Answer:
d. skewed right
Step-by-step explanation:
The shape of the given distribution is rightly skewed. For a symmetric distribution mean and median are equal and if mean is greater than median then the distribution is rightly skewed and if mean is less than median then the distribution is skewed left.
In the given distribution mean is greater than median and so the given distribution is skewed right.
APPLY IT Rate of Growth The area covered by a patch of moss is growing at a rate of
A'(t)=√ t ln t
cm^2 per dat, for t≥1. ≥ 1. Find the additional amount of area covered by the moss between 4 and 9 days.
Answer:
The additional amount of area covered between 4 to 9 days is 23.71 cm2
Step-by-step explanation:
As the relation is given as a combination of two functions so integration by parts is carried out thus
[tex]\int\limits^9_4 {\sqrt{t}\, ln t} \, dt[/tex]
In order to solve this integral, integration by parts is to be carried out which is given as
[tex]\int u v dx=u \int v dx -\int u'(\int vdx) dx[/tex]
Where
u(x) is a function of x
v(x) is a function of x
u' is the derivative of u wrt to x
Also u and v are defined on using the following sequence ILATE RULE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)
As here Logarithmic function is present which is taken as u and the algebraic function is taken as v so
[tex]u= ln t\\v=\sqrt{t}\\u'=\frac{1}{t}[/tex]
[tex]\int v dt =\int t^{1/2} dt =\frac{2}{3}t^{3/2}[/tex]
Substituting the values in equation gives
[tex]\int u v dt=u \int v dt -\int u'(\int vdt) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\int(\frac{1}{t} )(\frac{2}{3}t^{3/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\int(\frac{2}{3}t^{1/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{2}{3}\int(t^{1/2}) dt\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{2}{3}(\frac{2}{3}t^{3/2}) +C\\\int{\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2}) +C[/tex]
Now solving the definite integral
[tex]\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2}) +C\\\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=[ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2})]_{9} -[ln t (\frac{2}{3}t^{3/2}) -\frac{4}{9}(t^{3/2})]_{4}\\\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=18 ln (9)-\frac{16}{3} ln 4 -\frac{76}{9}\\A=\int\limits^9_4 {\sqrt{t}\, ln t} \, dt=23.71 cm^2\\[/tex]
So the additional amount of area covered between 4 to 9 days is 23.71 cm2
A food truck operator has traditionally sold 75 bowls of noodle soup each day. He moves to a new location and after a week sees that he has averaged 85 bowls of noodle soup sales each day. He runs a one-sided hypothesis test to determine if his daily sales at the new location have increased. The p-value of the test is 0.031. How should he interpret the p-value?
a. There is a 3.1% chance that the true mean of soup sales at the new location is 85 bowls a day.
b. There is a 96.9% chance that the true mean of soup sales at the new location is greater than 75 bowls a day.
c. There is a 96.9% chance that the sample mean of soup sales at the new location is 85 bowls a day.
d. There is a 3.1% chance of obtaining a sample with a mean of 85 or higher assuming that the true mean sales at the new location is still equal to or less than 75 bowls a day.
e. There is a 96.9% chance that the true mean of soup sales at the new location is within 3.1 bowls of 85 bowls a day.
Option d correctly interprets the p-value, signifying there is a 3.1% chance of observing an average sales of 85 or more daily bowls given the true mean is 75 or less. It indicates significant evidence against the null hypothesis, suggesting increased sales at the new location.
When interpreting the p-value of the hypothesis test conducted by the food truck operator, option d is the correct interpretation: There is a 3.1% chance of obtaining a sample with a mean of 85 or higher assuming that the true mean sales at the new location is still equal to or less than 75 bowls a day. The p-value in a one-sided hypothesis test indicates the probability of observing a result as extreme as, or more extreme than, the sample result, under the assumption that the null hypothesis is true. The null hypothesis in this case is that the true mean daily sales have not changed and remain at 75 bowls per day or less. Hence, with a p-value of 0.031, there is significant evidence against the null hypothesis, and the operator has reason to believe that the average sales have indeed increased at the new location.
The p-value of 0.031 means there's a 3.1% chance of obtaining a sample mean of 85 bowls or higher if the true mean remains 75 bowls per day. Hence option d is the correct option. This suggests sufficient evidence to reject the null hypothesis and conclude that soup sales at the new location have likely increased.
The food truck operator has conducted a one-sided hypothesis test to determine if his daily sales at the new location have increased from the traditional 75 bowls of noodle soup.
A p-value is the probability of obtaining a sample mean as extreme as 85 bowls of soup per day or higher, assuming the true mean is still 75 bowls per day.The p-value of 0.031 means there is a 3.1% chance of obtaining such a sample mean if the null hypothesis is true. Therefore, we interpret the p-value as follows:d. There is a 3.1% chance of obtaining a sample with a mean of 85 or higher assuming that the true mean sales at the new location is still equal to or less than 75 bowls a day.
Since the p-value of 0.031 is less than the typical significance level of 0.05, there is sufficient evidence to reject the null hypothesis and conclude that the daily sales at the new location have likely increased.
A company has fixed monthly costs of $100,000 and production costs on its product of $28 per unit. The company sells its product for $74 per unit. The cost function, revenue function and profit function for this situation are
Answer:
The cost function is [tex]C(x)=100000+x\cdot 28[/tex]
The revenue function is [tex]R(x)=x\cdot 74[/tex]
The profit function is [tex]P(x)=46x-100000[/tex]
Step-by-step explanation:
We have the following definitions:
The cost function consists of variable costs and fixed costs and is given by
[tex]C(x)=fixed\:costs+x\cdot variable\:costs[/tex]
The revenue function is given by
[tex]R(x)=x\cdot p(x)[/tex]
where x are the units sold and p(x) is the price per unit.
The profit function is given by
[tex]P(x)=R(x)-C(x)[/tex]
Given:
Fixed costs = $100,000
Variable costs = $28 per unit
Price per unit = $74 per unit
Applying the above definitions and the information given, we get that:
The cost function is [tex]C(x)=100000+x\cdot 28[/tex]
The revenue function is [tex]R(x)=x\cdot 74[/tex]
The profit function is [tex]P(x)=74x-(28x+100000)=46x-100000[/tex]
Final answer:
The total cost for producing 1,000 units of output, given average fixed costs of $100 and average variable costs of $50, is calculated to be $150,000.
Explanation:
The question asks us to calculate total cost of producing 1,000 units of output given the average fixed costs and average variable costs. To find the total cost, we need to add together the total fixed costs (average fixed cost × quantity) and the total variable costs (average variable cost × quantity).
The total fixed cost is $100,000 (since $100 × 1,000 units) and the total variable cost is $50,000 (since $50 × 1,000 units). Therefore, the total cost of producing 1,000 units of output is $150,000.
The negative sign in 15t-2t belongs to the term____?
The expression 15t – 2t not equivalent to 2t – 15t because the negative sign in 15t – 2t belongs to the term 2t.
Solution:
Given expressions are 15t – 2t and 2t – 15t.
To determine 15t – 2t is equivalent to 2t – 15t or not.
Substitute t = 2 in above two expressions.
15t – 2t = 15(2) – 2(2)
= 30 – 4
= 26
2t – 15t = 2(2) – 15(2)
= 4 – 30
= –26
The values of the expressions are different when t = 2.
So, 15t – 2t is not equivalent to 2t – 15t.
Hence the expression 15t – 2t not equivalent to 2t – 15t because the negative sign in 15t – 2t belongs to the term 2t.
See You Later Based on a Harris Interactive poll, 20% of adults believe in reincarnation. Assume that six adults are randomly selected, and find the indicated probability. a. What is the probability that exactly five of the selected adults believe in reincarnation? b. What is the probability that all of the selected adults believe in reincarnation? c. What is the probability that at least five of the selected adults believe in reincarnation? d. If six adults are randomly selected, is five a significantly high number who believe in reincarna
Answer:
a) There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.
b) 0.0064% probability that all of the selected adults believe in reincarnation.
c) There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.
d) Since [tex]P(X \geq 5) < 0.05[/tex], 5 is a significantly high number of adults who believe in reincarnation in this sample.
Step-by-step explanation:
For each of the adults selected, there are only two possible outcomes. Either they believe in reincarnation, or they do not. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 6, p = 0.2[/tex]
a. What is the probability that exactly five of the selected adults believe in reincarnation?
This is P(X = 5).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{6,5}.(0.2)^{5}.(0.8)^{1} = 0.0015[/tex]
There is a 0.15% probability that exactly five of the selected adults believe in reincarnation.
b. What is the probability that all of the selected adults believe in reincarnation?
This is P(X = 6).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{6,6}.(0.2)^{6}.(0.8)^{0} = 0.000064[/tex]
There is a 0.0064% probability that all of the selected adults believe in reincarnation.
c. What is the probability that at least five of the selected adults believe in reincarnation?
This is
[tex]P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564[/tex]
There is a 0.1564% probability that at least five of the selected adults believe in reincarnation.
d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation?
5 is significantly high if [tex]P(X \geq 5) < 0.05[/tex]
We have that
[tex]P(X \geq 5) = P(X = 5) + P(X = 6) = 0.0015 + 0.000064 = 0.001564 < 0.05[/tex]
Since [tex]P(X \geq 5) < 0.05[/tex], 5 is a significantly high number of adults who believe in reincarnation in this sample.
a. The probability that exactly five of the selected adults believe in reincarnation is approximately 0.00256.
b. The probability that all of the selected adults believe in reincarnation is approximately 0.000064.
c. The probability that at least five of the selected adults believe in reincarnation is approximately 0.002624.
d. To determine if five is significantly high, we need a significance level for comparison, which isn't provided in the question.
To solve this problem, we can use the binomial probability formula, where "n" is the number of trials, "p" is the probability of success (believing in reincarnation in this case), and "x" is the number of successes.
a. The probability that exactly five of the selected adults believe in reincarnation is calculated as follows:
P(X = 5) = C(6, 5) * (0.20)^5 * (0.80)^(6-5),
where C(6, 5) is the number of ways to choose 5 out of 6 adults, which equals 6.
P(X = 5) = 6 * (0.20)^5 * (0.80)^1 ≈ 0.00256
b. The probability that all of the selected adults believe in reincarnation is:
P(X = 6) = (0.20)^6 ≈ 0.000064
c. The probability that at least five of the selected adults believe in reincarnation is the sum of the probabilities from parts (a) and (b):
P(X ≥ 5) = P(X = 5) + P(X = 6) ≈ 0.00256 + 0.000064 ≈ 0.002624
d. To determine if five is a significantly high number who believe in reincarnation, we can compare the probability of getting at least five believers (from part c) to a significance level. If this probability is less than the significance level, it would be considered significant. The significance level would depend on the context and what is considered "significant" in the specific analysis.
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complete question should be :
See You Later Based on a Harris Interactive poll, 20% of adults believe in reincarnation. Assume that six adults are randomly selected, and find the indicated probability. a. What is the probability that exactly five of the selected adults believe in reincarnation? b. What is the probability that all of the selected adults believe in reincarnation? c. What is the probability that at least five of the selected adults believe in reincarnation? d. If six adults are randomly selected, is five a significantly high number who believe in reincarnation .
The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes.
The probability that she will finish her trip in 80 minutes or less is _____.
Answer:
0.8 or 80%
Step-by-step explanation:
Since the time is uniformly distributed, every possible travel time has the same likelihood of occurring.
Lower boundary (L) = 40 minutes
Upper boundary (U) = 90 minutes
The probability that a student finishes her trip in 80 minutes or less is:
[tex]P(t\leq 80) = \frac{80-L}{U-L}=\frac{80-40}{90-40}\\P(t\leq 80) = 0.8=80\%[/tex]
The probability is 0.8 or 80%.
Answer:
80%
Step-by-step explanation:
Researchers measure the body temperature of 52 randomly selected adults. They find a mean temperature of 98.2 degrees with a standard deviation of 0.682 degrees. Which of the following is the correct t-test statistic and p-value for a test of the following hypotheses?
H_o: mu = 98.6 degrees
H_a: mu notequalto 98.6 degrees
The test statistic is negative 1. 039.21.and the p-value is less than 0.000001.
The test statistic is negative 0.46.and the p-value is 2 times P(t_51 > -0.46).
The t-test statistic is negative 0.315.and the p-value is P(t_51 < -0.315).
The t-test statistic is negative 3.33.and the p-value is two times P(t_51 < -3.33).
Answer:
The t-test statistic is -3.33.and the p-value is two times P(t_51 < -3.33).
Step-by-step explanation:
Data given and notation
Assuming that the real sample mean is: [tex]\bar X=98.285[/tex] represent the sample mean
[tex]s=0.682[/tex] represent the sample standard deviation
[tex]n=52[/tex] sample size
[tex]\mu_o =98.6[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to apply a two tailed test.
What are H0 and Ha for this study?
Null hypothesis: [tex]\mu = 98.6[/tex]
Alternative hypothesis :[tex]\mu \neq 98.6[/tex]
Compute the test statistic
The statistic for this case is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{98.2-98.6}{\frac{0.682}{\sqrt{52}}}=-3.298[/tex]
Now we can calculate the degrees of freedom and we got:
[tex] df = n-1 = 52-1 = 51[/tex]
P value
Since is a two tailed test the p value would be:
[tex]p_v =2*P(t_{51}<-3.3)=0.0018[/tex]
So the most appropiate conclusion for this case would be:
The t-test statistic is -3.33.and the p-value is two times P(t_51 < -3.33).
Use the geometric definition of the cross product and the properties of the cross product to make the following calculations. (a) ((i⃗ +j⃗ )×i⃗ )×j⃗ = (b) (j⃗ +k⃗ )×(j⃗ ×k⃗ ) = (c) 4i⃗ ×(i⃗ +j⃗ ) = (d) (k⃗ +j⃗ )×(k⃗ −j⃗ ) =
Answer:
Step-by-step explanation:
we know
[tex]\vec{i}\times \vec{j}=\vec{k}[/tex]
[tex]\vec{j}\times \vec{k}=\vec{i}[/tex]
[tex]\vec{k}\times \vec{i}=\vec{j}[/tex]
(a)[tex]\left [ \left ( \hat{i}+\hat{j}\right )\times \hat{i}\right ]\times \hat{j}[/tex]
[tex]=\left [ \hat{i}\times \hat{i}+\hat{j}\times \hat{i}\right ]\times \hat{j}[/tex]
[tex]=\left [ 0-\hat{k}\right ]\times \hat{j}[/tex]
[tex]=\hat{i}[/tex]
(b)[tex]\left ( \hat{j}+\hat{k}\right )\times \left ( \hat{j}\times \hat{k}\right )[/tex]
[tex]=\left ( \hat{j}+\hat{k}\right )\times \left ( \hat{i}\right )[/tex]
[tex]=\hat{k}+\hat{j}[/tex]
(c)[tex]4\hat{i}\times \left ( \hat{i}+\hat{j}\right )[/tex]
[tex]=4\hat{i}\times \hat{i}+4\hat{i}\times \hat{j}[/tex]
[tex]=0+4\hat{k}[/tex]
(d)[tex]\left ( \hat{k}+\hat{j}\right )\times \left ( \hat{k}-\hat{j}\right )[/tex]
[tex]=\hat{k}\times \hat{k}-\hat{k}\times \hat{j}+\hat{j}\times \hat{k}-\hat{j}\times \hat{j}[/tex]
[tex]=0+\hat{i}+\hat{i}-0[/tex]
[tex]=2\hat{i}[/tex]
The direction of the cross product depends on the right-hand rule and the resulting cross-product is located on the plane that is perpendicular to the vectors undergoing the cross product.
Taking i, j, k as the unit vector along x, y, z-direction.
Since [tex]i \times i = 0[/tex] it implies the angle∠ between them is 0;
Then: sin(0) = 0
Also;
[tex]j \times j = 0 \\ \\ k\times k = 0[/tex]
Similarly, [tex]i \times j = k[/tex] which implies that the angle ∠ between them = 90°;
Then: sin (90°) = 1
Also;
[tex]j \times k= i \\ \\ k \times i = j[/tex]
[tex]j \times i = - k[/tex] which implies that the angle ∠ between them = -90° or 270°
Then; sin (-90) or Sin ( 270) = -1
Also;
[tex]i \times k = -j \\ \\ j \times i = -k[/tex]
As such i, j, k are unit vectors for x, y, and z-axis.
To determine the following calculations, we have;
(a)
[tex]\Big ( ( i^{\to }+ j^{\to })\times i^{\to }\Big)\times j^{\to }[/tex]
[tex]= ( i^{\to } \times i^{\to } + j^{\to }\times i^{\to }) \times j^{\to }[/tex]
[tex]=(0 - k^{\to })\times j^{\to }[/tex]
[tex]= - k^{\to } \times j^{\to }\\\\= -(-i^{\to }) \\ \\ \mathbf{= i^{\to }}[/tex]
(b)
[tex](j^{\to }+ k^{\to }) \times (j^{\to } + k^{\to })[/tex]
[tex]= j^{\to } \times (j^{\to } \times k ) + k^{\to } \times (j \times k)\\ \\ = j^{\to } \times i^{\to } + k^{\to } \times i^{\to } \\ \\ \mathbf{ = -k^{\to } + j^{\to }}[/tex]
(c)
[tex]4i^{\to } \times (j \times k^{\to }) \\ \\ = 4 (i^{\to } \times i^{\to }) \\ \\ \mathbf{= 0}[/tex]
(d)
[tex](k^{\to} + j^{\to}) \times (k^{\to} - j^{\to})) \\ \\ =(k^{\to} \times k^{\to}) - (k^{\to} \times j^{\to})+(j^{\to} + k^{\to}) - (j^{\to} \times j^{\to}) \\ \\ = 0 + i^{\to} + i^{\to} -0 \\ \\ \mathbf{ = 2 i^{\to}}[/tex]
Therefore, we can conclude that the calculations of the cross-product are well defined from the above explanations.
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Kayla set up an outdoor digital thermometer to record the temperature overnight as part of her science fair project. She began recording the temperature, in degrees Fahrenheit, at 10:00 p.m. Kayla modeled the overnight temperature with function t, where h represents the number of hours since 10:00 p.m. t(h) = 0.5h2 − 5h + 27.5 What is the lowest temperature and at what time did it occur? A. 5°F at 3:00 a.m. B. 15°F at 5:00 a.m. C. 15°F at 3:00 a.m. D. 5°F at 5:00 a.m.
Answer:
C. 15°F at 3:00 a.m
Step-by-step explanation:
We will start seeing the function they give us, as we can see it is of the form ax ^ 2 + bx + c, this means that it is a parabola.
First we will look the term a of the function
t(h) = 0.5h2 − 5h + 27.5
in this case a = 0.5 , is a positive number so we have a minimum, this point shows us when the temperature reaches its minimum at night.
To obtain it we will have to apply this parabola formula
x = -b / 2a
in this case h = -( -5) / 2(0.5)
h = 5
This 5 represents the hours that have passed since 10:00 p.m.
We add 5 to 10:00 p.m. and get the time that is 3:00 a.m.
Finally we replace the function t with this value, and obtain the value of the minimum temperature
t(h) = 0.5h2 − 5h + 27.5
t(5) = 0.5(5)^2 - 5(5) + 27.5
t = 12.5 - 25 + 27.5
t = 15
C. 15°F at 3:00 a.m
Answer:
C
Step-by-step explanation: because i take the test
Customers arrive at Rich Dunn’s Styling Shop at a rate of 3 per hour, distributed in a Poisson fashion. Rich’s service times follow a negative exponential distribution, and Rich can complete an average of 5 haircuts per hour. a) Find the average number of customers waiting for haircuts. b) Find the average number of customers in the shop. c) Find the average time a customer waits until it is his or her turn. d) Find the average time a customer spends in the shop. e) Find the percentage of time that Rich is busy.
Answer:
a) 0.9,b) 1.5,c) 0.3hrs, d) 0.5hrs,e) 60%
Step-by-step explanation:
Given Data:
rate of arrival = 3customers/hr ;
rate of service = 5 haircuts/hr ;
a)
Average number of customers = La = λ²/[μ(μ-λ)]
= 3²/[(5(5-3)]
Average number of customers = La = 0.9
b)
Number of customers in system = Ls = λ/(μ-λ)
= 3/(5-3)
Number of customers in system = Ls = 1.5
c)
Average waiting time = Ta = λ/[μ(μ-λ)]
= 3/[(5(5-3)]
Average waiting time = Ta =0.3hrs or 18mins
d)
Average time spent by customer = Ts = 1/(μ-λ)
= 1/(5-3)
Average time spent by customer = Ts = 0.5hrs or 30mins
e)
% of time = Tr = λ/μ
= 3/5
% of time = Tr = 0.6 or 60%
The arrival of customers follows a Poisson distribution
The average number of customers waiting for haircut is 0.9The average number of customers in the shop is 1.5The average time of waiting for haircut is 0.3 hourThe average time spent in the shop is 0.5 hourRich is busy 60% of the timeThe given parameters are:
[tex]\mathbf{\lambda = 3}[/tex] --- rate of arrival
[tex]\mathbf{\mu= 5}[/tex] ---- rate of service
(a) Average number of customers waiting
This is calculated using:
[tex]\mathbf{L_a = \frac{\lambda^2}{\mu(\mu - \lambda)}}[/tex]
So, we have:
[tex]\mathbf{L_a = \frac{3^2}{5(5 - 3)}}[/tex]
[tex]\mathbf{L_a = \frac{9}{5 \times 2}}[/tex]
[tex]\mathbf{L_a = \frac{9}{10}}[/tex]
[tex]\mathbf{L_a = 0.9}[/tex]
Hence, the average number of customers waiting for haircut is 0.9
(b) Average number of customers in the shop
This is calculated using:
[tex]\mathbf{L_s = \frac{\lambda}{\mu - \lambda}}[/tex]
So, we have:
[tex]\mathbf{L_s = \frac{3}{5 - 3}}[/tex]
[tex]\mathbf{L_s = \frac{3}{2}}[/tex]
[tex]\mathbf{L_s = 1.5}[/tex]
Hence, the average number of customers in the shop is 1.5
(c) Average time of waiting
This is calculated using:
[tex]\mathbf{T_a = \frac{\lambda}{\mu(\mu - \lambda)}}[/tex]
So, we have:
[tex]\mathbf{T_a = \frac{3}{5(5 - 3)}}[/tex]
[tex]\mathbf{T_a = \frac{3}{5 \times 2}}[/tex]
[tex]\mathbf{T_a = \frac{3}{10}}[/tex]
[tex]\mathbf{T_a = 0.3}[/tex]
Hence, the average time of waiting for haircut is 0.3 hour
(d) Average time spent in the shop
This is calculated using:
[tex]\mathbf{T_s = \frac{1}{\mu - \lambda}}[/tex]
So, we have:
[tex]\mathbf{T_s = \frac{1}{5 - 3}}[/tex]
[tex]\mathbf{T_s = \frac{1}{2}}[/tex]
[tex]\mathbf{T_s = 0.5}[/tex]
Hence, the average time spent in the shop is 0.5 hour
(e) Percentage of time Rich is busy
This is calculated as:
[tex]\mathbf{T = \frac{\lambda}{\mu}}[/tex]
So, we have:
[tex]\mathbf{T = \frac{3}{5}}[/tex]
Divide
[tex]\mathbf{T = 0.6}[/tex]
Express as percentage
[tex]\mathbf{T = 60\%}[/tex]
Hence, Rich is busy 60% of the time
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An ensemble of 100 identical particles is sent through a Stern-Gerlach apparatus and the z-component of spin is measured. 46 yield the value +\frac{\hbar}{2}+ ℏ 2 while the other 54 give -\frac{\hbar}{2}− ℏ 2. Compute the standard deviation of the measurements.
Answer:
The standard deviation is 0.4984 [tex]\hbar[/tex]
Step-by-step explanation:
In order to find standard deviation, The equation is given as
[tex]\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^{100} (\mu-x_i)^2[/tex]
Here μ is the mean which is calculated as follows
[tex]\mu=\frac{\sum_{i=1}^{100} x_i}{n}\\\mu=\frac{46\times \frac{\hbar}{2}+54\times \frac{-\hbar}{2}}{100}\\\mu=\frac{-4 \hbar}{100}\\\mu=-0.04 \hbar[/tex]
Now the standard deviation is given as
[tex]\sigma=\sqrt{\frac{1}{100} \sum_{i=1}^{100} (-0.04 \hbar-x_i)^2}\\\sigma=\sqrt{\frac{1}{100} [[46 \times(-0.04 \hbar-0.5 \hbar)^2]+[54 \times(-0.04 \hbar+0.5 \hbar)^2]}]\\\sigma=\sqrt{\frac{1}{100} [[46 \times(-0.54 \hbar)^2]+[54 \times(0.46 \hbar)^2]}]\\\sigma=\sqrt{\frac{1}{100} [[46 \times(0.2916 \hbar)]+[54 \times(0.2116 \hbar)]}]\\\sigma=\sqrt{\frac{1}{100} [13.4136 \hbar+11.4264 \hbar}]\\\sigma=\sqrt{\frac{24.84 \hbar}{100}}\\\sigma =0.4984 \hbar[/tex]
So the standard deviation is 0.4984 [tex]\hbar[/tex]
Final answer:
To calculate the standard deviation of the z-component of spin measurements from a Stern-Gerlach experiment, use the formula for standard deviation in a binomial distribution. With 46 particles showing spin up and 54 spin down, the standard deviation is found to be approximately 4.984.
Explanation:
The question involves calculating the standard deviation of the z-component of spin measurements in a Stern-Gerlach experiment. Given that 46 particles yielded a spin of +½ℏ and 54 particles yielded a spin of -½ℏ, we can use these values to compute the standard deviation. The formula for the standard deviation σ in this binomial distribution is σ = √(np(1-p)), where n is the total number of trials and p is the probability of success (getting a +½ℏ spin result).
Using these values, the standard deviation is:
σ = √(100 * (46/100) * (1 - 46/100))
σ = √(100 * 0.46 * 0.54)
σ = √(24.84)
σ = 4.984
The standard deviation of the z-component of spin measurements in this experiment is approximately 4.984.
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How do you convert $150,000 US dollars to $240,000 Australian dollars (1.31545 AU) using the exchange rate per US dollar?
Answer:
240,000 AUD to USD = 161,924.79 US Dollars or 150,000 USD to AUD = 222,316.63 Australian Dollars
Step-by-step explanation:
PLZ HELP A delivery truck is transporting boxes of two sizes: large and small. The large boxes weigh 55 pounds each, and the small boxes weigh 35 pounds each. There are 125 boxes in all. If the truck is carrying a total of 4950 pounds in boxes, how many of each type of box is it carrying?
Answer:
There were 28.75 large boxes and 96.25 small boxes.
Step-by-step explanation:
Create a system of equations to solve.
State your variables
let x be the number of small boxes
let y be the number of large boxes
35x + 55y = 4950 Equation for weight
x + y = 125 Equation for number of boxes
Rearrange equation for number of boxes to isolate "x".
x = 125 - y
Substitute the new equation into the equation for weight
35x + 55y = 4950
35(125 - y) + 55y = 4950
Expand the brackets
4375 - 35y + 55y = 4950 Combine like terms
4375 + 20y = 4950
Start isolating "y"
4375 + 20y = 4950
4375 - 4375 + 20y = 4950 - 4375 subtract 4375 from both sides
20y = 575
20y/20 = 575/20 divide both sides by 20
y = 28.75 number of large boxes
Calculate "x". Rearrange the equation for number of boxes to isolate "y".
y = 125 - x Substitute this expression
35x + 55y = 4950 equation for weight
35x + 55(125 - x) = 4950 expand the brackets
35x + 6875 - 55x = 4950 combine like terms
6875 - 20x = 4950 start isolating "x"
6875 - 6875 - 20x = 4950 - 6875 subtract 6875 on both sides
-20x = -1925
-20x/-20 = -1925/-20 divide both sides by -20
x = 96.25 number of small boxes
There is a mistake with this question because you should not be able to have partial boxes. However, the answer is:
There were 28.75 large boxes and 96.25 small boxes.
Using the bijection rule to count binary strings with even parity.
Let B = {0, 1}. Bn is the set of binary strings with n bits. Define the set En to be the set of binary strings with n bits that have an even number of 1's. Note that zero is an even number, so a string with zero 1's (i.e., a string that is all 0's) has an even number of 1's.
(a) Show a bijection between B9 and E10. Explain why your function is a bijection.
Answer:
Lets denote c the concatenation of strings. For a binary string a in B9, we define the element f(a) in E10 this way:
f(a) = a c {1} if a has an odd number of 1's f(a) = a c {0} if a has an even number of 1'sStep-by-step explanation:
To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.
f is well defined:
To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.
f is injective (or one on one):
If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.
f is surjective:
Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:
If x has an even number of 1's, then the last digit of y has to be 0, and f(x) = x c {0} = yIf x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = yThis shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.
I need a better explanation how to find length and area of a triangle
Answer:
Height(h) = 24in
Area = 204in²
Step-by-step explanation:
we would use the heron formula because we have all the sides to calulate the area
S = (A+B+C)/2 = (25 + 17 + 26)/2 = 68/2 = 34
area = √s(s-a)(s-b)(s-c) = √34(34-25)(34-17)(34-26) = √34(9)(17)(8) = √34x9x17x8 = √41616 = 204
area= (1/2)bh
204 = (1/2)bh b= 17, h =?
204 = (1/2)17h
204 x 2 = 17h
408 = 17h
h = 408/17 = 24in
Height(h) = 24in
Area = 204in²
Find the perimeter of the following shape, rounded to the nearest tenth:
Answer: the perimeter of the shape is 19.1
Step-by-step explanation:
To determine the length of each side of the quadrilateral, we would apply Pythagoras theorem which is expressed as
Hypotenuse² = opposite side² + adjacent side
For line AD,
AD² = 2² + 4² = 4 + 16 = 20
AD = √20 = 4.47
For line AB,
AB² = 1² + 5² = 1 + 25 = 26
AB = √26 = 5.1
For line BC,
BC² = 2² + 4² = 4 + 16 = 20
BC = √20 = 4.47
For line CD,
CD² = 1² + 5² = 1 + 25 = 26
CD = √26 = 5.1
The perimeter of a plane figure is the distance around the figure. Therefore
Perimeter = AB + AB + BC + CD
Perimeter = 4.47 + 5.1 + 4.47 + 5.1 =
19.1
Answer:
19.1
Step-by-step explanation:
Got it right on the test! <3
Suppose that we have a standard Solow model with a Cobb-Douglas production function. The central equation of the model is as follows: kt+1 = sAkα t + (1 − δ)kt . Consumption per worker is given by: ct = (1 − s)Akα t . (a) Solve for an expression for the steady state capital stock per worker. In doing so, assume that the level of productivity is fixed at some value A.
Answer:
The answer is shown in the attachment
Step-by-step explanation:
The detailed step by step and appropriate mathematical manipulation for the expression for the steady state capital stock per worker is as shown in the attachment.
Can u guys Pls help:((((
Answer: angle 8 = 118 degrees
Step-by-step explanation:
The sum of the angles on a straight line is 180 degrees. This means that
angle 1 + angle 3 = 180 degrees
Therefore,
118 + angle 3 = 180 degrees
Subtracting 118 from the left hand side and the right hand side of the equation, it becomes
118 - 118 + angle 3 = 180 - 118
Angle 3 = 62 degrees
Since line d is parallel to line e, then angle 3 = angle 6 because they are alternate angles. Therefore,
Angle 6 = 62 degrees
Since the sum of the angles in a straight line is 180 degrees,
angle 8 = 180 - angle 6
angle 8 = 180 - 62 = 118 degrees
Angela took a general aptitude test and scored in the 91st percentile for aptitude in accounting. (a) What percentage of the scores were at or below her score?(b) What percentage were above?
Answer:
a) 91% of the scores were at or below her score.
b) 9% of the scores were above her score.
Step-by-step explanation:
When a value V is said to be in the xth percentile of a set, x% of the values in the set are lower than V and (100-x)% of the values in the set are higher than V.
In this problem, we have that:
Angela scored on the 91st percentile.
(a) What percentage of the scores were at or below her score?
Since Angela scored on the 91st percentile, 91% of the scores were at or below her score.
(b) What percentage were above?
Since Angela scored on the 91st percentile, 9% of the scores were above her score.
Angela scored higher than or equal to 91% of her peers, meaning 9% scored above her.
Explanation:If Angela scored in the 91st percentile on her general aptitude test, that means she scored higher than or equal to 91% of the other scores. Therefore, (a) 91% of scores were at or below Angela's score. As for (b), if you subtract her percentile from 100%, you can determine that 9% of scores were above hers.
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Standby time is amount of time a phone can remain powered on while not being used. A cell phone company claims that the standby time of certain phone model is 16 days on average. A consumer report firm gathered a sample of 19 batteries and conducted tests on this claim. The sample mean was 15 days and 10 hours and the sample standard deviation was 30 hours. Assume that the standby time is distributed as normal. In testing if the average standby time is shorter than 16 days, compute the value of the test statistic (round off to second decimal place).
Answer:
[tex]t_{stat} = -2.03[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 16 days = 384 hours
Sample mean, [tex]\bar{x}[/tex] = 15 days 10 hours = 370 hours
Sample size, n = 19
Sample standard deviation, s = 30 hours
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 384\text{ hours}\\H_A: \mu < 384\text{ hours}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{370 - 384}{\frac{30}{\sqrt{19}} } = -2.03415 \approx -2.03[/tex]
The value of t-statistic is -2.03
The test statistic for the claim that the average standby time of the phone model is 16 days, with a sample mean of 15 days and 10 hours and standard deviation of 30 hours, is -2.50.
Explanation:The question requires computation of a test statistic for the claim that the average standby time of a certain phone model is 16 days, using a sample mean of 15 days and 10 hours. The sample standard deviation is given as 30 hours. The number of phones (or sample size) is 19.
First, convert the sample mean to the same unit as the standard deviation. In this case, convert 15 days and 10 hours to hours: (15 * 24) + 10 = 370 hours. The null hypothesis mean is also converted to hours (16 * 24 = 384 hours).
The formula for the test statistic in a one-sample z-test is z = (Xbar - μ) / (σ/√n), where Xbar is the sample mean, μ is the hypothesized population mean, σ is the sample standard deviation, and n is the sample size.
Substitute values into the formula to get: z = (370 - 384) / (30/√19) = -2.50 (rounded to the second decimal place). So the test statistic is -2.50.
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Consider the following hypothesis test: H0: LaTeX: \mu\leμ ≤ 12 Ha: LaTeX: \mu>μ > 12 A sample of 25 provided a sample mean LaTeX: \overline{x}x ¯ = 14 and a sample standard deviation s = 4.32. Use LaTeX: \alphaα = 0.05. a. Compute the value of the test statistic.
Answer:
[tex]t=\frac{14-12}{\frac{4.32}{\sqrt{25}}}=2.315[/tex]
[tex]p_v =P(t_{(24)}>2.315)=0.015[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 1% of signficance.
Step-by-step explanation:
Data given and notation
[tex]\bar X = 14[/tex] represent the sample mean
[tex]s=4.32[/tex] represent the sample standard deviation
[tex]n=25[/tex] sample size
[tex]\mu_o =12[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 12, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 12[/tex]
Alternative hypothesis:[tex]\mu > 12[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{14-12}{\frac{4.32}{\sqrt{25}}}=2.315[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=25-1=24[/tex]
Since is a one side right tailed test the p value would be:
[tex]p_v =P(t_{(24)}>2.315)=0.015[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 1% of signficance.
To calculate the test statistic for a given hypothesis test, use the formula for a one-sample z-test by taking the difference between the sample mean and the population mean under the null hypothesis, divided by the standard error of the mean. For the given values, the test statistic is approximately 2.315.
You asked how to compute the value of the test statistic for a hypothesis test with the following parameters:
Null hypothesis H0: μ ≤ 12
Alternative hypothesis Ha: μ > 12
Sample size n = 25
Sample mean μ = 14
Sample standard deviation s = 4.32
Significance level α = 0.05
To calculate the test statistic, we use the formula for a one-sample z-test since the sample size is large (n ≥ 30) or the population is normally distributed and σ is known.
The test statistic (z) is calculated as follows:
z = (μ - μ0) / (s / √n)
Where:
μ0 is the hypothesized population mean under the null hypothesis.
μ is the sample mean.
s is the sample standard deviation.
n is the sample size.
Substituting the given values:
z = (14 - 12) / (4.32 / √25)
z = 2 / (4.32 / 5)
z = 2 / 0.864
z ≈ 2.315
Therefore, This z value is the test statistic that you would then compare to the critical z value from the z-table that corresponds to the given significance level α = 0.05 for a right-tailed test.
What is the probability that one die has number ""5"" as the outcome and the other die has number ""1"" as the outcome?
Answer:
[tex]\frac{1}{36}[/tex]
Step-by-step explanation:
Probability refers to the chance of occurrence of some event.
Outcome refers to the result of the event that occurs.
When a die is thrown once, outcomes are [tex]\left \{ 1,2,3,4,5,6 \right \}[/tex]
Probability of occurrence of each of the events i.e. number appeared on the die when it is thrown is 1 or 2 or 3 or 4 or 5 or 6 = [tex]\frac{1}{6}[/tex]
To find: the probability that one die has the number ''5'' as the outcome and the other die has the number ''1'' as the outcome
Solution: the probability that one die has the number ''5'' as the outcome × the probability that the die has the number ''1'' as the outcome = [tex]\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}[/tex]
Let f be the function that determines the area of a circle (in square cm) given the radius of the circle in cm, r. That is, f(r) represents the area of a circle (in square cm) whose radius is r cm. Use function notation to complete the following tasks a. Represent the area (in square cm) of a circle whose radius is 4 cm. Preview syntax error b. Represent how much the area (in square cm) of a circle increases by when its radius increases from 10.9 to 10.91 cm. # Preview syntax error c. Represent the area of 5 circles that all have a radius of 12.7 cm *Preview syntax error d. A circle has a radius of 28 cm. Another larger circle has an area that is 59 square cm more than the first circle. Represent the area of the larger circle. # Preview) syntax error
Part(a):[tex]f(r)=f(4)[/tex]
Part(b):[tex]f(10.91)-f(10.9)[/tex]
Part(c):[tex]5 f(r)=5 f(12.7)[/tex]
Part(d):[tex]28+59 =f(28)+59[/tex]
Area of the circle:
The area of a circle is the region occupied by the circle in a two-dimensional plane. It can be determined easily using a formula,
[tex]A= \pi r^2[/tex]
where [tex]r[/tex] is the radius of the circle
The formula for the area of the circle is,
[tex]A=\pi r^2[/tex]
Part(a):
Given,
Radius([tex]r[/tex])=4 cm
So, the area is [tex]f(r)=f(4)[/tex]
Part(b):
Given,
[tex]r=10.91\\r=10.9[/tex]
The difference in area is,
[tex]f(10.91)-f(10.9)[/tex]
Part(c):
Area of 5 circles are,
[tex]5 f(r)=5 f(12.7)[/tex]
Part(d):
The area of the larger circle is,
Area of the circle of radius [tex]28+59 =f(28)+59[/tex]
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A glass of orange juice contains 3.2 dL of juice. How many milliliters of orange juice are in the glass?
Answer:
There are 320 milliliters of orange juice in the glass.
Step-by-step explanation:
Unit conversion problems can be solved by rules of three.
We have that each dL is worth 100 mL(milliliters).
How many milliliters of orange juice are in the glass?
How many ml are 3.2dL of juice?
1 dL - 100 mL
3.2 dL - x mL
[tex]x = 100*3.2[/tex]
[tex]x = 320[/tex]
There are 320 milliliters of orange juice in the glass.
The measure of juice is 320 milliliters.
Given that
A glass of orange juice contains 3.2 dL of juice.
We have to determine
How many milliliters of orange juice is in the glass?
According to the question
To determine the measure of juice in milliliters following all the steps given below.
A glass of orange juice contains 3.2 dL of juice.
Here 1 dL = 100 milliliters
Then,
The measure of juice in milliliters is,
[tex]\rm = 3.2 \times 100\\\\= 320 \ milliliters[/tex]
Hence, The measure of juice is 320 milliliters.
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This problem involves two distinct sets of events, which we label A1and A2 and B1 and B2. The events A1and A2 are mutually exclusive and collectively exhaustive within their set. The events B1and B2 are mutually exclusive and collectively exhaustive within their set. Intersections can occur between all events from the two sets.Given P(A1)= 0.8, P(B1|A1) = 0.6, and P(B1|A2) = 0.2, what is P(A1|B1)?
Answer:
0.42
Step-by-step explanation:
Let the events be given as:
For twp mutually exclusive events, the probability of A1 is given as follows:
P (B1A1) = [tex]\frac{ P(B1) x P (BA1)}{P(A1)}[/tex]
= 0.6
Solving the equation above to get B1:
P (B1) = [tex]\frac{(0.8)x (0.6)}{(0.6)}[/tex]
= 0.8
Therefore, computing P (A1B1) gives P(A1) × P (B1)
= (0.8) × (0.6)
= 0.42 Ans
please help. i have no idea where to start
Step-by-step explanation:
[tex] \because \triangle \: DEF \sim \triangle NPQ \\ \\ \therefore \: \frac{EF}{PQ} = \frac{DF}{NQ}... (csst) \\ \\ \therefore \: \frac{d}{ \frac{11}{2} } = \frac{7}{9} \\ \\ \therefore \: \frac{2d}{ 11 } = \frac{7}{9} \\ \\ \therefore \: d = \frac{11 \times 7}{2 \times 9} \\ \\ \therefore \: d = \frac{77}{18} \\ [/tex]
Answer:
77/18
Step-by-step explanation:
Similar figures have sides in the same ratio.
Ratio can be obtained using the ratio of DF/NQ
DF/NQ = 7/9
EF/PQ = 7/9
d/(11/2) = 7/9
d = 7/9 × 11/2
d = 77/18
The approximate heights of two of the world's tallest mountains are given. Determine which is the taller mountain and by how many meters. Round to the nearest meter. (1 in equals 2.54 cm; 1 ft equals 30.48 cm; 1 yd almost equals 0.9144 m; 1 mi almost equals 1.6 km1 in = 2.54 cm; 1 ft = 30.48 cm; 1 yd ≈ 0.9144 m; 1 mi ≈ 1.6 km) Lhotse: 8516 meters and Ngadi Chuli: 25,866 feet
Answer:
Height of Lhotse mountain is greater than Ngadi Chuli mountain.
Height of Lhotse mountain is 632.0432 meter more than height of Ngadi Chuli mountain.
Step-by-step explanation:
Height of Lhotse mountain = 8516 meters
Height of Ngadi Chuli mountain = 25,866 feet
1 ft = 30.48 cm
[tex]25,866 feet =25,866\times 30.48 m=788,395.68 cm[/tex]
1 cm = 0.01 m
[tex]788,395.68 cm=788,395.68\times 0.01 m=7883.9568 m[/tex]
8516 meters > 7883.9568 m
Height of Lhotse mountain is greater than Ngadi Chuli mountain.
Difference in their height :
8516 meters - 7883.9568 m = 632.0432 m
Height of Lhotse mountain is 632.0432 meter more than height of Ngadi Chuli mountain.
The height of Ngadi Chuli is approximately 7871 meters, while the height of Lhotse is 8516 meters. Therefore, Lhotse is the taller mountain and is about 645 meters taller than Ngadi Chuli.
Explanation:The height of Ngadi Chuli is given in feet, so we must first convert that to meters using the conversion factor 1 ft = 30.48 cm = 0.3048 m. Multiplying 25,866 feet by 0.3048 m/ft gives us an approximate height of 7871 meters for Ngadi Chuli.
Lhotse is given to be 8516 meters tall. Therefore, by comparing these two heights, we can determine that Lhotse is the taller mountain.
The difference in height is found by subtracting the height of Ngadi Chuli from that of Lhotse: 8516 m - 7871 m = 645 m. So, Lhotse is approximately 645 meters taller than Ngadi Chuli.
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Question 5: A recent CNN News survey reported that 76% of adults think the U.S. pennies should still be made. Suppose we select a sample of 20 people.
How many of the 20 would you expect to indicate that the Treasury should continue making pennies?
What is the standard deviation?
What is the likelihood that exactly eight people would indicate the Treasury should continue making pennies?
What is the likelihood that 10 to 15 adults would indicate the Treasury should continue making pennies?
Answer:
a) [tex] E(X) =np = 20*0.76=15.2[/tex]
b) [tex] Sd(X) = \sqrt{3.648}=1.910[/tex]
c) [tex]P(X=8)=(20C8)(0.76)^8 (1-0.76)^{20-8}=0.000512[/tex]
That correspond to approximately 0.0512%, so then we can conclude that is very unlikely since is <1%
d) [tex] P(10 \leq X \leq 15)=0.541[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=20, p=0.76)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Part a
For this case the expected value for the binomial distribution is given by:
[tex] E(X) =np = 20*0.76=15.2[/tex]
Part b
The variance for the binomial distribution is given by:
[tex] Var(X) = np(1-p) = 20*0.76*(1-0.76) =3.648[/tex]
And the deviation would be ust the square root of the variance and we got:
[tex] Sd(X) = \sqrt{3.648}=1.910[/tex]
Part c
For this case we want this probability:
[tex]P(X=8)=(20C8)(0.76)^8 (1-0.76)^{20-8}=0.000512[/tex]
That correspond to approximately 0.0512%, so then we can conclude that is very unlikely since is <1%
Part d
For this case we want this probability:
[tex] P(10 \leq X \leq 15)=P(X=10)+....+P(X=15)[/tex]
If we find the individual probabilities we got:
[tex]P(X=10)=(20C10)(0.76)^{10} (1-0.76)^{20-10}=0.0075[/tex]
[tex]P(X=11)=(20C11)(0.76)^{11} (1-0.76)^{20-11}=0.0217[/tex]
[tex]P(X=12)=(20C12)(0.76)^{12} (1-0.76)^{20-12}=0.0515[/tex]
[tex]P(X=13)=(20C13)(0.76)^{13} (1-0.76)^{20-13}=0.100[/tex]
[tex]P(X=14)=(20C14)(0.76)^{14} (1-0.76)^{20-14}=0.159[/tex]
[tex]P(X=15)=(20C15)(0.76)^{15} (1-0.76)^{20-15}=0.201[/tex]
And if we add the values we got:
[tex] P(10 \leq X \leq 15)=0.541[/tex]
The response provides the expected number of people in the sample supporting the production of pennies, calculates the standard deviation, evaluates the probability of exactly eight respondents, and determines the likelihood of 10 to 15 adults favoring the production of pennies.
Expectation: Out of 20 people, you would expect 76% to indicate that the Treasury should continue making pennies. So, 20 x 0.76 = 15.2 people.
Standard Deviation: To find the standard deviation, use the formula: sqrt(n x p x (1 - p)), where n = 20 and p = 0.76. So, sqrt(20 x 0.76 x 0.24) = 1.95.
Probability: To find the probability of exactly 8 people indicating they should continue making pennies, use the binomial probability formula: C(20, 8) x (0.76⁸) x (0.24¹²) ≈ 0.029.
Likelihood (10 to 15 adults): To find the likelihood of 10 to 15 adults wanting pennies made, sum the probabilities of 10, 11, 12, 13, 14, and 15 people: P(10) + P(11) + P(12) + P(13) + P(14) + P(15).
If a student determined that the volume of a sample of water is 11.4 mL, and its mass is 10.98 g, what should the student record as the experimental density of the water sample, with the correct number of significant figures
Answer:1.04 ml/g.
Step-by-step explanation:
The formula we use to find the density :
[tex]\text{Density}=\dfrac{\text{Volume}}{\text{Mass}}[/tex]
We are given that ,
Volume of a sample of water = 11.4 mL
Mass of sample of water = 10.98 g
Then , the experimental density of the water sample will be :-
[tex]\text{Density}=\dfrac{11.4}{10.98}=1.03825136\approx1.04[/tex]
Hence, the experimental density of the water sample is 1.04 ml/g.
You have decided that you want to be a millionaire when you retire in 44 years. If you can earn an annual return of 11.06 percent, how much do you have to invest today? What if you can earn an annual return of 5.53 percent?
Answer:
1. Let amount deposited now be P.
So P*(1.1106^44) = 1,000,000, which means P=9896.06
2. Let amount deposited now be P.
So P*(1.0553^44) = 1,000,000, which means P=93639.42
Step-by-step explanation:
Answer:
Step-by-step explanation:
Let P represent the amount that you need to invest today. Thus, principal = $P
The return would be compounded annually. So
n = 1
The rate at which the principal would be compounded is 11.06%. So
r = 11.06/100 = 0.1106
The investment would be for 44 years. So
t = 44
The formula for compound interest is
A = P(1+r/n)^nt
A = total amount in the account at the end of t years.
A = $1000000
Therefore
1000000 = P(1+0.1106/1)^1×44
1000000 = P(1.1106)^44
P = 1000000/101.05
P = $9896.1
2) if you can earn an annual return of 5.53 percent, then
r = 5.53/100 = 0.053
1000000 = P(1+0.53/1)^1×44
1000000 = P(1.053)^44
P = 1000000/9.7
P = $103093