Answer: The point estimate of the population mean is 23. The margin of error for the confidence interval is 2.
Step-by-step explanation:
The confidence interval for the mean is of the form is given by :-
[tex]\overline{x}-\text{error}\leq\mu\leq\overline{x}+\text{error}[/tex]
Given : Lower bound: [tex]\overline{x}-\text{error}=21[/tex]
Upper bound :[tex]\overline{x}+\text{error}= 25[/tex]
Then ,the sum of lower and upper bound will be :-
[tex]2\overline{x}=21+25=46\\\\\Rightarrow\ \overline{x}=23[/tex]
Since [tex]\overline{x}-\text{error}=21[/tex]
[tex]\Rightarrow\ \text{Error}=\overline{x}-21=23-21=2[/tex]
Hence, The point estimate of the population mean is 23. The margin of error for the confidence interval is 2.
The point estimate of the population mean is calculated as the mid-point of the interval, which is 23. The margin of error for the confidence interval is calculated as the distance between the point estimate and either end of the interval, which is 2.
Explanation:To determine the point estimate of the population mean and the margin of error for the confidence interval, we first need to understand the form of a confidence interval. A confidence interval generally has the form (lower bound, upper bound) = (point estimate – margin of error, point estimate + margin of error).
In this case, with a given lower bound as 21 and upper bound as 25, we can perform a simple calculation to find the point estimate which is the mid-point of the interval. It's given by (lower bound + upper bound) / 2, which is (21+25)/2 = 23.
The margin of error for the confidence interval is calculated as the distance between the point estimate and either end of the interval. This is given by (upper bound - point estimate) or (point estimate - lower bound). It is therefore, (25-23) = 2 or (23-21) = 2. So, the point estimate of the population mean is 23 and the margin of error for the confidence interval is 2.
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Suppose a system has two modules, A and B, that function independently. Module A fails with probability 0.24 and Module B fails with probability 0.38, when the system is executed. When the system is executed, find the probability that.
(a) only Module A fails (b) only Module B fails. (c) both modules fail. (d) neither module fails. (e) What do the probabilities add up to?
Let's denote the events as follows:
- A: Module A fails
- B: Module B fails
Given probabilities:
[tex]\[ P(A) = 0.24 \][/tex]
[tex]\[ P(B) = 0.38 \][/tex]
We can use the addition rule of probability for mutually exclusive events to calculate the probabilities:
(a) Probability that only Module A fails:
[tex]\[ P(A \text{ only}) = P(A) = 0.24 \][/tex]
(b) Probability that only Module B fails:
[tex]\[ P(B \text{ only}) = P(B) = 0.38 \][/tex]
(c) Probability that both modules fail:
[tex]\[ P(\text{both}) = P(A \cap B) = P(A) \times P(B) = 0.24 \times 0.38 = 0.0912 \][/tex]
(d) Probability that neither module fails:
[tex]\[ P(\text{neither}) = 1 - (P(A) + P(B) - P(A \cap B)) \][/tex]
[tex]\[ = 1 - (0.24 + 0.38 - 0.0912) \][/tex]
[tex]\[ = 1 - 0.5288 \][/tex]
[tex]\[ = 0.4712 \][/tex]
(e) The probabilities should add up to 1, representing the total probability of all possible outcomes:
[tex]\[ P(A \text{ only}) + P(B \text{ only}) + P(\text{both}) + P(\text{neither}) = 0.24 + 0.38 + 0.0912 + 0.4712 = 1 \][/tex]
So, the probabilities add up to 1, as expected.
a. The probability that only Module A fails is 0.1488
b. The probability that only Module B fails is 0.2888
c. The probability that both modules fail is 0.0912
d. The probability that neither module fails is 0.4712
e. The probabilities add up to is 1
Let's denote the events as follows:
Event A: Module A fails
Event B: Module B fails
Given probabilities:
P(A) = 0.24
P(B) = 0.38
(a) To find the probability that only Module A fails:
[tex]\[ P(A \cap \neg B) = P(A) \times P(\neg B) \][/tex]
To calculate [tex]\( P(\neg B) \)[/tex], we subtract the probability of B from 1:
[tex]\[ P(\neg B) = 1 - P(B) = 1 - 0.38 = 0.62 \][/tex]
[tex]\[ P(A \cap \neg B) = 0.24 \times 0.62 = 0.1488 \][/tex]
(b) Similarly, to find the probability that only Module B fails:
[tex]\[ P(\neg A \cap B) = P(\neg A) \times P(B) \][/tex]
[tex]\[ P(\neg A) = 1 - P(A) = 1 - 0.24 = 0.76 \][/tex]
[tex]\[ P(\neg A \cap B) = 0.76 \times 0.38 = 0.2888 \][/tex]
(c) To find the probability that both modules fail:
[tex]\[ P(A \cap B) = P(A) \times P(B) = 0.24 \times 0.38 = 0.0912 \][/tex]
(d) To find the probability that neither module fails:
[tex]\[ P(\neg A \cap \neg B) = P(\neg A) \times P(\neg B) \][/tex]
[tex]\[ P(\neg A \cap \neg B) = 0.76 \times 0.62 = 0.4712 \][/tex]
(e) The probabilities add up to :
[tex]\[ P(A \cap \neg B) + P(\neg A \cap B) + P(A \cap B) + P(\neg A \cap \neg B) = 0.1488 + 0.2888 + 0.0912 + 0.4712 = 1 \][/tex]
The Pew Internet and American Life Project finds that 95% of teenagers (12–17) use the Internet and that 81% of online teens use some kind of social media. Of online teens who use some kind of social media, 91% have posted a photo of themselves. STATE: What percent of all teens are online, use social media, and have posted a photo of themselves?
The percent of all teens are online, use social media, and have posted a photo of themselves are:
70% (approx)
Step-by-step explanation:It is given that:
95% of teenagers use the Internet this means that the teens are online.
This means that 95%=0.95.
and 81% of online teens use some kind of social media.
i.e. 81%=0.81
and Of online teens who use some kind of social media, 91% have posted a photo of themselves.
i.e. 91%=0.91
Now we are asked to find the percent of teens who are online, use social media, and have posted a photo of themselves
i.e. The percent is: (0.95×0.81×0.91)×100
= 70.0245%
which is approximately equal to 70%
Answer:
The percent of all teens are online, use social media, and have posted a photo of themselves 70% (approx)
Step-by-step explanation:
You want to endow a scholarship that will pay $10,000 per year forever, starting one year from now. If the school’s endowment discount rate is 7%, what amount must you donate to endow the scholarship? 13. How would your answer to Problem 12 change if you endow it now, but it makes the first award to a student 10 years from today?
Answer:
12. $142,857.14
13. $77,704.82 . . . it is changed by the accumulated interest on the amount
Step-by-step explanation:
12. You want one year's interest on the endowment to be equal to $10,000. The principal (P) can be found by ...
I = Prt
10,000 = P·0.07·1
10,000/0.07 = P ≈ 142,857.14
The endowment must be $142,857.14 to pay $10,000 in interest annually forever.
__
13. If the first award is in 10 years, we want the above amount to be the value of an account that has paid 7% interest compounded annually for 9 years. (The first award is 1 year after this amount is achieved.) Then we want the principal (P) to be ...
142,857.14 = P·(1 +0.07)^9
142,57.14/1.07^9 = P = 77,704.82
The endowment needs to be only $77,704.82 if the first award is made 10 years after the endowment date.
The initial amount required for endowment yielding $10,000 annually in perpetuity with a discount rate of 7% is $142,857.14. If the first payment is postponed to 10 years later, the present value of this amount is $72,975.85.
Explanation:The question pertains to figuring out the initial amount needed to fund an endowment that would yield $10,000 annually, indefinitely, given a discount rate of 7%. This is a case for the use of perpetuity, a financial concept in which an infinite amount of identical cash flows occur continually. The formula for perpetuity is: Perpetuity = Cash flow / Discount rate. Thus, in this case, the amount to endow is $10,000 / 0.07 = $142,857.14.
For the second part of the question, if the first award will be given 10 years from today, the present value of the perpetuity needs to be factored in. The present value of a perpetuity starting at a future point is: Present Value of Perpetuity = Perpetuity / (1 + r)^n where 'r' is the discount rate and 'n' is the number of periods before the perpetuity starts. Here, it will be $142,857.14 / (1 + 7%)^10 = $72,975.85
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The mean and the standard deviation of a sampled population are, respectively, 113.9 and 32.1. Find the mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64. Round your answers to one decimal place.
Answer: The mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 is
[tex]113.9\text{ and }4.0[/tex] respectively.
Step-by-step explanation:
Given : The mean of sampled population : [tex]\mu = 113.9[/tex]
The standard deviation of sampled population : [tex]\sigma = 32.1[/tex]
We know that the mean and standard deviation of the sampling distribution of the sample mean for samples of size n is given by :_
[tex]\mu_s=\mu\\\\\sigma_s=\dfrac{\sigma}{\sqrt{n}}[/tex]
Now, the mean and standard deviation of the sampling distribution of the sample mean for samples of size n=64 will be :-
[tex]\mu_s=113.9\\\\\sigma_s=\dfrac{32.1}{\sqrt{64}}=4.0125\approx4.0[/tex]
The mean of the sample is 113.9, and the standard deviation of the sample is 4.0
The given parameters are:
[tex]\mu = 113.9[/tex] --- the population mean
[tex]\sigma = 32.1[/tex] --- the population standard deviation
[tex]n = 64[/tex] --- the sample size
The sample mean is calculated as:
[tex]\bar x = \mu[/tex]
So, we have:
[tex]\bar x = 113.9[/tex]
The sample standard deviation is calculated as:
[tex]\sigma_x = \frac{\sigma}{\sqrt n}[/tex]
This gives
[tex]\sigma_x = \frac{32.1}{\sqrt {64}}[/tex]
[tex]\sigma_x \approx 4.0[/tex]
Hence, the mean of the sample is 113.9, and the standard deviation of the sample is 4.0
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A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 1000 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions?
Answer:
The largest area is 125000 m²
The dimensions of the farmland are 250 m and 500 m
Step-by-step explanation:
* Lets pick the information from the problem
- The farmland is shaped a rectangle
- The farmland will be bounded on one side by a river
- The other three sides are bounded by a single-strand electric fence
- The length of wire is 1000 m
- Lets consider the width of the rectangle is x and the length is y
- The side which will be bounded by the river is y
∴ The perimeter of the farmland which will be bounded by the electric
fence = x + x + y = 2x + y
- We will use the wire to fence the farmland
∵ The length of the wire is 1000 m
∵ The perimeter of the farmland is equal to the length of the wire
∴ 2x + y = 1000
- Lets find y in term of x
∵ 2x + y = 1000 ⇒ subtract 2x from both sides
∴ y = 1000 - 2x
- Now lets find the area can enclose by the wire
∵ The area of the rectangle = length × width
∵ The width of the farmland is x and its length is y
∴ The area of the farmland (A) = x × y = xy ⇒ (2)
- Use equation (1) to substitute the value of y in equation (2)
∴ A = x (1000 - 2x) ⇒ simplify
∴ A = 1000 x - 2 x²
- To find the maximum area we will differentiate A with respect to x
and equate the answer by zero to find the value of x which will make
the enclosed area largest
* Lets revise the rule of differentiation
- If y = ax^n, then dy/dx = a(n) x^(n-1)
- If y = ax, then dy/dx = a
- If y = a, then dy/dx = 0 , where a is a constant
∵ A = 1000 x - 2 x² ⇒ (3)
- Differentiate A with respect to x using the rules above
∴ dA/dx = 1000 - 2(2) x^(2-1)
∴ dA/dx = 1000 - 4x
- Put dA/dx = 0 to find the value of x
∵ 1000 - 4x = 0 ⇒ add 4x to both sides
∴ 1000 = 4x ⇒ divide both sides by 4
∴ 250 = x
∴ The value of x is 250
- Lets substitute this value in equation 3 to find the largest area
∵ A = 1000 x - 2 x²
∴ A = 1000 (250) - 2(250)² = 125000 m²
* The largest area is 125000 m²
∵ The width of the farmland is x
∵ x = 250
∴ The width of the farmland = 250 m
- Substitute the value of x in the equation (1) to find y
∵ y = 1000 - 2x
∵ x = 250
∴ y = 1000 - 2(250) = 1000 - 500 = 500
∵ The length of the farm lend is y
∴ The length of the farm land = 500 m
* The dimensions of the farmland are 250 m and 500 m
To get the maximum area from a rectangular farmland bounded on one side by a river and on the other three sides by a single-strand electric fence with 1000 meters of wire, the dimensions should be 250m x 500m yielding a maximum area of 125,000 square meters.
Explanation:This problem can be approached as a classic calculus maximization problem. The scenario mentioned in your question is about maximizing the area of a rectangle with a constant perimeter, this occurs when the rectangle is a square.
The three sides of your plot will consume the 1000 m of wire, if the length of each adjacent sides are x and y (where x is the length of the fence along the river, and y is the length of the other two fences), then
y + 2x = 1000
Substituting y from this equation into the area equation, A = xy(which results in A = x(1000-2x)), you can differentiate this equation with regards to x and set the derivative equal to zero to find the x-value that will give the maximum area. This happens when x = 250m, y = 500m then the maximum area is 125,000 m².
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Which of the following is a solution of 4x + 5y ≤ 20?
A. (5, 7)
B. (1, 1)
C. (0, 5)
D. (8, 0)
Answer:
B.
Step-by-step explanation:
Plug in and see:
Let's try the first (x,y)=(5,7)
4(5)+5(7) <=20
20+35<=20 not true so not A
Let's try (1,1) for (x,y)
4(1)+5(1)<=20
4+5<=20 that is true so choice B looks good
For this case we must replace each of the points and verify if the inequality is met:
Point: (5,7)
[tex]4 (5) +5 (7) \leq20\\20 + 35 \leq20[/tex]
It is not fulfilled!
Point: (1,1)
[tex]4 (1) +5 (1) \leq20\\9 \leq20[/tex]
Is fulfilled!
Point: (0,5)
[tex]4 (0) +5 (5) \leq20\\0 + 25 \leq20[/tex]
It is not fulfilled!
Point: (8,0)
[tex]4 (8) +5 (0) \leq20\\32 \leq20[/tex]
It is not fulfilled!
Answer:
(1,1)
identify the type of data that would be used to describe a response (A. Quantitative discrete B. Quantitative continuous, or C. Qualitative), and give an example of the data. (a) Number of tickets sold to a concert (b) Percent of body fat (c) Favorite baseball team (d) Time in line to buy groceries (e) Number of students enrolled at Evergreen Valley College (f) Most-watched television show (g) Brand of toothpaste (h) Distance to the closest movie theatre (i) Age of executives in Fortune 500 companies (j) Number of competing computer spreadsheet software packages.
Answer:
a) Quantitative discrete
b) Quantitative continuous
c) Qualitative
d) Quantitative continuous
e) Quantitative discrete
f) Qualitative
g) Qualitative
h) Quantitative continuous
i) Quantitative continuous
j) Quantitative discrete
Step-by-step explanation:
Okay, so an easy way to remember the difference between qualitative data and quantitative data:
Quantitative is measurable, it measures "QUANTITY."
Qualitative is categorical, meaning it has numeric value.
Now what about the difference between quantitative discrete and quantitative continuous.
Quantitative discrete there is NO IN BETWEEN. It counts things as whole numbers. The example would be your problem:
Number of tickets sold to a concert. Tickets are sold WHOLE, because you can't sell half a ticket.
Quantitative continuous has numbers in between, specifically, there could be decimals or fractions in between. Like in our problem:
Distance to the closest movie theater.
It can be 10.5m, 1.6 km. It does not necessarily have to be EXACTLY 1 Km away or 10 m away. So this is a continuous.
The type of data (Quantitative discrete, Quantitative continuous, or Qualitative) depends on whether a response involves counting, measuring, or categorizing. Examples were given to illustrate these different types of data, from counting tickets (Quantitative discrete) to measuring time (Quantitative continuous), to choosing a preferred product (Qualitative).
Explanation:The type of data used to describe a response differs depending on the nature of the response. Let's identify them:
(a) Number of tickets sold to a concert - Quantitative discrete (because we count the number of tickets)(b) Percent of body fat - Quantitative continuous (since it involves a measurement that can fall anywhere within a certain range)(c) Favorite baseball team - Qualitative (this is categorical data)(d) Time in line to buy groceries - Quantitative continuous (it can be measured precisely)(e) Number of students enrolled at Evergreen Valley College - Quantitative discrete (because students can be counted)(f) Most-watched television show - Qualitative (it's a category, not a numerical value)(g) Brand of toothpaste - Qualitative (it's categorical data)(h) Distance to the closest movie theatre - Quantitative continuous (since it's a measurable distance)(i) Age of executives in Fortune 500 companies - Quantitative continuous (age can fall anywhere within a certain range)(j) Number of competing computer spreadsheet software packages - Quantitative discrete (since we count the number of software packages).Learn more about Data Types here:https://brainly.com/question/37915128
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Factor the higher degree polynomial
5y^4 + 11y^2 + 2
[tex]5y^4 + 11y^2 + 2=\\5y^4+10y^2+y^2+2=\\5y^2(y^2+2)+1(y^2+2)=\\(5y^2+1)(y^2+2)[/tex]
For this case we must factor the following expression:
[tex]5y ^ 4 + 11y ^ 2 + 2[/tex]
We rewrite[tex]y^ 4[/tex] as[tex](y^ 2) ^ 2[/tex]:
[tex]5 (y ^ 2) ^ 2 + 11y ^ 2 + 2[/tex]
We make a change of variable:
[tex]u = y ^ 2[/tex]
So, we have:
[tex]5u ^ 2 + 11u + 2[/tex]
we rewrite the term of the medium as a sum of two terms whose product is 5 * 2 = 10 and whose sum is 11. Then:[tex]5u ^ 2 + (1 + 10) u + 2\\5u ^ 2 + u + 10u + 2[/tex]
We factor the highest common denominator of each group:
[tex]u (5u + 1) +2 (5u + 1)[/tex]
We factor [tex](5u + 1):[/tex]
[tex](5u + 1) (u + 2)[/tex]
Returning the change:
[tex](5y ^ 2 + 1) (y ^ 2 + 2)[/tex]
ANswer:
[tex](5y ^ 2 + 1) (y ^ 2 + 2)[/tex]
A card is selected at random from a standard deck of playing cards.
Compute the probability that the card is a 7. (Enter the probability as a fraction.)
[tex]|\Omega|=52\\|A|=4\\\\P(A)=\dfrac{4}{52}=\dfrac{1}{13}[/tex]
The probability of getting the card of 7 in a deck will be 1 / 13.
What is probability?Probability is defined as the ratio of the number of favourable outcomes to the total number of outcomes in other words the probability is the number that shows the happening of the event.
Probability = Number of favourable outcomes / Number of sample
Given that a card is selected at random from a standard deck of playing cards. The probability of getting a number 7 will be,
There are four 7 cards in the deck of 52 cards.
Number of favourable outcomes = 4
Number of sample = 52
The probability is,
P = 4 / 52
P = 1 / 13
Therefore, the probability of getting the card of 7 in a deck will be 1 / 13.
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Question 6 3 pts Joshua's diastolic blood pressure reading had been 80. After several weeks of medication, his reading is now 63. By what percent did he reduce his blood pressure? Round to two decimals.
Answer: There is 21.25% of reduction in blood pressure.
Step-by-step explanation:
Since we have given that
Diastolic blood pressure reading previously = 80
Diastolic blood pressure reading now = 63
Decrement is given by
[tex]80-63\\\\=17[/tex]
So, percentage he reduced his blood pressure is given by
[tex]\dfrac{17}{80}\times 100\\\\=\dfrac{1700}{80}\\\\=21.25\%[/tex]
Hence, there is 21.25% of reduction in blood pressure.
i need geometry help pleas will give brainliest
Area of the shaded segment=
6π-9√3un^2
12π-9√3un^2
9π-9√3un^2
Answer:
6π - 9√3 unit^2
Step-by-step explanation:
Area of sector
= 1/6(π (6)^2
= 6π
Area of triangle = √3/4 (6)^2 = 9√3
Area of the shaded segment = Area of sector - Area of triangle
= 6π - 9√3 unit^2
to the risk of sounding redundant.
[tex]\bf \textit{area of a segment of a circle}\\\\ A=\cfrac{r^2}{2}\left(\cfrac{\pi \theta }{180}-sin(\theta ) \right)~~ \begin{cases} r=&radius\\ \theta =&angle~in\\ °rees\\ \cline{1-2} r=&6\\ \theta =&60 \end{cases}\implies A=\cfrac{6^2}{2}\left(\cfrac{\pi 60}{180}-sin(60^o ) \right) \\\\\\ A=18\left( \cfrac{\pi }{3}-\cfrac{\sqrt{3}}{2} \right)\implies \boxed{A=6\pi -9\sqrt{3}}\implies \implies A\approx 3.26[/tex]
Use the ratio test to determine whether ∑n=14∞n+2n! converges or diverges. (a) Find the ratio of successive terms. Write your answer as a fully simplified fraction. For n≥14, limn→∞∣∣∣an+1an∣∣∣=limn→∞.
Answer:
The sum [tex]\displaystyle \sum^{\infty}_{n = 14} n + 2n![/tex] diverges ∵ of the Ratio Test.
General Formulas and Concepts:
Calculus
Limits
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]Special Limit Rule [Coefficient Power Method]: [tex]\displaystyle \lim_{x \to \pm \infty} \frac{ax^n}{bx^n} = \frac{a}{b}[/tex]Series Convergence Tests
Ratio Test: [tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{a_{n + 1}}{a_n} \bigg|[/tex]Step-by-step explanation:
Step 1: Define
Identify.
[tex]\displaystyle \sum^{\infty}_{n = 14} n + 2n![/tex]
Step 2: Find Convergence
[Series] Define: [tex]\displaystyle a_n = n + 2n![/tex][Series] Set up [Ratio Test]: [tex]\displaystyle \sum^{\infty}_{n = 14} n + 2n! \rightarrow \lim_{n \to \infty} \bigg| \frac{n + 1 + 2(n + 1)!}{n + 2n!} \bigg|[/tex][Ratio Test] Evaluate Limit [Coefficient Power Method]: [tex]\displaystyle \lim_{n \to \infty} \bigg| \frac{n + 1 + 2(n + 1)!}{n + 2n!} \bigg| = \infty[/tex][Ratio Test] Define conclusiveness: [tex]\displaystyle \infty > 1[/tex]Since infinity is greater than 1, the Ratio Test defines the sum [tex]\displaystyle \sum^{\infty}_{n = 14} n + 2n![/tex] to be divergent.
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Topic: AP Calculus BC (Calculus I + II)
Unit: Taylor Series
Which number is irrational?
A. 0.14
B.1/3
C. Square root 4
D. Square root 6
Answer:
[tex]\sqrt{6}[/tex]
Step-by-step explanation:
Let's define irrational by stating what rational means first, and the use the process of elimination.
A rational number is one that can be expressed as a ratio. When dividing the numerator of this ratio by the denominator, we will either get a whole number, a decimal that repeats a pattern, or we will get a decimal that terminates.
.14 terminates, so it is rational
1/3 divides to .333333333333333333333333333333333333333 indefinitely, so it is rational
square root of 4 is 2, a whole number, so it is rational
the square root of 6, to 9 decimal places, is 2.449489743... and it goes on without ending and without repeating. So this is the only irrational number of the bunch.
Answer:
D
Step-by-step explanation:
How to integrate with steps:
(4x2-6)/(x+5)(x-2)(3x-1)
[tex]\displaystyle\int\frac{4x^2-6}{(x+5)(x-2)(3x-1)}\,\mathrm dx[/tex]
You have a rational expression whose numerator's degree is smaller than the denominator's. This tells you you should consider a partial fraction decomposition. We want to rewrite the integrand in the form
[tex]\dfrac{4x^2-6}{(x+5)(x-2)(3x-1)}=\dfrac a{x+5}+\dfrac b{x-2}+\dfrac c{3x-1}[/tex]
[tex]\implies4x^2-6=a(x-2)(3x-1)+b(x+5)(3x-1)+c(x+5)(x-2)[/tex]
You can use the "cover-up" method here to easily solve for [tex]a,b,c[/tex]. It involves fixing a value of [tex]x[/tex] to make 2 of the 3 terms on the right side disappear and leaving a simple algebraic equation to solve for the remaining one.
If [tex]x=-5[/tex], then [tex]94=112a\implies a=\dfrac{47}{56}[/tex]If [tex]x=2[/tex], then [tex]10=35b\implies b=\dfrac27[/tex]If [tex]x=\dfrac13[/tex], then [tex]-\dfrac{50}9=-\dfrac{80}9c\implies c=\dfrac58[/tex]So the integral we want to compute is the same as
[tex]\displaystyle\frac{47}{56}\int\frac{\mathrm dx}{x+5}+\frac{10}{35}\int\frac{\mathrm dx}{x-2}+\frac58\int\frac{\mathrm dx}{3x-1}[/tex]
and each integral here is trivial. We end up with
[tex]\displaystyle\int\frac{4x^2-6}{(x+5)(x-2)(3x-1)}\,\mathrm dx=\frac{47}{56}\ln|x+5|+\frac27\ln|x-2|+\frac5{24}\ln|3x-1|+C[/tex]
which can be condensed as
[tex]\ln\left|(x+5)^{47/56}(x-2)^{2/7}(3x-1)^{5/24}\right|+C[/tex]
2007 Federal Income Tax Table Single: Over But not over The tax is $0 $7,825 10% of the amount over $0 $7,825 $31,850 $788 + 15% of the amount over $7,825 $31,850 $77,100 $4,386 + 25% of the amount over $31,850 $77,100 $160,850 $15,699 + 28% of the amount over $77,100 $160,850 $349,700 $39,149 + 33% of the amount over $160,850 $349,700 And Over $101,469 + 35% of the amount over $349,700 Mr. Profit had a taxable income of $35,000. He figured his tax from the table above. 1. Find his earned income level. 2. Enter the base amount. = $ 3. Find the amount over $ = $ 4. Multiply line 3 by % = $ 5. Add Lines 2 and 4 = $ 6. Compute his monthly withholding would be = $
The total tax comes to $5,173.50, with a monthly withholding of approximately $431.13.
To calculate Mr. Profit's federal income tax based on a taxable income of $35,000 in 2007, we need to find the correct tax bracket and perform a series of calculations:
Earned income level: Based on the provided information, an income of $35,000 falls into the third bracket (over $31,850 but not over $77,100).Enter the base amount: The base amount for this bracket is $4,386.Find the amount over: The amount over $31,850 is $35,000 - $31,850, which equals $3,150.Multiply line 3 by the percentage: Multiply the amount over $31,850 by the rate of 25%, which is $3,150 * 25% = $787.50.Add lines 2 and 4: Adding the base amount to the above calculation gives us the total tax: $4,386 + $787.50 = $5,173.50.Compute his monthly withholding: To find the monthly withholding amount, we divide the total tax by 12 months: $5,173.50 / 12 = $431.125.Therefore, Mr. Profit's federal income tax for 2007 would be $5,173.50, and his monthly withholding would be approximately $431.13.
Mr. Profit's earned income level is $35,000. The base tax is $4,386, with an additional tax of $787.50 for amounts over $31,850. Monthly withholding comes to approximately $431.13.
To calculate Mr. Profit's tax liability based on his taxable income, we need to identify which tax bracket his income falls into and then compute the tax accordingly.
Given that Mr. Profit has a taxable income of $35,000, he falls into the 25% tax bracket, where the taxable income is over $31,850 but not over $77,100.
The tax table for this bracket is:
Base tax for incomes from $7,825 to $31,850: $788
The marginal tax rate for income over $31,850: 25%
Let's compute the tax following the provided steps.
Earned Income Level
Mr. Profit's earned income level is his taxable income:
Earned Income Level = $35,000
Base Amount
The base tax amount for his tax bracket:
Base Amount = $4,386
Amount Over
The amount over the lower bound of his tax bracket:
Amount Over = $35,000 - $31,850 = $3,150
Multiply by Percentage
Multiply the amount over by the tax rate for his bracket (25%):
Multiply Line 3 by 25% = $3,150 \times 0.25 = $787.50
Add Base and Multiplication Result
Add the base amount and the calculated tax from step 4:
Add Lines 2 and 4 = $4,386 + $787.50 = $5,173.50
Compute Monthly Withholding
Given that there are 12 months in a year, compute the monthly withholding:
Compute monthly withholding [tex]= $5,173.50 \div 12 \approx $431.13[/tex]
Thus, rounding to two decimal places, we have :
Earned Income Level: $35,000
Base Amount: $4,386
Amount Over: $3,150
Multiply Line 3 by Percentage: $787.50
Add Lines 2 and 4: $5,173.50
Compute monthly withholding: $431.13
The complete question is : 2007 Federal Income Tax Table Single: Over But not over The tax is $0 $7,825 10% of the amount over $0 $7,825 $31,850 $788 + 15% of the amount over $7,825 $31,850 $77,100 $4,386 + 25% of the amount over $31,850 $77,100 $160,850 $15,699 + 28% of the amount over $77,100 $160,850 $349,700 $39,149 + 33% of the amount over $160,850 $349,700 And Over $101,469 + 35% of the amount over $349,700 Mr. Profit had a taxable income of $35,000. He figured his tax from the table above. 1. Find his earned income level. 2. Enter the base amount. = $ 3. Find the amount over $ = $ 4. Multiply line 3 by % = $ 5. Add Lines 2 and 4 = $ 6. Compute his monthly withholding would be = $
in what form is the following linear equation written y=9x+2
The linear equation can be written as; 9x - y = -2
What is a linear equation?A linear equation is an equation that has the variable of the highest power of 1. The standard form of a linear equation is of the form Ax + B = 0.
We are given the linear equation as;
y = 9x + 2
-9x + y = 2
A cannot be a negative:
-1(-9x + y = 2)
9x - y = -2
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Write the Ratio 32 :42 as a Fraction in Simplest Form
Divide by 2 for both of the numbers
32/2, 42/2
16/21
Answer is 16/21
Answer:
Divide both numbers by 2
32÷2 ,42÷2
Or, 16,21
answer is 16,21
Their perimeter of a triangle is 82' one side of the triangle is 2 times the 2nd side the 3rd side is 2' longer than the 2nd side find the length of each side
Answer:
side 1: 40; side 2: 20; side 3: 22
Step-by-step explanation:
It appears that sides 1 and 3 are based on side 2. So side 2 is our "control" and we'll call it x.
If side 1 is 2 times side 2, then
side 1 = 2x.
If side 3 is 2 feet longer than side 2, then
side 3 = x + 2
We are given the perimeter value. The perimeter of any shape is a measure around the outside of the shape. Since a triangle has 3 sides, we add them together and set their sum equal to the perimeter we were given. That looks like this:
x + 2x + x + 2 = 82
Combine like terms to get 4x + 2 = 82
Solve for x to get 4x = 80 and x = 20. That means that side 2 measures 20 feet; side 1 measures twice that, so side 1 = 40 feet, and side 3 measures 2 more than side 2, so side 3 measures 22. Let's add them all together and make sure the numbers work:
20 + 40 + 22 = 82
It works! So we're all done!
Which is an equation of a circle with center ( -5, -7) that passes through the point ( 0, 0 ) brainly
Answer:
(x+5)^2+(y+7)^2=74
Step-by-step explanation:
So we are given this (x+5)^2+(y+7)^2=r^2
We need to find r so just plug in your point (0,0) for (x,y) and solve for r :)
(0+5)^2+(0+7)^2=r^2
25+49=r^2
74=r^2
So the answer is (x+5)^2+(y+7)^2=74
Answer:
(x+5)^2+(y+7)^2=74
Step-by-step explanation:
Solve the following problem. PV=$24,122; n = 70; i = 0.024; PMT ?; PMT = $ (Round to two decimal places.)
Answer:
The payment per period is $714.82.
Step-by-step explanation:
Given information: PV=$24,122; n = 70; i = 0.024
The formula for payment per period is
[tex]PMT=\frac{PV\times i}{1-(1+i)^{-n}}[/tex]
Substitute PV=$24,122; n = 70; i = 0.024 in the above formula.
[tex]PMT=\frac{24122\times 0.024}{1-(1+0.024)^{-70}}[/tex]
[tex]PMT=\frac{578.928}{0.80989084337}[/tex]
[tex]PMT=714.822256282[/tex]
[tex]PMT\approx 714.82[/tex]
Therefore the payment per period is $714.82.
Fast! Which of the following are characteristics of the graph of the square root
parent function? Check all that apply.
Answer:
Only choices B and C are correct.
Step-by-step explanation:
The square root parent function is [tex]f(x)= \sqrt{x}.[/tex]
The function [tex]f(x)[/tex] is not a linear function, therefore it's graph cannot be a straight line. This rules out choice A.
The function [tex]f(x)[/tex] is defined at [tex]x=0[/tex] because [tex]f(0)= \sqrt{0} =0[/tex]. This means that choice B is correct.
The function [tex]f(x)[/tex] is only real for values [tex]x\geq 0[/tex], because negative values of [tex]x[/tex] give complex values for [tex]f(x)[/tex]. This means that choice C is correct.
The function [tex]f(x)[/tex] can take only positive values which means it is confined to only the I quadrant, and is not defined in quadrants II, III, and IV. This rules out Choice D.
Therefore only choices B and C are correct.
Answer:
B & C
Step-by-step explanation:
B and C are the answers
Evaluate the function at the given value of the independent variable. Simplify the results. (If an answer is undefined, enter UNDEFINED.) f(x) = x3 − 49 x f(x) − f(7) x − 7 =
To evaluate the function f(x) = 5x² + 7 at a given value, substitute that value into the function and simplify. The resulting answer is the function's value at that point. For the difference quotient f(x) - f(7)/(x - 7), the value is undefined if x = 7.
To evaluate the function f(x) = 5x² + 7 at a given value of the independent variable and simplify the results, we simply plug in the value of the independent variable into the function. For example, if we want to evaluate f at x = 3, we would calculate f(3) = 5(3)² + 7 = 5(9) + 7 = 45 + 7 = 52. Thus, the evaluated function at x = 3 is 52.
If we were to find the difference quotient f(x) - f(7)/(x - 7), we would need to plug in a value for x that is not 7, as the expression is undefined for x = 7. For any other value of x, we substitute x into the function, subtract f(7), and divide by (x - 7). If x = 7, the result is UNDEFINED.
Solve the following system of equations
3x - 2y =55
-2x - 3y = 14
Answer:
The solution is:
[tex](\frac{137}{13}, -\frac{152}{13})[/tex]
Step-by-step explanation:
We have the following equations
[tex]3x - 2y =55[/tex]
[tex]-2x - 3y = 14[/tex]
To solve the system multiply by [tex]\frac{3}{2}[/tex] the second equation and add it to the first equation
[tex]-2*\frac{3}{2}x - 3\frac{3}{2}y = 14\frac{3}{2}[/tex]
[tex]-3x - \frac{9}{2}y = 21[/tex]
[tex]3x - 2y =55[/tex]
---------------------------------------
[tex]-\frac{13}{2}y=76[/tex]
[tex]y=-76*\frac{2}{13}[/tex]
[tex]y=-\frac{152}{13}[/tex]
Now substitute the value of y in any of the two equations and solve for x
[tex]-2x - 3(-\frac{152}{13}) = 14[/tex]
[tex]-2x +\frac{456}{13} = 14[/tex]
[tex]-2x= 14-\frac{456}{13}[/tex]
[tex]-2x=-\frac{274}{13}[/tex]
[tex]x=\frac{137}{13}[/tex]
The solution is:
[tex](\frac{137}{13}, -\frac{152}{13})[/tex]
Answer:
x = 411/39 and y = -152/13
Step-by-step explanation:
It is given that,
3x - 2y = 55 ----(1)
-2x - 3y = 14 ---(2)
To find the solution of given equations
eq(1) * 2 ⇒
6x - 4y = 110 ---(3)
eq(2) * 3 ⇒
-6x - 9y = 42 ---(4)
eq(3) + eq(4) ⇒
6x - 4y = 110 ---(3)
-6x - 9y = 42 ---(4)
0 - 13y = 152
y = -152/13
Substitute the value of y in eq (1)
3x - 2y = 55 ----(1)
3x - 2*(-152/13) = 55
3x + 304/13 = 55
3x = 411/13
x = 411/39
Therefore x = 411/39 and y = -152/13
Convert ln x = y to exponential form.
A test consists of 10 multiple choice questions, each with 5 possible answers, one of which is correct. To pass the test a student must get 60% or better on the test. If a student randomly guesses, what is the probability that the student will pass the test?
Try this option:
if only one answer of five is correct, it means, the probability to choose it P=1/5=0.2.
If the student guesses randomly, it means, using the probability 0.2, he(she) can choose only 10*0.2=2 correct answers. To pass the test, the student must get 0.6*10=6 correct answers or more.
The student business club on your campus has decided to hold a pizza fund raiser. The club plans to buy 50 pizzas from Dominos and resell them in the student center. Based upon the specials advertised on the Dominos website, what will you need to charge per slice (assume 8 slices per pizza) in order to break even? Since this is a fund raiser, what would you suggest charging for each slice and, based on this, what would the net profit be to this club? Why do you feel breakeven analysis is so crucial in the development of new products for businesses?
Answer:
2.50 per slice would be okay because most people would order 2 slices and that would have them give you an even 5.00. Net profit for this would be 464.00
Step-by-step explanation:
The specials advertised on Domino's website is 5.99 for each pizza, 8 slices.
The break even price is 1.34 per slice. This is important to know because a business never wants to take a loss.
To break even, the club would need to charge $2 per slice. However, for fundraising, they might charge $3 per slice, giving a net profit of $400. Breakeven analysis is crucial in business as it helps in decision making and financial planning.
Explanation:Based on the information given, we see that the business club wants to purchase 50 pizzas from Domino's and resell them on campus. Assuming each pizza costs $16 (as per the example given for Authentic Chinese Pizza), the initial cost for the club would be $800 (50 pizzas x $16).
To calculate the break-even price per slice, we would need to divide the total cost by the total number of slices: $800 / (8 slices x 50 pizzas) = $2 per slice. However, since this is a fundraiser, the club might want to add in a margin to generate profit, so they might charge for example $3 per slice, resulting in a profit of $1 per slice, or $400 total.
The breakeven analysis is crucial in the development of new products because it helps businesses determine the minimum production and sales levels they must achieve to avoid losing money. This is useful for decision making and financial planning in any business operation.
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Find a parametrization for the curve「and determine the work done on a particle moving along Γ in R3 through the force field F:R^3--R^3'where F(x,y,z) = (1,-x,z) and (a) Im (Γ) is the line segment from (0,0,0) to (1,2,1) (b) Im (Γ) is the polygonal curve with successive vertices (1,0,0), (0,1,1), and (2,2,2) (c) Im (Γ) is the unit circle in the plane z = 1 with center (0,0,1) beginning and ending at (1,0,1), and starting towards (0,1,1)
a. Parameterize [tex]\Gamma[/tex] by
[tex]\vec r(t)=(t,2t,t)[/tex]
with [tex]0\le t\le1[/tex]. The work done by [tex]\vec F[/tex] along [tex]\Gamma[/tex] is
[tex]\displaystyle\int_\Gamma\vec F\cdot\mathrm d\vec r=\int_0^1(1,-t,t)\cdot(1,2,1)\,\mathrm dt=\int_0^1(1-t)\,\mathrm dt=\boxed{\frac12}[/tex]
b. Break up [tex]\Gamma[/tex] into each component line segment, denoting them by [tex]\Gamma_1[/tex] and [tex]\Gamma_2[/tex], and parameterize each respectively by
[tex]\vec r_1(t)=(1-t,t,t)[/tex] and[tex]\vec r_2(t)=(2t,1+t,1+t)[/tex]both with [tex]0\le t\le1[/tex]. Then the work done by [tex]\vec F[/tex] along each component path is
[tex]\displaystyle\int_{\Gamma_1}\vec F\cdot\mathrm d\vec r_1=\int_0^1(1,t-1,t)\cdot(-1,1,1)\,\mathrm dt=\int_0^1(2t-2)\,\mathrm dt=-1[/tex]
[tex]\displaystyle\int_{\Gamma_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(1,-2t,1+t)\cdot(2,1,1)\,\mathrm dt=\int_0^1(3-t)\,\mathrm dt=\frac52[/tex]
giving a total work done of [tex]-1+\dfrac52=\boxed{\dfrac32}[/tex].
c. Parameterize [tex]\Gamma[/tex] by
[tex]\vec r(t)=(\cos t,\sin t,1)[/tex]
with [tex]0\le t\le2\pi[/tex]. Then the work done by [tex]\vec F[/tex] is
[tex]\displaystyle\int_\Gamma\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}(1,-\cos t,1)\cdot(-\sin t,\cos t,0)\,\mathrm dr=-\int_0^{2\pi}(\sin t+\cos^2t)\,\mathrm dt=\boxed{-\pi}[/tex]
Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. lim x → (π/2)+ cos(x) 1 − sin(x)
Answer:
The limit of the function at x approaches to [tex]\frac{\pi}{2}[/tex] is [tex]-\infty[/tex].
Step-by-step explanation:
Consider the information:
[tex]\lim_{x \to \frac{\pi}{2}}\frac{cos(x)}{1-sin(x)}[/tex]
If we try to find the value at [tex]\frac{\pi}{2}[/tex] we will obtained a [tex]\frac{0}{0}[/tex] form. this means that L'Hôpital's rule applies.
To apply the rule, take the derivative of the numerator:
[tex]\frac{d}{dx}cos(x)=-sin(x)[/tex]
Now, take the derivative of the denominator:
[tex]\frac{d}{dx}1-sin(x)=-cos(x)[/tex]
Therefore,
[tex]\lim_{x \to \frac{\pi}{2}}\frac{-sin(x)}{-cos(x)}[/tex]
[tex]\lim_{x \to \frac{\pi}{2}}\frac{sin(x)}{cos(x)}[/tex]
[tex]\lim_{x \to \frac{\pi}{2}}tan(x)}[/tex]
Since, tangent function approaches -∞ as x approaches to [tex]\frac{\pi}{2}[/tex]
, therefore, the original expression does the same thing.
Hence, the limit of the function at x approaches to [tex]\frac{\pi}{2}[/tex] is [tex]-\infty[/tex].
The function cos(x)/(1-sin(x)) approaches an indeterminate form as x approaches π/2 from the right. By applying L'Hopital's Rule, we find that the limit is equivalent to the limit of tan(x) as x approaches π/2 from the right. However, tan(π/2) is undefined, so the limit does not exist.
Explanation:In this question, we are asked to find the limit of the function cos(x)/(1-sin(x)) as x approaches π/2 from the right. We can't directly substitute x = π/2 because it makes the denominator zero, yielding an indeterminate form of '0/0'.
So, we use L'Hopital's Rule, which states that the limit of a ratio of two functions as x approaches a particular value is equal to the limit of their derivatives.
The derivative of cos(x) is -sin(x) and the derivative of (1-sin(x)) is -cos(x). Using L'Hopital's Rule, we can now re-evaluate our limit substituting these derivatives.
lim [x → (π/2)+] (-sin(x)/-cos(x)) = lim [x → (π/2)+] tan(x)
When substituting x = π/2 into tan(x), we realize that the tan(π/2) is undefined, so the answer is the limit does not exist.
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The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 5 minutes and a standard deviation of 2 minutes. Find the probability that it takes at least 8 minutes to find a parking space. (Round your answer to four decimal places.)
Answer: 0.0668
Step-by-step explanation:
Given: Mean : [tex]\mu = 5\text{ minutes}[/tex]
Standard deviation : [tex]\sigma = 2\text{ minutes}[/tex]
The formula to calculate z is given by :-
[tex]z=\dfrac{X-\mu}{\sigma}[/tex]
To check the probability it takes at least 8 minutes (X≥ 8) to find a parking space.
Put X= 8 minutes
[tex]z=\dfrac{8-5}{2}=1.5[/tex]
The P Value =[tex]P(z\geq1.5)=1-P(z\leq1.5)=1- 0.9331927=0.0668073\approx0.0668[/tex]
Hence, the probability that it takes at least 8 minutes to find a parking space = 0.0668
The probability that it takes atleast 8 minutes to find a parking space is 0.0668
Given the Parameters :
Mean = 5 minutes Standard deviation = 2 minutes X ≥ 8First we find the Zscore :
Zscore = (X - mean) / standard deviationZscore = (8 - 5) / 2 = 1.5
The probability of taking atleast 8 minutes can be expressed thus and calculated using the normal distribution table :
P(Z ≥ 1.5) = 1 - P(Z ≤ 1.5)
P(Z ≥ 1.5) = 1 - 0.9332
P(Z ≥ 1.5) = 0.0668
Therefore, the probability, P(Z ≥ 1.5) is 0.0668
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1) Find the lump that must be deposited today to have a future value of $ 25,000 in 5 years if funds earn 6 % componded annually.
Answer: $ 18681.45
Step-by-step explanation:
Given: Future value : [tex]FV=\$25,000[/tex]
The rate of interest : [tex]r=0.06[/tex]
The number of time period : [tex]t=5[/tex]
The formula to calculate the future value is given by :-
[tex]\text{Future value}=P(1+i)^n[/tex], where P is the initial amount deposited.
[tex]\Rightarrow\ 25000=P(1+0.06)^5\\\\\Rightarrow\ P=\dfrac{25000}{(1.06)^5}\\\\\Rightarrow\ P=18681.4543217\approx=18681.45[/tex]
Hence, the lump that must be deposit today : $ 18681.45