To maximize the area, a 2m length of wire should be used for the square and the rest for the triangle. To minimize the area, nearly all the wire should be used for the triangle, leaving a negligible amount for the square.
Explanation:The problem described is a classic example of Mathematics optimization. In this case, we have two geometric shapes, a square and an equilateral triangle. To answer this question effectively, one needs to understand the relationship between the perimeter and area of these two shapes.
For the square, the area is given by A=s2, where s is the length of a side. For the equilateral triangle, the area is given by A=0.433*s2, where s is the length of a side. We want to understand how to divide the 6m wire so that we either maximize or minimize the total area of these two shapes.
The total length of wire used is fixed at 6m. Let's designate x as the length of wire used for the square. This means the length for the triangle would be 6-x. For the maximum area, the result generally comes around 2m for the square and 4m for the triangle. However, for the minimum area, the answer would be essentially 0m for the square and 6m for the triangle.
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(a) For maximizing total area, use [tex]\( s \approx 3.76 \)[/tex] meters in square.
(b) For minimizing area, use [tex]\( s = 0 \)[/tex] for maximum utilization of wire to the triangle, least yielding area near zero or minimal non-zero.
(a) Maximizing the Total Area
To maximize the total area, we need to determine the length of the wire to be used for the square [tex](\( s \))[/tex] and the length used for the equilateral triangle [tex](\( t \))[/tex].
(b) Minimizing the Total Area
To minimize the total area, confirm whether interiors of boundary values might be critical points. Specifically, see if using all the wire for one shape minimizes the area.
You just purchased two coins at a price of $670 each. Because one of the coins is more collectible, you believe that its value will increase at a rate of 7.1 percent per year, while you believe the second coin will only increase at 6.5 percent per year. If you are correct, how much more will the first coin be worth in 15 years?
Answer:
The value of first coin will be $151.51 more than second coin in 15 years.
Step-by-step explanation:
You have just purchased two coins at a price of $670 each.
You believe that first coin's value will increase at a rate of 7.1% and second coin's value 6.5% per year.
We have to calculate the first coin's value after 15 years by using the formula
[tex]A=P(1+\frac{r}{100})^{n}[/tex]
Where A = Future value
P = Present value
r = rate of interest
n = time in years
Now we put the values
[tex]A=670(1+\frac{7.1}{100})^{15}[/tex]
[tex]A=670(1+0.071)^{15}[/tex]
[tex]A=670(1.071)^{15}[/tex]
A = (670)(2.797964)
A = 1874.635622 ≈ $1874.64
Now we will calculate the value of second coin.
[tex]A=670(1+\frac{6.5}{100})^{15}[/tex]
[tex]A=670(1+0.6.5)^{15}[/tex]
[tex]A=670(1.065)^{15}[/tex]
A = 670 × 2.571841
A = $1723.13
The difference of the value after 15 years = 1874.64 - 1723.13 = $151.51
The value of first coin will be $151.51 more than second coin in 15 years.
To find out how much more the first coin will be worth compared to the second coin in 15 years, we can apply the formula for compound interest. The formula is:
\[ A = P(1 + r)^t \]
where:
- \( A \) is the future value of the investment/loan, including interest
- \( P \) is the principal investment amount (the initial deposit or loan amount)
- \( r \) is the annual interest rate (decimal)
- \( t \) is the number of years the money is invested or borrowed for
We will apply this formula separately for each coin and then find the difference between the two future values.
First, let's calculate the future value of the more collectible coin:
\( P = \$670 \) (the initial value of the coin)
\( r = 7.1\% = 0.071 \) (the annual increase rate converted to decimal)
\( t = 15 \) years
So the future value \( A_1 \) for the first coin is calculated as:
\[ A_1 = \$670 \times (1 + 0.071)^{15} \]
Now, let's calculate the future value of the less collectible coin:
\( P = \$670 \) (the initial value of the coin)
\( r = 6.5\% = 0.065 \) (the annual increase rate converted to decimal)
\( t = 15 \) years
So the future value \( A_2 \) for the second coin is calculated as:
\[ A_2 = \$670 \times (1 + 0.065)^{15} \]
Finally, to find out how much more the first coin will be worth, we subtract the future value of the second coin from the future value of the first coin:
\[ \text{Difference} = A_1 - A_2 \]
Let's compute this step by step.
First, calculate \( A_1 \):
\[ A_1 = \$670 \times (1 + 0.071)^{15} \]
\[ A_1 = \$670 \times (1.071)^{15} \]
\[ A_1 = \$670 \times 2.83844... \] (using a calculator)
\[ A_1 \approx \$1901.75 \] (rounded to two decimal places)
Next, calculate \( A_2 \):
\[ A_2 = \$670 \times (1 + 0.065)^{15} \]
\[ A_2 = \$670 \times (1.065)^{15} \]
\[ A_2 = \$670 \times 2.64716... \] (using a calculator)
\[ A_2 \approx \$1775.63 \] (rounded to two decimal places)
Now calculate the difference:
\[ \text{Difference} = A_1 - A_2 \]
\[ \text{Difference} = \$1901.75 - \$1775.63 \]
\[ \text{Difference} \approx \$126.12 \]
Therefore, if your expectations are correct, in 15 years, the first coin will be worth approximately $126.12 more than the second coin.
Seven times the first (smaller) of two consecutive odd integers is equal to five times the second (larger) integer. Find each integers.
Let the smaller odd integer be: a
then the larger odd integer which is consecutive to a will be: a+2
It is given that:
Seven times the first (smaller) of two consecutive odd integers is equal to five times the second (larger) integer.
This means that:
7 times of a is equal to 5 times of (a+2)
i.e.
[tex]7a=5(a+2)\\\\i.e.\\\\7a=5\times a+5\times 2[/tex]
( Since, by using the distributive property of multiplication)
i.e.
[tex]7a=5a+10\\\\i.e.\\\\7a-5a=10\\\\i.e.\\\\2a=10\\\\i.e.\\\\a=5[/tex]
Hence, the smaller number is: 5
and the larger number is: 7
( Since a+2=5+2=7 )
The population of a town with a 20162016 population of 87 comma 00087,000 grows at a rate of 1.91.9% per year. a. Find the rate constant k and use it to devise an exponential growth function that fits the given data. b. In what year will the population reach 145 comma 000145,000?
Answer:
Given,
The initial population, P = 87,000,
Annual rate of increasing, r = 1.9 %,
a) Thus, the population after t years,
[tex]A=P(1+\frac{r}{100})^t[/tex]
[tex]A=87000(1+\frac{1.9}{100})^{t}[/tex]
[tex]=87000(1+0.019)^t[/tex]
[tex]=87000(1.019)^t[/tex]
[tex]87000(1.019)^t=87000e^{kt}[/tex]
Where, k is the rate constant,
By comparing,
[tex]\implies k=log(1.019)=0.00817418400643\approx 0.00817[/tex]
Hence, the approximate value of k is 0.00817.
And, the exponential growth function would be,
[tex]A=87000e^{0.00817t}[/tex]
b) If A = 145,000,
[tex]145000=87000(1.019)^t[/tex]
By using graphing calculator,
We get,
[tex]t=27.14\approx 27[/tex]
The year after 27 years from 2016 is 2043.
Hence, in 2043 the population reach 145,000.
The rate constant k in the given exponential growth problem is 1.9% per year. The exponential growth function for this problem is P = 87,000 * e^(0.019t). The year in which the population reaches 145,000 can be found by solving the equation 145,000 = 87,000 * e^(0.019t) for t, which will be counted from 2016.
Explanation:The given problem is an example of an exponential growth problem. The formula for exponential growth is P = P0 * e^(kt), where P is the future population, P0 is the initial population (87,000 in this case), e is the natural base (approximately equal to 2.71828), k is the growth rate (1.9% or 0.019 in decimal), and t is the time (in years).
a. Finding the rate constant k
The rate constant k is given in the problem as 1.9% per year or 0.019 per year when expressed in decimal form.
The exponential growth function that fits the given data would therefore be P = 87,000 * e^(0.019t).
b. When will the population reach 145,000?
To find out when the population will reach 145,000, we need to set P equal to 145,000 and solve for t, giving us the equation 145,000 = 87,000 * e^(0.019t). Solving this equation using logarithms will yield the year, which will be counted from the base year 2016.
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Theorem: Let a and b be positive integer, if gcd(a,b)=1, then exist positive integer x and y such that ax+by=c for any integer greater than ab-a-b.
Prove the above theorem.
Answer with explanation:
It is given that , a and b are positive integers.
gcd(a,b)=1
We have to prove for any positive integer x and y ,
a x + by =c, for any integer greater than ab-a-b.
Proof:
GCD of two numbers is 1, when two numbers are coprime.
Consider two numbers , 9 and 7
GCD (9,7)=1
So, we have to calculate positive integers x and y such that
⇒ 9 x +7 y > 9×7-9-7
⇒9x +7 y> 47
To prove this we will draw the graph of Inequality.
So the ordered pair of Integers are
x>5 and y>6.
So, for any integers , a and b ,
→ax+ by > a b -a -b, if
[tex]\Rightarrow \frac{x}{\frac{ab-a-b}{a}}+ \frac{y}{\frac{ab-a-b}{b}}>1,\frac{ab-a-b}{a},\text{and},\frac{ab-a-b}{b}[/tex]
⇒Range of x for which this inequality hold
[tex]=[\frac{ab-a-b}{a},\infty)[/tex]
if,
[tex]\frac{ab-a-b}{a}[/tex]
is an Integer ,otherwise range of x
[tex]=(\frac{ab-a-b}{a},\infty)[/tex]
⇒Range of y for which this inequality hold
[tex]=[\frac{ab-a-b}{b},\infty)[/tex]
if,
[tex]\frac{ab-a-b}{b}[/tex]
is an Integer ,otherwise range of y
[tex]=(\frac{ab-a-b}{b},\infty)[/tex]
Enter your PIN: The technology company DataGenetics suggests that % of all four-digit personal identification numbers, or PIN codes, have a repeating digits format such as . Assuming this to be true, if the PIN codes of seven people are selected at random, what is the probability that at least one of them will have repeating digits? Round your answer to four decimal places.
Answer: 0.9917
Step-by-step explanation:
If repetition is allowed , then the total number of possible four digits pin codes = [tex]10^4=10,000[/tex]
Number of ways to make for digit code without repetition of digits =
[tex]10\times9\times8\times7=5040[/tex]
Number of ways to make for digit codes having repetition =
[tex]10,000-5040=4960[/tex]
Probability that a person has pin code that has repetition:-
[tex]\dfrac{4960}{10,000}=0.496[/tex]
Let x be number of pin codes with repeating digits.
If the PIN codes of seven people are selected at random, then the probability that at least one of them will have repeating digits:-
[tex]P(x\geq1)=1-(P(0))\\\\=1-(^7C_0(0.496)^0(1-0.496)^7)[/tex] (By Binomial distribution)
[tex]=1-((0.496)^0(0.504)^7)=0.991739358875\approx0.9917[/tex]
Hence, the probability that at least one of them will have repeating digits = 0.9917
Find a polynomial f(x) of degree 5 that has the following zeros.
4, 9, -5 (multiplicity 2), -7
Leave your answer in factored form.
Answer:
[tex]f (x) = (x-4)(x + 5)^2(x-9)(x+7)[/tex]
Step-by-step explanation:
The zeros of the polynomial are all the values of x for which the function [tex]f (x) = 0[/tex]
In this case we know that the zeros are:
[tex]x = 4,\ x-4 =0[/tex]
[tex]x = 9,\ x-9=0[/tex]
[tex]x = -5[/tex], [tex]x + 5 = 0[/tex] (multiplicity 2)
[tex]x = -7,\ x+7=0[/tex]
Now we can write the polynomial as a product of its factors
[tex]f (x) = (x-4)(x + 5)^2(x-9)(x+7)[/tex]
Note that the polynomial is of degree 5 because the greatest exponent of the variable x that results from multiplying the factors of f (x) is 5
The polynomial of degree 5 with the zeros 4, 9, -5 (multiplicity 2), and -7 is given by f(x) = (x - 4)(x - 9)(x + 5)² (x + 7).
Explanation:To find a polynomial f(x) of degree 5 that has the zeros 4, 9, -5 (multiplicity 2), -7, you would begin by setting up a polynomial that has these zeros. From the given roots, we can form the polynomial in factored form as follows:
f(x) = (x - 4)(x - 9)(x + 5)²(x + 7).
Multiplicity refers to how many times a root is repeated. In this case, -5 is repeated twice, hence the exponent of 2.
So, the polynomial in factored form with the zeros provided is:
f(x) = (x - 4)(x - 9)(x + 5)² (x + 7).
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what is the solution of the inequality shown below? c-7>2
Answer:
c > 9
Step-by-step explanation:
Isolate the variable, c. Treat the greater than sign as an equal, what you do to one side, you do to the other. Add 7 to both sides:
c - 7 > 2
c - 7 (+7) > 2 (+7)
c > 2 + 7
c > 9
c > 9 is your answer.
~
Answer:
[tex]\Huge \boxed{C>9}\checkmark[/tex]
Step-by-step explanation:
Add by 7 from both sides.
[tex]\displaystyle c-7+7>2+7[/tex]
Simplify, to find the answer.
[tex]\displaystyle 2+7=9[/tex]
[tex]\huge \boxed{c>9}[/tex], which is our answer.
Shown work to help me make sense would be greatly appreciated.
1. Convert the 8-binary binary expansion (1010 0110)2 to a decimal expansion.
2. Convert the following decimal expansion (145)10 to an 8-bit binary expansion.
3. Convert the following hexadecimal expansion (A3C)16 to an octal expansion.
4. Convert the following binary expansion (1111 1100 0011 0110)2 to a hexadecimal expansion.
1. (1010 0110)₂ = (166)₁₀
2. (145)₁₀ = (1001 0001)₂
3. (101 000 111 100)₂ = (5074)₈
4. (1111 1100 0011 0110)₂ = (FC36)₁₆
1. Convert the 8-binary binary expansion (1010 0110)₂ to a decimal expansion.
In order to solve this problem we have to use the expansion:
n = aₓbˣ + aₓ₋₁bˣ⁻¹ + ... + a₁b¹ + a₀
where b = 2, x = 8 - 1 = 7 due is a 8-binary
(1010 0110)₂ = 1 x 2⁷ + 0 x 2⁶ + 1 x 2⁵ + 0 x 2⁴ + 0 x 2³ + 1 x 2² + 1 x 2¹ + 0 x 2⁰
(1010 0110)₂ = 128 + 0 + 32 + 0 + 0 + 4 + 2 + 0
(1010 0110)₂ = (166)₁₀
2. Convert the following decimal expansion (145)₁₀ to an 8-bit binary expansion.
To solve this problem we have to use the divide by 2 process.
Since we are dividing by 2, when the dividend is an even number, the remainder will be 0, and when the dividend is an odd number the binary residual will be 1.
145 ---------------> 1 Less significant bit
145/2 = 72 -----> 0
72/ 2 = 36 -----> 0
36/2 = 18 ------> 0
18/2 = 9 -------> 1
9/2 = 4 --------> 0
4/2 = 2 ---------> 0
2/2 = 1 ----------> 1 Most significant bit
Then we order from the most significant bit to the less significant bit (from the bottom to the top) to obtain the 8-binary number:
(145)₁₀ = (1001 0001)₂
3. Convert the following hexadecimal expansion (A3C)₁₆ to an octal expansion.
To convert a hexadecimal expansion to an octal expansion we have to convert from hexadecimal to binary and then to octal using the table hexadecimal to binary and binary to octal.
Converting from hexadecimal to binary:
(A3C)₁₆
A = 1010, 3 = 0011 and C = 1100
(A3C)₁₆ = (1010 0011 1100)₂
Converting from binary to octal:
To convert binary to octal we have to order the binary expansion into group of 3-bits and use the table to convert binary to octal.
(1010 0011 1100)₂ = (101 000 111 100)₂
101 = 5, 000 = 0, 111 = 7 and 100 = 4
(101 000 111 100)₂ = (5074)₈
4. Convert the following binary expansion (1111 1100 0011 0110)₂ to a hexadecimal expansion.
To solve this exercise we have to use the binary to hexadecimal table.
(1111 1100 0011 0110)₂
1111 = F, 1100 = C, 0011 = 3 and 0110 = 6
(1111 1100 0011 0110)₂ = (FC36)₁₆
Use an integrating factor to solve the following first order linear ODE. xy' + 2y = 3x, y(1) = 3 Find the end behavior of y as x rightarrow infinity.
Answer:
Solution --> [tex] y(x) = x + \frac{2}{x^{2}} [/tex]
when x --> infinity the y goes to infinity too.
Step-by-step explanation:
We have the eq.
[tex] x y'+2 y = 3 x [/tex]
with y(1) = 3. So a reconfiguration o this last one equation can be like:
[tex] y' + \frac{2}{x} y = 3 [/tex]
so is of the form y'+p(x) y = f(x), where the integral factor is given by:
[tex] \mu = \exp[ \int p(x) dx] [/tex]
is,
[tex] \mu = \exp[ \int \frac{2}{x} dx] = x^{2} [/tex]
Multiplying the whole equation with this integral factor we can write the expression:
[tex] \int \frac{d}{dx}[y x^{2}] dx = \int 3x^{2} dx [/tex]
and from this we obtain,
[tex] y x^{2} = x^{3} + c,[/tex]
So when y(x=1) = 3, c = 2 and the complete solution can be writen as:
[tex] y(x) = x - \frac{2}{x^{2}} [/tex].
And we can see that when x --> infinity the second term of the solution is practically zero and the first is infinity so y also goes to infinity when x does.
The first-order linear ODE xy' + 2y = 3x is y = x + 2/x² and the end behavior of y is approximately as x.
To solve the first-order linear ordinary differential equation (ODE) of the form xy' + 2y = 3x using an integrating factor, follow these steps:
1. Rewrite the Equation
First, rewrite the ODE in the standard form:y' + (2/x)y = 32. Identify the Integrating Factor
The standard form is y' + p(x)y = q(x), where p(x) = 2/x and q(x) = 3. The integrating factor (μ) is given by:μ = e^(∫p(x) dx) = e^(∫2/x dx) = e^(2ln|x|) = x²3. Multiply by the Integrating Factor
Multiply through by the integrating factor x2:x²y' + 2xy = 3x²4. Simplify and Integrate
This can be written as (x²y)' = 3x².Integrate both sides with respect to x:∫(x²y)' dx = ∫3x² dxx²y = x³ + C5. Solve for y
Divide through by x²:y = x + C/x²6. Use Initial Condition
Use the initial condition y(1) = 3 to find C:3 = 1 + C/12C = 2So, the solution is:y = x + 2/x²7. Find the End Behavior
To find the end behavior of y as x → ∞:As x → ∞, the term 2/x² → 0.Thus, y ≈ x.Therefore, as x → ∞, y behaves approximately as x.
Complete Question:
Use an integrating factor to solve the following first order linear ODE. xy' + 2y = 3x, y(1) = 3 Find the end behavior of y as x rightarrow infinity.
A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 48.0 and 53.0 minutes. Find the probability that a given class period runs between 51.5 and 51.75 minutes. Find the probability of selecting a class that runs between 51.5 and 51.75 minutes.
Step-by-step answer:
Given:
class time uniformly distributed between 48 and 53 minutes (range 5 minutes)
Find probability that the class period runs between 51.5 and 51.75 minutes.
Solution:
Uniformly distributed means that the probability that the class time is the same for any minute (out of 5), i.e. 20%, or probability density is 0.2 / minutes, no matter which minute.
The interval between 51.5 and 51.75 is 0.25 minutes, so the probability of class period within that range is 0.25 minutes * 0.2 / minute = 0.05.
Final answer:
The probability that a class period runs between 51.5 and 51.75 minutes, when class lengths are uniformly distributed between 48.0 and 53.0 minutes, is 0.05 or 5%.
Explanation:
To find the probability that a given class period runs between 51.5 and 51.75 minutes when class lengths are uniformly distributed between 48.0 and 53.0 minutes, we utilize the properties of the uniform distribution. The probability is the area under the uniform probability density function (PDF) between the two specified lengths. For a uniform distribution, this area corresponds to the length of the interval divided by the length of the entire interval in which the variable is uniformly distributed.
Here, we're looking for the probability between 51.5 minutes and 51.75 minutes. The total length of the available interval is 53.0 - 48.0 = 5 minutes. The length of the desired interval is 51.75 - 51.5 = 0.25 minutes. So, we divide the length of the desired interval by the total interval to get the probability.
Probability = (51.75 - 51.5) / (53.0 - 48.0) = 0.25 / 5 = 0.05
Therefore, the probability that a class period runs between 51.5 and 51.75 minutes is 0.05 or 5%.
Suppose that a poll finds that 31.9% of taxpayers who filed their tax return electronically self-prepared their taxes. If three tax returns submitted electronically are randomly selected, what is the probability that all three were self-prepared?
Answer: 0.0325
Step-by-step explanation:
Binomial probability distribution formula for x successes in n trials:-
[tex]P(X=x)=^nC_x\ p^x\ (1-p)^{n-x}[/tex], where n is the number of trials , p is the probability of success.
Given : The probability that taxpayers who filed their tax return are electronically self-prepared their taxes : [tex]p= 0.319[/tex]
If three tax returns submitted electronically are randomly selected, then the probability that all three were self-prepared is given by :-
[tex]P(X=3)=^3C_3\ (0.319)^3\ (1-0.319)^{3-3}\\\\=(1) (0.319)^3(1)\\=0.032461759\approx0.0325[/tex]
Hence, the probability that all three were self-prepared =0.0325
Final answer:
The probability that all three tax returns submitted electronically are self-prepared, when each has a 31.9% chance of being self-prepared, is the product of their individual probabilities, resulting in approximately 3.24%.
Explanation:
To compute this, we will use the property that the probability of independent events all occurring is the product of their individual probabilities.
The probability of one taxpayer self-preparing is 0.319. Since the question implies that the taxpayers are chosen independently, the probability that all three returns are self-prepared is:
P(First self-prepared) = 0.319,
P(Second self-prepared) = 0.319,
P(Third self-prepared) = 0.319.
To find the combined probability, we multiply these probabilities together:
P(All three self-prepared) = 0.319 × 0.319 × 0.319,
This calculation results in approximately 0.0324, or 3.24%.
A model for the basal metabolism rate, in kcal/h, of a young man is R(t) = 95 − 0.18 cos(πt/12), where t is the time in hours measured from 5:00 AM. What is the total basal metabolism of this man, 24 R(t) dt, 0 over a 24-hour time period?
Answer:
2280 kcal
Step-by-step explanation:
The given function R(t) is periodic with a period of 24 hours, so the integral is the product of the average value (95 kcal/h) and the 24-hour interval:
(95 kcal/h)(24 h) = 2280 kcal
To calculate the total basal metabolism over 24 hours for a young man modeled by R(t) = 95 − 0.18 cos(πt/12), you need to apply integral calculus. First, integrate the function R(t) and then apply a definite integral over the period from 0 to 24 hours. The result will be the total basal metabolism over this time period.
Explanation:To solve this problem, you need to integrate the function R(t) = 95 − 0.18 cos(πt/12) over the time interval of 0 to 24 hours. In calculus, such an operation is represented as ∫R(t) dt from 0 to 24. This operation will give you the total basal metabolism during these 24 hours.
The first thing you need to do in this case is to integrate the function. As this function is a composition of functions, you would need to use the basics of integral calculus, specifically the trigonometric substitution method or table of integrals for cosine function.
Afterward, apply a definite integral over the period from 0 to 24 hours. This will give you the total basal metabolism for 24 hours, expressed in kcal.
The steps presented above represent a simplification of the process. The actual calculations might be quite complex, so please don't worry if they seem overwhelming - it's a normal part of learning integral calculus.
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2. In 1973, Secretariat set the current record of 1 minute 59.40 seconds for the Kentucky Derby. The length of this race is 10 furlongs1 furlong1/8 mile). What was Secretariat's average speed in this race in mi/hr?
Answer:
Secretariat's average speed in the Kentucky Derby of 1973 was of 47.05 miles per hour.
Step-by-step explanation:
Secretariat set the current record of 1 minute 59.40 seconds for the Kentucky Derby, running 10 furlongs, which are 1.25 miles.
To determine the speed at which the horse ran, we must first isolate the time in which he ran, to determine the fraction of time in which he traveled the 1.25 mile.
An hour has 60 minutes. To determine the speed, whe have to divide 60 by the minutes the horse ran, and then multiply it by the number of miles he ran: 60 / 1.5940 = 37,64 x 1.25 = 47,05.
The horse ran at 47.05 miles per hour.
Secretariat's record average speed during the Kentucky Derby was an impressive 37.68 mi/hr.
The question asks to calculate Secretariat's average speed during the Kentucky Derby in miles per hour (mi/hr). First, we need to express the distance of the Kentucky Derby in miles. The Derby is 10 furlongs long, and since 1 furlong is 1/8 mile, this translates to 1.25 miles (10 furlongs × 1/8 mile/furlong). Understanding this conversion is crucial because speed is typically measured in miles per hour in the United States.
Next, Secretariat's record-setting time for completing this distance is 1 minute and 59.40 seconds, which is equivalent to 119.40 seconds. To find the average speed, we use the formula:
Speed = Distance / Time
Hence, Secretariat's average speed is:
1.25 miles / (119.40 seconds × (1 hour / 3600 seconds)) = 1.25 miles / 0.03317 hours = 37.68 mi/hr, when the time is converted from seconds to hours.
Therefore, Secretariat's record average speed during the Kentucky Derby was an impressive 37.68 mi/hr.
Solve the following ODE's: c) y* - 9y' + 18y = t^2
Answer:
y = [tex]C_1e^{3t}+C_2e^{6t}[/tex] + [tex]\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})[/tex]
Step-by-step explanation:
y''- 9 y' + 18 y = t²
solution of ordinary differential equation
using characteristics equation
m² - 9 m + 18 = 0
m² - 3 m - 6 m+ 18 = 0
(m-3)(m-6) = 0
m = 3,6
C.F. = [tex]C_1e^{3t}+C_2e^{6t}[/tex]
now calculating P.I.
[tex]P.I. = \frac{t^2}{D^2 - 9D +18}[/tex]
[tex]P.I. = \dfrac{t^2}{(D-3)(D-6)}\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1-\frac{D}{6})^{-1}(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1+\frac{D}{6}+\frac{D^2}{36}+....)(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(1+\frac{D}{3}+\frac{D^2}{9}+....)(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})[/tex]
hence the complete solution
y = C.F. + P.I.
y = [tex]C_1e^{3t}+C_2e^{6t}[/tex] + [tex]\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})[/tex]
In a recent year, companies spent a total of $88.9 billion on newspaper, television, and radio ads. The total amount spent on television and radio ads was only $3.1 billion more than the amount spent on newspaper ads alone. The amount spent on newspaper ads was $5.4 billion more than what was spent on television ads. How much was spent on each form of advertising? (Hint: Let the variables represent numbers of billions of dollars.) How much was spent on newspaper ads? $ nothing billion How much was spent on television ads? $ nothing billion How much was spent on radio ads? $ nothing billion
[tex]\bf \begin{cases} t=\textit{tv ads}\\ r=\textit{radio ads}\\ n=\textit{newspaper ads} \end{cases}~\hspace{7em}t+r+n=88.9 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{tv, radio ads 3.1 more than newspaper}}{t+r=n+3.1}~\hfill \stackrel{\textit{newspaper ads 5.4 more than tv's}}{\boxed{n} = t+5.4}[/tex]
[tex]\bf \stackrel{\textit{substituting on the 2nd equation}}{t+r=\boxed{t+5.4}+3.1}\implies ~~\begin{matrix} t \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+r=~~\begin{matrix} t \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+8.5\implies \blacktriangleright r=8.5 \blacktriangleleft \\\\\\ \stackrel{\textit{we know that}}{t+r+n=88.9}\implies t+8.5+n=88.9\implies t+n=80.4[/tex]
[tex]\bf \boxed{n}=80.4-t~\hspace{7em}\stackrel{\textit{substituting on the 3rd equation}}{n=t+5.4\implies \boxed{80.4-t}=t+5.4} \\\\\\ 75-t=t\implies 75=2t\implies \cfrac{75}{2}=t\implies \blacktriangleright 37.5=t \blacktriangleleft \\\\\\ \stackrel{\textit{since we know that}}{n = t+5.4}\implies n=37.5+5.4\implies \blacktriangleright n=42.9 \blacktriangleleft[/tex]
The amounts spent on Newspaper, Television and Radio ads are $40.3 billion, $34.9 billion, and $13.7 billion, respectively.
Explanation:Let us represent the three advertising channels - newspaper, television, and radio - as N, T, and R respectively. Given that total spending on advertising is $88.9 billion, we can write this formally as N + T + R = 88.9 (equation 1). The problem also mentions that total spending on television and radio ads is $3.1 billion more than spending on newspaper ads. This can be written formally as T + R = N + 3.1 (equation 2). Lastly, it is given that spending on newspaper ads is $5.4 billion more than what was spent on television ads. Writing this we get N = T + 5.4 (equation 3). Substituting equation 3 into equation 2, we get (T + 5.4) + R = T + R + 3.1. This simplifies to T - R = 2.3. Solving these equations, we find that $40.3 billion is spent on Newspaper ads, $34.9 billion is spent on Television ads, and $13.7 billion is spent on Radio ads.
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M1Q13.) Find the relative frequency of the 2nd class to the nearest tenth of a percent.
Answer:
9.1%
Step-by-step explanation:
The total number of observations is 33. The number in the second class is 3, so the relative frequency is ...
3/33 × 100% = (100/11)% ≈ 9.1%
Add everything in the frequency column.
= 33
Take class two (3).
Solution:
3/33 * 100%
1/11 * 100%
100/11%
9.1%
Best of Luck!
Evaluate the line integral, where C is the given curve. C (x + yz) dx + 2x dy + xyz dz, C consists of line segments (1, 0, 1) to (2, 4, 1) and from (2, 4, 1) to (2, 6, 4)
Parameterize the first line segment [tex]C_1[/tex] by
[tex]\vec r(t)=(1,0,1)(1-t)+(2,4,1)t=(1+t,4t,1)[/tex]
and the second line segment [tex]C_2[/tex] by
[tex]\vec s(t)=(2,4,1)(1-t)+(2,6,4)t=(2,4+2t,1+3t)[/tex]
both with [tex]0\le t\le1[/tex]. Then
[tex]\displaystyle\int_{C_1}(x+yz)\,\mathrm dx+2x\,\mathrm dy+xyz\,\mathrm dz[/tex]
[tex]\displaystyle=\int_0^1\bigg((1+5t)\cdot1+2(1+t)\cdot4+(1-t)4t\cdot0\bigg)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1(9+13t)\,\mathrm dt=\frac{31}2[/tex]
and
[tex]\displaystyle\int_{C_2}(x+yz)\,\mathrm dx+2x\,\mathrm dy+xyz\,\mathrm dz[/tex]
[tex]\displaystyle=\int_0^1\bigg((2+(4+2t)(1+3t))\cdot0+2(2)\cdot2+2(4+2t)(1+3t)\cdot3\bigg)\,\mathrm dt[/tex]
[tex]\displaystyle=\int_0^1(32+84t+36t^2)\,\mathrm dt=86[/tex]
Then
[tex]\displaystyle\int_C(x+yz)\,\mathrm dx+2x\,\mathrm dy+xyz\,\mathrm dz=\frac{31}2+86=\boxed{\frac{203}2}[/tex]
To evaluate the line integral, parameterize each segment of the curve separately, substitute the parameterization into the line integral expression, evaluate the integral for each segment, and sum up the results.
Explanation:To evaluate the line integral, we need to first parameterize the curve C. The given curve consists of two line segments, so we will parameterize each segment separately.
For the first segment from (1, 0, 1) to (2, 4, 1), we can parameterize it as r(t) = (1 + t, 4t, 1), where t ranges from 0 to 1.
For the second segment from (2, 4, 1) to (2, 6, 4), we can parameterize it as r(t) = (2, 4 + 2t, 1 + 3t), where t ranges from 0 to 1.
Next, we substitute the parameterization into the line integral expression and evaluate the integral for each segment separately. Finally, we sum up the individual results to get the value of the line integral.
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A certain field is a rectangle with a perimeter of 918 feet. The length is 181 feet more than the width. Find the width and length of the rectangular field The width is? feet. The length is ? feet.
The width of the rectangular field is 139 feet and its length is 320 feet.
Explanation:To solve this problem, we can use the formula for the perimeter of a rectangle, which is 2*(length + width). We know that the perimeter is 918 feet, and it's given that the length is 181 feet more than the width.
Let's denote the width as w, therefore the length would be w + 181.
By substituting these expressions into the formula for the perimeter, we get:
2*(w + w + 181)=918,
which simplifies to 2*(2w + 181)=918,
further simplifies to 4w + 362 = 918.
To isolate 4w, subtract 362 from both sides:
4w = 918 - 362 = 556.
Finally, divide both sides by 4 to solve for w:
w = 556/4 = 139 feet.
Substitute w = 139 into the length equation to get the length:
length = w + 181 = 139 + 181 = 320 feet.
So the width is 139 feet, and the length is 320 feet.
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Length is 320 feet.
Let's define the width of the rectangle as w feet and the length as l feet.
According to the problem,
the perimeter of the rectangle is 918 feet and the length is 181 feet more than the width.
We can express these conditions using the following equations:
Perimeter equation: 2l + 2w = 918
Length equation: l = w + 181
First, we can solve the perimeter equation for l+w:
2l + 2w = 918
Divide both sides by 2:
l + w = 459
Next, we substitute the length equation l = w + 181 into the perimeter equation:
w + 181 + w = 459 Simplify:
2w + 181 = 459
Subtract 181 from both sides:
2w = 278
Divide by 2:
w = 139 feet
Now substitute the width back into the length equation:
l = w + 181
l = 139 + 181
l = 320 feet
So, the width is 139 feet, and the length is 320 feet.
A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 75%. Refer to Exhibit 9-6. At a .05 level of significance, it can be concluded that the proportion of the population in favor of candidate A is
Answer:
it would be A 1251
Step-by-step explanation:
The question involves hypothesis testing of proportions. Based on a sample where 80% favour Candidate A, we may infer the true population proportion is significantly greater than 75%, however, actual mathematical calculations are needed for confirmation.
Explanation:The question examines whether the proportion of the population favouring Candidate A is significantly more than 75% based on a sample of 100 people where 80 favoured Candidate A. This is a problem of hypothesis testing for proportions. The null hypothesis (H₀) is that the true population proportion is 75% (p = 0.75), versus the alternative hypothesis (H₁) stating the true population proportion is more than 75%. Using a significance level of 0.05, we examine the data.
With a sample proportion of 80 out of 100 (p' = 0.80) and given the large sample size, we apply the Normal approximation to the Binomial distribution, followed by a one-sample z-test. If the resulting p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that the true population proportion of individuals favouring Candidate A is significantly greater than 75%.
However, without performing the actual calculations, we cannot definitively determine the conclusion. From a practical perspective, an 80% sample proportion showing favour in a sample as large as 100 might indicate a significantly higher proportion than 75%, but an exact mathematical test should be done to confirm this.
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Based on a Comcast survey, there is a 0.8 probability that a randomly selected adult will watch prime-time TV live, instead of online, on DVR, etc. Assume that seven adults are randomly selected. Find the probability that fewer than three of the selected adults watch prime-time live.
Answer: Our required probability is 0.004672.
Step-by-step explanation:
Since we have given that
Number of adults = 7
Probability of getting adult will watch prime time TV live = 0.8
We need to find the probability that fewer than 3 of the selected adults watch prime time live.
We will use "Binomial Distribution":
here, n = 7
p = 0.8
So, P(X<3)=P(X=0)+P(X=1)+P(X=2)
So, it becomes,
[tex]P(X=0)=(1-0.8)^7=0.2^7=0.0000128[/tex]
[tex]P(X=1)=^7C_1(0.8)(0.2)^6=0.0003584\\\\P(X=2)=^7C_2(0.8)^2(0.2)^5=0.0043[/tex]
So, probability that fewer than 3 of the selected adult watch prime time live is given by
[tex]0.0000128+0.0003584+0.0043=0.004672[/tex]
Hence, our required probability is 0.004672.
This problem relates to the binomial distribution and requires us to find the sum of binomial probabilities for 0, 1, and 2 successes (adults watching live TV) out of seven trials (the seven randomly selected adults).
Explanation:This question is utilizing the concept of binomial distribution. The probability of a randomly selected adult watching prime-time TV live is 0.8. We want to find the probability that fewer than 3 out of 7 randomly selected adults watch prime-time live.
We find this by adding up the probabilities for 0, 1, and 2 adults watching live TV using the binomial distribution formula: P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)), where C(n, k) denotes the number of combinations of n items taken k at a time, p is the probability of success, and n is the number of trials.
We get:
P(X=0) = C(7, 0) * (0.8^0) * ((1-0.8)^(7-0))P(X=1) = C(7, 1) * (0.8^1) * ((1-0.8)^(7-1))P(X=2) = C(7, 2) * (0.8^2) * ((1-0.8)^(7-2))Adding these up will give the total probability that fewer than three adults out of seven watch prime-time live.
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On the day that his first child was born, Ezio Auditore de Firenze deposited $3,000 into an investment account. The only purpose for the account was to pay for his son’s first year of college tuition. Assume that his son, Flavia, started college on his 18t h birthday and his first year tuition payment had to be made that day. The amount needed on that day was $26,000. If that was indeed the amount of money in the account on Flavia’s 18t h birthday, what annual rate of return did Ezio earn on his investment account?
Answer:
about 12.75%
Step-by-step explanation:
Let r represent the annual rate of return. Compounded annually for 18 years, the account multiplier is (1+r)^18. Then we have ...
26,000 = 3,000(1+r)^18
(26,000/3,000)^(1/18) = 1+r . . . . . . divide by 3000, take the 18th root
(26/3)^(1/18) -1 = r ≈ 12.7465%
Ezio's account earned about 12.75% annually.
Ezio's $3,000 investment grew to $26,000 over 18 years at this rate.
To calculate the annual rate of return Ezio earned on his investment for his son, Flavia's, college tuition, we'll need to use the compound interest formula:
A = P(1 + r)^n
Where:
A is the amount of money accumulated after n years, including interest.
P is the principal amount (the initial amount of money).
r is the annual interest rate (decimal).
n is the number of years the money is invested.
We know that Ezio deposited $3,000 (P = 3000), the amount in the account after 18 years is $26,000 (A = 26000), and that the time period n is 18 years. We need to find the annual interest rate r.
Let's rearrange the formula to solve for r:
(1 + r)^n = A / P
(1 + r)^18 = 26000 / 3000
(1 + r)^18 = 8.6667
Now we need to take the 18th root of 8.6667 to find (1 + r):
1 + r = (8.6667)^(1/18)
The 18th root of 8.6667 is approximately 1.1225. This means:
1 + r = 1.1225
Subtract 1 from both sides to find r:
r = 1.1225 - 1
r = 0.1225 or 12.25%
The annual rate of return Ezio earned on his investment is approximately 12.25%.
47. An honest coin is tossed 10 times in a row. The result of each toss (H or T) is observed. Find the probability of the event E= “a T comes up at least once.” (Hint: Find the probability of the complementary event.)
Answer:
0.0488%
Step-by-step explanation:
Here we have the probability of different independent events, which means that none of the previous ones affect the next. For this kind of events, the probability 'P' that a series of events occur is the multiplication of the probability of each singular event.
p1: Probability that an H be obtained: 50% (is always 50% because it is independent of the previous results)
P: The probability that H be obtained all of the 10 times. This is the complementary probability to E:(a T comes up at least once).
By the first definition given
[tex]P=0.5^{11}=0.00048828=0.0488%[/tex]
The complementary probability 'P' is the probability that 'E' does not happen, so the probability that E happen is: [tex]P_{E} =1-P[/tex]
The last makes sense if we think about the fact that for the experiment there are just two possibilities, 'E' happen, or 'E' does not happen. Then,
[tex]P_{E} =1-0.000488=99.95[/tex]%
Several psychology students are unprepared for a surprise true/false test with 18 questions, and all of their answers are guesses. a. Find the mean and standard deviation for the number of correct answers for such students. b. Would it be unusual for a student to pass by guessing (which requires getting at least 14 correct answers)? Why or why not?
In a true/false test where answers are guessed, the mean number of correct answers out of 18 would be 9 (50%) and the standard deviation would be 3. Passing by guessing with a minimum of 14 correct answers would be unusual as it is significantly above the mean and more than a single standard deviation away.
Explanation:This question is about applying the binary distribution, specifically dealing with a true/false test. Before diving into the question, let's understand what we're working with. In a true/false test where students are guessing, the probability of a correct answer (p) is 0.5, as the expected value (mean) is np (number of trials times the probability of success) and the standard deviation is sqrt(np(1-p)).
In this scenario, for 18 questions, the mean (expected value) would be 18*0.5 = 9 and the standard deviation would be sqrt(18*0.5*0.5) = 3. Therefore, the mean is 9 and the standard deviation is 3.
For a student to 'pass' by guessing, they would need at least 14 correct answers. This number of successes is much greater than our mean expectation of 9, and it's more than a single standard deviation away. Therefore, it would be unusual for a student to pass the test by simply guessing.
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Use Laplace transforms to solve the initial value problem. x" + 9x = 1;x (0) = 0 = x' (0)
Answer: The required solution of the given IVP is
[tex]x(t)=\dfrac{1}{9}(1-\cos3t).[/tex]
Step-by-step explanation: We are given to use Laplace transforms to solve the following initial value problem :
[tex]x^{\prime\prime}+9x=1,~~~x(0)=0=x^\prime(0).[/tex]
We will be using the following formula for the Laplace transform :
[tex](i)~L\{t^n\}=\dfrac{n!}{s^{n+1}},\\\\\\(ii)~L\{\coskt\}=\dfrac{s}{s^2+k^2}.[/tex]
Applying Laplace transform on both sides of the above equation, we have
[tex]L\{x^{\prime\prime}+9x\}=L\{1\}\\\\\\\Rightarrow s^2X(s)-sx(0)-x^\prime(0)+9X(s)=\dfrac{1}{s}\\\\\\\Rightarrow s^2X(s)-s\times0-0=\dfrac{1}{s}\\\\\\\Rightarrow (s^2+9)X(s)=\dfrac{1}{s}\\\\\\\Rightarrow X(s)=\dfrac{1}{s(s^2+9)}\\\\\\\Rightarrow X(s)=\dfrac{1}{9s}-\dfrac{s}{9(s^2+9)}.[/tex]
Taking inverse Laplace transform on both sides of the above equation, we get
[tex]L^{-1}\{X(s)\}=L^{-1}\left(\dfrac{1}{9s}\right)-L^{-1}\left(\dfrac{s}{9(s^2+9)}\right)\\\\\\\Rightarrow x(t)=\dfrac{1}{9}\times1-\dfrac{1}{9}\times \cos3t\\\\\\\Rightarrow x(t)=\dfrac{1}{9}(1-\cos3t).[/tex]
Thus, the required solution of the given IVP is
[tex]x(t)=\dfrac{1}{9}(1-\cos3t).[/tex]
To solve the given initial value problem using Laplace transforms, apply the transforms to the equation, simplify to get the equation in the s-domain, and then take the inverse Laplace transform to get the solution in the time domain.
Explanation:To solve the initial value problem x" + 9x = 1 where x(0) = 0 = x'(0) using Laplace transforms, we firstly apply it to the whole equation. The Laplace transform of a second derivative x''(t) is s^2 X(s) - s x(0) - x`(0) and that of x(t) is X(s), where X(s) is the Laplace transform Of x(t). Since the Laplace transform of a constant is the constant itself divided by s, the Laplace transform of the equation becomes s^2 X(s) - 0 - 0 + 9X(s) = 1/s. Simplifying the above gives X(s) = 1/(s^3 + 9s), as a solution in the s-domain. Apply a combination of basic formulas and decomposition techniques to convert this equation into a zero-state solution. After this, take the inverse Laplace transform to get the solution in the time domain. It should be simple given that the initial values are zero.
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A shipment of 30 inexpensive digital watches, including 6 that are defective, is sent to a department store. The receiving department selects 10 at random for testing and rejects the whole shipment if 1 or more in the sample are found defective. What is the probability that the shipment will be rejected?
Answer:
0.935
Step-by-step explanation:
Let's notice that finding 1 or more defective watches in the sample is the complement (exactly the opposite) of finding 0 defective watches. That means P (1 or more defective watches) + P (0 defective watches) = 1
Let's find out P (0 defective watches):
The chances of getting 1 defective watch is 6 in 30 (in a shipment of 30 there are only 6 defective).
Therefore, we have 24/30 chances of getting a non-defective watch.
Needing 0 defective watches is the same as having 10 non defective watches.
When the receiving department select the first one, the probability of getting a non-defective watch is 24/30.
In the second one, they have a probability of 23/29 (they already took a non-defective one).
In the third one, the probability is 22/28.
And so on, up to the last one where the probability is 15/21.
As we need this to happen all at the same time, we have to multiply it:
P (0 defective watches) = [tex]\frac{24}{30}*\frac{23}{29}*\frac{22}{28}*\frac{21}{27}*\frac{20}{26}*\frac{19}{25}*\frac{18}{24}*\frac{17}{23}*\frac{16}{22}*\frac{15}{21}[/tex] = 0.065
Therefore, P (1 or more defective watches) = 1 - P (0 defective watches) = 1 - 0.065 = 0.935
The probability that the shipment will be rejected is 0.8696 or 86.96%.
Explanation:To find the probability that the shipment will be rejected, we need to find the probability that 1 or more watches are defective in a sample of 10. This can be done using the complement rule, which states that the probability of an event happening is equal to 1 minus the probability of it not happening.
The probability of selecting a defective watch in the sample is given by:
P(defective) = 6/30 = 1/5
Therefore, the probability of not selecting a defective watch is:
P(not defective) = 1 - P(defective) = 1 - 1/5 = 4/5
The probability that none of the watches in the sample are defective is:
P(all not defective) = (4/5)^10
Finally, the probability that at least one watch is defective (and the shipment is rejected) is:
P(rejected shipment) = 1 - P(all not defective) = 1 - (4/5)^10 = 0.8696
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Write an equation of the circle with center (-8,4) and diameter 6 .
Answer:
(x+8)^2 + (y-4)^2 = 3^2
or
(x+8)^2 + (y-4)^2 = 9
Step-by-step explanation:
The diameter is 6, so the radius would be d/2 =3
The equation of a circle is
(x-h)^2 + (y-k)^2 = r^2
where (h,k) is the center and r is the radius
(x- -8)^2 + (y-4)^2 = 3^2
(x+8)^2 + (y-4)^2 = 3^2
or
(x+8)^2 + (y-4)^2 = 9
32. Quality Control A campus bookstore buys 100 calcula-
tors. Assume that 2 have a defect of some kind. The math-
ematics department buys 8 of these calculators from the
bookstore. What is the probability that, of those calcula-
tors bought by the mathematics department,
(a) All are free of defects?
(b) Exactly 1 will have a defect?
(c) Exactly 2 will have a defect?
Answer:
a) 0.8508
b) 0.1389
c) 0.0099
Step-by-step explanation:
Total number of calculators bought = 100
Number of calculators with defect = 2
Probability of selecting a calculator with defect = p = 2 out of 100 = [tex]\frac{2}{100}=0.02[/tex]
Probability of selecting a calculator without defect = q = 1 - p = 1 - 0.02 = 0.98
Part a)
The bookstore buys 8 calculators. This means the number of trials is fixed ( n =8). The calculator is either with defect or without defect. This means there are 2 outcomes only. The outcome of one selection is independent of other selections. This means all the events(selections) are independent of each other.
Hence, all the conditions of a Binomial Experiment are being satisfied. So we will use Binomial Probability to solve this problem.
We need to find the probability that all calculators are free of defects. i.e. number of success is 0 i.e. P( x = 0 )
The formula of Binomial Probability is:
[tex]P(x) =^{n}C_{x} (p)^{x} q^{n-x}[/tex]
Using the values, we get:
[tex]P(0)=^{8}C_{0}(0.02)^{0}(0.98)^{8-0}\\\\ P(0)=0.8508[/tex]
Thus, the probability that all calculators are free of defects is 0.8508
Part b) Exactly 1 calculator will have defect
This means the number of success is 1. So, we need to calculate ( x = 1)
Using the formula of binomial probability again and using x = 1, we get:
[tex]P(1)=^{8}C_{1}(0.02)^{1}(0.98)^{8-1}\\\\ P(1)=0.1389[/tex]
Thus, the probability that exactly two calculators will be having a defect is 0.1389
Part c) Exactly 2 calculator will have defect
This means the number of success is 2. So, we need to calculate ( x = 2 )
Using the formula of binomial probability again and using x = 2, we get:
[tex]P(2)=^{8}C_{2}(0.02)^{2}(0.98)^{8-2}\\\\ P(2)=0.0099[/tex]
Thus, the probability that exactly two calculators will be having a defect is 0.0099
3) An oil prospector will drill a succession of holes in a given area to find a productive well. The probability that he is successful on a given trial is 0.2. What is the probability that the third hole drilled is the first to yield a productive well?
Answer: 0.128
Step-by-step explanation:
The geometric probability of getting success on nth trial is given by :-
[tex]P(n)=p(1-p)^{n-1}[/tex]
Given : The probability that he is successful on a given trial[tex]p= 0.2[/tex].
Then , the probability that the third hole drilled is the first to yield a productive well is given by :-
[tex]P(3)=0.2(1-0.2)^{3-1}=0.128[/tex]
Hence, the probability that the third hole drilled is the first to yield a productive well = 0.128
Barney is the owner and insured of a $200,000 life insurance policy. Upon learning that he has a terminal illness, he sells the policy to a third party for $140,000, and the third party becomes the owner and beneficiary of the policy. Barney entered into what type of arrangement? a. life endowment b. accelerated death benefit arrangement c. qualified policy surrender d. viatical settlement
Answer:
Viatical settlement
Step-by-step explanation:
Here Barney sold his life insurance to 3rd party for less net death benefit value.
He entered into the arrangement called - viatical settlement
A viatical settlement is the sale of a life insurance policy to a third party. The owner of the life insurance policy sells the policy for cash benefit, thus making the buyer the new owner of the policy who will get death benefits.
Find the area and the circumference of a circle with radius 3 m. Use the value 3.14 for π, and do not round your answers. Be sure to include the correct units in your answers.
Circumference: 18.84 m
Area: 28.26[tex]m^{2}[/tex]
Explanation
For circumference, use the formula [tex]\pi[/tex]*d, where d=the diameter(the radius times 2).
For area, use the formula [tex]\pi[/tex][tex]r^{2}[/tex], where r=the radius. Make sure to square the radius first before multiplying by pi!
A good way to remember these formulas is Cherry(circumference) Pi([tex]\pi[/tex]) is Delicious(diameter) and Apple(area) [tex]\pi[/tex] [tex]r^{2}[/tex]("are too")
The circumference of the circle with radius 3 meters is 18.84 meters and the area is 28.26 square meters.
Given that the radius of the circle is 3 meters, we can find the circumference and the area using the respective formulae.
Circumference of a circle with radius r (C) is given by the formula C = 2πr. Substituting π = 3.14 and r = 3 meters, the circumference C = 2*3.14*3 = 18.84 meters.
The Area of a circle with radius r (A) is given by the formula A = πr². Substituting π = 3.14 and r = 3 meters, the area A = 3.14*3² = 28.26 square meters.
So, the circumference of the circle with radius 3 meters would be 18.84 meters and the area would be 28.26 square meters.
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