Joan’s finishing time for the Bolder Boulder 10k race was 1.84 standard deviations faster than women’s average for her age group There were 395 women who ran in her age group assuming a normal distribution how many women ran faster than Joan(round down your answer to the nearest whole number)
Answers
Rounding down to the nearest whole number, we can conclude that about 12 women in her age group ran faster than Joan.
What is a whole number?A whole number is a number that is not a fraction, decimal or negative. Whole numbers include all positive integers (1, 2, 3, ...), as well as zero (0). They are also sometimes referred to as counting numbers, as they are commonly used for counting objects or items.
What is standard normal distribution?A standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is a specific type of normal distribution that has been standardized to have these specific parameters, which makes it easier to calculate probabilities and perform statistical analyses.
In the given question,
We can use the standard normal distribution to solve this problem. Since Joan's finishing time was 1.84 standard deviations faster than the mean for her age group, we know that her finishing time is in the top (100% - 1.84% = 98.16%) of the distribution.
To find the number of women who ran faster than Joan, we need to find the area under the standard normal distribution curve to the right of her z-score. Using a standard normal distribution table or calculator, we can find that the area to the right of a z-score of 1.84 is 0.0322.
Therefore, approximately 395 x 0.0322 = 12.729 women in Joan's age group ran faster than her. Rounding down to the nearest whole number, we can conclude that about 12 women in her age group ran faster than Joan.
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Related Questions
What is the difference between the questionnaire and an interview?
Answers
Answer: Questionnaire refers to a research instrument, in which a series of question, is typed or printed along with the choice of answers, expected to be marked by the respondents, used for survey or statistical study. It consists of aformalisedd set of questions, in a definite order on a form, which are mailed to the respondents or manually delivered to them for answers. The respondents are supposed to read, comprehend and give their responses, in the space provided.
A ‘Pilot Study’ is advised to be conducted to test the questionnaire before using this method. A pilot survey is nothing but a preliminary study or say rehearsal to know the time, cost, efforts, reliability and so forth involved in it.
The interview is a data collection method wherein a direct, in-depth conversation between interviewer and respondent takes place. It is carried out with a purpose like a survey, research, and the like, where both the two parties participate in the one to one interaction. Under this method, oral-verbal stimuli are presented and replied by way of oral-verbal responses.
It is considered as one of the best methods for collecting data because it allows two way exchange of information, the interviewer gets to know about the respondent, and the respondent learns about the interviewer. There are two types of interview:
Personal Interview: A type of interview, wherein there is a face to face question-answer session between the interviewer and interviewee, is conducted.
Telephonic Interview: This method involves contacting the interviewee and asking questions to them on the telephone itself.
A pipe with the diameter of 2.4 cm discharges water a rate 2.8 m per second. find the volume of water discharges one and a half hour, giving the answer in litres?
Answers
Answer: approximately 68,404 liters.
Step-by-step explanation:
To solve the problem, we first need to find the cross-sectional area of the pipe, which we can calculate using the formula for the area of a circle:
A = πr^2
where r is the radius of the pipe, which is half the diameter. So, in this case, the radius is 2.4 cm / 2 = 1.2 cm.
A = π(1.2 cm)^2
A ≈ 4.5239 cm^2
Next, we can use the formula for volume flow rate to find the volume of water that is discharged per second:
Q = Av
where Q is the volume flow rate, A is the cross-sectional area of the pipe, and v is the velocity of the water. In this case, we have:
Q = (4.5239 cm^2)(2.8 m/s)
Q ≈ 0.01266 m^3/s
To find the volume of water discharged in one and a half hours (which is 5400 seconds), we can multiply the volume flow rate by the time:
V = Qt
V = (0.01266 m^3/s)(5400 s)
V ≈ 68.404 m^3
Finally, to convert the volume from cubic meters to liters, we can multiply by 1000:
V = 68.404 m^3 × 1000 L/m^3
V ≈ 68,404 L
Therefore, the volume of water discharged in one and a half hours is approximately 68,404 liters.
Simplify 3(x+2) + 2x + 5
Answers
Answer:
[tex]5x+11[/tex]
Step-by-step explanation:
Step 1: Distribute
[tex]3x+6+2x+5[/tex]
Step 2: Add like terms
[tex]5x+11[/tex] < your answer
t/f the mean and standard deviation are more accurate measures of center and spread when the data is skewed
Answers
When the data is skewed, it is untrue that the mean and standard deviation are better indicators of the centre and spread.
The median is a better tool to use to locate the centre when it is skewed right or left with high or low outliers. The IQR is the most accurate indicator of spread when the median is the centre. When the mean is the centre, the standard deviation should be utilised because it gauges how far a data point is from the mean.
The standard deviation will be greatly overstated in cases when the distribution of the data is highly skewed, making it a poor choice as a measure of variability.
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This is accounting. Please answer
Answers
All answers are mentioned below.
Describe balance sheet?A balance sheet is a financial statement that provides a snapshot of a company's financial position at a specific point in time. It presents the company's assets, liabilities, and equity, and shows how these are financed. The balance sheet follows the accounting equation, which states that the total assets must equal the sum of liabilities and equity.
(a) Trading, Profit & Loss and Profit & Loss Appropriation Account for the year ended December 31, 1996
Trading Account
$
Sales (51,550 - 5,550 + 5,385) 51,385
Less: Cost of Goods Sold
Opening Stock 5,550
Add: Purchases 29,525
35,075
Less: Closing Stock 5,385
Cost of Goods Sold 29,690
Gross Profit 21,695
Profit & Loss Account
$
Gross Profit 21,695
Less: Expenses
Carriage Inward 100
Discount Allowed 1,000
Wages & Salaries 17,650
Motor Vehicle Repairs 5,850
Telephones 350
Rent & Rates 1,800
Advertising 2,750
Carriage Outward 1,000
Bank Charges 550
Depreciation
Land & Building (30,000/50 years) 600
Motor Vehicle (20% x 20,000) 4,000
34,550
Net Loss (12,855)
Profit & Loss Appropriation Account
$
Net Loss (12,855)
Add: Interest on Capital (5% x $90,000) 4,500
(8,355)
To be shared equally between Knox and Cox
Knox (4,178)
Cox (4,178)
Additional Salary for Cox (5,000)
Carried Forward (13,356)
(b) Partners' Current Account
Knox Cox
$ $
To Opening Balance 8,185 4,400
By Share of Net Loss (4,178) (4,178)
By Interest on Capital 2,500 2,000
By Drawings (1,200) (500)
By Salary 0 (5,000)
To Closing Balance 5,307 (3,278)
(c) Balance Sheet as at December 31, 1996
Fixed Assets
Land & Building (100,000 - 30,000) 70,000
Motor Vehicles (20,000 - 4,000) 16,000
86,000
Current Assets
Stock 5,385
Debtors 7,250
Bank 3,200
Cash 510
16,345
Current Liabilities
Creditors (7,650)
Net Current Assets 8,695
94,695
Less: Partners' Capital and Current Accounts
Knox Capital (50,000 + 5,307) 55,307
Cox Capital (40,000 - 3,278) 36,722
92,029
2,666
Notes:
Depreciation of Land and Building: 30,000/50 years = 600 per year
Depreciation of Motor Vehicles: 20% x 20,000 = 4,000
Interest on Capital: Knox - 5% x 50,000 = 2,500, Cox - 5% x 40,000 = 2,000
Interest on Drawings: Knox - 10% x 1,200 = 120, Cox - 10% x 500 = 50
Additional Salary for Cox is treated as an expense and charged to the Profit & Loss Account.
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(a) Trading, Profit & Loss and Profit for the year ended December 31, 1996 is mentioned below. (b) Knox Cox, Opening Balance 8,185 4,400, respectively.(c) Interest on Drawings: Knox - 10% x 1,200 = 120, Cox - 10% x 500 = 50
Describe balance sheet?A balance sheet is a financial statement that provides a snapshot of a company's financial position at a specific point in time. It presents the company's assets, liabilities, and equity, and shows how these are financed. The balance sheet follows the accounting equation, which states that the total assets must equal the sum of liabilities and equity.
(a) Trading, Profit & Loss and Profit & Loss Appropriation Account for the year ended December 31, 1996
Trading Account
$
Sales (51,550 - 5,550 + 5,385) 51,385
Less: Cost of Goods Sold
Opening Stock 5,550
Add: Purchases 29,525
35,075
Less: Closing Stock 5,385
Cost of Goods Sold 29,690
Gross Profit 21,695
Profit & Loss Account
$
Gross Profit 21,695
Less: Expenses
Carriage Inward 100
Discount Allowed 1,000
Wages & Salaries 17,650
Motor Vehicle Repairs 5,850
Telephones 350
Rent & Rates 1,800
Advertising 2,750
Carriage Outward 1,000
Bank Charges 550
Depreciation
Land & Building (30,000/50 years) 600
Motor Vehicle (20% x 20,000) 4,000
34,550
Net Loss (12,855)
Profit & Loss Appropriation Account
Net Loss (12,855)
Add: Interest on Capital (5% x $90,000) 4,500
(8,355)
To be shared equally between Knox and Cox
Knox (4,178)
Cox (4,178)
Additional Salary for Cox (5,000)
Carried Forward (13,356)
(b) Partners' Current Account
Knox Cox
$ $
To Opening Balance 8,185 4,400
By Share of Net Loss (4,178) (4,178)
By Interest on Capital 2,500 2,000
By Drawings (1,200) (500)
By Salary 0 (5,000)
To Closing Balance 5,307 (3,278)
(c) Balance Sheet as at December 31, 1996
Fixed Assets
Land & Building (100,000 - 30,000) 70,000
Motor Vehicles (20,000 - 4,000) 16,000
86,000
Current Assets
Stock 5,385
Debtors 7,250
Bank 3,200
Cash 510
16,345
Current Liabilities
Creditors (7,650)
Net Current Assets 8,695
94,695
Less: Partners' Capital and Current Accounts
Knox Capital (50,000 + 5,307) 55,307
Cox Capital (40,000 - 3,278) 36,722
92,029
2,666
Notes:
Depreciation of Land and Building: 30,000/50 years = 600 per year
Depreciation of Motor Vehicles: 20% x 20,000 = 4,000
Interest on Capital: Knox - 5% x 50,000 = 2,500, Cox - 5% x 40,000 = 2,000
Interest on Drawings: Knox - 10% x 1,200 = 120, Cox - 10% x 500 = 50
Additional Salary for Cox is treated as an expense and charged to the Profit & Loss Account.
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(a) If a is a zero of the polynomial P(x), then must be a factor of P(x). (b) If a is a zero of multiplicity m of the polynomial P(x), then must be a factor of P(x) when we factor P completely.
(a) If a is a zero of the polynomial P(x), then must be a factor of P(x).
(b) If a is a zero of multiplicity m of the polynomial P(x), then must be a factor of P(x) when we factor P completely.
Answers
If a is a zero of the polynomial P(x), then (x-a) must be a factor of P(x) and [tex](x-a)^m[/tex] be a factor of P(x) when we factor P completely.
The values of x that fulfil the formula f(x) = 0 are the zeros of a polynomial. The polynomial's zeros are the x values for which the function's value, f(x), equals zero in this case. The degree of the equation f(x) = 0 determines how many zeros a polynomial has.
The locations when a polynomial equals 0 overall are known as its zeros. In layman's terms, we may state that a polynomial's zeros are variable values at which the polynomial equals 0. The zeros of a polynomial are often referred to as the equation's roots and are frequently written as,, and. A few techniques for locating polynomial zeros include grouping, factoring, and employing algebraic expressions.
(a) if we have zero at x=a of polynomial P(x)
then, (x-a) must be factor of P(x).
(b) if we have zero at x=a of polynomial P(x)
with multiplicity=m
then, [tex](x-a)^m[/tex] must be factor of P(x).
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for each polynomial in factored form show the leading term, the zeros on the x-axis, and the general shape of the polynomial
Answers
For the given polynomials in factored form;
(a) the leading term is x³, the zeroes are -2,3 and 5 , the graph will be a cubic function passing through x-axis at -2, 3 and 5.
(b) the leading term is 2x², the zeros are -1, 4 the graph will be a quadratic function passing through x-axis at -1 and 4.
Part(a) : The polynomial in factored form is : f(x) = (x + 2)(x - 3)(x - 5)
The Leading Term is : x³; The Zeros are : -2, 3, 5.
The General Shape: The graph of the polynomial will be a cubic function that passes through the x-axis at -2, 3, and 5.
The function will approach negative infinity as x approaches negative infinity and positive infinity as x approaches positive infinity.
Part(b) : The Polynomial in factored form is : f(x) = 2(x + 1)(x - 4)
The Leading Term is : 2x²; The Zeros are : -1, 4.
The General Shape: The graph of the polynomial will be a quadratic function that passes through the x-axis at -1 and 4. The function will open upwards since the leading coefficient is positive.
The function will approach negative infinity as x approaches negative infinity and positive infinity as x approaches positive infinity.
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The given question is incomplete, the complete question is
For each polynomial in factored form show the leading term, the zeros on the x-axis, and the general shape of the polynomial.
(a) f(x) = (x + 2)(x - 3)(x - 5)
(b) f(x) = 2(x + 1)(x - 4).
Fill in the table using this function rule.
f(x)=√x+5
Simplify your answers as much as possible.
Click "Not a real number" if applicable.
x
-4
0
36
64
f(x)
11
0
0
0
Not a
real
number
Start over
5
Answers
Answer:
4
Step-by-step explanation:
What’s the answer???
Answers
The difference in price between the two shops for 300 cm of ribbon is £3.60 - £2.40 = £1.20. Therefore, the answer is £1.20.
How to solve and what is Selling?
We can use proportions to find the cost of 300 cm of ribbon at each shop, and then subtract the cost at Shop B from the cost at Shop A to find the difference in price:
For Shop A:
140 cm of ribbon cost £1.68, so 1 cm of ribbon cost £1.68/140 = £0.012.
Therefore, 300 cm of ribbon would cost £0.012 x 300 = £3.60.
For Shop B:
215 cm of ribbon cost £1.72, so 1 cm of ribbon cost £1.72/215 = £0.008.
Therefore, 300 cm of ribbon would cost £0.008 x 300 = £2.40.
The difference in price between the two shops for 300 cm of ribbon is £3.60 - £2.40 = £1.20. Therefore, the answer is £1.20.
Selling is the process of exchanging goods or services for money or other valuable consideration. In business, selling is an essential part of the marketing and sales process, and involves identifying potential customers or clients, communicating with them about the features and benefits of the product or service being sold, and negotiating a price or other terms of the sale.
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You have a map that is missing a scale. The distance from Point A to Point B is
five inches on the map, and after driving it, you know it is 250 miles in reality.
Answers
The scale of the map in the question is 1 inch = 50 miles.
What is the scale of the map?A scale in a map is a relation that tells us how many units each unit in the map represents. In this case, we know that the distance between two points A and B on the map is 5 inches, while the actual distance between these two places is 250 miles.
Then we start with the relation:
5 inches = 250 miles.
But to get the scale of the map we need to see how many miles one inch represents in the map, then we can divide both sides of the equation by 5 to geT:
5 in = 250 mi
1 in = 250mi/5
1 in = 50 mi
The scale of the map is 1 inch to 50 miles.
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factorise completely[tex]3x²-12xy
Answers
Answer:
3x(x - 4y)
Step-by-step explanation:
3x² - 12xy ← factor out 3x from each term
= 3x(x- 4y)
Write (28)to the power of 3 as a power of 2
Answers
The expression (2^8) to the power of 3 as a power of 2 is 4096^2
Rewritting the expression as a power of 2Given the following expression
(2^8) to the power of 3
To write (2^8) to the power of 3 as a power of 2, we can use the rule of exponents that says:
(a^b)^c = (a^c)^b
Applying this rule to (2^8)^3, we get:
(2^8)^3 = (2^3)^8
Simplifying the expression, we get:
(2^8)^3 = (8)^8
So, we have
(2^8)^3 = (8^4)^2
Simplify
(2^8)^3 = 4096^2
Therefore, (2^8)^3 can be written as 4096^2.
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Given that (1, 2, 3] System{1, 4, 7,6] for a system known to be LTI, compute the system's impulse response h[n] without using z-transforms.
Answers
Given that (1, 2, 3] System{1, 4, 7,6] for a system known to be LTI, the impulse response of the system: h[n] = (1/2)*δ[n] + δ[n-1] + (3/2)*δ[n-2]
To compute the impulse response h[n] of a linear time-invariant (LTI) system given its input-output relationship, we can use the convolution sum:
y[n] = x[n] * h[n]
y[n] = (1/2)*(x[n] + 2x[n-1] + 3x[n-2])
y[n] = (1/2)*(δ[n] + 2δ[n-1] + 3δ[n-2])
y[n] = (1/2)*δ[n] + δ[n-1] + (3/2)*δ[n-2]
Thus, the impulse response of the system is:
h[n] = (1/2)*δ[n] + δ[n-1] + (3/2)*δ[n-2],where δ[n] is the impulse signal.
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For which pair of functions is the exponential consistently growing at a faster rate than the quadratic over the interval 0 ≤ x ≤ 5?
Answers
One pair of functions that satisfies the given condition is:
Exponential function: [tex]f(x) = 1.46^x,[/tex] Quadratic function: [tex]g(x) = x^2[/tex]
What is expression ?In mathematics, an expression is a combination of numbers, variables, and mathematical operations such as addition, subtraction, multiplication, and division. Expressions can also include functions, brackets, and other symbols.
According to the given information:Let's consider the two functions:
Exponential function: [tex]f(x) = a^x, where a > 1[/tex]
Quadratic function: [tex]g(x) = x^2[/tex]
We want to find the pair of functions for which the exponential is consistently growing at a faster rate than the quadratic over the interval 0 ≤ x ≤ 5.
To determine this, we can compare the growth rates of the two functions by looking at their derivatives.
The derivative of the exponential function is:[tex]f'(x) = a^x * ln(a)[/tex]
The derivative of the quadratic function is: [tex]g'(x) = 2x[/tex]
To compare the growth rates of the two functions, we need to compare their derivatives. We want to find the value of x for which the exponential function is growing faster than the quadratic function, i.e., where f'(x) > [tex]g'(x).\\f'(x) > g'(x)\\a^x * ln(a) > 2x[/tex]
Now, we can solve for x:
[tex]a^x * ln(a) > 2xln(a)/2 * a^x > x[/tex]
Since we want to find the pair of functions for which the exponential is consistently growing at a faster rate than the quadratic over the interval 0 ≤ x ≤ 5, we need to find a value of a such that the inequality ln(a)/2 * [tex]a^5 > 5[/tex] is true for all values of a > 1.
We can use a graphing calculator or a numerical solver to find the value of a that satisfies this inequality. One possible solution is a ≈ 1.46.
Therefore, one pair of functions that satisfies the given condition is:
Exponential function: [tex]f(x) = 1.46^x,[/tex] Quadratic function: [tex]g(x) = x^2[/tex]
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The figure below displays the SAT scores of three students, but each chart looks different. The two charts have the same data, but the difference seems larger for the graph on the left. Why?
Answers
Answering the presented question, we may conclude that This greater expressions scale makes it easier to see the variations between the ratings of the three college students in a extra correct and informative way.
what is expression ?In mathematics, an expression is a collection of integers, variables, and complex mathematical (such as arithmetic, subtraction, multiplication, division, multiplications, and so on) that describes a quantity or value. Phrases can be simple, such as "3 + 4," or complicated, such as They may also contain functions like "sin(x)" or "log(y)". Expressions can be evaluated by swapping the variables with their values and performing the arithmetic operations in the order specified. If x = 2, for example, the formula "3x + 5" equals 3(2) + 5 = 11. Expressions are commonly used in mathematics to describe real-world situations, construct equations, and simplify complicated mathematical topics.
The difference in look between the two charts is due to the choice of the scales on the x and y-axes. In the left chart, the y-axis starts offevolved at 800 and has a range of only 200 points, whilst the x-axis starts offevolved at 1300 and has a vary of 200 points. This compressed scale makes the variations between the ratings of the three students appear larger than they absolutely are.
On the other hand, the proper chart has a y-axis that starts at zero and has a vary of 800 points, whilst the x-axis begins at 1200 and has a range of 800 points. This greater expanded scale makes it easier to see the variations between the ratings of the three college students in a extra correct and informative way.
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In the inequality, x (< with line underneath) 8, represents the number of books on a shelf.
How do you know if 8 is a possible value of x?
Answers
Answer:
The numbers 8 and lower are possible values of x.
Step-by-step explanation:
The inequality [tex]x\leq 8[/tex] means x is less than or equal to 8. Therefore, 8 is a possible value of x.
Mear's Taxi charges a $ 4.25 flat rate for a ride in the cab. In addition to that, they charge $ 0.47 per mile. Kevin has no more than $ 25 to spend on a ride. At most, how many miles can Kevin travel without exceeding his spending limit?
Answers
Answer: 44 miles
Step-by-step explanation:
Let's call the number of miles Kevin can travel "m". We can set up an equation using the given information:
4.25 + 0.47m ≤ 25
Solving for "m", we can begin by subtracting 4.25 from both sides:
0.47m ≤ 20.75
Then, divide both sides by 0.47:
m ≤ 44.15
Therefore, Kevin can travel at most 44 miles without exceeding his spending limit.
b is the set of odd positive integers less than 11.
(a)list all the elements of b in set notation.
b)state whether each of the followning statements is true or false.
(i)1∈b
Answers
Answer:
Step-by-step explanation:
True. 1 is an odd positive integer less than 11 and is an element of set B.
2 Sasha believes her soccer team plays better at away games than at home games. Her team played 12 games at home and 12 games away. She recorded the wins over this season. Based on these results, what is the probability for home and away wins? Is she right about her team playing better away? Show your work. Place: Home Away
Frequency: 8 7
Answers
Probability of winning at home is 0.67, Probability of winning away is 0.58.
How to calculate the probabilitiesTo calculate the probability of home and away wins, we need to divide the number of wins by the total number of games played at home and away respectively.
Probability of winning at home = (Number of wins at home) / (Total number of games played at home)
= 8 / 12
= 0.67
Probability of winning away = (Number of wins away) / (Total number of games played away)
= 7 / 12
= 0.58
Based on these results, Sasha's team has a higher probability of winning at home (0.67) compared to away games (0.58).
Therefore, her belief that her team plays better away is not supported by the data.
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Asaad invests $6800 in two different accounts. The first account paid 14 %, the second account paid 11 % in interest. At the end of the first year he had earned $856 in interest. How much was in each account?
Answers
Answer:
Step-by-step explanation:
Let x be the amount invested in the first account, which pays 14% interest. Then the amount invested in the second account, which pays 11% interest, is 6800 - x.
The interest earned on the first account is 0.14x, and the interest earned on the second account is 0.11(6800 - x). The total interest earned is the sum of these two amounts, so we have:
0.14x + 0.11(6800 - x) = 856
Simplifying and solving for x, we get:
0.14x + 748 - 0.11x = 856
0.03x = 108
x = 3600
Therefore, Asaad invested $3600 in the first account and $3200 (6800 - 3600) in the second account.
Mr. Chand is one of the landlords of his town. He buys a land for his daughter spanning over a
area of 480m². He fences the dimensions of the land measuring (x+12) mx (x+16) m. Now he
plans to erect a house with a beautiful garden in the ratio 5:3 respectively. A total of Rs. 5,00,000 is estimated as the budget for the expenses.
1)Give the area of the land purchased in linear polynomial form using algebraic expression
2)Mr. Chand's daughter is ready to share 3/5" of the expenses by her earnings. Express the
fraction in amount.
3)Can you solve the linear equation/polynomial of the area into different factors?
Answers
The required answers are 1) [tex]$$A = x^2 + 28x + 192$$[/tex] 2) 300000 3) [tex]$$x^2 + 28x + 192 = (x + 14 - 2\sqrt{19})(x + 14 + 2\sqrt{19})$$[/tex].
How to deal with area and fractions?area of the land purchased is given as 480m², and the dimensions of the land are (x+12)mx(x+16)m. Therefore, the area of the land can be expressed as:
[tex]$$A = (x+12)(x+16)$$[/tex]
Expanding this expression, we get:
[tex]$$A = x^2 + 28x + 192$$[/tex]
Hence, the area of the land purchased is given by the polynomial expression [tex]$x^2 + 28x + 192$[/tex].
The total budget for the expenses is Rs. 5,00,000. If Mr. Chand's daughter is ready to share 3/5 of the expenses, then the fraction of the expenses she will pay is:
[tex]$\frac{3}{5}=\frac{x}{500000}$$[/tex]
Simplifying this expression, we get:
[tex]$x = \frac{3}{5}\times 500000 = 300000$$[/tex]
Therefore, Mr. Chand's daughter will pay Rs. 3,00,000 towards the expenses.
We can solve the polynomial [tex]$x^2 + 28x + 192$[/tex] into different factors by using the quadratic formula:
[tex]$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$[/tex]
Here, the coefficients of the polynomial are:
[tex]$$a = 1, \quad b = 28, \quad c = 192$$[/tex]
Substituting these values in the quadratic formula, we get:
[tex]$x = \frac{-28 \pm \sqrt{28^2 - 4\times 1 \times 192}}{2\times 1}$$[/tex]
Simplifying this expression, we get:
[tex]$$x = -14 \pm 2\sqrt{19}$$[/tex]
Therefore, the polynomial [tex]$x^2 + 28x + 192$[/tex] can be factored as:
[tex]$$x^2 + 28x + 192 = (x - (-14 + 2\sqrt{19}))(x - (-14 - 2\sqrt{19}))$$[/tex]
or
[tex]$$x^2 + 28x + 192 = (x + 14 - 2\sqrt{19})(x + 14 + 2\sqrt{19})$$[/tex]
So, we have factored the polynomial into two factors.
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Please help with hard polynomial problem
Let [tex]$f(x)=(x^2+6x+9)^{50}-4x+3$[/tex], and let [tex]$r_1,r_2,\ldots,r_{100}$[/tex] be the roots of [tex]$f(x)$[/tex].
Compute [tex]$(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}$[/tex].
Answers
The value of [tex](r_1+3)^{100}+(r_2+3)^{100}+...+(r_{100}+3)^{100}[/tex] is -1500.
What is function?
A mathematical phrase, rule, or law that establishes the link between an independent variable and a dependent variable (the dependent variable). In mathematics, functions exist everywhere, and they are crucial for constructing physical links in the sciences.
Here the given function is,
[tex]f(x)=(x^2+6x+9)^{50}-4x+3[/tex]
For any r , [tex](x^2+6x+9)^{50}-4x+3[/tex] is satisfy, Then take f(x)=0 then
=> [tex](x^2+6x+9)^{50}-4x+3=0[/tex]
Take x=r then
=> [tex](r^2+6r+9)^{50}-4r+3=0[/tex]
=> [tex](r^2+6r+9)^{50}=4r-3[/tex]
=> [tex]((r+3)^2)^{50}=4r-3[/tex]
=> [tex](r+3)^{100}=4r-3[/tex]
Then,
=> [tex]\sum_{i=1}^{100} (r_i+3)^{100}=\sum_{i=1}^{100} (4r_i-3)[/tex] = 4 × sum of roots - 300
Expanding [tex](x+3)^{100}-4x+3[/tex] using the binomial theorem, we get
=> f(x) = [tex]x^{100}+300x^{99}+....[/tex]
So sum of roots = -300 then
=> [tex](r_1+3)^{100}+(r_2+3)^{100}+...+(r_{100}+3)^{100} =[/tex] 4*(-300)-300=-1200-300
=> -1500.
Hence , The value of [tex](r_1+3)^{100}+(r_2+3)^{100}+...+(r_{100}+3)^{100}[/tex] is -1500.
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3. Each sample of water from a river has a 10% chance of contamination by a particular heavy metal. Find the probability that in 18 independent samples taken from the same river, only two samples were contaminated. [3 marks]
Answers
The probability that, out of 18 independent samples received from one river, just two were contaminated is 0.8438.
Explain about the independent samples?Randomly chosen samples are known as independent samples since their results are independent of other observations' values. The premise that sampling are independent underlies many statistical analysis.When each trial possesses the same probability of achieving a given value, the number of trials or observations is represented using the binomial distribution.In the following 18 samples to be evaluated,
Let X = the number of samples that now the pollutant is present in.
Thus, with p = 0.10 and n = 18, X is a binomial random variable.
Using the binomial theorem:
[tex](^{n} _{r} ) p^{x} q^{n-x}[/tex]
p = 0.10
q = 1 - 0.10 = 0.9
n = 18
The likelihood that only two samples out of 18 obtained in different ways from the same river were polluted
P(x = 2) = [tex](^{18} _{2} ) (0.1)^{2} (0.9)^{18-2}[/tex]
= [tex](^{18} _{2} ) (0.1)^{2} (0.9)^{16}[/tex]
= 153 x 0.01 x 0.1853
= 0.8438
Thus, the probability that, out of 18 separate samples received from one river, just two were contaminated is 0.8438.
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A cyclist rides off from rest, accelerating at a constant rate for 3 minutes until she reaches 40 kmh-1. She then maintains a constant speed for 4 minutes until reaching a hill. She slows down at a constant rate over one minute to 30 kmh-1. then continues at this rate for 10 minutes.
At the top of the hill she reduces her speed uniformly and is stationary 2 minutes later.
b
How far has the cyclist travelled? Its 9.75 km, but I don't understand how to get there
PLEASE SHOW YOUR WORK
Answers
Answer:
Step-by-step explanation:
To solve this problem, we need to use the equations of motion for constant acceleration, constant velocity, and constant deceleration. We'll break the problem into several parts and use these equations to find the distance traveled in each part. Then, we'll add up the distances to get the total distance traveled.
First, we need to convert the units of speed from km/h to m/s, since the equations of motion use meters per second. We have:
Initial speed (u) = 0 km/h = 0 m/s
Final speed (v) = 40 km/h = 11.11 m/s
Constant speed = 40 km/h = 11.11 m/s (for 4 minutes)
Final speed before hill = 30 km/h = 8.33 m/s
Speed at top of hill = 0 m/s
Acceleration (a) = (v-u)/t = (11.11-0)/(3*60) = 0.0611 m/s^2
PART 1: ACCELERATION PHASE
Time taken (t) = 3 minutes = 180 seconds
Distance traveled (s) = ut + (1/2)at^2
s = 0 + (1/2)0.0611(180^2) = 331.83 meters
PART 2: CONSTANT SPEED PHASE
Time taken (t) = 4 minutes = 240 seconds
Distance traveled (s) = vt
s = 11.11*240 = 2666.4 meters
PART 3: DECELERATION PHASE
Time taken (t) = 1 minute = 60 seconds
Deceleration (a) = (v-u)/t = (8.33-11.11)/60 = -0.0461 m/s^2 (negative since it's deceleration)
Distance traveled (s) = vt + (1/2)at^2
s = 8.3360 + (1/2)(-0.0461)*(60^2) = 494.7 meters
PART 4: CONSTANT SPEED PHASE
Time taken (t) = 10 minutes = 600 seconds
Distance traveled (s) = vt
s = 8.33*600 = 4998 meters
PART 5: DECELERATION PHASE TO STOP
Time taken (t) = 2 minutes = 120 seconds
Initial speed (u) = 8.33 m/s
Final speed (v) = 0 m/s
Deceleration (a) = (v-u)/t = (0-8.33)/120 = -0.0694 m/s^2
Distance traveled (s) = vt + (1/2)at^2
s = 8.33120 + (1/2)(-0.0694)*(120^2) = 733.3 meters
TOTAL DISTANCE TRAVELED:
Adding up the distances from each part, we get:
Total distance = 331.83 + 2666.4 + 494.7 + 4998 + 733.3 = 9184.23 meters = 9.18 km (rounded to two decimal places)
Therefore, the cyclist has traveled approximately 9.18 km.
Consider the line that passes through the point and is parallel to the given vector. (4, -1, 9) ‹-1, 4, -2› symmetric equations for the line. -(x - 4) = y+1/ 4 = − z−9 /2 . (b) Find the points in which the line intersects the coordinate planes.
Answers
The symmetric equations of the line passing through a point and parallel to a vector are -(x - 4) = y + 1/4 = -(z - 9)/2. The line intersects the xy-, xz-, and yz-planes at (5, -9/4, 0), (15/4, 0, 23/2), and (0, -17/4, 11/2), respectively.
To find the symmetric equations of the line, we first need to find the direction vector of the line. Since the line is parallel to the vector <4, -1, 9>, any scalar multiple of this vector will be a direction vector of the line. So, let's choose the parameter t and write the vector equation of the line:
r = <4, -1, 9> + t<-1, 4, -2>
Expanding this vector equation component-wise, we get:
x = 4 - t
y = -1 + 4t
z = 9 - 2t
These equations can be rearranged to get the symmetric equations of the line:
-(x - 4) = y + 1/4 = -(z - 9)/2
To find the points in which the line intersects the coordinate planes, we substitute the corresponding variables with 0 in the equations for the line.
For the xy-plane, we set z = 0 and solve for x and y:
-(x - 4) = y + 1/4 = -(-9)/2
x = 5, y = -9/4
So, the line intersects the xy-plane at the point (5, -9/4, 0).
For the xz-plane, we set y = 0 and solve for x and z:
-(x - 4) = 0 + 1/4 = -(z - 9)/2
x = 15/4, z = 23/2
So, the line intersects the xz-plane at the point (15/4, 0, 23/2).
For the yz-plane, we set x = 0 and solve for y and z:
-(-4) = y + 1/4 = -(z - 9)/2
y = -17/4, z = 11/2
So, the line intersects the yz-plane at the point (0, -17/4, 11/2).
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Prove that for every real number if c is the root of a polynomial with rational coefficients then root ofa polynomial with integer coefficients: It may be helpful to suppose that is a solution to the polynomial equation: GnXn An-1xn-1_ +q1* + 4o Where qi € Q
Answers
Every real root of a polynomial with rational coefficients is also a root of a polynomial with integer coefficients.
Suppose that c is a root of the polynomial equation:
[tex]q_n[/tex] × [tex]x^n[/tex] + q_{n-1} × [tex]x^{n-1}[/tex] + ... + q1 × x + q0 = 0
where [tex]q_i[/tex] are rational coefficients. Since c is a root of this polynomial equation, we have:
[tex]q_n[/tex] × [tex]c^n[/tex] + [tex]q_{n-1}[/tex] × [tex]c^{n-1}[/tex] + ... + q1 × c + q0 = 0
Multiplying both sides of the equation by the common denominator of the coefficients [tex]q_i[/tex], we can obtain an equation with integer coefficients. Let d be the least common multiple of the denominators of the coefficients [tex]q_i[/tex]. Then we can write:
d × ([tex]q_n[/tex] × [tex]c^n[/tex] + [tex]q_{n-1}[/tex] × [tex]c^{n-1}[/tex] + ... + q1 × c + q0) = 0
Expanding the left-hand side of the equation, we obtain a polynomial with integer coefficients:
d × [tex]q_n[/tex] × [tex]x^n[/tex] + d × [tex]q_{n-1}[/tex] × [tex]x^{n-1}[/tex] + ... + d × q1 × x + d × q0 = 0
Since c is a root of the original polynomial equation, it is also a root of this polynomial with integer coefficients. Therefore, we have shown that if c is a root of a polynomial with rational coefficients, then it is also a root of a polynomial with integer coefficients.
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In May 2022, Reginald graduated from the Naval Academy with a degree in aeronautical engineering and was assigned to Pensacola, Florida as a permanent duty station. In his move to Pensacola, Reginald incurred the following costs: $450 in gasoline. $250 for renting a truck from UPAYME rentals. $100 for a tow trailer for his car. $90 in food. $35 in double espressos from Starbucks. $300 for motel lodging on the way to Pensacola. $475 for a previous plane trip to Pensacola to look for an apartment. $175 in temporary storage costs for his collection of sports memorabilia. Required: If the government reimburses him $900, how much, if any, may Reginald take as a moving expense deduction on his 2022 tax return?
Answers
Therefore , the solution of the given problem of unitary method comes out to be $500.
An unitary method is what?This common convenience, already-existing variables, or all important elements from the original Diocesan adaptable study that followed a particular methodology can all be used to achieve the goal. Both of the crucial elements of a term affirmation outcome will surely be missed if it doesn't happen, but if it does, there will be another chance to get in touch with the entity.
Here,
Reginald's total moving costs must first be determined in order to determine the moving expense deduction he may claim on his 2022 tax return.
The full cost of relocating is:
=> $450 (gasoline) + $250 (truck rental) + $100 (tow trailer) + $90 (food) + $35 (Starbucks) + $300 (motel lodging) + $175 (storage)
=> $1,400
Reginald's moving expenditures were covered by the government for $900, so the following are his actual moving costs:
=> $1,400 - $900 = $500
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the area of the parallelogram is 40 in2. one base of the parallelogram is 5 in long. the other base is 10 in long find its 2 heights
Answers
Step-by-step explanation:
Area of parallelogram = base × height or say b×h
Given: h=8 inches
Area=120 sq.inches
⇒120=8×b
⇒120/8=b
⇒120/8=b = 15
Answer = 15 inches b
Hmm, try solving 13 to the power of 16 do it the long way :0
and just in case you do
2(2 + ab) + b(r + 3)
Answers
The answer of the given question based on solving 13 to the power of 16 is 13 to the power of 16 is 3,947,868,257,259,789. and the simplified expression is 7 + 2ab + br.
What is Expression?A expression is a combination of symbols or values that represents a particular concept or computation.
In mathematics, an expression is a combination of numbers, variables, operators, and/or functions that can be evaluated to produce a numerical result.
an expression is a way to represent an idea, computation, or meaning using a set of symbols or words.
To solve 13 to the power of 16, we can start by multiplying 13 by itself 16 times:
13 × 13 = 169
169 × 13 = 2197
2197 × 13 = 28,561
28,561 × 13 = 371,293
371,293 × 13 = 4,826,389
4,826,389 × 13 = 62,748,857
62,748,857 × 13 = 815,730,721
815,730,721 × 13 = 10,604,807,473
10,604,807,473 × 13 = 137,858,491,849
137,858,491,849 × 13 = 1,792,160,390,737
1,792,160,390,737 × 13 = 23,303,986,079,681
23,303,986,079,681 × 13 = 303,305,489,096,753
303,305,489,096,753 × 13 = 3,947,868,257,259,789
Therefore, 13 to the power of 16 is 3,947,868,257,259,789.
As for the second expression, we can simplify it using the distributive property of multiplication:
2(2 + ab) + b(r + 3) = 4 + 2ab + br + 3b
Simplifying further, we can combine the constant terms:
2(2 + ab) + b(r + 3) = 7 + 2ab + br
So the simplified expression is 7 + 2ab + br.
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Suppose I go on a fishing trip where I visit 4 lakes, lakes L1, L2, L3, and L4. Let C1 be the event that I catch a fish from lake L1. Let C2 be the event that I catch a fish from lake L2. Let
C3
be the event that I catch a fish from lake L3. Let
C4
be the event that I catch a fish from lake L4. I am a poor fisherman, so I am happy if I catch at least one fish. The lakes are far enough apart so that whether I catch a fish in any lake is independent from catching a fish in any other lake. There is a
.3
probability that C1 happens, a .4 probability that C2 happens, a .2 probability that C3 happens and a
.2
probability that C4 happens a. What is the probability I catch fish in all 4 lakes? b. What is the probability I do not catch any fish at all? c. What is the probability that I catch at least one fish? (I am happy.) d. What is the probability that I catch fish in Lake L1 and lake L2? e. What is the probability that I catch fish in lake L1 or lake L2? f. What is the probability that I catch fish in exactly one lake? Add any comments below.
Answers
The required probabilities of catching or not catching a fish is given by,
Catching fish in all 4 lakes = 0.0048
Not catching any fish at all = 0.2688
Catching at least one fish = 0.7312
Catching fish in Lake L1 and lake L2 = 0.12
Catching fish in lake L1 or lake L2 = 0.58
Catching fish in exactly one lake = 1.1
Probability of catching fish in all 4 lakes,
Simply multiply the probabilities of catching fish in each lake,
since the events are independent,
P(C1 and C2 and C3 and C4)
= P(C1) x P(C2) x P(C3) x P(C4)
= 0.3 x 0.4 x 0.2 x 0.2
= 0.0048
= 0.48%
Probability of not catching any fish at all,
Probability of the complement of the event of catching at least one fish.
Happy if we catch at least one fish, the probability of not catching any fish is the probability that is not happy,
P(not happy)
= P(not C1 and not C2 and not C3 and not C4)
= (1 - P(C1)) x (1 - P(C2)) x (1 - P(C3)) x (1 - P(C4))
= 0.7 x 0.6 x 0.8 x 0.8
= 0.2688
= 26.88%
Probability of catching at least one fish,
Probability of the complement of the event of not catching any fish and subtract it from 1,
P(at least one fish)
= 1 - P(not happy)
= 1 - 0.2688
= 0.7312
= 73.12%
Probability of catching fish in Lake L1 and lake L2,
Simply multiply the probabilities of catching fish in each lake,
P(C1 and C2)
= P(C1) x P(C2)
= 0.3 x 0.4
= 0.12
= 12%
Probability of catching fish in lake L1 or lake L2,
Add the probabilities of catching fish in each lake,
And then subtract the probability of catching fish in both lakes to avoid double counting,
P(C1 or C2)
= P(C1) + P(C2) - P(C1 and C2)
= 0.3 + 0.4 - 0.12
= 0.58
= 58%
Probability of catching fish in exactly one lake can be broken down into four mutually exclusive events,
Catching fish in L1 only, catching fish in L2 only, catching fish in L3 only, or catching fish in L4 only.
Probabilities of each of these events is,
P(C1 or C2 or C3 or C4)
= P(C1) + P(C2) + P(C3) + P(C4)
= 0.3 + 0.4 + 0.2 + 0.2
= 1.1
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