amount borrowed and the number of payments per year.
To compare the two installment factors, we would look at the relative slopes and intercepts of the two lines. If one line has a steeper slope than the other, it would indicate that the total amount repaid increases more rapidly over time for that installment factor. If one line has a higher intercept than the other, it would indicate that the total amount repaid is higher at the beginning of the repayment period for that installment factor.
Overall, by plotting the two graphs on the same set of axes, we can visually compare how the total amount repaid (T) depends on the number of months (N) for both installment factors.
in a 100-meter race, the runner travels the 1st and 2nd 50-meters in time t1 and t2. usually: group of answer choices do not know t1 or t2 is longer. t1
In a 100-meter race, the runner travels the 1st and 2nd 50-meters in time t1 and t2. Then t1>t2. So, the correct option is D.
During the first half , the runner has to start from the rest and will gain some velocity as he move further from start point, where as for the second half , as he already has the velocity required to run, he will complete the second half faster than first half.
Let us assume that V1 is the velocity , that the runner is having during the first half. V1 doesn't start as soon as he starts running, at first the body is in rest and velocity V1 starts increasing after he travels few distance.
And Let V2 be the velocity for the runner in second half , as he already have the velocity V1, there will be no delay to start the second half. So, second half takes less time than first half. This can be expressed as:
t2>t1
where t2 is second half time and t1 is first half time.
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Complete question is:
In a 100-meter race, the runner travels the 1st and 2nd 50-meters in time t1 and t2. usually: group of answer choices
a) do not know t1 or t2 is longer.
b) t1 = t2
c) t1<t2
d) t1>t2
please help. . . . . . . .
Answer:
3. [tex]y=(x-3)(x-2)[/tex]
4. [tex]y = (x + 2)^2[/tex]
Step-by-step explanation:
In this problem, we are asked to find the equations in their factored form of the graphed parabolas.
3. We can see that the parabola's vertex is at (3, -4). We can plug the coordinates of that point into the vertex form equation:
[tex]y=(x-a)^2 + b[/tex]
where [tex](a,b)[/tex] is the vertex of the parabola.
[tex]y=(x-3)^2 -4[/tex]
Then, we can expand the right side of the equation to an unfactored form.
[tex]y=x^2-6x+9 -4[/tex]
[tex]y=x^2-6x-5[/tex]
Finally, we can factor the right side of the equation.
[tex]\boxed{y=(x-3)(x-2)}[/tex]
4. First, input the vertex's coordinates into the vertex form equation.
[tex]y=(x - (-2))^2 + 0[/tex]
Then, simplify.
(Remember that subtracting a negative is the same as adding the positive)
[tex]\boxed{y=(x+2)^2}[/tex]
5
6
Check
7
8
10
11
12
13
A-, B-, C- 51.1
14
15
Consider a triangle ABC like the one below. Suppose that C-96°, a-33, and b=39. (The figure is not drawn to scale.) Solve the triangle.
Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.
If there is more than one solution, use the button labeled "or".
DSD
X
16
No
solution
5
Save For Later
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The value of C is 108°
How to solve for thisGiven [tex]a=33,\ \ b=26, \ \ \angle B=31^0 .[/tex]
We have to find [tex]\angle C, \ \angle A \ and\ c[/tex]
We can use the cosine formula to find these:
[tex]cosB=\frac{a^2+c^2-b^2}{2ac}\\cos31^o=\frac{33^2+c^2-26^2}{2(33)(c)}\\0.8572=\frac{413+c^2}{66c}\\56.5752c=413+c^2\\i.e. c^2-56.5752c+413=0\\\Rightarrow c=47.9647, \ 8.6104\\So, \mathbf{c=48 \ or \ c=9}\\Now if c=48:\\cosA=\frac{b^2+c^2-a^2}{2bc}\\=\frac{26^2+48^2-33^2}{2(26)(48)}\\[/tex]
[tex]=\frac{1891}{2496}\\=0.75761\\\therefore A=cos^{-1}(0.75761)=40.746^0\approx 41^0\\i.e. \mathbf{\angle A= 41^0}\\cosC=\frac{a^2+b^2-c^2}{2ab}\\=\frac{33^2+26^2-48^2}{2(33)(26)}\\=\frac{-539}{1716}\\=-0.3141\\\therefore C=cos^{-1}(-0.3141)=108.306^0\approx 108^0[/tex]
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( 7 x + 5 ) ( 2 x 3 − 4 x 2 + 9 x − 3 ) A. 14 x 4 − 18 x 3 + 43 x 2 + 24 x − 15 B. 9 x 4 − 11 x 3 + 16 x 2 + 11 x − 15 C. 9 x 4 − 11 x 3 + 16 x 2 + 11 x − 8 D.
Expanding the polynomial, (7x + 5)(2x³ − 4x² + 9x − 3), we have: A. 14x^4 - 18x³ + 43x² + 24x - 15.
How to Expand Expressions?By applying the distribution property, we can expand a given polynomial expression like the one given above.
Given, (7x + 5)(2x³ − 4x² + 9x − 3), distribute to eliminate the parentheses:
7x(2x³ − 4x² + 9x − 3) + 5(2x³ − 4x² + 9x − 3)
14x^4 - 28x³ + 63x² - 21x + 10x³ - 20x² + 45x - 15
Combine like terms:
14x^4 - 18x³ + 43x² + 24x - 15
The correct solution is option A.
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help please see photo
The solution of the inequality using the coordinate pair is 5.
What is Linear Inequality?Linear inequalities are those expressions which are connected by inequality signs like >, <, ≤, ≥ and ≠ and the value of the exponent of the variable is 1.
Given linear inequality is,
y ≤ -2x + 3
We have a coordinate pair (-1, 5).
So substitute x = -1 and y = 5 in the given inequality.
5 ≤ (-2 × -1) + 3
5 ≤ 2 + 3
5 ≤ 5
Hence the missing numbers are found.
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An entry-level civil engineer earns an average bi-weekly net pay of $2,250.65. The engineer has created a monthly budget using the following percentages for expenses: Percent Housing 25% Food/Household 20% Savings 10% Transportation 5% Debt 15% Entertainment 5% Medical/Personal Care 5% Giving 5% Clothing 2% Miscellaneous 8% Which balance sheet correctly represents the engineer's income, expenses, and balance?
The engineer's balance sheet would show a monthly income of $9,002.60 ($4,501.30 x 2), total expenses of $4,001.36, and a balance of $499.94.
What is a balance sheet?Balance sheet is a financial statement that contains details of a company's assets or liabilities at a specific point in time.
First we can use the given information to create the following balance sheet for the civil engineer:
INCOME:
Bi-weekly net pay: $2,250.65 x 2 = $4,501.30 (this assumes two pay periods per month)
EXPENSES:
Housing: 25% of $4,501.30 = $1,125.33Food/Household: 20% of $4,501.30 = $900.26Savings: 10% of $4,501.30 = $450.13Transportation: 5% of $4,501.30 = $225.07Debt: 15% of $4,501.30 = $675.20Entertainment: 5% of $4,501.30 = $225.07Medical/Personal Care: 5% of $4,501.30 = $225.07Giving: 5% of $4,501.30 = $225.07Clothing: 2% of $4,501.30 = $90.03Miscellaneous: 8% of $4,501.30 = $360.10TOTAL EXPENSES: $4,001.36
BALANCE: $4,501.30 - $4,001.36 = $499.94
Therefore, the engineer's balance sheet would show a monthly income of $9,002.60 ($4,501.30 x 2), total expenses of $4,001.36, and a balance of $499.94.
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I NEED HELP PLEASE HELP ME
Cecily purchases a box of 100 paper clips. She puts 37 /100 of the paper clips in a jar on her desk and puts another 6 /10 in her drawer at home. Shade a grid that shows how many of the paper clips are in Cecily's jar and drawer, then write the fraction tbe grid represents.
The fraction of the total of paper clips Cecily put in the jar and drawer is [tex]\frac{97}{100}[/tex] (see the attachment below).
How many clips did Cecily put in the jar and drawer?Paper clips in the jar: 37/100, which means Cecily put 37 clips in the jar.
Paper clips in the drawer: 6/10, now let's find out the number of clips this fraction is equivalent to:
100 (total clips) / 10 x 6 = 60
Total clips: 60 +37= 97. This number of clips can be expressed as [tex]\frac{97}{100}[/tex] .
How to represent this in a grid?To represent this in a grid color a total of 97 squares in a grid with 100 squares as it is shown below.
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Accounting /
The following list of accounts was drawn for Tile, Etc., Inc. on December 31, Year 1, after the closing entries were posted:
Account Title
Cash
Accounts receivable
Allowance for doubtful accounts.
Inventory
Accounts payable
Common stock
Retained earnings
$110,000
125,000
18,000
425,000
95,000
450,000
97,000
Tile, Etc. had the following transactions in Year 2:
1. Purchased merchandise on account for $580,000.
2. Sold merchandise that cost $420,000 for $890,000 on account.
3. Sold for $245,000 cash merchandise that had cost $160,000.
4. Sold merchandise for $190,000 to credit card customers. The merchandise had cost $96,000. The credit card company charges a 4
percent fee.
5. Collected $620,000 cash from accounts receivable.
6. Paid $610,000 cash on accounts payable.
7. Paid $145,000 cash for selling and administrative expenses.
8. Collected cash for the full amount due from the credit card company (see item 4).
9. Loaned $60,000 to J. Parks. The note had an 8 percent interest rate and a one-year term to maturity.
10. Wrote off $7,500 of accounts as uncollectible.
11. Made the following adjusting entries:
(a) Recorded uncollectible accounts expense estimated at 1 percent of sales on account.
(b) Recorded seven months of accrued interest on the note at December 31, Year 2 (see item 9).
Required
a. Organize the transaction data in accounts under an accounting equation.
b. Prepare an income statement, a statement of changes in stockholders' equity, a balance sheet, and a statement of cash flows for
Year 2.
The Total Stockholder's Equity is $1,266,290.00
How to represent the income statementIncome Statement Amount $ Amount $
Sales Revenue 1,585,000.00
Interest Revenue 2,940.00
Total Revenue 1,587,940.00
Less: Expenses
Cost of merchandise sold 793,000.00
Selling and admin expense 158,000.00
Credit card expense 7,650.00
Bad debt reinstated (5,500.00)
Total Operating expense 953,150.00
Net Income 634,790.00
Statement of changes in Stockholder's Equity Amount $ Amount $
Beginning Common Stock 515,000.00
Add: Common Stock issued -
Ending Common Stock 515,000.00
Beginning Retained Earnings 116,500.00
Add: Net Income 634,790.00
Ending Retained Earnings 751,290.00
Total Stockholder's Equity 1,266,290.00
Balance Sheet Amount $ Amount $
Assets
Cash 586,350.00
Inventory 303,000.00
Accounts Receivable 399,200.00
Allowance for uncollectible accounts (10,200.00) 389,000.00
Notes Receivable 63,000.00
Interest Receivable 2,940.00
Total Assets 1,344,290.00
Liabilities & Stockholder's Equity
Liabilities
Current Liabilities
Accounts Payable 78,000.00
Current Liabilities 78,000.00
Stockholder's Equity
Common Stock 515,000.00
Retained Earnings 751,290.00
Total Stockholder's Equity 1,266,290.00
Total Liabilities & Stockholder's Equity 1,344,290.00
Statement of cash flows Amount $ Amount $
Cash flow from operating activities
Cash sales 310,000.00
Inflow from customers 997,350.00
Payment for inventory (675,000.00)
Payment for expenses (158,000.00)
Net Cash flow from operating activities 474,350.00
Cash flow from investing activities
Notes receivable provided (63,000.00)
Cash flow from financing activities
Cash flow from financing activities -
Net Change in cash 411,350.00
Add: Beginning cash balance 175,000.00
Ending cash balance 586,350.00
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Please solve within 30 minutes! Thanks!
Please solve for the variable indicated.
20.) A=1/2bh, solve for b
If you could break it down step by step that would be super helpful! I’m very confused. Thank you!
Step-by-step explanation:
A = 1/2 bh
divide by 1/2
2A = bh
divide by h
2A/h = b
Hope this helps.
Answer:
[tex] b = \dfrac{2A}{h} [/tex]
Step-by-step explanation:
[tex] A = \dfrac{1}{2}bh [/tex]
[tex] 2A = 2 \times \dfrac{1}{2}bh [/tex]
[tex] 2A = bh [/tex]
[tex] \dfrac{2A}{h} = \dfrac{bh}{h} [/tex]
[tex] \dfrac{2A}{h} = b [/tex]
[tex] b = \dfrac{2A}{h} [/tex]
Deduct 20% for taxes from the monthly gross income to get the net income (the amount made after taxes).
What is Andrew's monthly net income?
What is Lisa's monthly net income?
What is their total monthly net income?
Andrew's monthly net income is $65280.
Lisa's monthly net income is $60960.
The total monthly net income is $126240.
What is a percentage?The percentage is calculated by dividing the required value by the total value and multiplying by 100.
Example:
Required percentage value = a
total value = b
Percentage = a/b x 100
Example:
50% = 50/100 = 1/2
25% = 25/100 = 1/4
20% = 20/100 = 1/5
10% = 10/100 = 1/10
We have,
Andrew's monthly gross income.
= $81, 600
Lisa's monthly gross income.
= $76,200
Now,
Tax = 20%
So,
Andrew's monthly net income.
= 80/100 x 81, 600
= $65280
Lisa's monthly net income.
= 80/100 x 76,200
= $60960
Now,
Total monthly net income.
= 65280 + 60960
= $126240
Thus,
Andrew's monthly net income = $65280
Lisa's monthly net income = $60960
Total monthly net income = $126240
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Maximize Z = 7X1 + 5X2
Subject to X1 + 2X2 ≤ 6
4X1 + 3X2 ≤ 12
X1, X2 ≥ 0
The solution is [tex]x_1[/tex] = 3 and [tex]x_2[/tex]= 0.
What is Simplex Method?Simplex method is an approach to solving linear programming models by hand using slack variables, tableaus, and pivot variables as a means to finding the optimal solution of an optimization problem.
Given:
We transfer the row with the resolving element from the previous table into the current table, elementwise dividing its values into the resolving element:
[tex]P_2[/tex] = [tex]P_2[/tex] / [tex]x_{2,1[/tex] = 12 / 4 = 3;
[tex]x_{2,1[/tex] = [tex]x_{2,1[/tex] / [tex]x_{2,1[/tex] = 4 / 4 = 1;
[tex]x_{2,2[/tex] = [tex]x_{2,2[/tex] / [tex]x_{2,1[/tex] = 3 / 4 = 0.75;
[tex]x_{2,3[/tex] = [tex]x_{2,3[/tex] / [tex]x_{2,1[/tex] = 0 / 4 = 0;
[tex]x_{2,4[/tex] = [tex]x_{2,4[/tex] / [tex]x_{2,1[/tex] = 1 / 4 = 0.25;
The remaining empty cells, except for the row of estimates and the column Q, are calculated using the rectangle method, relative to the resolving element:
[tex]P_1[/tex] = ([tex]P_1[/tex] . [tex]x_{2,1[/tex]) - ([tex]x_{1,1[/tex] . [tex]P_2[/tex]) / [tex]x_{2,1[/tex]= ((6 x 4) - (1 x 12)) / 4 = 3
[tex]x_{1,1[/tex]= (([tex]x_{1,1[/tex]. [tex]x_{2,1[/tex]) - ([tex]x_{1,1[/tex] . [tex]x_{2,1[/tex])) / [tex]x_{2,1[/tex]= ((1 x 4) - (1 x 4)) / 4 = 0
[tex]x_{1,3[/tex]= (([tex]x_{1, 3[/tex]. [tex]x_{2,1[/tex]) - ([tex]x_{1,1[/tex] . [tex]x_{2,3[/tex])) / [tex]x_{2,1[/tex]= ((1 x 4) - (1 x 0)) / 4 = 1
[tex]x_{1,4[/tex] = (([tex]x_{1, 4[/tex]. [tex]x_{2,1[/tex]) - ([tex]x_{1,1[/tex]. [tex]x_{2,4[/tex])) / [tex]x_{2,1[/tex]= ((0 x 4) - (1 x 1)) / 4 = -0.25
Objective function value
We calculate the value of the objective function by elementwise multiplying the column [tex]C_b[/tex] by the column P, adding the results of the products.
Max P = (C[tex]b_1[/tex] x [tex]P_{01[/tex]) + (C[tex]b_{11[/tex] x [tex]P_2[/tex])= (0 x 3) + (7 x 3) = 21;
Evaluated Control Variables
We calculate the estimates for each controlled variable, by element-wise multiplying the value from the variable column, by the value from the Cb column, summing up the results of the products, and subtracting the coefficient of the objective function from their sum, with this variable.
Max [tex]x_1[/tex]= ((C[tex]b_1[/tex] . [tex]x_{1,1[/tex]) + (C[tex]b_2[/tex] . [tex]x_{2,1[/tex]) ) - k[tex]x_1[/tex] = ((0 x 0) + (7 x 1) ) - 7 = 0
Max [tex]x_2[/tex]= ((C[tex]b_1[/tex] . [tex]x_{1,2[/tex]) + (C[tex]b_2[/tex] . [tex]x_{2,2[/tex]) ) - k[tex]x_2[/tex] = ((0 x 1.25) + (7 x 0.75) ) - 5 = 0.25
Max [tex]x_3[/tex]= ((C[tex]b_1[/tex] . [tex]x_{1,3[/tex]) + (C[tex]b_2[/tex] . [tex]x_{2,3[/tex]) ) - k[tex]x_3[/tex] = ((0 x 1) + (7 x 0) ) - 0 = 0
Max [tex]x_4[/tex]= ((C[tex]b_1[/tex] . [tex]x_{1,4[/tex]) + (C[tex]b_2[/tex] . [tex]x_{2,4[/tex]) ) - k[tex]x_4[/tex] = ((0 x -0.25) + (7 x 0.25) ) - 0 = 1.75
Since there are no negative values among the estimates of the controlled variables, the current table has an optimal solution.
The value of the objective function:
F = 21
The variables that are present in the basis are equal to the corresponding cells of the column P, all other variables are equal to zero:
[tex]x_1[/tex] = 3;
[tex]x_2[/tex]= 0;
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An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.
Compute the probability of each of the following events:
Event A: The sum is greater than 6. Event B: The sum is divisible by 3 or 5 (or both).
Write your answers as exact fractions.
P(a)
P(b)
The final answers are:
a) P(A) = 7/12
b) P(B) = 11/36
To compute the probabilities of Events A and B, we can use a table to list all possible outcomes and their corresponding sums:
Die 1 Die 2 Sum
1 1 2
1 2 3
1 3 4
1 4 5
1 5 6
1 6 7
2 1 3
2 2 4
2 3 5
2 4 6
2 5 7
2 6 8
3 1 4
3 2 5
3 3 6
3 4 7
3 5 8
3 6 9
4 1 5
4 2 6
4 3 7
4 4 8
4 5 9
4 6 10
5 1 6
5 2 7
5 3 8
5 4 9
5 5 10
5 6 11
6 1 7
6 2 8
6 3 9
6 4 10
6 5 11
6 6 12
a) Event A: The sum is greater than 6. From the table, we can see that there are 21 outcomes where the sum is greater than 6 (highlighted in bold). Thus, the probability of Event A is:
P(A) = 21/36 = 7/12
b) Event B: The sum is divisible by 3 or 5 (or both). From the table, we can see that the sums that are divisible by 3 or 5 (or both) are:
3, 5, 6, 9, 10, 12
There are 11 outcomes with these sums (highlighted in bold). Thus, the probability of Event B is:
P(B) = 11/36
Therefore, the final answers are:
a) P(A) = 7/12
b) P(B) = 11/36
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Help me please.
Very urgently
The answers to each part is given below.
What is function?A function is a elation between a dependent and independent variable.
Given is the function as -
f(x) = x² + 10x + 12
{ 1 } -
Using the quadratic formula, we can write -
x = - 10 ± √(100 - 48)/2
x = {- 10 ± √52}/2
{ 2 } -
We can write the area as -
1/2 x b x h = 96
1/2 x 2(x + 4) x {x} = 96
x(x + 4) = 96
x² + 4x - 96 = 0
Using the quadratic formula, we can write -
x = - 4 ± √(16 + 384)/2
x = -4 ± 20/2
Therefore, the answers to each part is given above.
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A total of 14 different people were randomly surveyed and asked how many hours per day they worked the week before Their answers are included below. Construct a box and whisker plot using a TI-83. TI-83 Plus, or TI-84 graphing calculator. Au plots below have the following window settings: Xmin = 7.95. Xmax = 8.25, Xsel = 0.01, Ymin = -0.5, Ymax = 3.5, Ysel = 1, Xres=1 Average Work Hours Per Day 8.08 8.08 8.04 8.12 8.05 8.05 7.97 8.00 8.10 8.22 8.09 8.00 815 D ex 70 6 t y u g LLLL wa Lose Calculate measures of center and spread using Technology Calculator Med=8.08 P1L1 Med=8.065 Med=8.14 P1:11 Med=8.09
The boxplot for the provide data is present in above figure. Median or central tendency of data is equals to 8.08. The lower and upper limits of outliers are 7.928 and 8.248 respectively.
The first task is to compute the median and the quartiles. And, in order to compute the median and the quartiles, the data needs to be put into the ascending order, as shown in the above table. Since the sample size n = 14 is even, we have that (n+1)/2 = (14+1)/2 =7.5, is not an integer value, the median is computing by taking the average of the values at the positions 7ᵗʰ and 8ᵗʰ, as shown below Median = (8.08 + 8.08)/2 = 8.08
Quartiles : The quartiles are computed using the table with the data in the asencending order. For Q₁ we have to compute the following position:
pos(Q) = (n+1)× 25/100 = 15×25/100 = 3.75
Since 3.75 is not an integer number, Q₁ is computed by interpolating between the
values located in the 3ᵗʰ and 4ᵗʰ positions, as shown in the formula above
Q₁ = 8.04+ (3.75 - 3) (8.05 - 8.04) = 8.0475
For Q3 we have to compute the following position:pos(Q₃) = (n + 1)× 75/100
= (14+1)×75/100 = 11.25
Since 11.25 is not an integer number, Q₃ is computed by interpolating between
the values located in the 11ᵗʰ and 12ᵗʰ positions, as shown in the formula below
Q₃ = 8.12 + (11.25 - 11) × (8.15 8.12) = 8.1275
The interquartile range is therefore
IQR = Q₃ - Q₁ = 8.1275 - 8.0475 = 0.08
Now, we can compute the lower and upper limits for outliers:
Lower = Q₁ - 1.5 x IQR = 8.0475 - 1.5×0.08 =7.928
Upper = Q₃ + 1.5 x IQR = 8.1275 + 1.5× 0.08 =8.248
and then, an outcome X is an outlier if X < 7.928, or if X > 8.248.
In this case since all the outcomes X are within the values of Lower = 7.9275 and Upper = 8.2475, then there are no outliers. The above boxplot is obtained.
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Classify the following functions on R as even, odd or neither: (Problem 1.4A) (a) fi(t)=I (b) f2(t) = t3-2t
For each of the functions in problem (1), find the even and odd parts of the function. (Prob- lem 1.4C)
a) f1(t) = 1 is a even function, the even part is simply 1/2, and the odd part is 0.
b) f2(t) = t^3 - 2t is odd function, the even part is t^3 - t, and the odd part is -t.
Part (a) asks us to classify the function f1(t) = I as even, odd, or neither. Here, I represents a constant, so f1(t) is a constant function.
A function f(t) is even if f(-t) = f(t) for all t in the domain of f. In other words, the function has symmetry about the y-axis.
A function f(t) is odd if f(-t) = -f(t) for all t in the domain of f. In other words, the function has rotational symmetry about the origin.
For a constant function, we have f(-t) = I = f(t) for all values of t. Therefore, f1(t) is even.
Part (b) asks us to classify the function f2(t) = t^3 - 2t as even, odd, or neither.
To determine whether f2(t) is even, odd, or neither, we need to apply the definitions of even and odd functions.
If f2(t) is even, then we have f2(-t) = f2(t) for all t.
If f2(t) is odd, then we have f2(-t) = -f2(t) for all t.
We can check these conditions as follows:
f2(-t) = (-t)^3 - 2(-t) = -t^3 + 2t
Since f2(-t) is not equal to f2(t) for all t, we know that f2(t) is not even.
We can also check if f2(t) is odd:
f2(-t) = (-t)^3 - 2(-t) = -t^3 + 2t = -f2(t)
Since f2(-t) = -f2(t) for all t, we know that f2(t) is odd.
To find the even and odd parts of a function, we use the following formulas:
even part of f(t) = (f(t) + f(-t))/2
odd part of f(t) = (f(t) - f(-t))/2
For f1(t) = I, the even part is (I + I)/2 = I/2, and the odd part is (I - I)/2 = 0.
For f2(t) = t^3 - 2t, the even part is (t^3 - 2t + (-t)^3 - 2(-t))/2 = t^3 - t, and the odd part is (t^3 - 2t - (-t)^3 + 2(-t))/2 = -t.
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The Florida lottery has a game called Pick 4 where a player pays $1 and then picks a sequence of four
numbers 0-9 (e.g., 1234, 0389, 9020). If the four numbers come up in the order you picked, then you win
$3700.
a) The probability distribution for a player's winnings is shown in table.
b) The expected winnings for a player are $3.70.
What is probability distribution?
A probability distribution is a mathematical function that describes the likelihood of various variable values. Probability distributions are frequently represented using graphs or probability tables.
a) Since there are 10 possible digits for each of the four positions, there are a total of 10^4 = 10,000 possible sequences of numbers that could be drawn.
Only one of these sequences will be the winning sequence, corresponding to a probability of 1/10,000. If a player wins, the prize is $3,700. Therefore, the probability distribution for a player's winnings is shown in below table.
b) To find the expected winnings, we multiply each possible winning amount by its probability and add the products together:
E(Winnings) = (0) * (9999/10000) + (3700) * (1/10000)
= 37/10
= $3.70 (rounded to two decimal places)
Therefore, the expected winnings for a player are $3.70.
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Please help !! I need help I can’t figure it out
Answer:
i cant see
Step-by-step explanation:
it, its to tiny
For 3y-2x=-18 determine the value of y when x = 0, and the value of x when y = 0
26°
onotzib loinosho
er
11
Answer:chicken nuggets
Step-by-step explanation:
chickemn nuggets
Limosine Company P charges a rental fee plus an
additional charge per hour. The equation y = 50 +
30x shows the total cost, y of renting the limosine,
and the x represents the hours. The cost to rent a
limosine from Company R is shown below in the
table. Which statement is true?
Time (hours)
Total Cost
O
A
B
C
Company R
1
2
3
4
$100 $125 $150 $175
5
$200
Company P has a lower rate of
change than Company R.
Company P charges $25 more
for its rental fee compared to
Company R.
Company R charges less money
for 1 hour of service compared
to Company P.
The correct statement is -
company {R} charges less money for 1 hour of service compared to company {P}.
What is division?Division is used to split a number into equal parts. It is one of the four arithmetic operations.
Given is that Limousine Company {P} charges a rental fee plus an additional charge per hour. The equation {y} = 50 + 30x shows the total cost. The cost to rent a limousine from Company {R} is shown below in the table as -
Time : 1 2 3 4 5
Cost : $100 $125 $150 $175 $200
We can write the unit charge for company {P} as -
{r₁} = $30
We can write the unit charge for company {R} can be calculated as -
{r₂} = (125 - 100) ÷ (2 - 1)
{r₂} = $25
We can conclude -
company {R} charges less money for 1 hour of service compared to company {P}.
Therefore, the correct statement is -
company {R} charges less money for 1 hour of service compared to company {P}.
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1) Consider the second-order linear DE (1- tcot t) y - ty y 0 for 0 < t < T- Let Y1 (t)-t and y2 (t) sin t. a) Are y1 and y2 both solutions to this DE? b) Are yi and y2 linearly independent? If so, then find the general solution of the DE. If not, then find constants A and B, not both zero, such that Ay? By 0. 2) Suppose that is a constant and {31 , y2} is a set of solutions to the DE Find the Wronskian W(x) of {31 , y, given that W(0) = 1.
The solution of second order linear equation DE (1- tcot t) y - ty y 0 for 0 < t < T is y2(t) = sin(t). There cannot be exist a constant k such that {x^2, e^x} is a set of solutions to the given differential equation.
To check if y1(t) = t is a solution to the given differential equation, we substitute it into the DE and simplify:
(1 - t cot(t))y - t(dy/dt) = (1 - t cot(t))t - t(1) = t - t^2 cot(t) - t = t(1 - t cot(t))
So y1(t) = t is a solution to the DE.
To check if y2(t) = sin(t) is a solution, we substitute it into the DE and simplify:
(1 - t cot(t))y - t(dy/dt) = (1 - t cot(t))sin(t) - t(cos(t)) = sin(t) - t sin(t) (csc(t)) - t(cos(t)) = sin(t) - t tan(t) sin(t) - t(cos(t))
So y2(t) = sin(t) is also a solution to the DE.
To check if y1(t) and y2(t) are linearly independent, we can check if their Wronskian is nonzero. The Wronskian of two solutions y1(t) and y2(t) of a second-order linear differential equation is given by:
W(y1, y2) = y1(t) (dy2/dt) - y2(t) (dy1/dt)
Substituting y1(t) = t and y2(t) = sin(t), we get:
W(y1, y2) = t cos(t) - sin(t)
Since W(y1, y2) is not identically zero (it is zero only at certain values of t), we can conclude that y1(t) and y2(t) are linearly independent.
The general solution to the given differential equation is given by:
y(t) = c1 t + c2 sin(t)
where c1 and c2 are constants determined by the initial or boundary conditions.
Let W(x) be the Wronskian of the solutions y1(x) and y2(x) of the differential equation given by:
(1 - x^2) y'' - 2x y' + k y = 0
Then, we have:
W(x) = y1(x) y2'(x) - y1'(x) y2(x)
Substituting y1(x) = x^2 and y2(x) = e^x, we get:
W(x) = x^2 e^x - 2x e^x - x^2 e^x = -2x e^x
Since we are given that W(0) = 1, we have:
W(0) = -2(0) e^0 = 1
This gives us a contradiction, so there cannot exist a constant k such that {x^2, e^x} is a set of solutions to the given differential equation.
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Kareem is a software salesman. Let y represent his total pay (in dollars). Let x represent the number of copies of History is Fun he sells. Suppose that x and
y are related by the equation 90x+2500=y.
Answer the questions below.
Note that a change can be an increase or a decrease.
For an increase, use a positive number. For a decrease, use a negative number.
What is Kareem's total pay if he doesn't sell any copies of History is Fun?
What is the change in Kareem's total pay for each copy of History is Fun he sells?
?
70 8
The change in Kareem's total pay for each copy of Math is fun he sells is $90. Kareem's total pay if he doesn't sell any copies of Math is fun is $2500.
What is total pay?Total Pay means regular straight-time earnings or base salary, plus payments for overtime, shift differentials, incentive compensation, bonuses, and other special payments, fees, allowances or extraordinary compensation.
Given that, the equation 90x+2500 = y, represents the relation between number of copies (x) Kareem sells and total pay (y) he receives,
We are asked to find, Kareem's total pay if he doesn't sell any copies of History is Fun and the change in Kareem's total pay for each copy of History is Fun he sells,
So, from the equation, 90x+2500 = y,
If we compare this equation with the general equation of the line, i.e. y = mx+c, where m is the slope, which shows 'change', we will get m = 90.
Therefore, the change in Kareem's total pay is $90.
Now, the number of copies he is selling is shown by x, if he sells no copy, then x = 0,
Therefore, Kareem's total pay, for x = 0,
y = 2500+90(0)
y = 2500
Hence, the change in Kareem's total pay for each copy of Math is fun he sells is $90. Kareem's total pay if he doesn't sell any copies of Math is fun is $2500.
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Explain how to simplify an expression with like terms, such as 6g + 5g.
Answer: 11g
Step-by-step explanation:
This is so because 6g and 5g are both like terms
They both have the variable g which makes them like terms
U can just add like terms straight away
To simplify an expression with like terms, you add their coefficients. In this case, we have [tex]6g+5g[/tex].
Combine Like Terms:
[tex]6g+5g[/tex]
[tex]=\fbox{11g}[/tex]
If the mean of 4,6,9,y,16 and 19 is 13. What is the value of y.
The missing term y in the given data having mean is 33.
What is mean?The Arithmetic Mean is the average of the numbers: a calculated "central" value of a set of numbers.
Given that, the mean = 13,
we need to find the value of y.
4+6+9+y+16+19 / 6 = 13
y = 33
Hence, the missing term y in the given data having mean is 33.
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1.28 Ф Vector operations and units. (Chapter 2) Circle the vector operations below (scalar multiplication, addition, dot-product, etc.) that are defined for a position vector a (with units of m) and a velocity vector b (with units of P. a+b 1.29 Using vector identities to simplify expressions (refer to Homework 1.13). One reason to treat vectors as basis-independent quantities is to simplify vector expressions with- out resolving the vectors into orthogonal x, y, z or i k components. Simplify the following Vector expression (3u-29) x (u + v) vector expressions using various properties of dot-products and cross-products Simplified vector expression Express your results in terms of dot-products - and cross- products x of the arbitrary vectors υ. v. w (ie.. υ. v, w are not orthogonal). 2 1.30 Vector concepts: Solving a vector equation. (Section 2.10.5) Consider the vector equation to the right and the process that follows that solves for θ (a, is a unit vector and v,, θ, R are scalars). This process is a valid way to solve for θ. True/False Explain: vr ax 0
The vector operations defined for a position vector a and a velocity vector b are vector addition and scalar multiplication.
Simplified Vector expression is (3u - 29) x u + (3u - 29) x v.
False, as the process that follows, solving for θ by dividing by the magnitude of v and taking the inverse sine, is only valid if v x a = 0.
1.28: The vector operations defined for a position vector a and a velocity vector b are vector addition and scalar multiplication.
1.29: Using the properties of the cross-product, the simplified expression is:
3u x u + 3u x v - 29 x u - 29 x v
Using the fact that the cross-product is distributive over vector addition, we can write this as:
3u x u + 3u x v - 29 x u - 29 x v = (3u - 29) x u + (3u - 29) x v
So the simplified vector expression is (3u - 29) x u + (3u - 29) x v.
1.30: False.
The given equation,
v x (a x R) = 0,
can be simplified to a scalar triple product,
(v x a) · R = 0.
Since the scalar triple product is equal to the determinant of a matrix formed by the three vectors,
this equation implies that either v x a = 0 or R = 0.
Therefore, the process that follows, solving for θ by dividing by the magnitude of v and taking the inverse sine, is only valid if v x a = 0.
The vector operations defined for a position vector a and a velocity vector b are vector addition and scalar multiplication.
Simplified Vector expression is (3u - 29) x u + (3u - 29) x v.
False as the process that follows, solving for θ by dividing by the magnitude of v and taking the inverse sine, is only valid if v x a = 0.
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What is the algebraic relationship between the resonant wavelength, λ, and length of an open-closed tube, L, for the n=1,3,5, and 7 harmonics. The options below refer to the following expressions 1. λ= 2L/52. λ= 2L/73. λ=5L4. λ=3L5. λ = 4L/36. λ=4L7.λ = L8. λ= 4L/79. λ=7L10. λ= 4L/5
The relationship between the resonant wavelength λ and the length of an open-closed tube L for the n=1,3,5, and 7 harmonics follows the general equation λ=(2n+1)L/2n. This means that the nth harmonic has a wavelength that is (2n+1) times the length of the tube divided by 2n.
For example, the wavelength of the 1st harmonic is λ= (21+1) L/21=3L/2, the wavelength of the 3rd harmonic is λ= (23+1) L/23=7L/4, the wavelength of the 5th harmonic is λ= (25+1) L/25=11L/4, and the wavelength of the 7th harmonic is λ= (27+1) L/27=15L/4. Therefore, the algebraic relationship between the resonant wavelength and length of an open-closed tube follows the general equation λ= (2n+1) L/2n, where n is the harmonic number. This equation can be used to calculate the resonant wavelength for any harmonic number.
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Recall the equation for simple interest, A = P(1 + rt). Compare the amount of interest earned in 10 years for an investment of $3700 with a 7% annual simple interest rate and a 7% annual interest rate compounded monthly.
simple interest: $______
compound interest: $______
The compound interest investment earned $______more than the simple interest investment.
Hence, or otherwise, find the coordinates of the minimum point of the curve y =
x² + 4x – 7.
Answer: To find the minimum point of the curve y = x^2 + 4x - 7, we need to find the vertex of the parabolic function. The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
where a = 1 (coefficient of x^2), b = 4 (coefficient of x), and c = -7 (constant term). So, we have:
x = -4 / (2 * 1) = -2
The y-coordinate of the vertex can be found by substituting the x-coordinate into the original equation:
y = -2^2 + 4 * -2 - 7 = 4 - 8 - 7 = -11
So, the minimum point of the curve y = x^2 + 4x - 7 is located at the point (-2, -11).
Step-by-step explanation: