The expected value of the business return is RM112.50. Whether you should invest in the business venture depends on your risk tolerance and the potential return you expect. If the potential profit of RM112.50 is attractive to you and aligns with your investment goals, you might consider investing.
The expected value of a business return is calculated by multiplying each possible outcome by its respective probability and summing them up. In this case, the probability of making a profit of RM250 is 0.75, and the probability of making a loss of RM300 is 0.25.
i. To calculate the expected value of the business return, we can use the following formula:
Expected Value = (Probability of Profit * Profit) + (Probability of Loss * Loss)
Expected Value = (0.75 * RM250) + (0.25 * RM(-300))
Expected Value = RM187.50 + RM(-75)
Expected Value = RM112.50
Therefore, the expected value of the business return is RM112.50.
ii. Whether you should invest in the business venture depends on your risk tolerance and the potential return you expect. The expected value of the business return is RM112.50, which means, on average, you can expect to make RM112.50 from the venture.
If the potential profit of RM112.50 is attractive to you and aligns with your investment goals, you might consider investing in the business venture. However, it's important to note that the expected value is just an average, and individual outcomes can vary. There is a 0.75 probability of making a profit of RM250 and a 0.25 probability of making a loss of RM300.
It is recommended to consider other factors such as the initial investment required, the level of risk you are comfortable with, and the overall viability of the business venture before making a decision.
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Consider the following series and level of accuracy. ∑ n=0
[infinity]
(−1) n
6 n
+2
1
(10 −4
) Determine the least number N such that ∣R N
∣ is less than the given level of accuracy. N= 0. [-/8 Points] Consider the following series and level of accuracy. ∑ n=1
[infinity]
7 n
n
(−1) n
(10 −4
) Determine the least number N such that ∣R N
∣ is less than the given level of accuracy. N=
The least number N such that |[tex]R_N[/tex]| is less than the given level of accuracy is N = 4.
To determine the least number N such that the remainder term |[tex]R_N[/tex]| is less than the given level of accuracy, we need to apply the alternating series remainder theorem.
For the series Σₙ₌₀(-1)ⁿ × [tex]6^{(n+2)[/tex]/(10⁴), the remainder term [tex]R_N[/tex] is given by:
|[tex]R_N[/tex]| ≤ |[tex]a_{(N+1)[/tex]|,
where [tex]a_{(N+1)[/tex] is the absolute value of the (N+1)-th term of the series.
To find N, we need to find the term that satisfies |[tex]a_{(N+1)[/tex]| < 10⁻⁸. Let's calculate the terms of the series:
a₁ = (-1)¹ × [tex]6^{(1+2)[/tex]/(10⁴)
= -6³/(10⁴)
= -216/10000
a₂ = (-1)² × [tex]6^{(2+2)[/tex]/(10⁴)
= 6⁴/(10⁴)
= 1296/10000
a₃ = (-1)³ × [tex]6^{(3+2)[/tex]/(10⁴)
= -6⁵/(10⁴)
= -7776/10000
a₄ = (-1)⁴ × [tex]6^{(4+2)[/tex]/(10⁴)
= 6⁶/(10⁴)
= 46656/10000
We can observe that the magnitude of the terms alternates between increasing and decreasing.
Checking the magnitude of the terms:
|a₁| = |216/10000| ≈ 0.0216
|a₂| = |1296/10000| ≈ 0.1296
|a₃| = |7776/10000| ≈ 0.7776
|a₄| = |46656/10000| ≈ 4.6656
We see that |a₄| ≈ 4.6656 > 10⁻⁸.
Therefore, we need to find the least number N such that N ≥ 4.
Hence, the least number N such that |[tex]R_N[/tex]| is less than the given level of accuracy is N = 4.
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If f(x)={ e x
+5
sinx+6cosx+tanx
if x<0
if x≥0
Is f(x) Continuous at x
˙
=0 ? Explain Is f(x) Differentiable at x=0 ? Explain
Given function is f(x)={ ex+5sinx+6cosx+tanxif x<0if x≥01. To determine if the function is continuous at x = 0, we have to determine if the function exists at that point and if the left-hand limit is equal to the right-hand limit at x = 0.2. To determine if the function is differentiable at x = 0, we have to determine if the derivative exists at that point.
Is f(x) Continuous at x = 0?Answer: Yes. f(x) is continuous at x = 0. We can write f(x) as shown below:⇒f(x)={ ex+5sinx+6cosx+tanxif x<0if x≥0⇒f(x)={ ex+5sinx+6cosx+tanxif x≤0if x>0Since the expression of the function is different for x < 0 and x > 0, we have to calculate both the left and right limits of the function to check if f(x) is continuous at x = 0.Let's first calculate the left-hand limit. So, as x approaches 0 from the left-hand side, x takes on negative values. Therefore, we have,⇒
limx→0−f(x)=limx→0−(ex+5sinx+6cosx+tanx)=limx→0−ex+5sinx+6cosx+tanx
Let's now calculate the right-hand limit. As x approaches 0 from the right-hand side, x takes on positive values. Therefore, we have,⇒
limx→0+f(x)=limx→0+(ex+5sinx+6cosx+tanx)=limx→0+ex+5sinx+6cosx+tanx
We can now evaluate the limits. We know that,⇒
limx→0sinx/x=1⇒limx→0cosx−1/x=0⇒limx→0tanx/x=1
Thus,⇒
limx→0−f(x)=1+6+1+0=8⇒limx→0+f(x)=1+6+1+0=8
Since both the left and right-hand limits exist and are equal, we can say that the limit of the function f(x) exists and is equal to 8. Thus, f(x) is continuous at x = 0. 2. Is f(x) Differentiable at x = 0?Answer: No. f(x) is not differentiable at x = 0. We can use the definition of the derivative to calculate the left and right-hand derivatives of the function at x = 0. Let's first calculate the left-hand derivative. As x approaches 0 from the left-hand side, x takes on negative values. Therefore, we have,⇒
f′(0−)=limh→0−f(0+h)−f(0)h=limh→0−[(e0+h+5sin(0+h)+6cos(0+h)+tan(0+h))−(e0+5sin0+6cos0+tan0)]h⇒f′(0−)=limh→0−[eh+5sinh+6cosh+tanh−12]h
Using the limit properties, we can simplify the expression. Therefore,⇒
f′(0−)=limh→0−[eh−12h+h(sinh+6cosh+tanh)]
By taking the limit of the second term, we get,⇒
limh→0−h(sinh+6cosh+tanh)=0
Therefore,⇒
f′(0−)=limh→0−[eh−12h]=limh→0−(eh)e−h2
This limit does not exist. Therefore, the left-hand derivative does not exist at x = 0. Now, let's calculate the right-hand derivative. As x approaches 0 from the right-hand side, x takes on positive values. Therefore, we have,⇒
f′(0+)=limh→0+f(0+h)−f(0)h=limh→0+[(e0+h+5sin(0+h)+6cos(0+h)+tan(0+h))−(e0+5sin0+6cos0+tan0)]h⇒f′(0+)=limh→0+[eh+5sinh+6cosh+tanh−12]h
Using the limit properties, we can simplify the expression. Therefore,⇒
f′(0+)=limh→0+[eh−12h+h(sinh+6cosh+tanh)]
By taking the limit of the second term, we get,⇒
limh→0+h(sinh+6cosh+tanh)=0
Therefore,⇒
f′(0+)=limh→0+[eh−12h]=limh→0+(eh)e−h2
This limit does not exist. Therefore, the right-hand derivative does not exist at x = 0. Since both the left and right-hand derivatives do not exist, we can say that the derivative of f(x) does not exist at x = 0.
Conclusion: Therefore, f(x) is continuous but not differentiable at x = 0.
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Evaluate the given equation using integration by parts. ∫tan−γdγ
The integration of tan (x) can be done by using the integration by parts. Integration by parts is used when the function is expressed as a product of two different functions.
It can also be used to integrate the functions that cannot be integrated directly. Let's use this method to evaluate the given equation using integration by parts
∫tan(−γ)dγ.Integration by parts formula is given as;
\int udv= uv - \
int vdu
Let u be tan(−γ) and dv be dγ
We have;
du/dγ = sec²(−γ)
dv/dγ = 1
By substituting u and v into the formula we get;
\int \tan (-γ) dγ = \int u dv
= uv - \int v du
= -\tan (-γ)γ - \int (-\sec^2(-γ))(-dγ)
= -\tan (-γ)γ + \int \sec^2(-γ)dγ
We know that\int \sec^2 (-γ)dγ
= -\tan (-γ) + C
Substituting it in the above equation;
\int \tan (-γ) dγ
= -\tan (-γ)γ + (-\tan (-γ) + C)
=-\tan (-γ) (\gamma + 1) + C
Therefore, ∫tan (−γ)dγ using integration by parts is:
\boxed{\int \tan (-γ) dγ = -\tan (-γ) (\gamma + 1) + C}$$
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Find the average value of the function f(x)= e 2x
x 2
+3
over x∈[0,2].
The average value of the given function f(x) = e²ˣ + x² + 3 for the interval [0, 2] is equal to (1/4) × e⁴ + 11/12.
To find the average value of a function over an interval,
Evaluate the definite integral of the function over that interval and divide it by the length of the interval.
find the average value of the function f(x) = e²ˣ + x²+ 3 over the interval [0, 2].
The average value, Avg, is ,
Avg = (1 / (b - a))× ∫[a, b] f(x) dx
Plugging in the values for a = 0 and b = 2, we have,
Avg = (1 / (2 - 0))× ∫[0, 2] (e²ˣ + x² + 3) dx
Simplifying further,
Avg = (1 / 2) × ∫[0, 2] (e²ˣ + x² + 3) dx
Now, let's integrate each term separately,
∫ e²ˣ dx
= (1/2) × e²ˣ| [0, 2]
= (1/2) × (e⁴ - e⁰)
= (1/2) × (e⁴- 1)
∫ x² dx
= (1/3) × x³ | [0, 2]
= (1/3) ×(2³ - 0³)
= (1/3) × 8
= 8/3
∫ 3 dx
= 3x | [0, 2]
= 3 × (2 - 0)
= 6
Now, we can substitute these values into the expression for Avg,
Avg = (1 / 2) × [(1/2) × (e⁴ - 1) + 8/3 + 6]
= (1/4) ×(e⁴ - 1) + 4/3 + 3
= (1/4) × e⁴ - 1/4 + 4/3 + 3
= (1/4) × e⁴ + 11/12
Therefore, the average value of the function f(x) = e²ˣ + x² + 3 over the interval [0, 2] is (1/4) × e⁴ + 11/12.
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The above question is incomplete, the complete question is:
Find the average value of the function f(x)= e²x + x² +3 over x∈[0,2].
Show that "if n is an integer and (n2 + 3) is odd, then n is an
even integer," by applying (a) Proof by Contraposition. (b) Proof
by Contradiction. Show every step of the proof process.
Using proof by contraposition and proof by contradiction, it has been shown that if n is an integer and (n² + 3) is odd, then n must be an even integer.
(a) To prove the statement "if n is an integer and (n² + 3) is odd, then n is an even integer" using proof by contraposition, we need to show that the negation of the statement implies the negation of the original statement.
Negation of the original statement:
"If n is an integer and n is not even, then (n² + 3) is not odd."
Let's assume that n is an integer but not even. Then, n must be an odd integer.
If n is an odd integer, it can be expressed as n = 2k + 1, where k is an integer.
Substituting n = 2k + 1 into the expression (n² + 3), we get:
(n² + 3) = (2k + 1)²2 + 3
= 4k² + 4k + 1 + 3
= 4k² + 4k + 4
= 4(k² + k + 1)
Since k is an integer, k² + k + 1 is also an integer. Let's denote it as m.
So, we have (n² + 3) = 4m.
Since 4 is divisible by 2, we can rewrite 4m as 2(2m), where 2m is an integer.
Therefore, (n² + 3) is divisible by 2, and it is not odd.
Hence, we have proved that if n is an integer and (n² + 3) is odd, then n is an even integer.
(b) To prove the statement "if n is an integer and (n² + 3) is odd, then n is an even integer" using proof by contradiction, we assume the opposite of the statement and show that it leads to a contradiction.
Assume that n is an integer and (n² + 3) is odd, but n is an odd integer.
If n is an odd integer, we can write it as n = 2k + 1, where k is an integer.
Substituting n = 2k + 1 into the expression (n² + 3), we get:
(n² + 3) = (2k + 1)² + 3
= 4k² + 4k + 1 + 3
= 4k² + 4k + 4
= 4(k² + k + 1)
Since k is an integer, k² + k + 1 is also an integer. Let's denote it as m.
So, we have (n² + 3) = 4m.
Since 4 is divisible by 2, we can rewrite 4m as 2(2m), where 2m is an integer.
Therefore, (n² + 3) is divisible by 2, and it is not odd.
This leads to a contradiction since we assumed that (n² + 3) is odd.
Hence, our assumption that n is an odd integer must be false.
Therefore, n must be an even integer.
Thus, we have proved that if n is an integer and (n² + 3) is odd, then n is an even integer.
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ms. smythe and her husband wanted to take the neighborhood children to see a play. there were 14 children and the cost of their tickets was $6 each. the adult tickets were three times that amount. how much did the smythe's spend? a. $36 b. $84 c. $120 d. $130
To calculate the total amount spent by the Smythes, we need to consider the cost of the children's tickets and the adult tickets the correct answer is c. $120.
Given that there are 14 children, each ticket costs $6. Therefore, the total cost of the children's tickets is 14 * $6 = $84.The cost of the adult tickets is three times the cost of a child's ticket. So each adult ticket costs $6 * 3 = $18.Assuming there are two adults (Ms. Smythe and her husband), the total cost of the adult tickets is 2 * $18 = $36.
To find the total amount spent, we add the cost of the children's tickets and the adult tickets: $84 + $36 = $120.Therefore, the Smythes spent a total of $120.In summary, the correct answer is c. $120.
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If a population has 5261 members and you select a sample of 45 using systematio sampoling, what is the value of m ? 83 Calculate the IHD for each of the following molecular formulas: (a) C 6
H 6
; (b) C 6
H 5
NO 2
; (c) C 8
H 13
F 2
NO ;
(d) C 4
H 12
Si; (e) C 8
H 5
BrO; (f) C 4
H 6
O 3
S
Answer:
ok, here is you answer
Step-by-step explanation:
I'm sorry, but the first part of your question about systematic sampling is incomplete and unclear. Can you please provide more information or clarify the question?
For the second part of your question, to calculate IHD (index of hydrogen deficiency) for a molecular formula, you can use the formula:
IHD = (2n + 2 - x)/2
Where n is the number of carbons and x is the number of hydrogens and halogens (excluding fluorine).
Using this formula, we can calculate the IHD for each of the given molecular formulas:
(a) C6H6: n = 6, x = 6, IHD = 4
(b) C6H5NO2: n = 6, x = 7, IHD = 3
(c) C8H13F2NO: n = 8, x = 16, IHD = 4
(d) C4H12Si: n = 4, x = 12, IHD = 0
(e) C8H5BrO: n = 8, x = 6, IHD = 4
(f) C4H6O3S: n = 4, x = 6, IHD = 3
Therefore, the IHD for each of the given molecular formulas are: (a) 4, (b) 3, (c) 4, (d) 0, (e) 4, (f) 3.
mark me as brainliestSection 6 Student Loan Repayment (15 marks)
Shekha Bhalla completed his college program in December 2016 with a $11,600 Canada Student Loan.
He selected the fixed rate option (prime +2. 5%) and agreed to make end-of-month payments of $120
beginning July 31 2017. The Prime Rate began the six-month grace period at 6% and rose by 0. 5% effective
March 29 2017. On June 30 2017, he paid all the interest that had accrued (at prime +2. 5%) during the
six-month grace period. On July 1 2017, the prime rate rose to 9%. On August 13, the prime rate rose by
another 0. 5%.
Required:
מחרוזורח
Shekha Bhalla paid a total of $2350 in interest during the repayment period.
To calculate the total amount of interest paid by Shekha Bhalla during the repayment period, we need to consider the different interest rates and payment dates.
Grace Period Interest:
During the six-month grace period (December 2016 - June 2017), the prime rate was 6%. However, Shekha paid all the accrued interest at prime +2.5% on June 30, 2017. So there was no interest paid during the grace period.
Interest from July 1, 2017, to March 28, 2018:
Starting July 1, 2017, the prime rate was 9%.
The repayment period lasted from July 1, 2017, to March 28, 2018 (8 months).
Using the formula: Interest = Loan Amount x Interest Rate x Time
Interest = $11,600 x (9% + 2.5%) x (8/12) = $11,600 x 11.5% x 2/3 = $1090
Interest from March 29, 2018, to August 13, 2018:
On March 29, 2018, the prime rate increased by 0.5% to 9.5%.
The repayment period lasted from March 29, 2018, to August 13, 2018 (4.5 months).
Using the same formula: Interest = $11,600 x (9.5% + 2.5%) x (4.5/12) = $11,600 x 12% x 3/8 = $1260
Total Interest Paid = Grace Period Interest + Interest July 1, 2017 - March 28, 2018 + Interest March 29, 2018 - August 13, 2018
= $0 + $1090 + $1260
= $2350
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What is the GCF of 28 and 42
GCF of 28 and 42: 14.
To find the greatest common factor (GCF) of 28 and 42, we need to determine the largest number that can evenly divide both 28 and 42.
1. List the factors of each number:
The factors of 28 are: 1, 2, 4, 7, 14, 28.
The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, 42.
2. Identify the common factors:
The common factors of 28 and 42 are: 1, 2, 7, 14.
3. Determine the greatest common factor:
The greatest common factor among the common factors is 14.
Therefore, the GCF of 28 and 42 is 14.
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estion 4: Below are the prices of the same car in different countries.
£1= €1.18 £1 = ¥140
£1 = $1.25
USA
$20000
Ireland
€17500
England
£15000
In which country is the car the best value?
Japan
¥3000000
From the given question the prices of the same car in different countries best value for the car is in the USA with a price of 16000 euros.
We can use the currency conversion rates to compare the prices of the same car in different countries.
The prices of the same car in different countries are as follows:
USA: $20000
Ireland: €17500
England: £15000
Japan: ¥3000000
To compare these prices, we need to convert them to a common currency.
Here, we can use the conversion rates given in the question:
1 pound = 1.18 euro
1 pound = 140 yen
1 pound = 1.25 dollars
Price of car in euros in Ireland: 17500 euros
Price of car in pounds in England: 15000 pounds = 15000 x 1.18
= 17700 euros
Price of car in dollars in USA: 20000 dollars
= 20000 / 1.25 = 16000 euros
Price of car in yen in Japan: 3000000 yen = 3000000 / 140
= 21428.57 euros
Best value for the car is in the USA with a price of 16000 euros.
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A boat heads
37°,
propelled by a force of
650
lb. A wind from
326°
exerts a force of
200
lb on the boat. How large is the resultant force
F,
and in what direction is the boat moving?
Given data:A boat heads at 37° and is propelled by a force of 650 lb. A wind from 326° exerts a force of 200 lb on the boat.
To find:How large is the resultant force F, and in what direction is the boat moving?Solution:Firstly, we need to make a rough sketch for the given scenario as given below: [tex]AO = 650lb [/tex] is the force which boat is propelled with and [tex] OB = 200 lb[/tex] is the force of wind blowing from 326 degrees.
[tex]OC[/tex] is the resultant force and the angle formed by this force with the x-axis is [tex] \theta [/tex] to be found.Now we can see the triangle [tex] OAB[/tex] forms a scalene triangle so, it's tough to get any angle directly. Let's break the vectors into their components form and solve the problem.
Let, A be the angle made by the boat's force and x-axis, thenA = 90 - 37° = 53°Hence, the x-component of force due to the boat [tex]OA[/tex] is:[tex] OA = 650 cos 53° = 408.53[/tex]and, the y-component of force due to the boat [tex]OA[/tex] is:[tex] OA = 650 sin 53° = 527.39[/tex]Let, B be the angle made by the wind's force and x-axis, thenB = 360° - 326° = 34°
Hence, the x-component of force due to the wind [tex]OB[/tex] is:[tex] OB = 200 cos 34° = 165.65[/tex]and, the y-component of force due to the wind [tex]OB[/tex] is:[tex] OB = 200 sin 34° = 113.57[/tex]Now we can find out the resultant force acting on the boat i.e [tex] OC [/tex] which is the vector sum of [tex]OA[/tex] and [tex] OB[/tex].
Now applying the Pythagorean theorem we can find the magnitude of the resultant force.Finally, to find the direction of the resultant force, we use the below formula :[tex] \theta = arctan (\frac{527.39 + 113.57}{408.53 + 165.65}) = arctan (\frac{640.96}{574.18}) = 50.3 [/tex]degree (approx.)
Resultant Force [tex]OC[/tex] :[tex]OC = \sqrt{(527.39+113.57)^2 + (408.53+165.65)^2}[/tex][tex]OC = 846.56 lb[/tex]Direction of boat = 50.3 degrees (approx.)
Therefore, the magnitude of the resultant force acting on the boat is 846.56 lb and the direction of the boat is 50.3 degrees.
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Verify the identity:
2cos2(x/2)=(sin2 x)/(1-cos x)
The given trigonometric identity cannot be verified for all values of x.
Given the trigonometric identity to verify:2 cos(x/2) = sin(x)/1-cos(x)We know the following trigonometric identities: Cosine double-angle identity:cos(2x) = cos²(x) - sin²(x)Cosine half-angle identity:cos(x/2) = ±√(1 + cos(x)) / 2Sine double-angle identity:sin(2x) = 2sin(x)cos(x).
Let us convert the left-hand side of the given equation to sin(x)/1-cos(x) by using the half-angle identity:2 cos(x/2) = 2(√(1 + cos(x)) / 2) = √(1 + cos(x))Next, let us square the right-hand side of the given equation using the double-angle identity:sin²(x) = 2sin(x)cos(x) / (1 - cos²(x))Therefore,2 cos(x/2) = sin(x)/1-cos(x)2(√(1 + cos(x)) / 2) = √(1 - cos²(x)) / (1 - cos(x)) = sin(x) / (1 - cos(x))2√(1 + cos(x)) = sin(x)Multiply both sides by 2 to obtain:4(1 + cos(x)) = sin²(x)Use the identity sin²(x) + cos²(x) = 1 to substitute cos²(x) with 1 - sin²(x):4(1 + (1 - sin²(x))) = sin²(x)5sin²(x) + 8 = 4sin²(x)5sin²(x) - 4sin²(x) + 8 = 04sin²(x) = -8sin²(x) = -2.
Hence, sin²(x) = -2 which is not possible as the square of a sine function cannot be negative. Therefore, the given trigonometric identity cannot be verified for all values of x.
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Find the solution of the following polynomial inequality.
Express your answer in interval notation.
x2≤−x+12
The solution of the polynomial inequality x^2 ≤ −x + 12 is x ∈ [−4, 3].
To solve this inequality, we need to bring all the terms to one side and then factorize it. After this, we can find the roots of the quadratic equation and then use test points to see which interval(s) satisfy the inequality. Let's solve this inequality step by step.
Step 1: Write the given inequality in standard form. We get:
x^2 + x - 12 ≤ 0
Step 2: Factorize the quadratic equation. We get:
(x + 4)(x - 3) ≤ 0
Step 3: Find the roots of the quadratic equation. We get:
x = -4 and x = 3.
Step 4: Plot these roots on the number line. This divides the number line into three intervals. They are: (−∞, −4], [−4, 3], and [3, ∞).
Step 5: Now, we need to find the sign of the inequality in each of these intervals. We can do this by taking a test point from each of these intervals and substituting it into the inequality. For example, let's take x = −5. Substituting this into the inequality, we get(−5)^2 + (−5) - 12 ≤ 0⟹ 25 − 5 - 12 ≤ 0⟹ 8 ≤ 0. This is false.
Hence, the sign of the inequality in the interval (−∞, −4] is negative. Let's take x = 0. Substituting this into the inequality, we get0^2 + 0 - 12 ≤ 0⟹ -12 ≤ 0. This is true. Hence, the sign of the inequality in the interval [−4, 3] is positive. Let's take x = 4. Substituting this into the inequality, we get:
4^2 + 4 - 12 ≤ 0⟹ 12 ≤ 0.
This is false. Hence, the sign of the inequality in the interval [3, ∞) is negative. The following table summarizes the signs of the inequality in each interval. Interval(x + 4)(x - 3)x^2 + x - 12. Sign of x^2 + x - 12(−∞, −4](−)(−)+Negative[−4, 3](+)(−)Negative[3, ∞)(+)(+)Negative.
Step 6: From the above table, we see that the inequality is true only in the interval [−4, 3]. Therefore, the solution of the inequality x^2 ≤ −x + 12 is x ∈ [−4, 3].
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GIVING 25 POINTS FOR WHOEVER HELPS!! Find the solution to the system of equations, represented by the matrix shown below.
Answer:
A. x = 3; y = 5; z = 9
Step-by-step explanation:
7 4 -1 32
4 1 -3 -10
1 2 1 22
Switch rows 1 and 3.
1 2 1 22
4 1 -3 -10
7 4 -1 32
R2 --> -4 × R1 + R2
1 2 1 22
0 -7 -7 -98
7 4 -1 32
R3 --> -7 × R1 + R3
1 2 1 22
0 -7 -7 -98
0 -10 -8 -122
R2 --> -1/7 × R2
1 2 1 22
0 1 1 14
0 -10 -8 -122
R1 --> -2 × R2 + R1
1 0 -1 -6
0 1 1 14
0 -10 -8 -122
R3 --> 10 × R2 + R3
1 0 -1 -6
0 1 1 14
0 0 2 18
R3 --> 1/2 × R2
1 0 -1 -6
0 1 1 14
0 0 1 9
R2 --> -R3 + R2
1 0 -1 -6
0 1 0 5
0 0 1 9
R1 --> R1 + R3
1 0 0 3
0 1 0 5
0 0 1 9
Answer: A. x = 3; y = 5; z = 9
**Please, Solve the Math problem properly.**
Find all value (s) of x where the tangent line to the graph of f(x) = 3. x-6 3x-2 is perpendicular to the line y = -6+ 16
Given f(x) = 3x - 6 - 3x - 2 Find all the values of x where the tangent line to the graph of f(x) is perpendicular to the line y = -6x + 16. We have to find the derivative of the function f(x) first. f(x) = 3x - 6 - 3x - 2f(x) = -5x - 8f'(x) = -5Now, the slope of the tangent line to the graph of f(x) is -5.
Since the line y = -6x + 16 is perpendicular to the tangent line, its slope will be the negative reciprocal of -5.Now, the slope of the line y = -6x + 16 is -6. Therefore, -6 = -1/m, where m is the slope of the tangent line. m = 1/6 Now, the tangent line has a slope of 1/6 and passes through a point (x, f(x)). The equation of the tangent line is given by:y - (3x - 6 - 3x - 2) = (1/6)(x - x)y - 5x + 8 = (1/6)x - (1/6)x + C.
where C is the y-intercept. To find C, we use the point-slope form of the equation and substitute x = 1:y - 5(1) + 8 = (1/6)(1) + C= -17/6 + C= y - (3x - 6 - 3x - 2) = (1/6)(x - 1)y - 5x + 8 = (1/6)x - (1/6)(1) - 17/6y - 5x = (1/6)x - 23/6y = (1/6)x - 23/6 + 5xy = (1/6)x + (23 - 5x)/6 The slope of the tangent line is 1/6, and it passes through a point (x, f(x)).Therefore, the equation of the tangent line is:y = (1/6)x + (23 - 5x)/6To find all the values of x where the tangent line is perpendicular to the line y = -6x + 16, we need to solve the following equation:-6 = 1/6x + (23 - 5x)/6-36 = x + 23 - 5x-36 = -4x + 23-4x = -13x = 13/4Therefore, the only value of x where the tangent line is perpendicular to the line y = -6x + 16 is x = 13/4.Answer: x = 13/4
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1d
(d) Describe three situations in which a function fail to be differentiable. Support your answer with sketches.
A function f(x) is said to be non-differentiable at a given point x0 if it is discontinuous at x0 or it has a cusp at x0 or it has a vertical tangent at x0.
Below are the three situations in which a function fails to be differentiable:1. Discontinuity: A function is non-differentiable at a point where it has a sharp bend or a vertical tangent or where it is discontinuous. When a function has a point of discontinuity, it cannot have a derivative at that point. The derivative does not exist at discontinuous points.2. Cusps: A function is non-differentiable at the cusp point.
A cusp is a point where the slope of the function changes abruptly. The derivative of the function at a cusp point does not exist.3. Vertical Tangent: A function is non-differentiable at a point where it has a vertical tangent. A vertical tangent is a tangent that is parallel to the y-axis. The derivative of the function does not exist at points where the function has a vertical tangent. Below are the sketches that support the above three situations:Image of the Discontinuity function is given below: Image of the Cusp function is given below: Image of the Vertical Tangent function is given below:
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Find f such that f ′
(x)= x
5
,f(16)=55. f(x)= Find all antiderivatives of the following function. f(x)=e −15x
∫f(x)dx=
We are required to find f such that f ′(x) = x5, f(16)=55 and find all antiderivatives of the following function, f(x) = e^(-15x).
The required function is f(x) = (x^6/6) - 11453246068.5.
So, we can solve these two problems separately.
Solution:
I. Integration of f(x)
= e^(-15x):
Let ∫f(x) dx
= F(x)So, F'(x)
= f(x) = e^(-15x)
∴ F(x)
= ∫e^(-15x) dx
= (-1/15) * e^(-15x) + C
Where C is an arbitrary constant II.
Finding f such that
f ′(x)= x5, f(16)
=55
:Integrating the given function, we have f(x)
= (x^6/6) + C
Where C is a constant
.Now,
f(16) = 55
∴ (16^6/6) + C
= 55
∴ 68719476737/6 + C
= 55
∴ C
= 55 - 11453246123.5
= -11453246068.5
So, the required function is
f(x) = (x^6/6) - 11453246068.5.
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A brand of laptop has a lifetime that is normally distributed with a mean of 6 years and a standard deviation of 1.5 years. (i) What is the probability that a randomly chosen laptop will last more than 8 years? (ii) If the manufacturer wishes to guarantee the laptop for 5 years, what percentage of the laptops will they have to replace under the guarantee?
A brand of laptop has a lifetime that is normally distributed with a mean of 6 years and a standard deviation of 1.5 years. So,
(i) The probability that a randomly chosen laptop will last more than 8 years is approximately 90.88%.(ii) If the manufacturer wishes to guarantee the laptop for 5 years, they will have to replace approximately 25.14% of the laptops under the guarantee.Now, let's calculate these probabilities step by step:
(i) To find the probability that a randomly chosen laptop will last more than 8 years, we need to calculate the z-score first. The z-score measures the number of standard deviations a particular value is from the mean. It is calculated as:
z = (x - μ) / σ
where x is the value we are interested in, μ is the mean, and σ is the standard deviation.
In this case, we want to find the probability of the laptop lasting more than 8 years, so x = 8.
Plugging in the values, we get:
z = (8 - 6) / 1.5 = 2 / 1.5 ≈ 1.33
Next, we can use a standard normal distribution table or a calculator to find the probability corresponding to this z-score. The probability of the laptop lasting more than 8 years is the area under the normal distribution curve to the right of the z-score.
By looking up the z-score of 1.33 in a standard normal distribution table or using a calculator, we find that the probability is approximately 0.9088, or 90.88%.
(ii) To determine the percentage of laptops that will require replacement under the 5-year guarantee, we need to calculate the probability of a laptop failing before the 5-year mark.
Using the same formula as above, we calculate the z-score for x = 5:
z = (5 - 6) / 1.5 = -1 / 1.5 ≈ -0.67
Again, we can use a standard normal distribution table or a calculator to find the probability corresponding to this z-score. The probability of the laptop failing before 5 years is the area under the normal distribution curve to the left of the z-score.
By looking up the z-score of -0.67 in a standard normal distribution table or using a calculator, we find that the probability is approximately 0.2514, or 25.14%.
Therefore, the manufacturer will need to replace approximately 25.14% of the laptops under the 5-year guarantee.
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(2+ 2 pts) How many integer solutions are there to the equation x+y+z= 8 such that (i) x, y, z> 0? (ii) x, y, z ≥ 0?
(i)The three integer solutions are:(x, y, z) = (1, 1, 6), (1, 2, 5) and (1, 3, 4).
x, y, z > 0.To find the integer solutions to the equation `x+y+z= 8` such that `x, y, z> 0`, we can make use of the concept of the stars and bars, where n stars are placed in m bins, i.e., the variables `x, y and z` in our case, which are separated by `m - 1` bars and each bin has at least one star.
So, for the given equation `x+y+z= 8`, the number of integer solutions is 2^2 - 1 = 3.
We use 2^2 because we have 3 variables x, y and z, and we need 2 bars to separate them, and 1 less is subtracted to eliminate the case where any variable is equal to 0.
The three integer solutions are:(x, y, z) = (1, 1, 6), (1, 2, 5) and (1, 3, 4).
(ii)There are 45 integer solutions to the equation `x+y+z= 8` such that `x, y, z ≥ 0`.
x, y, z ≥ 0.Similarly, for the equation `x+y+z= 8` such that `x, y, z ≥ 0`, we can make use of the same concept of the stars and bars, where n stars are placed in m bins, i.e., the variables `x, y and z` in our case, which are separated by `m - 1` bars and any bin may have 0 stars.
So, for the given equation `x+y+z= 8`, the number of integer solutions is 10C2 = 45.
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Find the infinite sum of the following series ∑ k=1
[infinity]
3 k+1
(−1) k
The infinite sum of the infinite series is (3k + 1)/2
Finding the infinite sum of the infinite seriesFrom the question, we have the following parameters that can be used in our computation:
∑k=1[infinity] (3k + 1)(−1)k
From the above, we have the following
First term, a = 3k + 1
Common ratio, r = -1
The infinite sum of the series is then calculated as
[tex]Sum = \frac{a}{1 - r}[/tex]
substitute the known values in the above equation, so, we have the following representation
[tex]Sum = \frac{3k + 1}{1 + 1}[/tex]
Evaluate
Sum = (3k + 1)/2
Hence, the infinite sum of the infinite series is (3k + 1)/2
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Simplify the following by combining all constants and combining the \( a \) and \( b \) terms using exponential notation. \[ -2 a a a a a a b b b b b= \]
The given expression to simplify by combining all constants and combining the a and b terms using exponential notation is,-2 a a a a a a b b b b b For this expression, we can combine the constants and a terms using exponential notation in the following manner,-2 * (a⁶) * (b⁵)Therefore, the main answer of the given question is, -2 * (a⁶) * (b⁵).
We have to simplify the given expression by combining all constants and combining the a and b terms using exponential notation. The given expression is -2 a a a a a a b b b b b.In order to solve the expression, we need to simplify the constant terms and combine the a and b terms in exponential notation form.Constant terms are those that are multiplied by the variables and have a constant value. In this case, the constant is -2. Therefore, we only have one constant to simplify.For the a and b terms, we can see that the a variable is repeated six times, whereas the b variable is repeated five times. Hence, we can combine these variables using exponential notation by multiplying a⁶ with b⁵.So, the simplified form of the expression is -2 * (a⁶) * (b⁵). Therefore, this is the final answer.
The given expression -2 a a a a a a b b b b b is simplified by combining all constants and combining the a and b terms using exponential notation, which results in -2 * (a⁶) * (b⁵).
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If \( f(x)=\frac{5 x^{2} \tan x}{\sec x} \), find \[ f^{\prime}(x)=\frac{5 x(x+2 \tan (x))}{\sec (x)} \] Find \( f^{\prime}(3) \)
According to the question on evaluating [tex]\[f'(3) = \frac{5(3)(3 + 2(-0.14254654))}{3.44122254}\][/tex] we get [tex]\(f'(3)[/tex] [tex]\approx 14.21684169\).[/tex]
To find the derivative of the function [tex]\(f(x) = \frac{5x^2 \tan x}{\sec x}\)[/tex] , we can use the quotient rule and simplify the expression.
Using the quotient rule, the derivative [tex]\(f'(x)\)[/tex] is given by:
[tex]\[f'(x) = \frac{(\sec x) \cdot \frac{d}{dx}(5x^2 \tan x) - (5x^2 \tan x) \cdot \frac{d}{dx}(\sec x)}{(\sec x)^2}\][/tex]
To differentiate [tex]\(5x^2 \tan x\)[/tex], we can apply the product rule:
[tex]\[\frac{d}{dx}(5x^2 \tan x) = 5x^2 \frac{d}{dx}(\tan x) + \tan x \frac{d}{dx}(5x^2)\][/tex]
The derivative of tangent function is given by:
[tex]\[\frac{d}{dx}(\tan x) = \sec^2 x\][/tex]
Differentiating [tex]\(5x^2\)[/tex] gives:
[tex]\[\frac{d}{dx}(5x^2) = 10x\][/tex]
Substituting these derivatives back into the expression for [tex]\(f'(x)\):[/tex]
[tex]\[f'(x) = \frac{(\sec x) \cdot (5x^2 \sec^2 x + 10x \tan x) - (5x^2 \tan x) \cdot (\sec^2 x)}{(\sec x)^2}\][/tex]
Simplifying this expression, we get:
[tex]\[f'(x) = \frac{5x(x + 2\tan x)}{\sec x}\][/tex]
To find [tex]\(f'(3)\)[/tex], we substitute [tex]\(x = 3\)[/tex] into the derivative expression:
[tex]\[f'(3) = \frac{5(3)(3 + 2\tan 3)}{\sec 3}\][/tex]
To find [tex]\(f'(3)\)[/tex], we substitute [tex]\(x = 3\)[/tex] into the derivative expression:
[tex]\[f'(3) = \frac{5(3)(3 + 2\tan 3)}{\sec 3}\][/tex]
First, let's evaluate [tex]\(\tan 3\) and \(\sec 3\)[/tex] separately.
Using a calculator or trigonometric table, we find:
[tex]\(\tan 3 \approx -0.14254654\)[/tex] (rounded to eight decimal places)
[tex]\(\sec 3 \approx 3.44122254\)[/tex] (rounded to eight decimal places)
Now we substitute these values back into the derivative expression:
[tex]\[f'(3) = \frac{5(3)(3 + 2(-0.14254654))}{3.44122254}\][/tex]
Simplifying further:
[tex]\[f'(3) = \frac{5(3)(3 - 0.28509308)}{3.44122254}\][/tex]
[tex]\[f'(3) = \frac{5(3)(2.71490692)}{3.44122254}\][/tex]
[tex]\[f'(3) = \frac{48.8978606}{3.44122254}\][/tex]
[tex]\[f'(3) \approx 14.21684169\][/tex]
Therefore, [tex]\(f'(3)[/tex] [tex]\approx 14.21684169\).[/tex]
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Solve the initial value problem: y(x) = dy dx || -0.8 cos(y)' y(0) = K|4
The solution of the given initial value problem is x + ln|K/4 + 0.8sin(K/4)|
The given initial value problem is:y(x) = dy/dx - 0.8cos(y).....(1)y(0) = K/4
We are to find the solution of this initial value problem.
Step 1: Separating the variables:Separating the variables in equation (1), we get:dy / (y + 0.8cos(y)) = dx.....(2)
Step 2: Integrating both sides:Integrating both sides of equation (2), we get:∫ dy / (y + 0.8cos(y)) = ∫ dxOr, ln|y + 0.8sin(y)| = x + c_1 , where c_1 is the constant of integration.
Step 3: Applying initial condition:Given that y(0) = K/4, we have:ln|K/4 + 0.8sin(K/4)| = 0 + c_1Or, c_1 = ln|K/4 + 0.8sin(K/4)|
The solution of the given initial value problem is:ln|y + 0.8sin(y)| = x + ln|K/4 + 0.8sin(K/4)|
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Solve the following linear equations for the unknown x i) 2x - 1 = 5x + 11 ii) 4(3x + 1) = 7(x + 4) - 2(x + 5)
The solution to the equation is x = 4.67 (rounded to two decimal places).
i) 2x - 1 = 5x + 11To solve the linear equation for the unknown x, first, isolate the unknown term (x) to one side of the equation and the constant terms to the other side of the equation. To achieve this, we need to add the constant term -1 to both sides and simplify the equation. 2x - 1 + 1 = 5x + 11 + 1 2x = 5x + 12To isolate x, subtract 5x from both sides.2x - 5x = 5x - 5x + 12-3x = 12To get x alone, divide both sides by -3.-3x/-3 = 12/-3x = -4Therefore, the solution to the equation is x = -4.ii) 4(3x + 1) = 7(x + 4) - 2(x + 5)The equation has two pairs of brackets. First, simplify the brackets by multiplying out.4(3x) + 4(1) = 7(x) + 7(4) - 2(x) - 2(5)12x + 4 = 7x + 28 - 2x - 10Group the x terms and the constant terms.12x - 7x - 2x = 28 - 10 - 4Collect the x terms and constant terms12x - 9x = 14Simplify the equation3x = 14Divide by 3x = 4.67Therefore, the solution to the equation is x = 4.67 (rounded to two decimal places).
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Consider two populations for which H₁ = 32, 0₁ = 2, H₂= 27, and 0₂= 3. Suppose that two independent random samples of sizes n. 48 and n₂ = 57 are selected. Describe the approximate sampling distribution of x₁-x₂ (center, variability, and shape). What is the shape of the distribution? a. The distribution would be non-normal. b. The distribution is approximately normal. c. The shape cannot be determined. What is the mean of the distribution? (letter only) What is the standard deviation of the distribution? (Round your answer to three decimal places.)
The approximate sampling distribution of x₁ - x₂ has a mean of 5 and a standard deviation of approximately 0.492. The shape of the distribution can be considered approximately normal.
To describe the approximate sampling distribution of x₁ - x₂ (the difference between two sample means), we can use the following properties:
1. Center: The mean of the sampling distribution of x₁ - x₂ is equal to the difference between the population means, which is H₁ - H₂.
2. Variability: The standard deviation of the sampling distribution of x₁ - x₂ is obtained by the formula:
σ(x₁ - x₂) = sqrt((σ₁² / n₁) + (σ₂² / n₂))
where σ₁ and σ₂ are the population standard deviations, and n₁ and n₂ are the sample sizes.
3. Shape: The shape of the sampling distribution of x₁ - x₂ can be approximated as normal if the sample sizes are reasonably large (typically, n₁ ≥ 30 and n₂ ≥ 30) or if the populations are approximately normal.
We have:
H₁ = 32
σ₁ = 2
H₂ = 27
σ₂ = 3
n₁ = 48
n₂ = 57
The mean of the distribution is:
Mean = H₁ - H₂ = 32 - 27 = 5 (Answer: b)
The standard deviation of the distribution can be calculated using the formula:
σ(x₁ - x₂) = sqrt((σ₁² / n₁) + (σ₂² / n₂))
Substituting the values:
σ(x₁ - x₂) = sqrt((2² / 48) + (3² / 57))
= sqrt(0.0833 + 0.1586)
= sqrt(0.2419)
≈ 0.492 (Answer: 0.492)
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Given that: \( \cos \theta+\sin \theta=\frac{5}{4} \) where: \( 0
The value of theta that satisfies the equation is approximately 0.301 radians.
We have,
To solve the equation cos(theta) + sin(theta) = 5/4, where
0 < theta < π/2, we can use trigonometric identities and algebraic manipulation.
First, let's square both sides of the equation to eliminate the square root:
(cos(theta) + sin(theta))² = (5/4)²
Expanding the left side using the identity (a + b)² = a² + 2ab + b²:
cos²(theta) + 2cos(theta)sin(theta) + sin²(theta) = 25/16
Since cos²(theta) + sin²(theta) = 1 (trigonometric identity),
we can simplify further:
1 + 2cos(theta)sin(theta) = 25/16
Rearranging the equation:
2cos(theta)sin(theta) = 25/16 - 1
2cos(theta)sin(theta) = 9/16
Next, we can use the identity sin(2theta) = 2sin(theta)cos(theta):
sin(2theta) = 9/16
Now, we solve for 2theta:
2theta = arcsin(9/16)
Taking the arcsin of both sides gives us:
2theta = 0.601 radians (rounded to three decimal places)
Finally, divide by 2 to find theta:
theta = 0.601 / 2
theta = 0.301 radians (rounded to three decimal places)
Therefore,
The value of theta that satisfies the equation is approximately 0.301 radians.
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The complete question:
Given the equation cos(theta) + sin(theta) = 5/4, where 0 < theta < pi/2, determine the value of theta that satisfies the equation.
(20 points) Use the Laplace Transform to solve the following initial value problems (a,b). (a) \( y^{\prime \prime}+2 y^{\prime}+2 y=\delta(t-\pi) ; y(0)=1, y^{\prime}(0)=0 \)
The solution of initial value problem is y(t) = e^(-t)sin(t) + e^(-t)H(t-\pi), where H(t) is the Heaviside step function. To solve the initial value problem using the Laplace Transform, we first take the Laplace Transform of both sides of the given differential equation.
Let L{y(t)} = Y(s) denote the Laplace Transform of y(t).
Taking the Laplace Transform of the differential equation, we have s^2Y(s) - sy(0) - y'(0) + 2(sY(s) - y(0)) + 2Y(s) = e^(-\pi s).
Using the initial conditions y(0) = 1 and y'(0) = 0, we can simplify the equation to obtain (s^2 + 2s + 2)Y(s) - s - 1 + 2 = e^(-\pi s).
Rearranging the equation, we have (s^2 + 2s + 2)Y(s) = s + 3 - e^(-\pi s).
Solving for Y(s), we get Y(s) = (s + 3 - e^(-\pi s)) / (s^2 + 2s + 2).
Using partial fraction decomposition, we express Y(s) as Y(s) = A/(s+1) + (Bs + C)/(s^2 + 2s + 2).
Solving for the coefficients A, B, and C, we find A = 1, B = -1, and C = 2.
Thus, Y(s) = 1/(s+1) - (s - 2)/(s^2 + 2s + 2).
Taking the inverse Laplace Transform of Y(s), we obtain y(t) = e^(-t)sin(t) + e^(-t)H(t-\pi).
Therefore, the solution of the initial value problem is y(t) = e^(-t)sin(t) + e^(-t)H(t-\pi).
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If a seed is planted, it has a \( 75 \% \) chance of growing into a healthy plant. If 10 seeds are planted, what is the probability that exactly 4 don't grow?
The probability that exactly 4 seeds don't grow is approximately 0.250282 or about 25.03%.
To solve this problem, we will use the binomial probability formula. Let X be the number of seeds that don't grow, then X follows a binomial distribution with parameters n = 10 and p = 0.25. We want to find P(X = 4).
The binomial probability formula is:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where (n choose k) = n! / (k! * (n-k)!) is the binomial coefficient.
Substituting n = 10, p = 0.25, and k = 4, we get:
P(X = 4) = (10 choose 4) * 0.25^4 * (1-0.25)^(10-4)
P(X = 4) = (10! / (4! * 6!)) * 0.25^4 * 0.75^6
P(X = 4) = (10*9*8*7 / (4*3*2*1)) * 0.00390625 * 0.17850625
P(X = 4) ≈ 0.250282
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By applying the substitution t = tan² 0 to B(x, y)= I 25 (sin 0)2x-1 (cos 0)2y-1de, show that dt B(x, y) = √o (1+t)x+y tx-1
The substitution that has been applied to B(x, y) is dt B(x, y) = √(1+t) (1+t)x+y tx-1.
The substitution that has been applied to B(x, y) is t = tan² 0
The substitution for x and y using the trigonometric identities is given by,
x = sin² 0 / cos² 0 = tan² 0 …(1)
y = sin² 0 …(2)
Differentiating both sides of equation (1) with respect to θ, we get
dx / dθ = 2tan 0 sec² 0
Putting the values of x and y in B(x, y), we get
B(x, y) = I 25 (sin 0)² (sin² 0 / cos² 0)-1 (cos 0)² (sin² 0)-1 dθ
= I 25 (sin 0)² / cos² 0 * cos² 0 / sin² 0 dθ
= I 25 tan² 0 dθ
= 25
∫ t dt √1+t
Now, we need to find the value of dt B(x, y) in terms of t.
To find dt B(x, y), we use the chain rule of differentiation and get
dt B(x, y) = ∂B/∂x dx/dt + ∂B/∂y dy/dt
= 25(2 sin 0 cos 0 sin² 0 / cos³ 0) * 2 sin 0 cos 0 sin 0 cos 0 dθ
= 100 (sin 0)³ (cos 0)³ / cos⁴ 0 dθ
= 100 (sin 0)³ / cos 0 dθ
Now, putting the values of x, y, and t, we get
dt B(x, y) = 100 sin³ 0 / cos 0 dθ
= 100 sin³ θ / cos θ dθ
Using the identity 1+t = sec² 0 or cos² 0 = 1 / (1+t), we can rewrite the above integral as
∫ 100 sin³ θ / cos θ dθ
= ∫ (1+t) (1+t)½ dt
Substituting the limits, we get
∫ 100 sin³ θ / cos θ dθ
= ∫ √(1+t) (1+t)x+y tx-1 dt
Answer: dt B(x, y) = √(1+t) (1+t)x+y tx-1.
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Evaluate the infinite geometric series
1/2
3/4
1/4
the sum doenst exist
The given geometric series is: 1/2, 3/4, 9/16, 27/64, …The common ratio is 3/2
Since the common ratio is greater than 1, the series diverges.
A geometric series is a sequence of numbers in which each term is obtained by multiplying the preceding term by a fixed number.
The sum of the first n terms of a geometric series is given by the formula:
Sn = a(1 - rn) / (1 - r), where a is the first term, r is the common ratio, and n is the number of terms.
If the common ratio is greater than 1, the series is said to diverge to infinity, and the sum doesn’t exist.
If the common ratio is less than 1, the series converges to a finite value.
If the common ratio is equal to 1, the series either converges or diverges depending on the first term.
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