Answer:
a.[tex]w(t)=-12e^{2t}[/tex]
b.[tex] y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}[/tex]
Step-by-step explanation:
We have a differential equation
y''-2 y'-35 y=0
Auxillary equation
[tex](D^2-2D-35)=0[/tex]
By factorization method we are finding the solution
[tex]D^2-7D+5D-35=0[/tex]
[tex](D-7)(D+5)=0[/tex]
Substitute each factor equal to zero
D-7=0 and D+5=0
D=7 and D=-5
Therefore ,
General solution is
[tex]y(x)=C_1e^{7t}+C_2e^{-5t}[/tex]
Let [tex]y_1=e^{7t} \;and \;y_2=e^{-5t}[/tex]
We have to find Wronskian
[tex]w(t)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}[/tex]
Substitute values then we get
[tex]w(t)=\begin{vmatrix}e^{7t}&e^{-5t}\\7e^{7t}&-5e^{-5t}\end{vmatrix}[/tex]
[tex]w(t)=-5e^{7t}\cdot e^{-5t}-7e^{7t}\cdot e^{-5t}=-5e^{7t-5t}-7e^[7t-5t}[/tex]
[tex]w(t)=-5e^{2t}-7e^{2t}=-12e^{2t}[/tex]
a.[tex]w(t)=-12e^{2t}[/tex]
We are given that y(0)=-7 and y'(0)=23
Substitute the value in general solution the we get
[tex]y(0)=C_1+C_2[/tex]
[tex]C_1+C_2=-7[/tex]....(equation I)
[tex]y'(t)=7C_1e^{7t}-5C_2e^{-5t}[/tex]
[tex]y'(0)=7C_1-5C_2[/tex]
[tex]7C_1-5C_2=23[/tex]......(equation II)
Equation I is multiply by 5 then we subtract equation II from equation I
Using elimination method we eliminate[tex] C_1[/tex]
Then we get [tex]C_2=-\frac{5}{2}[/tex]
Substitute the value of [tex] C_2 [/tex] in I equation then we get
[tex] C_1-\frac{5}{2}=-7[/tex]
[tex] C_1=-7+\frac{5}{2}=\frac{-14+5}{2}=-\frac{9}{2}[/tex]
Hence, the general solution is
b.[tex] y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}[/tex]
Find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. About ______% of the area is between z equals minus 2 and z equals 2 (or within 2 standard deviations of the mean).
Final answer:
The area under the standard normal distribution curve between z = -2 and z = 2 corresponds to approximately 84.4 %.
Explanation:
To find the indicated area under the curve of the standard normal distribution between z = -2 and z = 2, we refer to a z-table that provides us with the area under the curve to the left of a given z-score.
First, we find the area under the normal curve to the left of z = 2, which typically is around 0.8672.
Since the normal distribution is symmetric about the mean, the area to the left of z = -2 is the same as the area to the right of z = 2, which is 1 - 0.8672 = 0.0228.
The total area between z = -2 and z = 2 is the area to the left of z = 2 minus the area to the left of z = -2, or 0.8672 - 0.0228.
The difference gives us approximately 0.8444, which represents the probability that a value falls within 2 standard deviations of the mean in a standard normal distribution.
Converting this to a percentage, we multiply by 100 to find that about 84.4 % of the area is within 2 standard deviations of the mean.
A thief steals an ATM card and must randomly guess the correct three-digit pin code from a 8-key keypad. Repetition of digits is allowed. What is the probability of a correct guess on the first try?
Answer:
[tex] \frac{1}{ {8}^{3} } [/tex]
Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. (If an answer does not exist, enter DNE.) x = t2 − t + 9, y = t3 − 3t
Answer:
The horizontal tangents occur at: (9,-2) and (11,2)
The vertical tangent occurs at (8.75,-1.375)
See attachment
Step-by-step explanation:
The given parametric equations are:
[tex]x=t^2-t+9[/tex] and [tex]y=t^3-3t[/tex]
The slope function is given by:
[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]
[tex]\frac{dy}{dx}=\frac{3t^2-3}{2t-1}[/tex]
The tangent is vertical when [tex]\frac{dx}{dt}=0[/tex]
[tex]\implies 2t-1=0[/tex]
[tex]t=\frac{1}{2}[/tex]
When [tex]t=\frac{1}{2}[/tex], [tex]x=(\frac{1}{2})^2-\frac{1}{2}+9=8.75[/tex], [tex]y=0.5^3-3(0.5)=-1.375[/tex]
The vertical tangent occurs at (8.75,-1.375)
The tangent is horizontal when [tex]\frac{dy}{dt}=0[/tex]
[tex]3t^2-3=0[/tex]
[tex]\implies t=\pm1[/tex]
When t=1, [tex]x=(1)^2-1+9=9[/tex], [tex]y=1^3-3(1)=-2[/tex]
When t=-1, [tex]x=(-1)^2+1+9=11[/tex], [tex]y=(-1)^3-3(-1)=2[/tex]
The horizontal tangents occur at: (9,-2) and (11,2)
Consider the experiment of rolling a dice. Find the probability of getting an even number 3) and a number that is multiple of 3.
The required probability of rolling a dice and getting an even number and a number that is a multiple of 3 is 1/6.
To find the probability of rolling a dice and getting an even number (3, 4, or 6) and a number that is a multiple of 3 (3 or 6), we need to consider the outcomes that satisfy both conditions.
There are three even numbers on a standard six-sided dice: 2, 4, and 6. Out of these three even numbers, two of them are multiples of 3 (3 and 6).
The probability of rolling an even number is 3/6 (3 even numbers out of 6 possible outcomes) or simply 1/2.
The probability of rolling a number that is a multiple of 3 is 2/6 (2 multiples of 3 out of 6 possible outcomes) or 1/3.
Now, to find the probability of both events occurring together (rolling an even number and a number that is a multiple of 3), we multiply the individual probabilities:
Probability of getting an even number AND a number that is a multiple of 3 = (1/2) * (1/3) = 1/6.
So, the probability of rolling a dice and getting an even number and a number that is a multiple of 3 is 1/6.
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The probability of rolling an even number is 1/2, while the probability of rolling a multiple of 3 is 1/3. However, the probability of rolling a number that is both an even number and a multiple of 3 (which is only the number 6) is 1/6 or approximately 16.67%.
Explanation:The experiment in question involves rolling a single die and finding the probability of getting an even number as well as a number that is a multiple of 3. A standard die has six faces, numbered from 1 to 6. An even number on a die could be 2, 4, or 6, while a multiple of 3 could be 3 or 6.
To find the probability of getting an even number, we count the favorable outcomes for an even number, which are three (2, 4, 6), and divide by the total number of outcomes possible on a die, which is six. This gives us a probability of ½ or 50% for rolling an even number.
To find the number that is a multiple of 3, we have two outcomes (3 and 6). Again, dividing by the total number of outcomes we get a probability of ⅓ or approximately 33.33%.
But in this case, we are interested in the occurrence of both events together, which means we are looking for the probability of getting a number that is both an even number and a multiple of 3. The only number on a die that is both even and a multiple of 3 is the number 6.
Since there is only one favorable outcome (6) and six possible outcomes in total, the probability of rolling a number that is both even and a multiple of 3 is ⅖ or approximately 16.67%.
7. What is the cardinality of each of the following sets?
a) { }
b) { { } }
c) {a, {a}, {a, {a}} }
NOTE: I need the answer type out NOT hand written, the last person to answer this question teriibly hand wrote the answer and I could not read it.
Answer:
The cardinality of set (a) is 0,
the cardinality of set (b) is 1
and
the cardinality of set (c) is 3.
Step-by-step explanation: We are given to find the cardinality of each of the following sets :
(a) { }.
(b) { { } }.
(c) {a, {a}, {a, {a}} }.
We know that
CARDINALITY of a set is the number of elements present in the set.
(a) The given set is A = { }.
The set A is an empty set, so it does not contain any element. Hence, the cardinality of the set A is 0.
(b) The given set is B = { { } }.
The set is B is singleton set, contains only one element (that is empty set). So, the cardinality of the set B is 1.
(c) The given set is C = {a, {a}, {a, {a}} }.
The set C has three elements, a, the set {a} and the set {a, {a}}. So, the the cardinality of set C is 3.
Thus,
the cardinality of set (a) is 0,
the cardinality of set (b) is 1
and
the cardinality of set (c) is 3.
Shawn is interested in purchasing a new computer system for $1,650.00 and would like to apply a down payment of 20%. Calculate the down payment amount. Round dollars to the nearest cent.
Answer: Down payment amount = $330
Step-by-step explanation:
Given in the question that Shawn is interested in purchasing a new computer system and he wants to to give a 20% down payment.
Cost of Computer system = $1650
He would like to made a 20% down payment
So, the down payment amount is as follows:
20% of $1650 = [tex]\frac{20}{100}[/tex] × 1650
= $ 330 ⇒ Down payment amount
Every cereal box has a gift inside, but you cannot tell from the outside what the gift is. The store manager assures you that 11 of the 54 boxes on the shelf have the secret decoder ring. The other 43 boxes on the shelf have a different gift inside. If you randomly select two boxes of cereal from the shelf to purchase, what is the probability that BOTH of them have the secret decoder ring?
Answer:
The probability that BOTH of them have the secret decoder ring is [tex]\frac{55}{1431}[/tex].
Step-by-step explanation:
From the given information it is clear that the total number of boxes is 54.
Total number of boxes that have the secret decoder ring = 11
Total number of boxes that have a different gift inside = 43
Total number of ways to select 2 boxes from the boxes that have the secret decoder ring is
[tex]\text{Favorable outcomes}=^{11}C_2=\frac{11!}{2!(11-2)!}=\frac{11\times 10\times 9!}{2!9!}=55[/tex]
Total number of ways to select 2 boxes from the total number of boxes is
[tex]\text{Total outcomes}=^{54}C_2=\frac{52!}{2!(52-2)!}=\frac{52\times 51\times 50!}{2!50!}=1431[/tex]
The probability that BOTH of them have the secret decoder ring is
[tex]P=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]
[tex]P=\frac{55}{1431}[/tex]
Therefore the probability that BOTH of them have the secret decoder ring is [tex]\frac{55}{1431}[/tex].
Probability = 5/131, which is the final answer.
To calculate the probability that both boxes contain the secret decoder ring, we need to consider that the boxes are selected one after the other without replacement. Since there are 11 boxes with secret decoder rings out of 54, the probability of picking a secret decoder ring on the first try is 11/54. If a secret decoder box is picked first, there will be 10 secret decoder rings left out of the remaining 53 boxes. So the probability of picking another secret decoder ring is 10/53.
The overall probability of both events happening is the product of the two probabilities because we need both events to occur. Therefore, we multiply the separate probabilities together:
Probability = (11/54) imes (10/53)
When we calculate this, we get:
Probability = 110/2862
This fraction can be simplified to:
Probability = 5/131, which is the final answer.
A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.
Answer:
Step-by-step explanation:
With the information you have provided, the volume can only be stated as a third degree polynomial. We do not know the measurements of the square cut out of each corner, we can only call it "x".
If each side of the square measures 3 feet wide and we cut out 2 squares from a side, the length of that side is 3 - 2x. Volume is length times width times height. The length and the width will both be 3 - 2x, and the height is the measurement of x. But since we don't know it, the height is just x. Multiplying the length times the width times the height looks like this:
(3 - 2x)(3 - 2x)(x)
When you FOIL all this together you get a third degree polynomial
[tex]4x^3-12x^2+9x[/tex]
If you know the measurement of the squares cut out, plug that value in for x and get the volume in a number.
John a US citizen, is a product manager who moved to Argentina to work at the South American office of Ridmore Corporation, arn American compary Based on this information, John is an)
A) expatriate manger
B) third-country manager
C)inpatriate manager
D) polycentric manager
E) virtual manager
Answer:
The correct answer is A. Being American but working in Argentina, John is an expatriate manager.
Step-by-step explanation:
An expatriate is a person who, temporarily or permanently, resides in a country different from the country in which he was born. The term is commonly used in the case where companies send their professionals or workers to their delegations abroad.
In a survey conducted by Helena, a financial consultant, it was revealed of her 426 clients
288 own stocks.
200 own bonds.
184 own mutual funds.
123 own both stocks and bonds.
106 own both stocks and mutual funds.
102 own both bonds and mutual funds.
How many of Helena's clients own stocks, bonds, and mutual funds? (Assume each client invested in at least one of the three types of funds.)
_______clients
Answer: There are 85 Helena's client own stocks, bonds and mutual funds.
Step-by-step explanation:
Since we have given that
Let A: who own stocks
B : who own bonds
C : who own mutual fund
So, According to question,
n(A) = 288
n(B) = 200
n(C) = 184
n(A∩B) = 123
n(B∩C) = 106
n( A∩C) = 102
n(A∪B∪C) = 426
As we know the formula :
[tex]n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\426=288+200+184-123-106-102+n(A\cap B\cap C)\\\\426-341=n(A\cap B\cap C)\\\\85=n(A\cap B\cap C)[/tex]
Hence, there are 85 Helena's client own stocks, bonds and mutual funds.
To determine the number of clients who own stocks, bonds, and mutual funds, we can use the principle of inclusion-exclusion. This principle allows us to properly account for overlap in the sets of clients for different investment types.
We were told the following:
- The total number of clients is 426.
- The number of clients who own stocks is 288.
- The number of clients who own bonds is 200.
- The number of clients who own mutual funds is 184.
- The number of clients who own both stocks and bonds is 123.
- The number of clients who own both stocks and mutual funds is 106.
- The number of clients who own both bonds and mutual funds is 102.
Now, when we sum up the number of clients who own stocks, bonds, and mutual funds individually, we're double-counting those clients who have investments in more than one of these. We need to subtract the clients who are counted twice.
So, let's add up all the individuals:
S + B + M = 288 + 200 + 184 = 672
Now, let's subtract the number of clients who were double-counted:
(S + B + M) - (SB + SM + BM) = 672 - (123 + 106 + 102) = 672 - 331 = 341
However, in this calculation, we've subtracted clients who own all three: stocks, bonds, and mutual funds, three times (once for each pair), and then added them back in only once, so we've subtracted them two times too many.
Therefore, we need to correct for this: to find the number of people who own all three, we add the total number of clients (since everyone owns at least one of the three) and then subtract the sum we have just calculated.
Total + All three (overcorrected) = Total clients
426 + All three (overcorrected) = 341
To solve for the overcorrection (the actual number of clients who own all three types), we can now rearrange the equation:
All three (overcorrected) = 341 - 426
All three (overcorrected) = -85
However, since the number of people cannot be negative, this outcome indicates a logical inconsistency. Such an inconsistency generally means there must have been a mistake in either the data provided or the calculations based on that data. Under normal circumstances, you would go back and verify the numbers. But given this answer, it would suggest that the data provided has some inconsistencies, and it is not possible for a negative number of clients to own all three funds.
Simplify the following. (a) 3(-3 + 5x) - 1 (4 - 4x) (b) 3 squareroot 64 x^15 y^3 -2(-15 e^5 t/30 e^-2 t^-3)^0
Answer:
a. 19x-13 b. [tex]2(32x^{15}y^{3}-1).[/tex]
Step-by-step explanation:
a. 3(-3+5x)-1(4-4x) = -9+15x-4+4x = 15x+4x-9-4 = 19x-13.
b. [tex]64x^{15}y^{3}-2(-15e^{5}\frac{t}{30}e^{-2}t^{-3})^{0}[/tex]
= [tex]64x^{15}y^{3}-2 = 2(32x^{15}y^{3}-1).[/tex]
Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare your answer with the value of the integral produced by a calculator. (Round your answers to the nearest whole number.) y = 1 5 x5, 0 ≤ x ≤ 5
[tex]\[ \int_{0}^{5} \frac{1}{5} x^5 \, dx \approx 520.8333 \][/tex]
Given:is picking out some movies to rent, primarily interested in horror films and comedies. He has narrowed down his selections to 15 horror films and 18 comedies. How many different combinations of 4 movies can he rent if he wants at least two comedies?
Answer: 31,365
Step-by-step explanation:
Given : The number of horror films = 15
The number of comedies = 18
Then , the number of different combinations of 4 movies can he rent if he wants at least two comedies is given by :-
[tex]^{18}C_2\times ^{15}C_2+^{18}C_3\times ^{15}C_1+^{18}C_4\times ^{15}C_0\\\\=\dfrac{18!}{2!(18-2)!}\times\dfrac{15!}{2!(15-2)!}+\dfrac{18!}{3!(18-3)!}\times\dfrac{15!}{1!(15-1)!}+\dfrac{18!}{4!(18-4)!}\times\dfrac{15!}{0!(15-0)!}\\\\=16065+12240+3060=31365[/tex]
Hence, the number of different combinations of 4 movies can he rent if he wants at least two comedies = 31,365
The student can calculate different combinations of 4 movies by considering the cases for two, three, or four comedies, and then adding up the combinations for each case.
Explanation:The student is interested in renting movies and wants to know how many different combinations of 4 movies he can rent if he requires at least two comedies from a selection of 15 horror films and 18 comedies. We can use the combination formula C(n, k) = n! / (k!(n - k)!), where n is the total number of items to choose from, k is the number of items to choose, and ! denotes factorial.
To calculate the combinations with at least two comedies, we must consider the following cases:
Two comedies and two horror filmsThree comedies and one horror filmFour comediesFor each case, we calculate the combinations separately for comedies and horror films and then multiply them:
Two comedies and two horror films: C(18, 2) * C(15, 2)Three comedies and one horror film: C(18, 3) * C(15, 1)Four comedies: C(18, 4)Finally, we add all these possibilities together to get the total number of combinations.
A coin is tossed 5 times. Find the probability that exactly 1 is a tail. Find the probability that at most 2 are tails.
Answer:
Step-by-step explanation:
First questionThe only possibilities where there is exactly 1 tail are:
(t,h,h,h,h)(h,t,h,h,h)(h,h,t,h,h)(h,h,h,t,h)(h,h,h,h,t)those are 5 favorable outcomes.
where h represent heads and t represent tails. There are [tex]2^5 32[/tex] total number of outcomes after tossing the coin 5 times. Because every time you toss the coin, you have 2 possibilities, and as you do it 5 times, those are [tex]2^5[/tex] options. We can conclude from this that
The probability that exactly 1 is a tail is [tex]5/32[/tex].
Second questionWe already know the total number of outcomes; 32. Now we need to find the number of favorable outcomes. In order to do that, we can divide our search in three cases: 1.-there are no tails, 2.-exactly 1 is a tail, 3.- exactly 2 are tails.
The first case is 1 when every coin is a head. The second case we already solved it, and there are 5. The third case is the interesting one, we can count the outcomes as we did in the previous questions, but that's only because there are not too many outcomes. Instead we are going to use combinations:
We need to have 2 tails, the other coins are going to be heads. We made 5 tosses, then the possible combinations are [tex]C_{5,2} = \frac{5!}{3!2!} = \frac{120}{6*2} = 10[/tex]
Finally, we conclude that there are 1 + 5 + 10 favorable outcomes, and this implies that
The probability that at most 2 are tails is [tex]\frac{16}{32} = \frac{1}{2}[/tex].
In a five-coin toss, the probability of getting exactly one tail is 5/32 and the probability of getting at most two tails is 0.5. These probabilities are calculated considering all possible outcomes and arranging the heads and tails in distinct manners.
Explanation:The question you've asked involves calculating the probabilities in coin flipping, a common concept in mathematics and particularly in statistics. This falls under the topic of probability theory.
When a fair coin is tossed 5 times, there are 2^5 or 32 equally likely outcomes. If we want exactly 1 tail, there are 5 ways this can happen (one for each position the tails can be in). Thus, the probability for this occurrence is 5/32.
To find out the probability of getting at most 2 tails, we need to calculate the probability for getting exactly 0, 1, or 2 tails. As we already know that the probability for 1 tail is 5/32 and for 0 tails is 1/32 (only 1 way to get this outcome, getting heads every time). The probability for exactly 2 tails can be found in the same manner as for 1 tail, now we have 2 tails and it can be arranged in 5C2 ways which is 10 ways. Therefore, the probability of 2 tails is 10/32. Hence, the probability of getting at most 2 tails is the sum of probabilities of 0,1 or 2 tails, which is (1 + 5 + 10 )/32 = 16/32 = 0.5.
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Using the simple random sample of weights of women from a data set, we obtain these sample statistics: nequals45 and x overbarequals148.79 lb. Research from other sources suggests that the population of weights of women has a standard deviation given by sigmaequals31.37 lb. a. Find the best point estimate of the mean weight of all women. b. Find a 90% confidence interval estimate of the mean weight of all women.
Answer: (141.1, 156.48)
Step-by-step explanation:
Given sample statistics : [tex]n=45[/tex]
[tex]\overline{x}=148.79\text{ lb}[/tex]
[tex]\sigma=31.37\text{ lb}[/tex]
a) We know that the best point estimate of the population mean is the sample mean.
Therefore, the best point estimate of the mean weight of all women = [tex]\mu=148.79\text{ lb}[/tex]
b) The confidence interval for the population mean is given by :-
[tex]\mu\ \pm E[/tex], where E is the margin of error.
Formula for Margin of error :-
[tex]z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}[/tex]
Given : Significance level : [tex]\alpha=1-0.90=0.1[/tex]
Critical value : [tex]z_{\alpha/2}=z_{0.05}=\pm1.645[/tex]
Margin of error : [tex]E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69[/tex]
Now, the 90% confidence interval for the population mean will be :-
[tex]148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)[/tex]
Hence, the 90% confidence interval estimate of the mean weight of all women= (141.1, 156.48)
The Earth is 4.6 billion years old, but human civilization (the building of cities) began only about 10,000 years ago. If you represent the history of Earth with a line one mile long (63,360 inches), how long must the line be to represent the history of human civilization?
Answer:
0.13774 inches ( approx ).
Step-by-step explanation:
Given,
The age of earth = 4.6 billion years = 4600000000 years,
The represented age of earth = 63,360 inches,
That is, scale factor in representation of age
[tex]=\frac{\text{Represented age}}{\text{Actual age}}[/tex]
[tex]=\frac{ 63,360}{4600000000}[/tex]
Now, the age of human civilization = 10000 years,
Thus, the represented age of human civilization = actual age × scale factor
[tex]=10000\times \frac{ 63,360}{4600000000}[/tex]
[tex]=\frac{633600000}{4600000000}[/tex]
[tex]=0.137739130435[/tex]
[tex]\approx 0.13774\text{ inches}[/tex]
Which of the following is true regarding the PTIN? A. A PTIN is required to prepare or sign most tax returns. B A PTIN is required to represent a taxpayer before IRS. C. The PTIN is renewed semi-annually. D. Your PTIN can be shared with other members of a firm
Answer:
The first option is the correct answer.
Step-by-step explanation:
A PTIN is required to prepare or sign most tax returns.
The PTIN or full form Preparer Tax Identification Number (PTIN) is an identification number, used by the pre parers to claim for refund or compensation during tax return filing.
So, a person who has to claim refund must have his or her own PTIN and each tax return pre parer may only obtain one PTIN.
A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall above10.95 ounces.
Answer:
6.68 %.
Step-by-step explanation:
The standardised z-score = ( 10.95 - 10.5) / 0.3
= 1.5.
Looking up the Normal Distribution tables ( area to the left) 1.5 corresponds to 0.93319 so for a weight above 10.95 the proportion is
1 - 0.93319 = 0.06681
= 6.68%.
We can use the z-score to find the proportion of jars that are above a certain weight in a normal distribution. The z-score for 10.95 ounces is 1.5. Using a standard normal distribution table, we find that about 6.68% of the jars weigh more than 10.95 ounces.
Explanation:In this problem, we are using the concept of normal distribution, specifically to find the proportion of jars that are above a certain weight. Given that the mean (average) weight of the jars is 10.5 ounces and the standard deviation (which measures the dispersion of the weights) is 0.3 ounce, we can calculate the z-score for 10.95 ounces.
The z-score is defined as the number of standard deviations a data point is from the mean. Compute it using the formula: Z = (X - μ) / σ where X is the value, μ is the mean and σ is the standard deviation.
Plugging into the formula we get: Z = (10.95 - 10.5) / 0.3 = 1.5
You can then look up this z-score in a standard normal distribution table (or use a calculator or computer software that calculates it), to find the proportion below this z-score. But we need the proportion above, so we subtract this from 1. Let's say the value from a z-table for 1.5 is 0.9332, the proportion of values above this would be 1 - 0.9332 = 0.0668 or about 6.68% of the jars weigh more than 10.95 ounces.
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Prove that if a is equivalent to 5 mod (8) and b is equivalent to 3 mod (8), then 8 divides ab+1
Answer:
Explanation contains the proof.
Step-by-step explanation:
[tex]a \equiv 5 (mod 8) \text{ means there is integer } k \text{ such that } a-5=8k[/tex].
[tex]b \eqiv 3 (mod 8) \text{ means there is integer } m \text{ such that } b-3=8m[/tex].
We want to show that [tex]8 \text{ divides } ab+1[/tex]. So we are asked to show that there exist integer [tex]n \text{ such that } 8n=ab+1 \text{ or 8n-1=ab[/tex]
So what is [tex]ab[/tex]?
[tex]a-5=8k \text{ gives us } a=8k+5[/tex].
[tex]b-5=8m \text{ gives us } b=8m+5[/tex].
So back to [tex]ab[/tex]....
[tex]ab[/tex]
[tex]=(8k+5)(8m+5)[/tex]
[tex]=64km+40k+40m+25[/tex] (I use foil to get this)
Factoring out 8 gives us:
[tex]=8(8km+5k+5m)+25[/tex]
Now I could have factored some 8's out of 25. There are actually three 8's in 25 with a remainder of 1.
[tex]=8(8km+5k+5m+3)+1[/tex]
We have shown that there is integer [tex]n \text{ such that } ab=8n-1[/tex].
The integer I found that is n is 8km+5k+5m+3.
Therefore [tex]8|(ab+1)[/tex].
//
Answer:
See below.
Step-by-step explanation:
If a = 5 mod 8 and b = 3 mod 8
then ab = 5*3 mod 8 = 15 mod 8 = 7 mod 8.
ab + 1 = 8 mod 8 = 0 mod 8 so it is divisible by 8.
2^30 + 2^30 + 2^30 + 2^30 =
a) 8^120 b) 8^30 c) 8^32 d) 2^32 e) 2^30
Answer:
D
Step-by-step explanation:
[tex]x+x+x+x=4 \cdot x \text{ or } 4x[/tex]
So [tex]2^{30}+2^{30}+2^{30}+2^{30}=4 \cdot 2^{30}[/tex].
Now 4 can be rewritten so that it is 2 to some power.
4 is actually 2 to the second.
That is, [tex]4=2^2[/tex].
So [tex]2^{30}+2^{30}+2^{30}+2^{30}=2^2 \cdot 2^{30}[/tex].
Now there is a law of exponents that says if the bases are the same and your multiplying add the exponents.
[tex]2^{30}+2^{30}+2^{30}+2^{30}=2^2 \cdot 2^{30}[/tex].
[tex]2^{30}+2^{30}+2^{30}+2^{30}=2^{32}[/tex].
Answer:
The answer is D! Hope this helped.
Step-by-step explanation:
9. If you have twice as many pennies as nickels, and the total value of these coins is the same as the value of 32 quarters, how many pennies do you have?
Answer:
The answers is 228 pennies.
Step-by-step explanation:
The values of the subunits of a dollar are:
1 penny = $0.01.
1 nickel = $0.05.
1 quarter = $0.25.
If all the pennies and nickels that you have are equal to 32 quarters:
32*$0.25=$8Now, we have twice pennies as many nickels, and x is the total amount of coins:
2x * $0.1 + $0.05x = $8Multiply and add the terms of the equation:
$0.02x + $0.05x = $8$0.07x = $8Divide by $0.07 in both terms of the equation:
x = $8/$0.07x ≅ 114 because you can have a half penny nor nickel.When x = 114 means that you have 114 nickels and 228 pennies.
Replacing x=114
$0.02*114 + $0.05*114 = $8Given P(A) = 0.37, P(B) 0.38, and P(A or B) 0.27, are events A and B mutually exclusive? Yes, they are mutually exclusive No, they are not mutually exclusive
Answer:
No, they are not mutually exclusive.
Step-by-step explanation:
Two events are called mutually exclusive they are independent to each other.
Also, If A and B are any two events,
Then they are called mutually exclusive,
If P(A ∪ B ) = P(A) + P(B)
Or P ( A or B ) = P(A) + P(B)
Here, P(A) = 0.37, P(B) = 0.38 and P(A or B) = 0.27
Since,
0.27 ≠ 0.37 + 0.38
⇒ P ( A or B ) ≠ P(A) + P(B)
Hence, A and B are not mutually exclusive.
Consider the following functions. f(x) = x − 3, g(x) = x2 Find (f + g)(x). Find the domain of (f + g)(x). (Enter your answer using interval notation.) Find (f − g)(x). Find the domain of (f − g)(x). (Enter your answer using interval notation.) Find (fg)(x). Find the domain of (fg)(x). (Enter your answer using interval notation.) Find f g (x). Find the domain of f g (x). (Enter your answer using interval notation.)
Answer:
[tex](f+g)(x)=x-3+x^2[/tex] ; Domain = (-∞, ∞)
[tex](f-g)(x)=x-3-x^2[/tex] ; Domain = (-∞, ∞)
[tex](fg)(x)=x^3-3x^2[/tex] ; Domain = (-∞, ∞)
[tex](\frac{f}{g})(x)=\frac{x-3}{x^2}[/tex] ; Domain = (-∞,0)∪(0, ∞)
Step-by-step explanation:
The given functions are
[tex]f(x)=x-3[/tex]
[tex]g(x)=x^2[/tex]
1.
[tex](f+g)(x)=f(x)+g(x)[/tex]
Substitute the values of the given functions.
[tex](f+g)(x)=(x-3)+x^2[/tex]
[tex](f+g)(x)=x-3+x^2[/tex]
The function [tex](f+g)(x)=x-3+x^2[/tex] is a polynomial which is defined for all real values x.
Domain of (f+g)(x) = (-∞, ∞)
2.
[tex](f-g)(x)=f(x)-g(x)[/tex]
Substitute the values of the given functions.
[tex](f-g)(x)=(x-3)-x^2[/tex]
[tex](f-g)(x)=x-3-x^2[/tex]
The function [tex](f-g)(x)=x-3-x^2[/tex] is a polynomial which is defined for all real values x.
Domain of (f-g)(x) = (-∞, ∞)
3.
[tex](fg)(x)=f(x)g(x)[/tex]
Substitute the values of the given functions.
[tex](fg)(x)=(x-3)x^2[/tex]
[tex](fg)(x)=x^3-3x^2[/tex]
The function [tex](fg)(x)=x^3-3x^2[/tex] is a polynomial which is defined for all real values x.
Domain of (fg)(x) = (-∞, ∞)
4.
[tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)}[/tex]
Substitute the values of the given functions.
[tex](\frac{f}{g})(x)=\frac{x-3}{x^2}[/tex]
The function [tex](\frac{f}{g})(x)=\frac{x-3}{x^2}[/tex] is a rational function which is defined for all real values x except 0.
Domain of (f/g)(x) = (-∞,0)∪(0, ∞)
[tex](f + g)(x) = x^2 + x - 3[/tex], domain: all real numbers.
[tex](f - g)(x) = -x^2 + x - 3[/tex], domain: all real numbers.
[tex](fg)(x) = x^3 - 3x^2[/tex], domain: all real numbers.
[tex]f(g(x)) = x^2 - 3[/tex], domain: all real numbers.
To find (f + g)(x), we need to add the functions f(x) and g(x).
The function f(x) = x - 3 and the function [tex]g(x) = x^2.[/tex]
So, [tex](f + g)(x) = f(x) + g(x) = (x - 3) + (x^2).[/tex]
Expanding this equation, we get [tex](f + g)(x) = x^2 + x - 3.[/tex]
To find the domain of (f + g)(x), we need to consider the domain of the individual functions f(x) and g(x).
Since both f(x) = x - 3 and [tex]g(x) = x^2[/tex] are defined for all real numbers, the domain of (f + g)(x) is also all real numbers.
To find (f - g)(x), we need to subtract the function g(x) from f(x).
So, [tex](f - g)(x) = f(x) - g(x) = (x - 3) - (x^2).[/tex]
Expanding this equation, we get [tex](f - g)(x) = -x^2 + x - 3.[/tex]
The domain of (f - g)(x) is also all real numbers, since both f(x) and g(x) are defined for all real numbers.
To find (fg)(x), we need to multiply the functions f(x) and g(x).
So, [tex](fg)(x) = f(x) * g(x) = (x - 3) * (x^2).[/tex]
Expanding this equation, we get [tex](fg)(x) = x^3 - 3x^2.[/tex]
The domain of (fg)(x) is all real numbers, since both f(x) and g(x) are defined for all real numbers.
To find f(g(x)), we need to substitute g(x) into the function f(x).
So, [tex]f(g(x)) = f(x^2) = x^2 - 3.[/tex]
The domain of f(g(x)) is also all real numbers, as [tex]g(x) = x^2[/tex] is defined for all real numbers, and f(x) = x - 3 is defined for all real numbers.
In summary:
- [tex](f + g)(x) = x^2 + x - 3[/tex], domain: all real numbers.
- [tex](f - g)(x) = -x^2 + x - 3[/tex], domain: all real numbers.
- [tex](fg)(x) = x^3 - 3x^2[/tex], domain: all real numbers.
- [tex]f(g(x)) = x^2 - 3[/tex], domain: all real numbers.
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If the costs (S and H) and demands (D) are the same, which of the following is not true with regard to the EPQ model as compared to the EOQ model?
a) the EPQ model produces a lower annual cost
b) the maximum inventory level is lower under the EPQ model than under the EOQ model
c) both models use the same formula to compute annual ordering cost
d) the inventory depletion rate is not the same for both models
e) the two models use different formulas to compute annual holding costs
Final answer:
The EPQ model differs from the EOQ model in terms of annual cost, maximum inventory level, inventory depletion rate, and formulas for computing holding costs.
Explanation:
In the EPQ (Economic Production Quantity) model, the costs and demands are assumed to be the same. Compared to the EOQ (Economic Order Quantity) model, there are a few differences:
The EPQ model does not necessarily produce a lower annual cost. It may or may not, depending on the specific parameters and assumptions.The maximum inventory level is typically higher under the EPQ model than under the EOQ model.Both models use the same formula to compute annual ordering cost.The inventory depletion rate is not the same for both models. In the EPQ model, the rate at which inventory is used depends on the production rate, whereas the EOQ model assumes a constant rate of usage.The two models use different formulas to compute annual holding costs. The EPQ model considers holding cost as a percentage of average inventory, while the EOQ model considers holding cost as a function of order quantity and holding cost per unit..A single card is randomly drawn from a deck of 52 cards. a, Find the probability that it is an ace. b. Find the probability that it is a number less than 5 (not including the ace).
Answer:
P( ace) = 1/13
P (< 5) = 3/13
Step-by-step explanation:
There are 4 Aces in a deck of cards, 1 in each suit
P (ace) = number of aces/ total cards
= 4/52 = 1/13
Numbers less than 5: 4 twos, 4 threes, 4 fours =12 cards less than 5 ( since we do not include aces) There are 4 of each card since there is one in each suit.
P ( card < 5) = cards less than 5/ total
=12/52 = 3/13
Answer:
A: 1/13 B: 3/13
Step-by-step explanation:
There are 4 aces in a deck. 4/52=1/13
If you are talking about 2,3,4 then 12/52=3/13
If you include the 5 it is 16/64= 4/13
The measurement of the height of 600 students of a college is normally distributed with a mean of
175 centimeters and a standard deviation of 5 centimeters.
What percent of students are between 180 centimeters and 185 centimeters in height?
12.5
13.5
34
68
Answer: Second Option
[tex]P(180<X <185)=13.5\%[/tex]
We know that the mean is:
[tex]\mu=175[/tex]
and the standard deviation is:
[tex]\sigma=5[/tex]
We are looking at the percentage of students between 180 centimeters and 185 centimeters in height.
This is:
[tex]P(180<X <185)[/tex]
We calculate the Z-score using the formula:
[tex]Z=\frac{X-\mu}{\sigma}[/tex]
For [tex]X=180[/tex]
[tex]Z_{180}=\frac{180-175}{5}[/tex]
[tex]Z_{180}=1[/tex]
For [tex]X=185[/tex]
[tex]Z_{185}=\frac{185-175}{5}[/tex]
[tex]Z_{185}=2[/tex]
Then we look at the normal table
[tex]P(1<Z<2)[/tex]
[tex]P(1<Z<2)=P(Z<2)-P(Z<1)[/tex]
[tex]P(1<Z<2)=0.9772-0.8413[/tex]
[tex]P(1<Z<2)=0.135[/tex]
[tex]P(180<X <185)=13.5\%[/tex]
Note: You can get the same conclusion using the empirical rule
Look at the attached image for [tex]\mu+ 1\sigma <\mu <\mu + 2\sigma[/tex]
The most popular mathematician in the world is throwing aparty for all of his friends. As a way to kick things off, they decidethat everyone should shake hands. Assuming all 10 people atthe party each shake hands with every other person (but notthemselves, obviously) exactly once, how many handshakes takeplace?
Answer:
The no. of possible handshakes takes place are 45.
Step-by-step explanation:
Given : There are 10 people in the party .
To Find: Assuming all 10 people at the party each shake hands with every other person (but not themselves, obviously) exactly once, how many handshakes take place?
Solution:
We are given that there are 10 people in the party
No. of people involved in one handshake = 2
To find the no. of possible handshakes between 10 people we will use combination over here
Formula : [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
n = 10
r= 2
Substitute the values in the formula
[tex]^{10}C_{2}=\frac{10!}{2!(10-2)!}[/tex]
[tex]^{10}C_{2}=\frac{10!}{2!(8)!}[/tex]
[tex]^{10}C_{2}=\frac{10 \times 9 \times 8!}{2!(8)!}[/tex]
[tex]^{10}C_{2}=\frac{10 \times 9 }{2 \times 1}[/tex]
[tex]^{10}C_{2}=45[/tex]
No. of possible handshakes are 45
Hence The no. of possible handshakes takes place are 45.
Type in only your numerical answer to the following problem; do not type any words or letters with your answer. The population of Neverland was 4.3 billion, one hundred years ago. Currently the population is 5.2 billion. What will this population be 100 years from now? NOTE: Round to the nearest tenth of a billion.
Answer:
6.3 billion ( approx )
Step-by-step explanation:
Given,
The population of Neverland = 4.3 billion,
Let it is increasing by the rate of r % per year,
So, the population after 100 years,
[tex]A=4.3(1+\frac{r}{100})^{100}[/tex]
According to the question,
A = 5.2 billion,
[tex]\implies 5.2 = 4.3(1+\frac{r}{100})^{100}[/tex]
By the graphing calculator,
[tex]\implies r = 0.19\%[/tex]
If the initial population is 5.2 billion,
Then the population after 100 years,
[tex]A=5.2(1+\frac{0.19}{100})^{100}[/tex]
[tex]=6.28696444394[/tex]
[tex]\approx 6.3\text{ billion}[/tex]
If a distribution of test scores is normal with a mean of 78 and a standard deviation of 11, calculate the z-score for the following scores X Z-score 60 70 80 90 60 65 70 80 99 89 75 Make sure to round up your answers to two digits after the decimal point.
To calculate the z-score, subtract the mean from the score and divide by the standard deviation. The calculated z-scores for the given test scores are -1.64, -0.73, 0.18, 1.09, -1.64, -1.18, -0.73, 0.18, 1.91, 1.00, and -0.27.
Explanation:The z-score is a standardized value that measures how many standard deviations a score is above or below the mean. To calculate the z-score, we use the formula z = (X - µ) / σ, where X is the score, µ is the mean, and σ is the standard deviation.
For the score 60, the z-score is (60 - 78) / 11 = -1.64.For the score 70, the z-score is (70 - 78) / 11 = -0.73.For the score 80, the z-score is (80 - 78) / 11 = 0.18.For the score 90, the z-score is (90 - 78) / 11 = 1.09.For the score 60, the z-score is (60 - 78) / 11 = -1.64.For the score 65, the z-score is (65 - 78) / 11 = -1.18.For the score 70, the z-score is (70 - 78) / 11 = -0.73.For the score 80, the z-score is (80 - 78) / 11 = 0.18.For the score 99, the z-score is (99 - 78) / 11 = 1.91.For the score 89, the z-score is (89 - 78) / 11 = 1.00.For the score 75, the z-score is (75 - 78) / 11 = -0.27.Which of the following is NOT a conclusion of the Central Limit Theorem? Choose the correct answer below. A. The mean of all sample means is the population mean mu. B. The standard deviation of all sample means is the population standard deviation divided by the square root of the sample size. C. The distribution of the sample data will approach a normal distribution as the sample size increases. D. The distribution of the sample means x overbar will, as the sample size increases, approach a normal distribution.
Answer:
The distribution of the sample data will approach a normal distribution as the sample size increases.
Step-by-step explanation:
Central limit theorem states that the mean of all samples from the same population will be almost equal to the mean of the population, if the large sample size from a population, is given with a finite level of variance.
So, here Option C is not correct conclusion of central limit theorem -The distribution of the sample data will approach a normal distribution as the sample size increases.
We can say that the average of sample mean tends to be normal but not the sample data.
The option that's is not a conclusion of the Central Limit Theorem is C. The distribution of the sample data will approach a normal distribution as the sample size increases.
According to the central limit theorem, the standard deviation of all sample means will be the population standard deviation divided by the square root of the sample size.
Also, the mean of all sample means is the population mean is equal to the mean of the population.
It should be noted that the distribution of the sample data will not approach a normal distribution as the sample size increases.
In conclusion, the correct option is C.
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