Answer: 0.5507
Step-by-step explanation:
The cumulative distribution function for exponential distribution:-
[tex]P(X\leq x)=1-e^{-\lambda x}[/tex]
Given : The life of a light bulb is exponentially distributed with a mean of 1,000 hours.
Then , [tex]\lambda=\dfrac{1}{1000}[/tex]
Then , the probability that the bulb will last less than 800 hours is given by :-
[tex]P(X\leq 800)=1-e^{\frac{-800}{1000}}\\\\=1-e^{-0.8}=0.5506710358\approx0.5507[/tex]
Hence, the probability that the bulb will last less than 800 hours = 0.5507
The probability that the bulb will last less than 800 hours is 0.550
Further explanationAn incandescent light bulb is an electric light with a wire filament heated to such a high temperature that it glows with visible light. The filament of an incandescent light bulb is protected from oxidation with a glass or fused quartz bulb that is filled with inert gas or a vacuum. The exponential distribution is the probability distribution of the time between events in a Poisson point process. It is a particular case of the gamma distribution. While the arithmetic mean is the central value of a discrete set of numbers: specifically, the sum of the values divided by the number of values
The life of a light bulb is exponentially distributed with a mean of 1,000 hours. What is the probability that the bulb will last less than 800 hours?
What is the probability that the light bulb will have to replaced within 800 hours?
[tex]\lambda = \frac{1}{1000}[/tex]
[tex]P(x<=800) = 1 - e^{(-\lambda*x)} = 1-e^{[(-\frac{1}{1000} )*800]} = 1-e^{-0.8} = 1-0.449 = 0.550[/tex]
Learn moreLearn more about bulb https://brainly.com/question/9584045Learn more about hours https://brainly.com/question/1032699Learn more about mean https://brainly.com/question/3777157Answer details
Grade: 5
Subject: math
Chapter: probability
Keywords: mean, bulb, hours, exponential, light
Theorem: Let a and b be positive integer, if gcd(a,b)=1, then exist positive integer x and y such that ax+by=c for any integer greater than ab-a-b.
Prove the above theorem.
Answer with explanation:
It is given that , a and b are positive integers.
gcd(a,b)=1
We have to prove for any positive integer x and y ,
a x + by =c, for any integer greater than ab-a-b.
Proof:
GCD of two numbers is 1, when two numbers are coprime.
Consider two numbers , 9 and 7
GCD (9,7)=1
So, we have to calculate positive integers x and y such that
⇒ 9 x +7 y > 9×7-9-7
⇒9x +7 y> 47
To prove this we will draw the graph of Inequality.
So the ordered pair of Integers are
x>5 and y>6.
So, for any integers , a and b ,
→ax+ by > a b -a -b, if
[tex]\Rightarrow \frac{x}{\frac{ab-a-b}{a}}+ \frac{y}{\frac{ab-a-b}{b}}>1,\frac{ab-a-b}{a},\text{and},\frac{ab-a-b}{b}[/tex]
⇒Range of x for which this inequality hold
[tex]=[\frac{ab-a-b}{a},\infty)[/tex]
if,
[tex]\frac{ab-a-b}{a}[/tex]
is an Integer ,otherwise range of x
[tex]=(\frac{ab-a-b}{a},\infty)[/tex]
⇒Range of y for which this inequality hold
[tex]=[\frac{ab-a-b}{b},\infty)[/tex]
if,
[tex]\frac{ab-a-b}{b}[/tex]
is an Integer ,otherwise range of y
[tex]=(\frac{ab-a-b}{b},\infty)[/tex]
Prove by induction that all of the hexagonal numbers are odd.
*(Problem from book is incorrect)
If [tex]H_n[/tex] denotes the [tex]n[/tex]-th hexagonal number, then this is indeed false because [tex]H_2=6[/tex].
You are laying 1.200 ft of pipe. After doing 900 ft. your vendor has run out of $15 pipe, and you have to buy more pipe at the store for $25 per foot. How much will it cost you in materials to install the 1,200 ft of pipe? Next
Answer:
The total cost to install the 1,200 ft of the pipe is $21000.
Step-by-step explanation:
Consider the provided information.
You buy 900 ft pipe with a cost of $15 per foot and the remaining 300 ft pipe with a cost of $25 per foot.
Therefore, the total cost is:
[tex]900 \times 15+300 \times 25[/tex]
[tex]13500+7500[/tex]
[tex]13500+7500[/tex]
[tex]21000[/tex]
Hence, the total cost to install the 1,200 ft of the pipe is $21000.
A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 48.0 and 53.0 minutes. Find the probability that a given class period runs between 51.5 and 51.75 minutes. Find the probability of selecting a class that runs between 51.5 and 51.75 minutes.
Step-by-step answer:
Given:
class time uniformly distributed between 48 and 53 minutes (range 5 minutes)
Find probability that the class period runs between 51.5 and 51.75 minutes.
Solution:
Uniformly distributed means that the probability that the class time is the same for any minute (out of 5), i.e. 20%, or probability density is 0.2 / minutes, no matter which minute.
The interval between 51.5 and 51.75 is 0.25 minutes, so the probability of class period within that range is 0.25 minutes * 0.2 / minute = 0.05.
Final answer:
The probability that a class period runs between 51.5 and 51.75 minutes, when class lengths are uniformly distributed between 48.0 and 53.0 minutes, is 0.05 or 5%.
Explanation:
To find the probability that a given class period runs between 51.5 and 51.75 minutes when class lengths are uniformly distributed between 48.0 and 53.0 minutes, we utilize the properties of the uniform distribution. The probability is the area under the uniform probability density function (PDF) between the two specified lengths. For a uniform distribution, this area corresponds to the length of the interval divided by the length of the entire interval in which the variable is uniformly distributed.
Here, we're looking for the probability between 51.5 minutes and 51.75 minutes. The total length of the available interval is 53.0 - 48.0 = 5 minutes. The length of the desired interval is 51.75 - 51.5 = 0.25 minutes. So, we divide the length of the desired interval by the total interval to get the probability.
Probability = (51.75 - 51.5) / (53.0 - 48.0) = 0.25 / 5 = 0.05
Therefore, the probability that a class period runs between 51.5 and 51.75 minutes is 0.05 or 5%.
21. A courier company has motorbikes which can travel 300 km starting with a full tank. Two couriers, Anna and Brian, set off from the depot together to deliver a letter to Connor's house. The only refuelling is when they stop for Anna to transfer some fuel from her tank to Brian's tank. She then returns to the depot while Brian keeps going, delivers the letter and returns to the depot. What is the greatest distance that Connor's house could be from the depot? (A) 180km (B) 200 km (C) 225 km (D) 250 km (E) 300 km
Answer:
(B) 200 km
Step-by-step explanation:
Let A represent the distance Anna goes before transferring fuel. Let C represent the distance to Connor's house. All distances are in km. Here, we will measure fuel quantity in terms of the distance it enables.
The total distance that can be driven by the two motorbikes is ...
2A +2C = 600
Anna can transfer to Brian an amount of fuel that is 300-2A, since she needs to get back to the depot from the stopping point. When they stop, the amount of fuel in Brian's tank is 300-A. After that transfer, the most fuel Brian can have is a full tank (300). Then ...
(300 -A) +(300 -2A) = 300 . . . . fuel in Brian's tank after the transfer
This second equation simplifies to ...
600 -3A = 300
300 = 3A . . . . . . add 3A-300
100 = A . . . . . . . . divide by 3
Using this in the first equation, we get ...
2·100 +2C = 600
2C = 400 . . . . . . . . subtract 200
C = 200 . . . . . . . . . .divide by 2
The distance from the depot to Connor's house can be at most 200 km.
Determine Whether the following function is even, odd, or neither
f(x) = x^4 + 7x^2 - 30
Answer:
even
Step-by-step explanation:
f(-x)=f(x) means f is even
f(-x)=-f(x) means f is odd
If you get neither f(x) or -f(x), you just say it is neither.
f(x)=x^4+7x^2-30
f(-x)=(-x)^4+7(-x)^2-30
f(-x)=x^4+7x^2-30
f(-x)=f(x)
so f is even.
Notes:
(-x)^even=x^even
(-x)^odd=-(x^odd)
Examples (-x)^88=x^88 and (-x)^85=-(x^85)
Answer: even
Step-by-step explanation:
By definition a function is even if and only if it is fulfilled that:
[tex]f(-x) = f(x)[/tex]
By definition, a function is odd if and only if it is true that:
[tex]f (-x) = -f(x)[/tex]
Then we must prove the parity for the function: [tex]f(x) = x^4 + 7x^2 - 30[/tex]
[tex]f(-x) = (-x)^4 + 7(-x)^2 - 30[/tex]
[tex]f(-x) = x^4 + 7x^2 - 30=f(x)[/tex]
Note that for this case it is true that: [tex]f(-x) = f(x)[/tex]
Finally the function is even
Find dx/dt when y=2 and dy/dt=1, given that x^4=8y^5-240
dx/dt=
Answer:
The value of [tex]\frac{dx}{dt}[/tex] is [tex]\frac{160}{x^3}[/tex].
Step-by-step explanation:
The given equation is
[tex]x^4=8y^5-240[/tex]
We need to find the value of [tex]\frac{dx}{dt}[/tex].
Differentiate with respect to t.
[tex]4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0[/tex] [tex][\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0][/tex]
[tex]4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}[/tex]
It is given that y=2 and dy/dt=1, substitute these values in the above equation.
[tex]4x^3\frac{dx}{dt}=40(2)^4(1)[/tex]
[tex]4x^3\frac{dx}{dt}=40(16)(1)[/tex]
[tex]4x^3\frac{dx}{dt}=640[/tex]
Divide both sides by 4x³.
[tex]\frac{dx}{dt}=\frac{640}{4x^3}[/tex]
[tex]\frac{dx}{dt}=\frac{160}{x^3}[/tex]
Therefore the value of [tex]\frac{dx}{dt}[/tex] is [tex]\frac{160}{x^3}[/tex].
32. Quality Control A campus bookstore buys 100 calcula-
tors. Assume that 2 have a defect of some kind. The math-
ematics department buys 8 of these calculators from the
bookstore. What is the probability that, of those calcula-
tors bought by the mathematics department,
(a) All are free of defects?
(b) Exactly 1 will have a defect?
(c) Exactly 2 will have a defect?
Answer:
a) 0.8508
b) 0.1389
c) 0.0099
Step-by-step explanation:
Total number of calculators bought = 100
Number of calculators with defect = 2
Probability of selecting a calculator with defect = p = 2 out of 100 = [tex]\frac{2}{100}=0.02[/tex]
Probability of selecting a calculator without defect = q = 1 - p = 1 - 0.02 = 0.98
Part a)
The bookstore buys 8 calculators. This means the number of trials is fixed ( n =8). The calculator is either with defect or without defect. This means there are 2 outcomes only. The outcome of one selection is independent of other selections. This means all the events(selections) are independent of each other.
Hence, all the conditions of a Binomial Experiment are being satisfied. So we will use Binomial Probability to solve this problem.
We need to find the probability that all calculators are free of defects. i.e. number of success is 0 i.e. P( x = 0 )
The formula of Binomial Probability is:
[tex]P(x) =^{n}C_{x} (p)^{x} q^{n-x}[/tex]
Using the values, we get:
[tex]P(0)=^{8}C_{0}(0.02)^{0}(0.98)^{8-0}\\\\ P(0)=0.8508[/tex]
Thus, the probability that all calculators are free of defects is 0.8508
Part b) Exactly 1 calculator will have defect
This means the number of success is 1. So, we need to calculate ( x = 1)
Using the formula of binomial probability again and using x = 1, we get:
[tex]P(1)=^{8}C_{1}(0.02)^{1}(0.98)^{8-1}\\\\ P(1)=0.1389[/tex]
Thus, the probability that exactly two calculators will be having a defect is 0.1389
Part c) Exactly 2 calculator will have defect
This means the number of success is 2. So, we need to calculate ( x = 2 )
Using the formula of binomial probability again and using x = 2, we get:
[tex]P(2)=^{8}C_{2}(0.02)^{2}(0.98)^{8-2}\\\\ P(2)=0.0099[/tex]
Thus, the probability that exactly two calculators will be having a defect is 0.0099
M1Q13.) Find the relative frequency of the 2nd class to the nearest tenth of a percent.
Answer:
9.1%
Step-by-step explanation:
The total number of observations is 33. The number in the second class is 3, so the relative frequency is ...
3/33 × 100% = (100/11)% ≈ 9.1%
Add everything in the frequency column.
= 33
Take class two (3).
Solution:
3/33 * 100%
1/11 * 100%
100/11%
9.1%
Best of Luck!
13. Imagine that you are taking a class and your chances of being asked a question in class are 1% during the first week of class and double each week thereafter (i.e., you would have a 2% chance in Week 2, a 4% chance in Week 3, an 8% chance in Week 4). What is the probability that you will be asked a question in class during Week 7?
Answer: There is probability that he will be asked a question in class during Week 7 is 64%.
Step-by-step explanation:
Since we have given that
Probability that question in class being asked during Week 1 = 1%
Probability that question in class being asked during Week 2 = 2%
Probability that question in class being asked during Week 3 = 4%
and so on.
So, we need to find the probability that question being asked in class during week 7.
Since it forms geometric series:
1%, 2%, 4%, ........
So, we need to find the 7 th term:
[tex]a_n=ar^{n-1}\\\\a_7=ar^{7-1}\\\\a_7=1(2)^6\\\\a_7=64\%[/tex]
Hence, there is probability that he will be asked a question in class during Week 7 is 64%.
Question 1: Find the distance between the points (1, 4) and (5, 1).
Question 1 options:
7
5
25
√7
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{(5-1)^2+(1-4)^2}\implies d=\sqrt{4^2+(-3)^2} \\\\\\ d=\sqrt{16+9}\implies d=\sqrt{25}\implies d=5[/tex]
Answer: second option.
Step-by-step explanation:
You need to use the formula for calculate the distance between two points. This is:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Given the point (1, 4) and the point (5, 1), you can say that:
[tex]x_2=5\\x_1=1\\y_2=1\\y_1=4[/tex]
Now you must substitute these values into the formula.
The distance between the points (1, 4) and (5, 1) is the following:
[tex]d=\sqrt{(5-1)^2+(1-4)^2}\\\\d=5[/tex]
This matches with the second option.
On the day that his first child was born, Ezio Auditore de Firenze deposited $3,000 into an investment account. The only purpose for the account was to pay for his son’s first year of college tuition. Assume that his son, Flavia, started college on his 18t h birthday and his first year tuition payment had to be made that day. The amount needed on that day was $26,000. If that was indeed the amount of money in the account on Flavia’s 18t h birthday, what annual rate of return did Ezio earn on his investment account?
Answer:
about 12.75%
Step-by-step explanation:
Let r represent the annual rate of return. Compounded annually for 18 years, the account multiplier is (1+r)^18. Then we have ...
26,000 = 3,000(1+r)^18
(26,000/3,000)^(1/18) = 1+r . . . . . . divide by 3000, take the 18th root
(26/3)^(1/18) -1 = r ≈ 12.7465%
Ezio's account earned about 12.75% annually.
Ezio's $3,000 investment grew to $26,000 over 18 years at this rate.
To calculate the annual rate of return Ezio earned on his investment for his son, Flavia's, college tuition, we'll need to use the compound interest formula:
A = P(1 + r)^n
Where:
A is the amount of money accumulated after n years, including interest.
P is the principal amount (the initial amount of money).
r is the annual interest rate (decimal).
n is the number of years the money is invested.
We know that Ezio deposited $3,000 (P = 3000), the amount in the account after 18 years is $26,000 (A = 26000), and that the time period n is 18 years. We need to find the annual interest rate r.
Let's rearrange the formula to solve for r:
(1 + r)^n = A / P
(1 + r)^18 = 26000 / 3000
(1 + r)^18 = 8.6667
Now we need to take the 18th root of 8.6667 to find (1 + r):
1 + r = (8.6667)^(1/18)
The 18th root of 8.6667 is approximately 1.1225. This means:
1 + r = 1.1225
Subtract 1 from both sides to find r:
r = 1.1225 - 1
r = 0.1225 or 12.25%
The annual rate of return Ezio earned on his investment is approximately 12.25%.
3) An oil prospector will drill a succession of holes in a given area to find a productive well. The probability that he is successful on a given trial is 0.2. What is the probability that the third hole drilled is the first to yield a productive well?
Answer: 0.128
Step-by-step explanation:
The geometric probability of getting success on nth trial is given by :-
[tex]P(n)=p(1-p)^{n-1}[/tex]
Given : The probability that he is successful on a given trial[tex]p= 0.2[/tex].
Then , the probability that the third hole drilled is the first to yield a productive well is given by :-
[tex]P(3)=0.2(1-0.2)^{3-1}=0.128[/tex]
Hence, the probability that the third hole drilled is the first to yield a productive well = 0.128
Based on a Comcast survey, there is a 0.8 probability that a randomly selected adult will watch prime-time TV live, instead of online, on DVR, etc. Assume that seven adults are randomly selected. Find the probability that fewer than three of the selected adults watch prime-time live.
Answer: Our required probability is 0.004672.
Step-by-step explanation:
Since we have given that
Number of adults = 7
Probability of getting adult will watch prime time TV live = 0.8
We need to find the probability that fewer than 3 of the selected adults watch prime time live.
We will use "Binomial Distribution":
here, n = 7
p = 0.8
So, P(X<3)=P(X=0)+P(X=1)+P(X=2)
So, it becomes,
[tex]P(X=0)=(1-0.8)^7=0.2^7=0.0000128[/tex]
[tex]P(X=1)=^7C_1(0.8)(0.2)^6=0.0003584\\\\P(X=2)=^7C_2(0.8)^2(0.2)^5=0.0043[/tex]
So, probability that fewer than 3 of the selected adult watch prime time live is given by
[tex]0.0000128+0.0003584+0.0043=0.004672[/tex]
Hence, our required probability is 0.004672.
This problem relates to the binomial distribution and requires us to find the sum of binomial probabilities for 0, 1, and 2 successes (adults watching live TV) out of seven trials (the seven randomly selected adults).
Explanation:This question is utilizing the concept of binomial distribution. The probability of a randomly selected adult watching prime-time TV live is 0.8. We want to find the probability that fewer than 3 out of 7 randomly selected adults watch prime-time live.
We find this by adding up the probabilities for 0, 1, and 2 adults watching live TV using the binomial distribution formula: P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)), where C(n, k) denotes the number of combinations of n items taken k at a time, p is the probability of success, and n is the number of trials.
We get:
P(X=0) = C(7, 0) * (0.8^0) * ((1-0.8)^(7-0))P(X=1) = C(7, 1) * (0.8^1) * ((1-0.8)^(7-1))P(X=2) = C(7, 2) * (0.8^2) * ((1-0.8)^(7-2))Adding these up will give the total probability that fewer than three adults out of seven watch prime-time live.
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the solution of the differential equation x^2y"-5xy'+5y=0 is select the correct answer. (a)y=c1x+c2x^2, (b)y=c1xcoslnx+c2x^2sinlnx, (c)y=c1xcos2lnx+c2x^2sin2lnx, (d) c1x^(5+rad5/2)+c2x^(5-rad5/2) , or (e)y=c1e^2x+c2xe^2xsinx
Answer:
[tex]y=c_1x+c_2x^5[/tex]
Step-by-step explanation:
The given second order homogeneous Cauchy-Euler ordinary differential equation is
[tex]x^2y''-5xy'+5y=0[/tex]
The corresponding auxiliary equation is given by:
[tex]a {m}^{2} + (b - a)m + c = 0[/tex]
where a=1, b=-5, c=5
We substitute the coefficients into the auxiliary equation to obtain:
[tex] {m}^{2} + ( - 5 - 1)m + 5= 0[/tex]
[tex] {m}^{2} - 6m + 5= 0[/tex]
[tex](m - 1)(m - 5) = 0[/tex]
[tex] \implies \: m = 1 \: or \: m = 5[/tex]
The auxiliary equation has two distinct real roots. The general solution to the corresponding differential equation is of the form:
[tex]y=c_1x ^{m1} +c_2 {x}^{m2} [/tex]
We substitute the values to get:
[tex]y=c_1x+c_2x^5[/tex]
The provided differential equation [tex]x^2y'' - 5xy' + 5y = 0[/tex] can be solved using the method for Cauchy-Euler equations, resulting in the general solution [tex]y = c_1 x + c_2 x^5.[/tex] Therefore, the correct answer is (d) [tex]y = c_1 x^{5 + \frac{\sqrt{5}}{2}} + c_2 x^{5 - \frac{\sqrt{5}}{2}}[/tex]
To solve the given differential equation [tex]x^2y'' - 5xy' + 5y = 0[/tex], we use the method for Cauchy-Euler equations:
1. Assume a solution of the form[tex]y = x^m.[/tex]
2. Substitute [tex]y = x^m[/tex] into the differential equation:
[tex]y' = m x^(m-1)[/tex][tex]y'' = m(m-1)x^(m-2)[/tex]3. Plug these into the equation:
[tex]x^2 [m(m-1)x^(m-2)] - 5x[mx^(m-1)] + 5[x^m] = 0[/tex]
4. Simplify to get the characteristic equation:
[tex]m(m-1) - 5m + 5 = 0[/tex]
[tex]m^2 - 6m + 5 = 0[/tex]
5. Solve the quadratic equation:
[tex]m = \frac{{6 \pm \sqrt{6^2 - 4 \times 5}}}{{2}}[/tex]
[tex]m = \frac{{6 \pm \sqrt{16}}}{{2}}[/tex]
[tex]m = \frac{{6 \pm 4}}{2}[/tex]
[tex]m = 5 \quad \text{or} \quad m = 1[/tex]
6. The general solution is then a linear combination of these solutions:
[tex]y = c_1 x^5 + c_2 x^1[/tex]
[tex]y = c_1 x + c_2 x^5[/tex]
[tex]y = c_1 x^{5 + \frac{\sqrt{5}}{2}} + c_2 x^{5 - \frac{\sqrt{5}}{2}}[/tex]
A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 431 gram setting. It is believed that the machine is underfilling the bags. A 23 bag sample had a mean of 430 grams with a standard deviation of 24. A level of significance of 0.025 will be used. Assume the population distribution is approximately normal. State the null and alternative hypotheses.
Answer: [tex]H_0:\mu\geq431\\\\H_a:\mu<431[/tex]
Step-by-step explanation:
Given : A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 431 gram setting. t is believed that the machine is underfilling the bags.
Claim : [tex]\mu<431[/tex]
We know that the null hypothesis always have equals sign , then the required set of hypothesis for the situation :-
[tex]H_0:\mu\geq431\\\\H_a:\mu<431[/tex]
A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 75%. Refer to Exhibit 9-6. At a .05 level of significance, it can be concluded that the proportion of the population in favor of candidate A is
Answer:
it would be A 1251
Step-by-step explanation:
The question involves hypothesis testing of proportions. Based on a sample where 80% favour Candidate A, we may infer the true population proportion is significantly greater than 75%, however, actual mathematical calculations are needed for confirmation.
Explanation:The question examines whether the proportion of the population favouring Candidate A is significantly more than 75% based on a sample of 100 people where 80 favoured Candidate A. This is a problem of hypothesis testing for proportions. The null hypothesis (H₀) is that the true population proportion is 75% (p = 0.75), versus the alternative hypothesis (H₁) stating the true population proportion is more than 75%. Using a significance level of 0.05, we examine the data.
With a sample proportion of 80 out of 100 (p' = 0.80) and given the large sample size, we apply the Normal approximation to the Binomial distribution, followed by a one-sample z-test. If the resulting p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that the true population proportion of individuals favouring Candidate A is significantly greater than 75%.
However, without performing the actual calculations, we cannot definitively determine the conclusion. From a practical perspective, an 80% sample proportion showing favour in a sample as large as 100 might indicate a significantly higher proportion than 75%, but an exact mathematical test should be done to confirm this.
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47. An honest coin is tossed 10 times in a row. The result of each toss (H or T) is observed. Find the probability of the event E= “a T comes up at least once.” (Hint: Find the probability of the complementary event.)
Answer:
0.0488%
Step-by-step explanation:
Here we have the probability of different independent events, which means that none of the previous ones affect the next. For this kind of events, the probability 'P' that a series of events occur is the multiplication of the probability of each singular event.
p1: Probability that an H be obtained: 50% (is always 50% because it is independent of the previous results)
P: The probability that H be obtained all of the 10 times. This is the complementary probability to E:(a T comes up at least once).
By the first definition given
[tex]P=0.5^{11}=0.00048828=0.0488%[/tex]
The complementary probability 'P' is the probability that 'E' does not happen, so the probability that E happen is: [tex]P_{E} =1-P[/tex]
The last makes sense if we think about the fact that for the experiment there are just two possibilities, 'E' happen, or 'E' does not happen. Then,
[tex]P_{E} =1-0.000488=99.95[/tex]%
The pmf of the amount of memory X (GB) in a purchased flash drive is given as the following. x 1 2 4 8 16 p(x) 0.05 0.10 0.30 0.45 0.10 (a) Compute E(X). (Enter your answer to two decimal places.) GB (b) Compute V(X) directly from the definition. (Enter your answer to four decimal places.) GB2 (c) Compute the standard deviation of X. (Round your answer to three decimal places.) GB (d) Compute V(X) using the shortcut formula. (Enter your answer to four decimal places.) GB2
a. Expected value is defined by
[tex]E[X]=\displaystyle\sum_xx\,p(x)[/tex]
so we get
[tex]E[X]=1\cdot0.05+2\cdot0.10+4\cdot0.30+8\cdot0.45+16\cdot0.10[/tex]
[tex]\boxed{E[X]=6.65}[/tex]
b. Variance is defined by
[tex]V[X]=E[(X-E[X])^2][/tex]
so with the expectation found above, we have
[tex]V[X]=E[(X-6.65)^2][/tex]
[tex]V[X]=\displaystyle\sum_x(x-6.65)^2\,p(x)[/tex]
(by definition of expectation)
[tex]V[X]=(1-6.65)^2\cdot0.05+(2-6.65)^2\cdot0.10+(4-6.65)^2\cdot0.30+(8-6.65)^2\cdot0.45+(16-6.65)^2\cdot0.10[/tex]
[tex]\boxed{V[X]=15.4275}[/tex]
c. Standard deviation is the square root of variance:
[tex]\boxed{\sqrt{V[X]}\approx3.928}[/tex]
d. I assume "shortcut formula" refers to
[tex]V[X]=E[X^2]-E[X]^2[/tex]
which is easily derived from the definition of variance. We have (by def. of expectation)
[tex]E[X^2]=\displaystyle\sum_xx^2\,p(x)[/tex]
[tex]E[X^2]=1^2\cdot0.05+2^2\cdot0.10+4^2\cdot0.30+8^2\cdot0.45+16^2\cdot0.10[/tex]
[tex]E[X^2]=59.65[/tex]
and so the variance is again
[tex]V[X]=59.65-6.65^2[/tex]
[tex]\boxed{V[X]=15.4275}[/tex]
as expected.
To compute E(X), multiply each outcome by its probability and sum them up. Compute V(X) directly from the definition and also using the shortcut formula. Compute the standard deviation of X.
Explanation:To compute E(X), we need to multiply each outcome x by its corresponding probability p(x) and sum them up. So, E(X) = 1(0.05) + 2(0.10) + 4(0.30) + 8(0.45) + 16(0.10) = 7.6 GB.
To compute V(X) directly from the definition, we need to first compute the squared deviations of each outcome from the expected value, which is 7.6 GB. Then, multiply each squared deviation by its corresponding probability, and sum them up. So, V(X) = (1 - 7.6)^2(0.05) + (2 - 7.6)^2(0.10) + (4 - 7.6)^2(0.30) + (8 - 7.6)^2(0.45) + (16 - 7.6)^2(0.10) ≈ 51.64 GB^2.
The standard deviation of X is the square root of the variance, which is SD(X) ≈ √(51.64) ≈ 7.19 GB.
To compute V(X) using the shortcut formula, we can use the formula: V(X) = E(X^2) - [E(X)]^2. First, we compute E(X^2) by multiplying each outcome squared by its corresponding probability and summing them up. Then, we subtract the square of E(X) to find V(X). This gives us V(X) = (1^2)(0.05) + (2^2)(0.10) + (4^2)(0.30) + (8^2)(0.45) + (16^2)(0.10) - [7.6]^2 ≈ 51.64 GB^2.
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Let X be a random variable with mean X = 25 and X = 6 and let Y be a random variable with mean Y = 30 and Y = 4. It is known that X and Y are independent random variables. Suppose the random variables X and Y are added together to create new random variable W (i.e., W = X + Y). What is the standard deviation of W?
I'm guessing you intended to say [tex]X[/tex] has mean [tex]\mu_X=E[X]=25[/tex] and standard deviation [tex]\sigma_x=\sqrt{\mathrm{Var}[X]}=6[/tex], and [tex]Y[/tex] has means [tex]\mu_Y=E[Y]=30[/tex] and standard deviation [tex]\sigma_Y=\sqrt{\mathrm{Var}[Y]}=4[/tex].
If [tex]W=X+Y[/tex], then [tex]W[/tex] has mean
[tex]E[W]=E[X+Y]=E[X]+E[Y]=55[/tex]
and variance
[tex]\mathrm{Var}[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2[/tex]
Given that [tex]\mathrm{Var}[X]=36[/tex] and [tex]\mathrm{Var}[Y]=16[/tex], we have
[tex]\mathrm{Var}[X]=E[X^2]-E[X]^2\implies E[X^2]=36+25^2=661[/tex]
[tex]\mathrm{Var}[Y]=E[Y^2]-E[Y]^2\implies E[Y^2]=16+30^2=916[/tex]
Then
[tex]E[W^2]=E[(X+Y)^2]=E[X^2]+2E[XY]+E[Y^2][/tex]
[tex]X[/tex] and [tex]Y[/tex] are independent, so [tex]E[XY]=E[X]E[Y][/tex], and
[tex]E[W^2]=E[X^2]+2E[X]E[Y]+E[Y^2]=661+2\cdot25\cdot30+916=3077[/tex]
so that the variance, and hence standard deviation, are
[tex]\mathrm{Var}[W]=3077-55^2=52[/tex]
[tex]\implies\sqrt{\mathrm{Var}[W]}=\sqrt{52}=\boxed{2\sqrt{13}}[/tex]
# # #
Alternatively, if you've already learned about the variance of linear combinations of random variables, that is
[tex]\mathrm{Var}[aX+bY]=a^2\mathrm{Var}[X]+b^2\mathrm{Var}[Y][/tex]
then the variance of [tex]W[/tex] is simply the sum of the variances of [tex]X[/tex] and [tex]Y[/tex], [tex]\mathrm{Var}[W]=36+16=52[/tex], and so the standard deviation is again [tex]\sqrt{52}[/tex].
The value is W = sqrt(52) = 7.211.
To find the standard deviation of the sum of two independent random variables X and Y, use the formula W = sqrt(X² + Y²), plugging in the given values to calculate the standard deviation of W as 7.211.
The standard deviation of the sum of two independent random variables X and Y is the square root of the sum of their variances:
W = sqrt(X² + Y²)
Substitute the given values to find the standard deviation of W:
Given: X = 6, Y = 4W = sqrt(6² + 4²) = sqrt(36 + 16) = sqrt(52)Therefore, W = sqrt(52) = 7.211A certain connected graph has 68 vertices and 72 edges. Does it have a circuit?
Answer:
Yes.
Step-by-step explanation:
If a graph G doesn't have a circuit, we must have that
[tex]|E(G)|=|V(G)|-1[/tex]
where [tex]|E(G)|[/tex] is the number of edges of the graph and [tex]|V(G)|[/tex] the number of vertices. However, in this case it holds that
[tex]|E(G)|=72>68=|V(G)|.[/tex]
Your instructor has 50 questions on a quiz and the are 2 points a peice. The maxium score is 100. and you miss 11 of these question what would my score be?
Answer:
78
Step-by-step explanation:
50 questions worth 2 a piece and you miss 11 questions so we are going to take 2(11) off of 50.
We are doing 100-2(11).
100-22=78.
Or if you miss 11 questions you get 39 right. So 39(2)=78.
ANSWER :
There are 50 questions, total marks are 100.
EXPLANATION:
Since each question is 2 marks, missing 11 questions is equal to missing 22 marks.
therefore, your marks will be 100-22=78 marks.
urgent help please!!!!!!!!!!!!!!!!!!
Answer:
B. AAS Congruence Theorem
Step-by-step explanation:
Previous steps showed congruence of a side of the designated triangle, and two angles that do not bracket that side. In short form, you have shown congruence of ...
Angle - Angle - Side
so the AAS congruence theorem applies.
A certain field is a rectangle with a perimeter of 918 feet. The length is 181 feet more than the width. Find the width and length of the rectangular field The width is? feet. The length is ? feet.
The width of the rectangular field is 139 feet and its length is 320 feet.
Explanation:To solve this problem, we can use the formula for the perimeter of a rectangle, which is 2*(length + width). We know that the perimeter is 918 feet, and it's given that the length is 181 feet more than the width.
Let's denote the width as w, therefore the length would be w + 181.
By substituting these expressions into the formula for the perimeter, we get:
2*(w + w + 181)=918,
which simplifies to 2*(2w + 181)=918,
further simplifies to 4w + 362 = 918.
To isolate 4w, subtract 362 from both sides:
4w = 918 - 362 = 556.
Finally, divide both sides by 4 to solve for w:
w = 556/4 = 139 feet.
Substitute w = 139 into the length equation to get the length:
length = w + 181 = 139 + 181 = 320 feet.
So the width is 139 feet, and the length is 320 feet.
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Length is 320 feet.
Let's define the width of the rectangle as w feet and the length as l feet.
According to the problem,
the perimeter of the rectangle is 918 feet and the length is 181 feet more than the width.
We can express these conditions using the following equations:
Perimeter equation: 2l + 2w = 918
Length equation: l = w + 181
First, we can solve the perimeter equation for l+w:
2l + 2w = 918
Divide both sides by 2:
l + w = 459
Next, we substitute the length equation l = w + 181 into the perimeter equation:
w + 181 + w = 459 Simplify:
2w + 181 = 459
Subtract 181 from both sides:
2w = 278
Divide by 2:
w = 139 feet
Now substitute the width back into the length equation:
l = w + 181
l = 139 + 181
l = 320 feet
So, the width is 139 feet, and the length is 320 feet.
what is the solution of the inequality shown below? c-7>2
Answer:
c > 9
Step-by-step explanation:
Isolate the variable, c. Treat the greater than sign as an equal, what you do to one side, you do to the other. Add 7 to both sides:
c - 7 > 2
c - 7 (+7) > 2 (+7)
c > 2 + 7
c > 9
c > 9 is your answer.
~
Answer:
[tex]\Huge \boxed{C>9}\checkmark[/tex]
Step-by-step explanation:
Add by 7 from both sides.
[tex]\displaystyle c-7+7>2+7[/tex]
Simplify, to find the answer.
[tex]\displaystyle 2+7=9[/tex]
[tex]\huge \boxed{c>9}[/tex], which is our answer.
Jason received an invoice for $5,000, and the invoice was dated on August 26 with terms of 2/10, n/45 EOM. Please answer the following questions:
(1) How much did Jason owe if the bill is paid by September 26?
(2) When will be the last day that Jason can receive the cash discount?
(3) When will be the last day that Jason has to pay for this bill?
Step-by-step explanation:
Consider the provided information.
Jason received an invoice for $5,000, and the invoice was dated on August 26.
Part 1)
We need to find the bill Jason needs to pay by September 26.
n/45 means invoice amount should be paid within 45 days after the august 26.
From August 26 to September 26 Jason complete a month.
Thus, Jason just need to pay only $5,000. Which is the invoice amount.
Part 2)
The last day of discount can be calculated by 2/10.
Here, the 2/10 represents that discount of 2% within 10 days.
Now, calculate the 10 days form August 26.
The 10 days from August 26 is September 5.
Hence, September 5 is the last day that Jason can receive the cash discount.
Part 3)
Now, we need to calculate the last day that Jason has to pay for the bill.
n/45 means invoice amount should be paid within 45 days after the august 26.
Hence, the sum of the invoice should be paid within 45 days of August 26, whatever the date.
Find a compact form for generating function of the sequence 4, 4, 4, 4, 1, 0, 1, 0, 1, 0, 1, 0,
The generating function for this sequence is
[tex]f(x)=4+4x+4x^2+4x^3+x^4+x^6+x^8+\cdots[/tex]
assuming the sequence itself is {4, 4, 4, 4, 1, 0, 1, 0, ...} and the 1-0 pattern repeats forever (as opposes to, say four 4s appearing after every four 1-0 pairs). We can make this simpler by "displacing" the odd-degree terms and considering instead the generating function,
[tex]f(x)=3+4x+3x^2+4x^3+\underbrace{(1+x^2+x^4+x^6+x^8+\cdots)}_{g(x)}[/tex]
where the coefficients of [tex]g(x)[/tex] follow a much more obvious pattern of alternating 1s and 0s. Let
[tex]g(x)=\displaystyle\sum_{n=0}^\infty a_nx^n[/tex]
where [tex]a_n[/tex] is recursively given by
[tex]\begin{cases}a_0=1\\a_1=0\\a_{n+2}=a_n&\text{for }n\ge0\end{cases}[/tex]
and explicitly by
[tex]a_n=\dfrac{1+(-1)^n}2[/tex]
so that
[tex]g(x)=\displaystyle\sum_{n=0}^\infty\frac{1+(-1)^n}2x^n[/tex]
and so
[tex]\boxed{f(x)=3+4x+3x^2+4x^3+\displaystyle\sum_{n=0}^\infty\frac{1+(-1)^n}2x^n}[/tex]
Final answer:
The generating function for the sequence is found by splitting it into two parts and expressing each as a series. The constant part can be expressed as a finite series, while the alternating sequence is a geometric series that can be simplified. Their sum yields the generating function.
Explanation:
The student has asked for a compact form for the generating function of the sequence 4, 4, 4, 4, 1, 0, 1, 0, 1, 0, 1, 0, ... .
To find the generating function for the given sequence, we can split it into two parts: The constant part (4, 4, 4, 4) and the alternating sequence (1, 0, 1, 0, ...).
The constant part can be represented as:
4 + 4x + 4x2 + 4x3 = 4(1 + x + x2 + x3)
The alternating sequence can be represented as a geometric series:
1 - x2 + x4 - x6 + ... = 1 / (1+x2)
The generating function G(x) would then be the sum of these two parts, simplifying by multiplication of the series and a fraction:
G(x) = 4(1 + x + x2 + x3) + x4 / (1 + x2)
The baseball team needs new equipment. Company A can provide 9 helmets, 6 bats, and 12 balls for $525. Company B can provide 10 helmets, 8 bats, and 10 balls for $600. Company C can provide 8 helmets, 5 bats, and 15 balls for $500. Which system of equations matches the equipment choices available for purchase? 9x + 6y + 12z = 525 10x + 8y + 10z = 600 8x + 5y + 15z = 500 9x + 12y + 6z = 525 10x + 8y + 10z = 600 8x + 5y + 15z = 500 9x + 6y + 12z = 525 10x + 10y + 8z = 600 8x + 5y + 15z = 500 9x + 6y + 12z = 525 10x + 8y + 10z = 600 8x + 15y + 5z = 500
Answer:
Choice A.
Step-by-step explanation:
Let x = price of a helmet, y = price of a bat, z = price of a ball.
Company A can provide 9 helmets, 6 bats, and 12 balls for $525.
9x + 6y + 12z = 525
Company B can provide 10 helmets, 8 bats, and 10 balls for $600.
10x + 8y + 10z = 600
Company C can provide 8 helmets, 5 bats, and 15 balls for $500.
8x + 5y + 15z = 500
Answer: Choice A.
Answer:
The correct option is A.
Step-by-step explanation:
Let the price of a helmet is x, the price of a bat is y and the price of a ball is z.
It is given that Company A can provide 9 helmets, 6 bats, and 12 balls for $525. The equation for Company A is
[tex]9x+6y+12x=525[/tex]
It is given that Company B can provide 10 helmets, 8 bats, and 10 balls for $600. The equation for Company B is
[tex]10x+8y+10x=600[/tex]
It is given that Company C can provide 8 helmets, 5 bats, and 15 balls for $500. The equation for Company C is
[tex]8x+5y+15x=500[/tex]
The system of equations is
[tex]9x+6y+12x=525[/tex]
[tex]10x+8y+10x=600[/tex]
[tex]8x+5y+15x=500[/tex]
Therefore the correct option is A.
Seven times the first (smaller) of two consecutive odd integers is equal to five times the second (larger) integer. Find each integers.
Let the smaller odd integer be: a
then the larger odd integer which is consecutive to a will be: a+2
It is given that:
Seven times the first (smaller) of two consecutive odd integers is equal to five times the second (larger) integer.
This means that:
7 times of a is equal to 5 times of (a+2)
i.e.
[tex]7a=5(a+2)\\\\i.e.\\\\7a=5\times a+5\times 2[/tex]
( Since, by using the distributive property of multiplication)
i.e.
[tex]7a=5a+10\\\\i.e.\\\\7a-5a=10\\\\i.e.\\\\2a=10\\\\i.e.\\\\a=5[/tex]
Hence, the smaller number is: 5
and the larger number is: 7
( Since a+2=5+2=7 )
Find the number z such that the proportion of observations that are less than z in a standard Normal distribution is 0.5414. (b) Find the number z such that 65.91% of all observations from a standard Normal distribution are greater than z. (a)
Answer:
Part (a) The value of Z is 0.10396. Part (b) The value of Z is 0.410008.
Step-by-step explanation:
Consider the provided information.
Part (a)
In order to find the number z such that the proportion of observations that are less than z in a standard Normal distribution is 0.5414, simply find 0.5414 in the table and search for the appropriate Z-value.
Now, observing the table it can be concluded that the value of Z is 0.10396.
Part (b)
Consider the number 65.91%
The above number can be written as 0.6591.
Now, find 0.6591 in the table and search for the appropriate Z-value.
By, observing the table it can be concluded that the value of Z is 0.410008.
Using a Z-table, we find that a z-score of approximately 0.1 will give us 0.5414 of observations less than z in a standard normal distribution. Similarly, for 65.91% of observations being greater than z, we subtract this from 1 and find z to be approximately 1.0.
Explanation:To find a number, z, such that a certain proportion of observations are less than z in a standard normal distribution, we use a Z-table. In the case where observations less than z comprise 0.5414 of the total, we cross reference this probability in the Z-table to find that z is approximately 0.1.
Similarly, when we need to find the number z where 65.91% of all observations from a standard normal distribution are greater than z, we subtract this percentage from 1, as we are interested in the observations to the left of z. Doing this, we get 0.3409. Checking the Z-table shows that the z-score that corresponds with this area under the curve (or probability) is approximately 1.0.
Remember that a standard normal distribution is denoted Z ~ N(0, 1), meaning it has a mean of 0 and a standard deviation of 1. When calculating z-scores, this allows us to see how many standard deviations a certain point is from the mean (µ).
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