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 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).

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

Answer:

a. Categorical

b. Quantitative

c. Categorical

d. Categorical

e. Quantitative

Step-by-step explanation:

a.

Month of year with most reservations is a qualitative or categorical variable because it can't be represented numerically in a meaningful way. For example, with most reservations month of a year can be June or July.

b.

Airbnb's  total annual profit is a quantitative variable because it can be presented in numerical form and mathematical operation can be meaningfully  interpreted.

c.

Type of rental on Airbnb is a qualitative or categorical variable because it can't be represented numerically in a meaningful way. Also, it can be divided into categories whole house, private room and shared room etc.

d.

Unique 10-digit reservation number is a qualitative or categorical variable as these exists in numerical form but these numbers are used only as identifiers. The  mathematical operation on these numbers can't be meaningfully be interpreted.

e.

Number of house rentals is quantitative variable because it can be presented in numerical form and mathematical operation can be meaningfully interpreted.

The qualitative, categorical, and quantitative statements of the above cases are:

a. Categorical

b. Quantitative

c. Categorical

d. Categorical

e. Quantitative

Quantitative is the term used mainly to describe the quantity of a particular case, but not describe it as an attribute.

Categorical means describing anything in a particular way or series.

a.

The month of the year with most reserves is a qualitative or categorical variable because it can not be equal numerically in a meaningful way.

For example, with most reserves the month of the year can be June or July.

b.

Airbnb's total annual profit is a quantitative variable because it can be shown in mathematical operations that can be meaningfully interpreted.

c.

The type of rental on Airbnb is a  categorical variable because it can't be represented numerically in a meaningful way. Also, it can be divided into categories whole house, private room, shared room etc.

d.

A unique 10-digit reservation number is a qualitative or categorical variable as these exist in numerical form, but these numbers are used only as identifiers. The mathematical operation on these numbers can't be meaningfully be interpreted.

e.

The number of house rentals is a quantitative variable because it can be presented in numerical form and mathematical operation can be meaningfully interpreted.

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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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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.

Answer:

1300 millimeters

Step-by-step explanation:

Answer:

1300 mm

Step-by-step explanation:

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.

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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Answer: The markup on the T-shirts is $ 13.50.

Step-by-step explanation:

Given : The cost price of each t-shirt = $1.50

The selling price of each t-shirt = $15

Then ,the markup on the T-shirts =  (Selling price of each t-shirt) -( Cost price of each t-shirt)

i.e. The markup on the T-shirts = $15- $1.50= $ 13.50

Hence, the markup on the T-shirts is $ 13.50.

Answer:

90%

Step-by-step explanation:

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

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

Full Question

The number of messages that arrive at a Web site is a Poisson distributed random variable with a mean of 6 messages per hour.

a. What is the probability that 6 messages are received in 1 hour?

b. What is the probability that 10 messages are received in 1.5 hours?

c. What is the probability that fewer than 2 messages are received in 0.5 hour?

Answer and Explanation

Given

λ = 6 per hour

Poisson Probability P(X = k) = (λ^k e^-λ)/k!

a. K = 6

P(X = 6) = (6^6 e^-6)/6!

P(X = 6) = 0.160623141047980

P(X = 6) = 0.1606--------- Approximated

b.

If 6 messages are received on average per hour then the number of messages received on average per 1.5 hours is

λ = 6 *1.5

λ = 9

For k = 10

P(X = 10) = (9^10 e^-9)/10!

P(X = 10) = 0.118580076008570

P(X=10) = 0.1186 ---------- Approximated

c.

If 6 messages are received on average per hour then the number of messages received on average per 0.5 hours is

λ = 6 *0.5

λ = 3

For messages fewer than 2 means than k = 0 or k = 1

For k = 0

P(X = 0) = (3^0 e^-3)/0!

P(X = 0) = 0.049787068367863

P(X = 0) = 0.0498 ------_--- Approximated

For X = 1

P(X = 1) = (3^1 e^-3)/1!

P(X = 1) = 0.149361205103591

P(X = 1) = 0.1494 ---------- Approximated

P(X <2) = P(X=0) + P(X=1)

P(X<2) = 0.0498 + 0.1494

P(X<2) = 0.1996

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:

Step-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:

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

Answer:

The person would end up with 8 Bitcoins.

Step-by-step explanation:

This question can be solved by consecutive rules of three.

If someone traded $83,000 US dollars for Japanese yen, then traded the yen for Bitcoin, how many Bitcoin would that person end up with?

Each US dollar is worth 107.35 yen. So how many yens are $83,000 US dollars worth?

$1 - 107.35 yen

$83,000 - x yen

[tex]x = 83000*107.35[/tex]

[tex]x = 8,910,050[/tex]

The person has 8,910,050 yens. Each bitcoin is worth 1,086,300 yens. How many bitcoins are worth 8,910,050 yens?

1 bitcoin - 1,086,300 yens

x bitcoins - 8,910,050 yens

[tex]1086300x = 8910050[/tex]

[tex]x = \frac{8910050}{1086300}[/tex]

[tex]x = 8.2[/tex]

Rouded to the nearest whole Bitcoin, is 8.

So the person would end up with 8 Bitcoins.

By first converting the US dollars to yen and then trading the yen for Bitcoin, using the provided exchange rates, we determine that the person would end up with roughly 8 Bitcoin.

To answer this exchange rate problem, we must first convert the US dollars to yen, then convert the yen to Bitcoin.

First, we multiply the amount of US dollars, $83,000 by the US dollar to yen exchange rate, which is 107.35 yen for every 1 US dollar. This gives us:

$83,000 * 107.35 yen/US dollar = 8,910,050 yen

Next, we trade the yen for Bitcoin by dividing by the yen to Bitcoin exchange rate. Our yen to Bitcoin rate is 1,086,300 yen for 1 Bitcoin:

8,910,050 yen ÷ 1,086,300 yen/Bitcoin ≈ 8.2 Bitcoin.

Rounding this to the nearest whole number, we find that the person ends up with approximately 8 Bitcoin.

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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 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.

Answer:

it is not affected by a change of units

Step-by-step explanation:

Since the correlation coefficient has no dimensions, it is not affected by a change of units. Then it will remain the same after the conversion

In fact, the linear correlation coefficient ρ ,where

ρ = Cov (X,Y) / (σx*σy)

then the units [ ] of ρ are

[ρ] = [ Cov (X,Y) ] / [σx]*[σy] = σ²/σ² = 1 → dimensionless

is more useful than using covariance [ Cov (X,Y) ]  , since dividing by the standard deviations eliminates the units and standardise the variable

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

Answer:

The solution to the differential equation

y' = (7cos5x)/(8 - 3y); y(0) = 3

is

16y - 3y² = 70sin5x + 21

Step-by-step explanation:

y' = (7cos5x)/(8 - 3y)

This can be written as

dy/dx = (7cos5x)/(8 - 3y)

Separate the variables

(8 - 3y)dy = (7cos5x)dx

Integrate both sides

8y - (3/2)y² = 35sin5x + C

Applying the initial condition y(0) = 3

8(3) - (3/2)(3)² = 35sin(5(0)) + C

24 - (27/2) = 0 + C

C = 21/2

Therefore,

8y - (3/2)y² = 35sin5x + 21/2

Or

16y - 3y² = 70sin5x + 21

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 .

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 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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Answer:

There is a 2.28% probability that the average weight of these 64 adult males is over 205 pounds.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s= \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 197, \sigma = 32, n = 64, s = \frac{32}{\sqrt{64}} = 4[/tex]

What is the probability that the average weight of these 64 adult males is over 205 pounds?

This is 1 subtracted by the pvalue of Z when X = 205.

So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{205 - 197}{4}[/tex]

[tex]Z = 2[/tex]

[tex]Z = 2[/tex] has a pvalue of 0.9772

1 - 0.9772 = 0.0228

There is a 2.28% probability that the average weight of these 64 adult males is over 205 pounds.

The probability that the average weight of 64 randomly selected adult males is over 205 pounds is approximately 2.28%.

This problem involves the concept of normal distribution and probability in statistics. Given the mean (μ) is 197 pounds and the standard deviation (σ) is 32 pounds, we want to find the probability that the average weight of 64 randomly selected adult males (n=64) is over 205 pounds.

Firstly, we need to calculate the standard error (SE), which is σ/√n, thus SE=32/√64= 4 pounds. Next, we calculate the Z-score, which is (X-μ)/SE, thus Z=(205-197)/4=2.

A Z-score of 2 refers to a value that is 2 standard deviations away from the mean. Looking this up on a Z-table or using statistical software, we can see that the area to the left of Z=2 is approximately 0.9772, meaning there is a 97.72% chance that a randomly selected adult male's weight is below 205 pounds. Hence the probability of the weight being over 205 pounds is 1-0.9772=0.0228 or 2.28%.

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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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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:

Answer: for 10th percentile, X = 60.53inches

for 80th percentile, X = 61.53 inches

Step-by-step explanation:

the relationship between the mean, standard deviation and the standard normal distribution is given as

X = μ + σZ

where μ is the mean and σ is the standard deviation of the variable X, and Z is the value from the standard normal distribution for the desired percentile.

Hence to determine the 10th and 80th percentile, we lookup the standard normal distribution table attached below,

from the table,

at 10th percentile Z = -1.282

at 80th percentile we interpolate between 75th percentile and 90th

(80 - 75)/(90 - 75) = (Z - 0.675)/(1.282 - 0.675)

5/15 = Z - 0.675/0.607

0.333*(0.607) = Z - 0.675

Z = 0.8771

hence the Z value for the 80th percentile is 0.8771

hence

X value for 10th percentile and 80th is calculated as

X = μ + σZ

since mean = 63.7 and standard deviation = 2.47

For 10th percentile

X = 63.7 + 2.47*(-1.282)

X = 60.53

for 80th percentile

X = 63.7 + 2.47*(0.8771)

X = 61.53

Final answer:

Height at 10th percentile: 61.53 inches

Height at 80th percentile: 65.78 inches

Explanation:

Given that the heights of women are normally distributed with a mean of 63.7 inches and a standard deviation of 2.47 inches, we can find the 10th and 80th percentiles using the Z-score formula in the context of a normal distribution.

The Z-score formula is given by: Z = (X - μ) / σ, where X is the value whose Z-score we're calculating, μ is the mean, and σ is the standard deviation.

To find the 10th and 80th percentiles, we first use Z-scores corresponding to these percentiles from a standard normal distribution table:

We then apply the formula for each Z-score to find the heights corresponding to these percentiles:

Answer:

Correct answer is (c). 5

Step-by-step explanation:

It is important to note that solving problem requires techniques and intelligent people most especially when problem are complex or hard in nature. It is therefore important to ensure the numbers of problem solving experts should not be undersized than required to avoid over burden them and should not be too large to avoid conflict in their individual resolutions. Hence, most scientific reports state that problem solving experts should be within 3 to 5 members and as for this question, the optimum is 5 members.

Answer:

B.) square root of 2

E.) -0.54

G.) 5/3

H.) 1.29

Step-by-step explanation:

A probability of an event is how likely the event is to occur. It is always positive values, between 0% and 100%, or as decimals, between 0 and 1.

A.) 1

Can be a probability

B.) square root of 2

The square root of 2 is 1.41. 1.41 is higher than 1, so square root of 2 cannot be a probability

C.) 0

Can be a probability

D.) 0.04

0.04 = 4%

Can be a probability

E.) -0.54

Negative values cannot be probabilities

F.) 3/5

3/5 = 0.6 = 60%

Can be a probability

G.) 5/3

5/3 = 1.67

Higher than 1, so cannot be a probability

H.) 1.29

Higher than 1, cannot be a probability

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]

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).

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.

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.

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.