Answer:
The length is 78 feet and the width is 30 feet.
Step-by-step explanation:
The perimeter of a rectangle can be calculated with this formula:
[tex]P=2l+2w[/tex]
Where "l" is the length and "w" is the width.
Since we know that the perimeter of the playing field is 216 feet and its length is 48 feet longer than the width ([tex]l=w+48[/tex]), we can substitute them into the formula and solve for "w":
[tex]216=2(w+48)+2w\\\\216=2w+96+2w\\\\216-96=4w\\\\\frac{120}{4}=w\\\\w=30\ ft[/tex]
Finally, substitute the width into [tex]l=w+48[/tex] to find the length. This is:
[tex]l=30+48\\\\l=78\ ft[/tex]
Final answer:
To find the dimensions of the playing field, we identify the width as w feet and the length as w + 48 feet. By using the perimeter formula and solving the resulting equation, we determine that the width is 30 feet and the length is 78 feet.
Explanation:
The student is asking to find the dimensions of a rectangle given its perimeter and the relationship between its length and width. The perimeter of the rectangle is known to be 216 feet, and the length is specified to be 48 feet longer than the width.
Let's call the width w feet. Then, the length would be w + 48 feet. Since the perimeter of a rectangle is given by the formula P = 2(l + w), where P is the perimeter, l is the length, and w is the width, we can set up the following equation:
2(w + w + 48) = 216
Solving this equation, we find:
4w = 120
Therefore, the width of the playing field is 30 feet. To find the length, add 48 feet to the width:
Length = w + 48 = 30 + 48 = 78 feet.
The dimensions of the playing field are 30 feet in width and 78 feet in length.
8.) Out of 600 employees, only 270 are happy with their pay. What percent is this? (Please show your work.) *
For this case we can raise a rule of three:
600 employees -------------> 100%
270 employees -------------> x
Where the variable "x" represents the percentage of employees who are satisfied with their salary. So, we have:
[tex]x = \frac {270 * 100} {600}\\x = 45[/tex]
Thus, 45% of employees are satisfied with their salary.
Answer:
45%
ANSWER
[tex]45\%[/tex]
EXPLANATION
The total number of employees is 600.
The number of employees who are happy with their pay is 270.
The percentage of employees who are happy with their pay is the number who are happy with their pay divided by total number of employees times 100%
[tex] \frac{270}{600} \times 100\%[/tex]
This simplifies to
[tex]45\%[/tex]
verify sin^4x-sin^2x=cos^4x-cos^2x is an identity
ANSWER
See below
EXPLANATION
We want to verify that,
[tex] { \sin ^{4} x} - { \sin^{2} x} = { \cos ^{4} x} - { \cos^{2} x}[/tex]
To verify this identity, we can take the left hand side simplify it to get the right hand side or vice versa.
[tex]{ \sin ^{4} x} - { \sin^{2} x} =( { \sin ^{2} x} )^{2} - { \sin^{2} x}[/tex]
[tex]{ \sin ^{4} x} - { \sin^{2} x} ={ \sin ^{2} x}({ \sin ^{2} x} - 1)[/tex]
[tex]{ \sin ^{4} x} - { \sin^{2} x} ={ \sin ^{2} x} \times - (1 - { \sin ^{2} x})[/tex]
[tex]{ \sin ^{4} x} - { \sin^{2} x} =({1 - \cos^{2} x} )\times - ({ \cos^{2} x})[/tex]
[tex]{ \sin ^{4} x} - { \sin^{2} x} =({ \cos^{2} x} - 1 )\times ({ \cos^{2} x})[/tex]
We now expand the right hand side to get:
[tex] { \sin ^{4} x} - { \sin^{2} x} = { \cos ^{4} x} - { \cos^{2} x}[/tex]
FInd the values and sketch the direction field for dy/dx = x^2 +y for x [-2,-1,0,1,2] and y[-2,-1,0,1,2]
You want to buy a $232,000 home. You plan to pay 20% as a down payment, and take out a 3.6% APR loan for the rest. a) How much is the loan amount going to be? b) What will your monthly payments be if the length of the loan is 10 years? c) What will your monthly payments be if the length of the loan is 20 years? d) Over the course of the entire loan, how much more do you end up paying with the longer loan? (Hint: take the difference of the total amounts paid)
Answer:
The price of the home = 232,000
20% is down payment.
Part A:
[tex]0.20\times232000=46400[/tex]
So, the loan amount will be =[tex]232000-46400=185600[/tex]
Loan amount or p = $185,600
Part B:
p = 185600
r = [tex]3.6/12/100=0.003[/tex]
n = [tex]10\times12=120[/tex]
The EMI formula is :
[tex]\frac{p\times r\times(1+r)^{n} }{(1+r)^{n}-1 }[/tex]
Now putting the values in formula we get
[tex]\frac{185600\times 0.003\times(1+0.003)^{120} }{(1+0.003)^{120}-1 }[/tex]
=> [tex]\frac{185600\times 0.003\times(1.003)^{120} }{(1.003)^{120}-1 }[/tex]
Monthly payments = $1844.02
Part C:
p = 185600
r = [tex]3.6/12/100=0.003[/tex]
n = [tex]20\times12=240[/tex]
Now putting the values in formula we get
[tex]\frac{185600\times 0.003\times(1+0.003)^{240} }{(1+0.003)^{240}-1 }[/tex]
=> [tex]\frac{185600\times 0.003\times(1.003)^{240} }{(1.003)^{240}-1 }[/tex]
Monthly payments = $1085.96
Part D:
For 10 year loan you have to pay = [tex]120\times1844.02=221282.40[/tex]
For 20 years loan you have to pay =[tex]240\times1085.96=260630.40[/tex]
So, you ended up paying [tex]260630.40-221282.40=39348[/tex] dollars more in longer loan.
The difference is $39,348.
In 1898, L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that the number of soldiers killed by horse kicks each year in each corps in the Prussian cavalry followed a Poisson distribution with a mean of 0.61. (a) What is the probability of more than 1 death in a corps in a year
Answer: Hence, Probability of more than 1 death in a corps in a year is 0.126.
Step-by-step explanation:
Since we have given that
Mean for a poisson distribution (λ) = 0.61
Number of years = 20 years
We need to find the probability of more than 1 death in a corps in a year.
P(X>1)=1-P(X=0)-P(X=1)
Here,
[tex]P(X=0)=\dfrac{e^{-0.61}(0.61)^0}{0!}=0.543\\\\and\\\\P(X=1)=\dfrac{e^{-0.61}(0.61)}{1}=0.331[/tex]
So,
P(X>1)=1-0.543-0.331=0.126
Hence, Probability of more than 1 death in a corps in a year is 0.126.
Using the Poisson distribution with a mean of 0.61, we calculate the probability of 0 or 1 death and subtract that from 1 to get the probability of more than 1 death in a Prussian cavalry corps in a year.
Explanation:Calculating Probability Using the Poisson DistributionBased on L. J. Bortkiewicz's study, the number of soldiers killed by horse kicks in the Prussian cavalry follows a Poisson distribution with a mean (λ) of 0.61. To calculate the probability of more than one death in a corps in a year, we use the Poisson probability formula:
P(X > k) = 1 - P(X ≤ k)
Where P(X > k) is the probability of having more than k events (in this case, deaths), and P(X ≤ k) is the probability of k or fewer events. In this scenario, k equals 1. So, we need to calculate the probability of 0 or 1 death and subtract from 1 to get the probability of more than 1 death.
Using the Poisson probability formula:
Calculate P(X = 0) and P(X = 1).Add P(X = 0) and P(X = 1) to get P(X ≤ 1).Subtract P(X ≤ 1) from 1 to obtain P(X > 1).Let's calculate:
P(X = 0) = (e^(-0.61))*(0.61^0)/(0!) = e^(-0.61)P(X = 1) = (e^(-0.61))*(0.61^1)/(1!) = 0.61*e^(-0.61)P(X ≤ 1) = P(X = 0) + P(X = 1)P(X > 1) = 1 - P(X ≤ 1)The resulting calculation will give us the probability of more than one death due to horse kicks in a Prussian cavalry corps within one year.
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Of the coffee makers sold in an appliance store, 4.0% have either a faulty switch or a defective cord, 2.5% have a faulty switch, and 0.1% have both defects. What is the probability that a coffee maker will have a defective cord? Express the answer as a percentage.
Final answer:
The probability that a coffee maker will have a defective cord is calculated by using the principle of inclusion-exclusion. Subtracting the percentage of coffee makers with faulty switches from the total percentage with any defect and adding the percentage with both defects, we find that the probability is 1.6%.
Explanation:
To calculate the probability that a coffee maker will have a defective cord, we can use the principle of inclusion-exclusion. According to the question, 4.0% of coffee makers have either a faulty switch or a defective cord. Of these, 2.5% have a faulty switch, and 0.1% have both defects. The probability that a coffee maker will have a defective cord is the total probability of any defect minus the probability of a faulty switch, plus the probability of having both defects, since those with both defects were counted in both the faulty switch and defective cord categories.
The formula we will use is: Probability(defective cord) = Probability(faulty switch or defective cord) - Probability(faulty switch) + Probability(both defects).
Plugging in the values we have: Probability(defective cord) = 4.0% - 2.5% + 0.1% = 1.6%
Therefore, the probability that a coffee maker will have a defective cord is 1.6%.
Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xe^y + ye^z + ze^x, (0, 0, 0), v = 6, 3, −3
Final answer:
The directional derivative of the function f(x, y, z) = [tex]xe^y + ye^z + ze^x[/tex] at the point (0, 0, 0) in the direction of the vector v = 6, 3, −3 is 0.
Explanation:
To find the directional derivative of the function f(x, y, z) = [tex]xe^y + ye^z + ze^x[/tex] at the point (0, 0, 0) in the direction of the vector v = 6, 3, −3, we first need to find the gradient of f. The gradient of f, denoted as ∇f, is a vector of partial derivatives with respect to each variable. We calculate the partial derivatives as follows:
∂f/∂x = [tex]e^y + ze^x[/tex]∂f/∂y = [tex]xe^y + e^z[/tex]∂f/∂z = [tex]ye^z + xe^x[/tex]At the point (0, 0, 0), the gradient ∇f is (0 + 0, 0 + 1, 0 + 1) = (0, 1, 1).
Next, we need to normalize the given vector v. The normalization process involves dividing v by its magnitude to obtain a unit vector u in the direction of v. The magnitude of v is √(6² + 3² + (-3)²) = √(36 + 9 + 9) = √54. Therefore, the unit vector u is (6/√54, 3/√54, -3/√54).
Finally, the directional derivative of f at (0, 0, 0) in the direction of v is the dot product of ∇f and u, which is (0, 1, 1) ⋅ (6/√54, 3/√54, -3/√54) = 0*6/√54 + 1*3/√54 + 1*(-3)/√54 = 0.
An owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number of key rings is given by , where n is the number of key rings in thousands. Find the number of key rings sold on a particular day when the total profit is $5,000.
n^2-2n-3
Answer:
The number of key rings sold on that day is 4000 key rings
Step-by-step explanation:
* Lets explain the information in the problem
- The profit earned (in thousands of dollars) per day by selling n number
of key rings is given by the function P(n) = n² - 2n - 3, where n is the
number of key rings in thousands and P is the profit in thousands
for one day
- On a particular day the total profit is $5,000
∵ 5000 = 5 in thousands
∵ The function P(n) is the profit of n key ring in thousands
∴ P(n) = 5
- Lets solve the function to find the number of key rings
∵ P(n) = n² - 2n - 3
∴ 5 = n² - 2n - 3 ⇒ subtract 5 from both sides
∴ 0 = n² - 2n - 8 ⇒ factorize it
∵ n² = n × n ⇒ 1st terms in the 2 brackets
∵ -8 = -4 × 2 ⇒ 2nd terms in the 2 brackets
∵ n × -4 = -4n ⇒ nears
∵ n × 2 = 2n ⇒ extremes
∵ -4n + 2n = -2n ⇒ the middle term
∴ (n - 4)(n + 2) = 0 ⇒ equate each bracket by 0 to find n
∴ n - 4 = 0 ⇒ add 4 to both sides
∴ n = 4 key ring in thousands = 4000 key rings
- OR
∴ n + 2 = 0 ⇒ subtract 2 from both sides
∴ n = -2 ⇒ we will refused this value because number of key rings
must be positive
∴ The number of key rings sold on that day is 4000 key rings
To find the number of key rings sold on a particular day when the total profit is $5,000, we need to solve the given equation for n.
Explanation:The owner of a key rings manufacturing company found that the profit earned (in thousands of dollars) per day by selling n number of key rings is given by the equation P(n) = n^2-2n-3. To find the number of key rings sold on a particular day when the total profit is $5,000, we need to solve the equation P(n) = 5000 for n.
Step 1: Set the equation equal to 5000: n^2-2n-3 = 5000.
Step 2: Rearrange the equation and set it equal to zero: n^2-2n-5003 = 0.
Step 3: Solve the quadratic equation using factoring, completing the square, or the quadratic formula to find the values of n.
Step 4: The solutions will give us the possible values of n, representing the number of key rings sold on the particular day when the total profit is $5,000.
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3×
__ -2=15
7
2×
__+1=7
5
7×
__-1=22
15
5×
__+10=2
8
×
__+4=20
6
×
__-4=10
3
×
__+2=8
6
×
__+8=8
9
×
__+7=20
6
Answer:
1. x = 39.67
2. x = 15
3. x = 49.29
4. x = -12.8
5. x = 96
6. x = 42
7. x = 36
8. x = 0
9. x = 78
Step-by-step explanation:
Just remember to always isolate the unknown. Here are the solutions to your problem. I will explain each step for the first for you to give you an idea how the others were worked out.
1. [tex]\dfrac{3x}{7}-2 = 15[/tex]
Add 2 to both sides to get rid of -2 on the left side.
[tex]\dfrac{3x}{7}-2+(2)=15+(2)///dfrac{3x}{7}=17[/tex]
Multiply both sides by 7 to get rid of 7 on the left side.
[tex]\dfrac{3x}{7}\times 7 = 17\times 7\\\\3x = 119[/tex]
Divide both sides by 3 to get rid of 3 on the left side.
[tex]\dfrac{3x}{3} = \dfrac{119}{3}\\\\x = 39.67[/tex]
You could also transpose everything by the x to the other side of the equation. Just remember that whatever OPERATION used on the original side, must be opposite on the other side. I'll use the second problem to show this.
[tex]\dfrac{2x}{5}+1=7[/tex]
Transpose 1 on the left to the right. It is addition on the left, then it would be subtraction on the other side.
[tex]\dfrac{2x}{5}+1=7\\\\\dfrac{2x}{5}=7-1\\\\\dfrac{2x}{5}=6[/tex]
Transpose 5 from the left side to the right. It is division on the left, then it would be multiplication on the right.
[tex]\dfrac{2x}{5}=6\\\\2x=6\times 5\\\\2x=30[/tex]
Transpose 2 from the left side to the right. It is multiplication on the left, then it would be division on the right.
[tex]2x=30\\\\x=\dfrac{30}{2}\\\\x=15[/tex]
Let's move on with the rest now.
3.
[tex]\dfrac{7x}{15}-1=22\\\\\dfrac{7x}{15}=22+1\\\\\dfrac{7x}{15}=23\\\\7x=23\times15\\\\7x=345\\\\x=\dfrac{345}{7}\\\\x=49.29[/tex]
4.
[tex]\dfrac{5x}{8}+10=2\\\\\dfrac{5x}{8}=2-10\\\\\dfrac{5x}{8}=-8\\\\5x=-8\times8\\\\5x=-64\\\\x=\dfrac{-64}{5}\\\\x=-12.8[/tex]
5.
[tex]\dfrac{x}{6}+4=20\\\\\dfrac{x}{6}=20-4\\\\\dfrac{x}{6}=16\\\\x=16\times6\\\\x=96[/tex]
6.
[tex]\dfrac{x}{3}-4=10\\\\\dfrac{x}{3}=10+4\\\\\dfrac{x}{3}=14\\\\x=3\times14\\\\x=42[/tex]
7.
[tex]\dfrac{x}{6}+2=8\\\\\dfrac{x}{6}=8-2\\\\\dfrac{x}{6}=6\\\\x=6\times6\\\\x=36[/tex]
8.
[tex]\dfrac{x}{9}+8=8\\\\\dfrac{x}{9}=8-8\\\\\dfrac{x}{9}=0\\\\x=0\times9\\\\x=0[/tex]
9.
[tex]\dfrac{x}{6}+7=20\\\\\dfrac{x}{6}=20-7\\\\\dfrac{x}{6}=13\\\\x=13\times6\\\\x=78[/tex]
The answers are as follows- 1. x = 17/3 or 5.67,2. x=3 . 3.x = 23/7 or about 3.29. 4.x = -8/5 or -1.6., 5.x=2, 6.x = 14/6 or about 2.33., 7. x=2, 8. x=0, 9. x = 13/9 or about 1.44.
Solving the Given Equations
3 __ - 2 = 15Light bulbs of a certain type are advertised as having an averagelifetime of 750 hours. The price of these bulbs is very favorable,so a potential customer has decided to go ahead with a purchasearrangement unless it can be conclusively demonstrated thatthe true average lifetime is smaller than what is advertised. Arandom sample of 50 bulbs was selected, the lifetime of each bulbdetermined, and the appropriate hypothesis were tested usingMINITAB, resulting in the accompanying output.Variable N Mean St Dev SEMean ZP -Valuelifetime 50 738.44 38.20 5.40-2.14 0.016 What conclusion would be appropriate for a significance level of.05 ? .A significance level of .01 ?. What significance level wouldyou recommend ?
answer 40
Step-by-step explanation:
because you added all together to make one
Can you divide 25 into 4 equal groups? Explain why or why not.
Answer:
No, not really.
dividing 25 into 4 equal groups, equals 6.25.
So no, you can not divide 25 into 4 equal group.
If if was 25 into 5 equal groups, then yes.
Hope this helps!!!
Please mark brainliest, if this helps. :)
Step-by-step explanation:
No, 25 cannot be divided into 4 equal groups without a remainder. Dividing 25 by 4 results in a quotient of 6 with a remainder of 1, meaning the groups would not be equal.
To determine if this is possible, we need to perform a division operation. Dividing 25 by 4 gives us a quotient of 6 with a remainder of 1. This means that 25 cannot be evenly divided into 4 equal groups, as one group would end up with one less or one more than the others.
Therefore, 25 cannot be divided into 4 equal groups .
If one card is drawn from a standard 52 card playing deck, determine the probability of getting a jack, a three, a club or a diamond. Round to the nearest hundredth.
Answer:
The probability of getting a jack, a three, a club or a diamond is 0.58.
Step-by-step explanation:
In a standard deck of 52 cards have 13 club, 13 spade, 13 diamond, 13 heart cards. Each suit has one jack and 3.
Number of club cards = 13
Number of diamond cards = 13
Number of jack = 4
Number of 3 = 4
Jack of club and diamond = 2
3 of club and diamond = 2
Total number of cards that are either a jack, a three, a club or a diamond is
[tex]13+13+4+4-2-2=30[/tex]
The probability of getting a jack, a three, a club or a diamond is
[tex]Probability=\frac{\text{A jack, a three, a club or a diamond}}{\text{Total number of cards}}[/tex]
[tex]Probability=\frac{30}{52}[/tex]
[tex]Probability=0.576923076923[/tex]
[tex]Probability\approx 0.58[/tex]
Therefore the probability of getting a jack, a three, a club or a diamond is 0.58.
The correct answer is 0.50.
To determine the probability of getting a jack, a three, a club, or a diamond from a standard 52-card deck, we can calculate the probability of each individual event and then combine them, taking care to avoid double-counting any cards.
First, let's calculate the probability of drawing a jack. There are 4 jacks in the deck (one for each suit). Since there are 52 cards in total, the probability of drawing a jack is:
[tex]\[ P(\text{jack}) = \frac{4}{52} \][/tex]
Next, we calculate the probability of drawing a three. There are also 4 threes in the deck, one for each suit. So, the probability of drawing a three is:
[tex]\[ P(\text{three}) = \frac{4}{52} \][/tex]
Now, let's calculate the probability of drawing a club. There are 13 clubs in the deck (since there are 13 cards in each suit). Thus, the probability of drawing a club is:
[tex]\[ P(\text{club}) = \frac{13}{52} \][/tex]
Similarly, there are 13 diamonds in the deck, so the probability of drawing a diamond is:
[tex]\[ P(\text{diamond}) = \frac{13}{52} \][/tex]
However, we must be careful not to double-count the cards that are both a jack or a three and a club or a diamond. There are 2 jacks and 2 threes that are also clubs or diamonds (one jack and one three of clubs, and one jack and one three of diamonds).
To find the total probability, we add the probabilities of each event and subtract the probabilities of the events that have been counted twice (the jack and three of clubs and diamonds):
[tex]\[ P(\text{total}) = P(\text{jack}) + P(\text{three}) + P(\text{club}) + P(\text{diamond}) - 2 \times P(\text{jack or three of clubs or diamonds}) \] \[ P(\text{total}) = \frac{4}{52} + \frac{4}{52} + \frac{13}{52} + \frac{13}{52} - 2 \times \frac{2}{52} \] \[ P(\text{total}) = \frac{4 + 4 + 13 + 13 - 4}{52} \] \[ P(\text{total}) = \frac{30}{52} \] \[ P(\text{total}) = \frac{15}{26} \] \[ P(\text{total}) = \frac{5}{8} \] \[ P(\text{total}) = 0.625 \][/tex]
Since we need to round to the nearest hundredth, the final answer is:
[tex]\[ P(\text{total}) \approx 0.63 \][/tex]
[tex]\[ P(\text{total}) = P(\text{jack}) + P(\text{three}) + P(\text{club}) + P(\text{diamond}) \] \[ P(\text{total}) = \frac{4}{52} + \frac{4}{52} + \frac{13}{52} + \frac{13}{52} \] \[ P(\text{total}) = \frac{4 + 4 + 13 + 13}{52} \] \[ P(\text{total}) = \frac{34}{52} \] \[ P(\text{total}) = \frac{17}{26} \] \[ P(\text{total}) = \frac{1}{2} \] \[ P(\text{total}) = 0.50 \][/tex]
A scientist mixes water (containing no salt) with a solution that contains 30% salt. She wants to obtain 180 ounces of a mixture that is 15% salt. How many ounces of water and how many ounces of the 30% salt solution should she use?
Answer:
water : 90 ounces
salt solution: 90 ounces
Step-by-step explanation:
Call w the amount of water and call the solution containing 30% salt.
We want to get 180 ounces of a mixture with 15% salt.
So:
The amount of mixture will be:
[tex]w + s = 180[/tex]
the amount of salt will be
[tex]0w + 0.3s = 180 * 0.15[/tex]
[tex]0.3s = 27[/tex]
[tex]s = 90\ ounces[/tex]
Now we substitute the value of s in the first equation and solve for w
[tex]w + 90 = 180[/tex]
[tex]w = 90\ ounces[/tex]
Mens' heights are normally distributed with u 68 in and o 6 in. Womens' heights have u 65 in with o 4 in. Dan is 70 in tall and Amy is 68 in tall. Who is relatively taller? Are they unusually tall?
Answer:
i say option (a) is the answer its correct.
Step-by-step explanation:
New Tab mect OUantit MATH Hosted by ALEKS Carp Homework #1 Prep Lesson 1 Previous 2 $3 Question 10 of 10 (1 point) Find the most likely next two numbers. 9, 7, 4, 0,-5,-11 andy The most likely next two numbers are
Answer:
-18, -26
Step-by-step explanation:
The differences are decreasing by 1 each time, so the next differences will be -7 and -8.
-11 -7 = -18
-18 -8 = -26
Is (2,3), a solution to the system of equation x 2 + y 2 = 13 and 2x - y = 4?
Answer:
No, the point [tex](2,3)[/tex] is not a solution to the system of equation [tex]x2+y2=13[/tex] and [tex]2x-y=4[/tex].
Step-by-step explanation:
To make sure it is, it has to work on both equation. Plugging this into the first and second equations respectivly gives us:
[tex]13=x2+y2=(2)2+(3)2=4+6=10[/tex]
[tex]4=2x-y=2(2)-(3)=4-3=1[/tex]
Since [tex]13\neq10[/tex] and [tex]4\neq 1[/tex], the point [tex](2,3)[/tex] is not a solution to the system of equation [tex]x2+y2=13[/tex] and [tex]2x-y=4[/tex].
Answer:
That is False.
Step-by-step explanation:
Find the solution of IVP for the differential equation (x+2)^2e^y when y(1) =0 y'= dy/dx=(x+2)^2.e^y; y(1)=0
Answer:
The solution is [tex]y\:=-\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]
Step-by-step explanation:
We need to find the solution of IVP for differential equation [tex]\frac{dy}{dx}=(x+2)^{2}e^{y}[/tex] when [tex]y(1)=0[/tex]
[tex]\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}[/tex]
[tex]\mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)[/tex]
[tex]\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}[/tex]
[tex]\frac{1}{e^y}y'\:=\left(x+2\right)^2[/tex]
[tex]N\left(y\right)\cdot y'\:=M\left(x\right)[/tex]
[tex]N\left(y\right)=\frac{1}{e^y},\:\quad M\left(x\right)=\left(x+2\right)^2[/tex]
[tex]\mathrm{Solve\:}\:\frac{1}{e^y}y'\:=\left(x+2\right)^2[/tex]
[tex]\frac{1}{e^y}y'\:=x^{2}+4+4x[/tex]
Integrate both the sides with respect to dx
[tex]\int\frac{1}{e^y}y'dx\:=\intx^{2}dx+4\int dx+4\int x dx[/tex]
[tex]\int e^{-y}dy\:=\intx^{2}dx+4\int dx+4\int x dx[/tex]
[tex]-\frac{1}{e^{y}}\:=\frac{x^{3}}{3}+4x+4\frac{x^{2}}{2}+c_1[/tex]
[tex]-\frac{1}{e^{y}}\:=\frac{x^{3}}{3}+4x+2x^{2}+c_1[/tex]
Since, IVP is y(1)=0
put x=1 and y=0 in above equation
[tex]-\frac{1}{e^{0}}\:=\frac{1^{3}}{3}+4(1)+2(1)^{2}+c_1[/tex]
[tex]-1\:=\frac{1}{3}+4+2+c_1[/tex]
[tex]-1\:=\frac{19}{3}+c_1[/tex]
add both the sides by [tex]-\frac{19}{3}[/tex]
[tex]-1-\frac{19}{3}\:=\frac{19}{3}-\frac{19}{3}+c_1[/tex]
[tex]-1-\frac{19}{3}\:=c_1[/tex]
[tex]-\frac{22}{3}\:=c_1[/tex]
so,
[tex]-\frac{1}{e^{y}}\:=\frac{x^{3}}{3}+4x+2x^{2}-\frac{22}{3}[/tex]
Multiply both the sides by '-1'
[tex]\frac{1}{e^{y}}\:=-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3}[/tex]
[tex]e^{-y}\:=-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3}[/tex]
Take natural logarithm both the sides,
[tex]-y\:=\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]
Multiply both the sides by '-1'
[tex]y\:=-\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]
Therefore, the solution is [tex]y\:=-\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]
A box in a supply room contains 22 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 5 are rated 23-watt. Suppose that three of these bulbs are randomly selected. (Round your answers to three decimal places.)(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?(b) What is the probability that all three of the bulbs have the same rating?(c) What is the probability that one bulb of each type is selected?(d) If bulbs are selected one by one until a 23-watt bulb is obtained, what is the probability that it is necessary to examine at least 6 bulbs?
Answer:
(a) 0.110 ⇒ rounded to three decimal places
(b) 0.097 ⇒ rounded to three decimal places
(c) 0.234 ⇒ rounded to three decimal places
(d) 0.235 ⇒ rounded to three decimal places
Step-by-step explanation:
* Lets solve the problem using the combination
- The order is not important in this problem, so we can use the
combination nCr to find the probability
- There are 22 compact fluorescent light bulbs
# 8 ⇒ rated 13 watt
# 9 ⇒ rated 18 watt
# 5 ⇒ rated 23 watt
- 3 bulbs are randomly selected
(a) Exactly two of the selected bulbs are rated 23 watt
∵ Exactly two of them are rated 23 watt
∵ The number of bulbs rated 23 watt is 5
∴ We will chose 2 from 5
∴ 5C2 = 10 ⇒ by using calculator or by next rule
# nCr = n!/[r! × (n-r)!]
- 5C2= 5!/[2! × (5 - 2)!] = (5×4×3×2×1)/[(2×1)×(3×2×1)] = 120/12 = 10
∴ There are 10 ways to chose 2 bulbs from 5
∵ The 3rd bulbs will chosen from the other two types
∵ The other two types = 8 + 9 = 17
∴ We will chose 1 bulbs from 17 means 17C1
∵ 17C1 = 17 ⇒ (by the same way above)
∴ There are 17 ways to chose 1 bulbs from 17
- 10 ways for two bulbs and 17 ways for one bulb
∴ There are 10 × 17 = 170 ways two chose 3 bulbs exact 2 of them
rated 23 watt (and means multiply)
∵ There are 22C3 ways to chose 3 bulbs from total 22 bulbs
∵ 22C3 = 1540 ⇒ (by the same way above)
∴ The probability = 170/1540 = 17/154 = 0.110
* The probability that exactly two of the selected bulbs are rated
23-watt is 0.110 ⇒ rounded to three decimal places
(b) All three of the bulbs have the same rating
- We can have either all 13 watt, all 18 watt or all 23 watt.
# 13 watt
∵ There are 8 are rated 13 watt
∵ We will chose 3 of them
∴ There are 8C3 ways to chose 3 bulbs from 8 bulbs
∵ 8C3 = 56
∴ There are 56 ways to chose 3 bulbs rated 13 watt
# 18 watt
∵ There are 9 are rated 18 watt
∵ We will chose 3 of them
∴ There are 9C3 ways to chose 3 bulbs from 9 bulbs
∵ 9C3 = 84
∴ There are 84 ways to chose 3 bulbs rated 18 watt
# 23 watt
∵ There are 5 are rated 23 watt
∵ We will chose 3 of them
∴ There are 5C3 ways to chose 3 bulbs from 5 bulbs
∵ 5C3 = 10
∴ There are 10 ways to chose 3 bulbs rated 23 watt
- 65 ways or 84 ways or 10 ways for all three of the bulbs have
the same rating
∴ There are 56 + 84 + 10 = 150 ways two chose 3 bulbs have the
same rating (or means add)
∵ There are 22C3 ways to chose 3 bulbs from total 22 bulbs
∵ 22C3 = 1540
∴ The probability = 150/1540 = 15/154 = 0.097
* The probability that all three of the bulbs have the same rating is
0.097 ⇒ rounded to three decimal places
(c) One bulb of each type is selected
- We have to select one bulb of each type
# 13 watt
∵ There are 8 bulbs rated 13 watt
∵ We will chose 1 of them
∴ There are 8C1 ways to chose 1 bulbs from 8 bulbs
∵ 8C1 = 8
∴ There are 8 ways to chose 1 bulbs rated 13 watt
# 18 watt
∵ There are 9 bulbs rated 18 watt
∵ We will chose 1 of them
∴ There are 9C1 ways to chose 1 bulbs from 9 bulbs
∵ 9C1 = 9
∴ There are 9 ways to chose 1 bulbs rated 18 watt
# 23 watt
∵ There are 5 bulbs rated 23 watt
∵ We will chose 1 of them
∴ There are 5C1 ways to chose 1 bulbs from 5 bulbs
∵ 5C1 = 5
∴ There are 5 ways to chose 1 bulbs rated 23 watt
- 8 ways and 9 ways and 5 ways for one bulb of each type is selected
∴ There are 8 × 9 × 5 = 360 ways that one bulb of each type is
selected (and means multiply)
∵ There are 22C3 ways to chose 3 bulbs from total 22 bulbs
∵ 22C3 = 1540
∴ The probability = 360/1540 = 18/77 = 0.234
* The probability that one bulb of each type is selected is 0.234
⇒ rounded to three decimal places
(d) If bulbs are selected one by one until a 23-watt bulb is obtained,
it is necessary to examine at least 6 bulbs
- We have a total of 22 bulbs and 5 of them are rate 23 watt
∴ There are 22 - 5 = 17 bulbs not rate 23 watt
∵ We examine at least 6 so the 6th one will be 23 watt
∴ There are 17C5 ways that the bulb not rate 23 watt
∵ 17C5 = 6188
∴ There are 6188 ways that the bulb is not rate 23 watt
∵ There are 22C5 ways to chose 5 bulbs from total 22 bulbs
∵ 22C5 = 26334
∴ The probability = 6188/26334 = 442/1881 = 0.235
* If bulbs are selected one by one until a 23-watt bulb is obtained,
the probability that it is necessary to examine at least 6 bulbs is
0.235 ⇒ rounded to three decimal places
OR
we can use the rule:
# P ( examine at least six ) = 1 − P ( examine at most five )
# P ( examine at least six ) = 1 − P (1) − P (2) − P (3) − P (4) − P (5)
where P is the probability
∵ P(1) = 5/22
∵ P(2) = (5 × 17)/(22 × 21) = 85/462
∵ P(3) = (5 × 17× 16)/(22 ×21 × 20) = 34/231
∵ P(4) = (5 × 17 × 16 × 15)/(22 ×21 × 20 ×19) = 170/1463
∵ P(5) = (5 × 17 × 16 × 15 × 14)/(22 × 21 × 20 × 19 × 18) = 170/1881
∴ P = 1 - 5/22 - 85/462 - 34/231 - 170/1463 - 170/1881 = 442/1881
∴ P = 0.235
These probabilities are calculated by considering the number of combinations of selections.
a) Probability of Exactly Two 23-Watt Bulbs= 0.110
b)Probability of All Bulbs Having the Same Rating = 0.097
c)Probability of One Bulb of Each Type = 0.234
d)Probability of At Least 6 Bulbs Needed to Get a 23-Watt Bulb =0.151
Given a box containing 22 lightbulbs with specific wattage ratings, we can calculate the probabilities of various selection events using combinatorial methods.
(a) Probability of Exactly Two 23-Watt Bulbs
We need the probability of selecting exactly two 23-watt bulbs out of three:
Number of ways to choose 2 out of 5 23-watt bulbs: C(5,2) = 10Number of ways to choose 1 out of the remaining 17 bulbs: C(17,1) = 17Total number of ways to choose any 3 bulbs out of 22: C(22,3) = 1540Thus, the probability is P = (10 * 17) / 1540 = 0.110.
(b) Probability of All Bulbs Having the Same Rating
Probability of all 13-watt: C(8,3) / C(22,3) = 56 / 1540 = 0.036.Probability of all 18-watt: C(9,3) / C(22,3) = 84 / 1540 = 0.055.Probability of all 23-watt: C(5,3) / C(22,3) = 10 / 1540 = 0.006.Total probability: 0.036 + 0.055 + 0.006 = 0.097.
(c) Probability of One Bulb of Each Type
Number of ways to select 1 bulb from each type: 8 * 9 * 5 = 360.
Probability: P = 360 / 1540 = 0.234.
(d) Probability of At Least 6 Bulbs Needed to Get a 23-Watt Bulb
Probability of the first 5 bulbs not being 23-watt: (17/22) * (16/21) * (15/20) * (14/19) * (13/18).
Calculation: P = (17*16*15*14*13) / (22*21*20*19*18) = 0.151.
<c is opposite
please help me find the answer thank you :)
Answer:
A) [tex]\angle C \text{ is opposite } \overline{AB}[/tex]
Step-by-step explanation:
We need to name the side of the triangle that is across from angle C. A side of a triangle, if not otherwise given, is named based on the two points that form it. In this case, the side is formed by points B and C, so the side is called [tex]\overline{BC}[/tex].
How do you solve this system by substitution?
Answer is 'a'.(4;8;8)
All the details are provided in the attachment; the answer is marked with green colour.
Evaluate the Expression : (3+3)^2 / 10-4×3
Simplify 3 + 3 to 6
6^2/10 - 4 × 3
Simplify 4 × 3 to 12
6^2/10 - 12
Simplify 10 - 12 to -2
6^2/-2
Simplify 6^2 to 36
36/-2
Move the negative sign to the left
-36/2
Simplify 36/2 to 18
= -18
Solve |2x - 6| > 10
{x|x < -8 or x > 2}
{x|x < -2 or x > 8}
{x|-2 < x < 8}
ANSWER
{x|x < -2 or x > 8}
EXPLANATION
The given absolute inequality is
[tex] |2x - 6| \: > \: 10[/tex]
By the definition of absolute value,
[tex] (2x - 6)\: > \: 10 \: or \: \: - (2x - 6)\: > \: 10[/tex]
Multiply through the second inequality by -1 and reverse the inequality sign
[tex]2x - 6\: > \: 10 \: or \: \: 2x - 6\: < \: - 10[/tex]
[tex]2x \: > \: 10 + 6\: or \: \: 2x \: < \: - 10 + 6[/tex]
Simplify
[tex]2x \: > \: 16\: or \: \: 2x \: < \: -4[/tex]
Divide through by 2
[tex]x \: > \: 8\: or \: \: x \: < \: -2[/tex]
Answer:
{x|x < -2 or x > 8}
Step-by-step explanation:
|2x - 6| > 10
We split the inequality into two functions, one positive and one negative. The negative one flips the inequality. since this is greater than, this is an or problem
2x-6 >10 or 2x-6 < -10
Add 6 to each side
2x-6+6 > 10+6 2x-6+6 < -10+6
2x > 16 2x < -4
Divide by 2
2x/2 > 16/2 2x/2 < -4/2
x >8 or x < -2
......Help Please.....
Answer:
A= 4 , B= 3
Step-by-step explanation:
Simply put, use systems of equations to solve for A and B. First make y equal 4, and to make A equal to four, while being multiplied by 1 it has to be 4. Then solve backwards and now you know that A equals four, what multiplied by 4 is equal to 36, 9. What is the square root of 9? 3!
Plz help out!! honest answer!
Answer:
2nd choice: Counterclockwise rotation about the origin by 180 degrees followed by a reflection about the y-axis
Step-by-step explanation:
A simple reflection across the x-axis will do.
A rotation of 180 degrees about the origin is equivalent to a reflection across both axes. Then a reflection back across the y-axis leaves the net effect being the desired reflection across the x-axis.
The toco toucan, the largest member of the toucan family, possesses the largest beak relative to body size of all birds. This exaggerated feature has received various interpretations, such as being a refined adaptation for feeding. However, the large surface area may also be an important mechanism for radiating heat (and hence cooling the bird) as outdoor temperature increases. Here are data for beak heat loss, as a percent of total body heat loss, at various temperatures in degrees Celsius: Temperature 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Percent heat loss from beak 32 34 35 33 37 46 55 51 43 52 45 53 58 60 62 62 The equation of the least-squares regression line for predicting beak heat loss, as a percent of total body heat loss from all sources, from temperature is y=? + ? x (Use decimal notation. Give your answer to four decimal places.) Use the equation to predict beak heat loss, as a percent of total body heat loss from all sources, at a temperature of 25 degrees Celsius. ? % What percent of the variation in beak heat loss is explained by the straight-line relationship with temperature? ? % Find the correlation ( 0.001) between beak heat loss and temperature: r=?
The equation of the least-squares regression line can be used to predict the beak heat loss at a specific temperature. The coefficient of determination indicates the proportion of variation explained by the relationship between beak heat loss and temperature. The correlation coefficient quantifies the strength and direction of the linear relationship.
Explanation:The given data represents the relationship between the beak heat loss of the toco toucan and the outdoor temperature. To find the equation of the least-squares regression line, we need to calculate the slope, which represents the change in beak heat loss for every degree Celsius increase in temperature, and the y-intercept, which represents the predicted beak heat loss at 0 degrees Celsius. Using these values, we can then predict the beak heat loss at a temperature of 25 degrees Celsius. To determine the percent of variation explained by the straight-line relationship, we can calculate the coefficient of determination (r^2). Finally, the correlation coefficient (r) represents the strength and direction of the linear relationship between beak heat loss and temperature.
Learn more about least-squares regression line here:https://brainly.com/question/34639207
#SPJ3
Factoring out the GCF
8m^2n^3-24m^2n^2 +4m^3 n
Answer:
[tex]\large\boxed{8m^2n^3-24m^2n^2+4m^3n=4m^2n(2n^2-6n+m)}[/tex]
Step-by-step explanation:
[tex]8m^2n^3-24m^2n^2+4m^3n\\\\8m^2n^3=\boxed{(2)}\boxed{(2)}(2)\boxed{(m)}\boxed{(m)}\boxed{(n)}(n)(n)\\\\24m^2n^2=\boxed{(2)}\boxed{(2)}(2)(3)\boxed{(m)}\boxed{(m)}\boxed{(n)}(n)\\\\4m^3n=\boxed{(2)}\boxed{(2)}\boxed{(m)}\boxed{(m)}(m)\boxed{(n)}\\\\8m^2n^3-24m^2n^2+4m^3n\\\\=\boxed{(2)}\boxed{(2)}\boxed{(m)}\boxed{(m)}\boxed{(n)}\bigg((2)(n)(n)-(2)(3)(n)+(m)\bigg)\\\\=4m^2n(2n^2-6n+m)[/tex]
Suppose more than kn marbles are distributed into n jars. Prove that one jar will contain at least ? k + 1 marbles.
Answer:yes
Step-by-step explanation:because there are more marbles than jars
IF YOU GUYS COULD ACTUALLY HELP ME FOR ONCE THAT WOULD BE GREAT
Students in a school are asked to pick either football, rugby or tennis to play in their lesson.
The number of students who picked football and rugby are in the ratio of 3:2
The number of students who picked rugby and tennis are in the ratio of 6:13
78 students picked tennis.Work out how many students picked football.
To solve this you must make a formula like so:
ratio of rugby over tennis = unknown over students that picked tennis
***Unknown will be x
use a proportion like so...
[tex]\frac{rugby}{tennis} = \frac{x}{tennis}[/tex]
[tex]\frac{6}{13} = \frac{x}{78}[/tex]
Now you must cross multiply
6 * 78 = 13*x
468 = 13x
To isolate x divide 13 to both sides
468/13 = 13x/13
x= 36
This means that 36 people choose rugby when there were 78 people choose tennis
To find how many people choose football you must make another proportion similar to the first proportion:
ratio of football over rugby = unknown over students that picked rugby
use a proportion like so...
[tex]\frac{3}{2} = \frac{x}{36}[/tex]
Now you must cross multiply
3* 36 = 2*x
108 = 2x
To isolate x divide 2 to both sides
108/2 = 2x/2
x= 54
This means that 54 people choose football
Hope this helped!
~Just a girl in love with Shawn Mendes
PROBLEM 1: Looking at different combinations of letters we can decide certain facts. Decide how many different "words" can be formed using all the letters from the word PARSLEY they do not have to mean anything) and how many'words" can be formed using all the letters from the word PEPPER (they do not have to mean anything) NOTE: the 2 answers are different because Parsley does not have repeating letters and pepper does. Give the number of words you can form from each and how you got that number HINT: If you have trouble getting started, do the excersize for the words WOW and WET to help see the pattern.
Answer:
The words from parsley are PAR, YES, RAP, PAY,and SLAP
The words from pepper are PEEP and that is it
Step-by-step explanation:
PLEASE HELP I REALLY DONT UNDERSTAND HOW TO DO THESE QUESTIONS!!!!! THANK YOU I APPRECIATE YOU!!!!
6) The mean salary of 5 employees is $42100. The employee salaries have 2 modes: $34000 and $50500. If the median salary gets a $2400 raise, then ...
a) What is the new mean?
New Mean = $
b) What is the new median?
New Median = $
Answer:
a) $42,580
b) $43,900
Step-by-step explanation:
Recall, "mode" refers to the value which occurs most frequently.
In this case, the question says that there are 2 modes,
this means $34,000 and $50,500 both share the spot for the most frequently appearing salary.
because there are only 5 employees (and hence 5 salaries), the only possible way that there are two modes is if there are two of each mode, leaving only the last salary unknown.
i.e if we list the 5 salaries (in no particular order)
$34,000 $34,000 $50,500 $50,500 $ x
where x is the 5th unknown salary.
Given that the mean is $42,100
Then (34,000 + 34,000 + 50,500 + 50,500 + x) / 5 = 42,100
solving for x gives x = $41,500
Now we know all the values, we can rearrange them in increasing value:
$34,000 $34,000 $41,500 $50,500 $50,500
from this, we can see that the median salary is $41,500
Given that the median salary gets a $2400 raise,
the new median salary = $41,500 + $2400 = $43,900 (Ans for part b)
new mean salary,
= ($34,000 + $34,000 + $43,900 + $50,500 + $50,500 ) / 5
= $42,580 (answer for part a)