In a statistics class of 42 students, 14 have volunteered for community service in the past. If two students are selected at random from this class, what is the probability that both of them have volunteered for community service? Round your answer to four decimal places. P(both students have volunteered for community service) the absolute tolerance is __________

By first reducing the following function f(x, y) to g(x)=x3 with y=0 and then determining its derivative at x=1, we can obtain fx(1, 0) equals 3.

You are required to determine the value of fx(1, 0) in the given function f(x, y) = x(x(x2 + y2)3/2 e)(sin(x2y)). The tip suggests that you can address this problem by computing g'(1) where g(x) = f(x, 0). When y = 0, the function changes to f(x,0) = x * x2 * e0, which is then expressed as x3. G(x)=x3 has a derivative, g(x)=3x2. Consequently, fx(1, 0) equals g'(1), which equals 3.

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Answer:

The probability that a randomly selected American adult takes multivitamins regularly and works out regularly is  0.103

Step-by-step explanation:

p(Americans adults take multivitamins regularly) = 50% = 0.50

p(American adults work out regularly) = 20.6 = 0.206

p(Americans adults take multivitamins regularly and American adults work out regularly) = 0.50* 0.206 = 0.103

Thus, p(Americans adults take multivitamins regularly and American adults work out regularly)  = 0.103

Using the combination formula, the student can make 165 different possible schedules for next semester if she is planning to take three classes from the eleven available that she has met the prerequisites for.

This problem is a combination problem where students need to select 3 classes from 11 available classes to plan her schedule. The combination formula is used when the order of election does not matter. Here, the formula to find a combination is written as C(n, r) = n! / [r!(n - r)!].

In this particular case, the student has 11 classes (n=11) and plans to take 3 classes (r=3). So, the combination becomes C(11,3) = 11! / [3!(11-3)!] = (11*10*9) / (3*2*1) = 165.

This means, the student can make 165 different possible schedules for next semester assuming she is planning to take three classes from the eleven available classes.

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The student could make 165 different schedules.

The formula for combinations is given by:

[tex]\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \][/tex]

where n! (n factorial) is the product of all positive integers up to n, and r is the number of items to choose.

 Given that the student has 11 classes left and she plans to take 3 classes next semester, we substitute n = 11 and r = 3 into the combination formula:

[tex]\[ \binom{11}{3} = \frac{11!}{3!(11-3)!} \][/tex]

 Now, we simplify the factorial expressions:

[tex]\[ \binom{11}{3} = \frac{11!}{3! \times 8!} \][/tex]

We can cancel out the common terms in the numerator and the denominator:

[tex]\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} \][/tex]

Now, we perform the arithmetic:

[tex]\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 \][/tex]

Thus, the student has 165 different possible schedules for next semester if she is to take 3 classes out of the remaining 11.

Answer:

a) c = 2

b) 0.99

c) 0.84

d) 0.8485

Step-by-step explanation:

We are given the following in the question:

[tex]f(x) = \dfrac{c}{x^3}, x > 1[/tex]

a) Value of c

Property of probability density function

[tex]\displaystyle\int^{\infty}_{-\infty} f(x) = 1[/tex]

Putting values, we get,

[tex]\displaystyle\int^{\infty}_{1} \frac{c}{x^3} = 1\\\\\Rightarrow -\frac{c}{2}\bigg[\frac{1}{x^2}\bigg]^{\infty}_{1} = 1\\\\\Rightarrow \frac{c}{2} = 1\\\\\Rightarrow c = 2[/tex]

Thus, the value of c is 2.

[tex]f(x) = \dfrac{2}{x^3}, x > 1[/tex]

b) proportion of contaminating particles are PM10

We have to evaluate

[tex]P( x \leq 10) =\displaystyle\int ^{10}_{1}\frac{2}{x^3}dx\\\\=\bigg(\frac{2}{-2x^2}\bigg)^{10}_{1}\\\\=-(\frac{1}{100}-1)\\\\=0.99[/tex]

c) proportion of contaminating particles are PM2.5

[tex]P( x \leq 2.5) =\displaystyle\int ^{2.5}_{1}\frac{2}{x^3}dx\\\\=\bigg(\frac{2}{-2x^2}\bigg)^{2.5}_{1}\\\\=-(\frac{1}{6.25}-1)\\\\=0.84[/tex]

d)  proportion of the PM10 particles are PM2.5

[tex]P(PM ~2.5|PM~10) = \dfrac{0.84}{0.99} = 0.8485[/tex]

Answer:

The probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.

Step-by-step explanation:

Given:

Weight of a given sample (x) = 2.33 oz

Mean weight (μ) = 1.75 oz

Standard deviation (σ) = 0.22 oz

The distribution is normal distribution.

So, first, we will find the z-score of the distribution using the formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Plug in the values and solve for 'z'. This gives,

[tex]z=\frac{2.33-1.75}{0.22}=2.64[/tex]

So, the z-score of the distribution is 2.64.

Now, we need the probability [tex]P(x\geq 2.33 )=P(z\geq 2.64)[/tex].

From the normal distribution table for z-score equal to 2.64, the value of the probability is 0.9959. This is the area to the left of the curve or less than z-score value.

But, we need area more than the z-score value. So, the area is:

[tex]P(z\geq 2.64)=1-0.9959=0.0041=0.41\%[/tex]

Therefore, the probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.

Answer:

Option B)  (197773, 228207)    

Step-by-step explanation:

We are given the following in the question:

Sample mean, [tex]\bar{x}[/tex] = 212990

Sample size, n = 6

Alpha, α = 0.05

Sample standard deviation = 14500

95% Confidence interval:

[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]  

Putting the values, we get,  

[tex]t_{critical}\text{ at degree of freedom 5 and}~\alpha_{0.05} = \pm 2.5705[/tex]  

[tex]212990 \pm 2.5705(\dfrac{14500}{\sqrt{6}} ) \\\\= 212990 \pm 15216.33 \\= (197773.67 ,228206.33)\\\approx (197773, 228207)[/tex]  

Option B)  (197773, 228207)

Using the t-distribution, it is found that the 95% confidence interval for the average price of a home in Gainesville of this size is (197773, 228207).

The first step is finding the number of degrees of freedom, which is the sample size subtracted by 1, so df = 6 - 1 = 5.

Now, we look at the t-table for the critical value for a 95% confidence interval with 5 df, which is t = 2.5706.

The margin of error is:

[tex]M = t\frac{s}{\sqrt{n}}[/tex]

For this problem, [tex]s = 14500, n = 6[/tex], thus:

[tex]M = 2.5706\frac{14500}{\sqrt{6}}[/tex]

[tex]M = 15217[/tex]

The confidence interval is:

[tex]\overline{x} \pm M[/tex]

For this problem, [tex]\overline{x} = 212990[/tex], then:

[tex]\overline{x} - M = 212990 - 15217 = 197773[/tex]

[tex]\overline{x} + M = 212990 + 15217 = 228207[/tex]

The correct option is (197773, 228207).

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Answer:

Step-by-step explanation:

Hello!

The definition of the Central Limi Theorem states that:

Be a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

X[bar]≈N(μ;σ²/n)

If the variable of interest is X: the number of accidents per week at a hazardous intersection.

There is no information about the distribution of this variable, but a sample of n= 52 weeks was taken, and since the sample is large enough you can approximate the distribution of the sample mean to normal. With population mean μ= 2.2 and standard deviation σ/√n= 1.1/√52= 0.15

I hope it helps!

Answer:

a) For this case the intercept of 2.52 represent a common effect of measure for any student without taking in count the other variables analyzed, and we know that if HSGPA=0, ACT= 0 and skip =0 we got [tex] colGPA=2.52[/tex]

b) This value represent the effect into the ACT scores in the GPA, we know that:

[tex]\hat \beta_{ACT} = 0.015[/tex]

So then for every unit increase in the ACT score we expect and increase of 0.015 in the GPA or the predicted variable

c) If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:

Null Hypothesis: [tex]\beta_i = 0[/tex]

Alternative hypothesis: [tex]\beta_i \neq 0[/tex]

Or in other wouds we want to check if an specific slope is significant.

The significance level assumed for this case is [tex]\alpha=0.05[/tex]

Th degrees of freedom for a linear regression is given by [tex]df=n-p-1 = 45-3-1 = 41[/tex], where p =3 the number of variables used to estimate the dependent variable.

In order to test the hypothesis the statistic is given by:

[tex]t=\frac{\hat \beta_i}{SE_{\beta_i}}[/tex]

And replacing we got:

[tex] t = \frac{-0.5}{0.0001}=-5000[/tex]

And for this case we see that if we find the p value for this case we will get a value very near to 0, so then we can conclude that this coefficient would be significant for the regression model .

Step-by-step explanation:

For this case we have the following multiple regression model calculated:

colGPA =2.52+0.38*HSGPA+0.015*ACT-0.5*skip

Part a

(a) Interpret the intercept in this model.

For this case the intercept of 2.52 represent a common effect of measure for any student without taking in count the other variables analyzed, and we know that if HSGPA=0, ACT= 0 and skip =0 we got [tex] colGPA=2.52[/tex]

(b) Interpret [tex]\hat \beta_{ACT}[/tex] from this model.

This value represent the effect into the ACT scores in the GPA, we know that:

[tex]\hat \beta_{ACT} = 0.015[/tex]

So then for every unit increase in the ACT score we expect and increase of 0.015 in the GPA or the predicted variable

(c) What is the predicted college GPA for someone who scored a 25 on the ACT, had a 3.2 high school GPA and missed 4 lectures. Show your work.

For this case we can use the regression model and we got:

[tex] colGPA =2.52 +0.38*3.2 +0.015*25 - 0.5*4 = 26.751[/tex]

(d) Is the estimate of skipping class statistically significant? How do you know? Is the estimate of skipping class economically significant? How do you know? (Hint: Suppose there are 45 lectures in a typical semester long class).

If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:

Null Hypothesis: [tex]\beta_i = 0[/tex]

Alternative hypothesis: [tex]\beta_i \neq 0[/tex]

Or in other wouds we want to check if an specific slope is significant.

The significance level assumed for this case is [tex]\alpha=0.05[/tex]

Th degrees of freedom for a linear regression is given by [tex]df=n-p-1 = 45-3-1 = 41[/tex], where p =3 the number of variables used to estimate the dependent variable.

In order to test the hypothesis the statistic is given by:

[tex]t=\frac{\hat \beta_i}{SE_{\beta_i}}[/tex]

And replacing we got:

[tex] t = \frac{-0.5}{0.0001}=-5000[/tex]

And for this case we see that if we find the p value for this case we will get a value very near to 0, so then we can conclude that this coefficient would be significant for the regression model .

Answer:

68% of the sample can be expected to fall between 28 and 32 cm

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 30

Standard deviation = 2

What proportion of the sample can be expected to fall between 28 and 32 cm

28 = 30 - 2

28 is one standard deviation below the mean

32 = 30 + 2

32 is one standard deviation above the mean.

By the Empirical Rule, 68% of the sample can be expected to fall between 28 and 32 cm

Answer:

(D) -48.193

Step-by-step explanation:

We know that regression equation is

y=a+bx where a is intercept and b is slope of the regression equation.

[tex]Slope=b=\frac{sum(x-xbar)(y-ybar)}{sum(x-xbar)^2}[/tex]

City            Price ($)(x)     Sales(Y)  

River Falls       1.30            100

Hudson           1.60            90

Ellsworth         1.80            90

Prescott           2.00          40

Rock Elm         2.40           38

Stillwater         2.90           32

xbar=sumx/n=12/6=2

ybar=sumy/n=390/6=65

x y x-xbar y-ybar (x-xbar)(y-ybar) (x-xbar)²

1.3 100   -0.7   35              -24.5          0.49

1.6 90  -0.4 25                    -10          0.16

1.8 90   -0.2 25                      -5          0.04

2 40     0         -25                       0           0

2.4 38    0.4 -27                    -10.8         0.16

2.9 32   0.9 -33                   -29.7         0.81

                 Total                     -80                  1.66

[tex]Slope=b=\frac{sum(x-xbar)(y-ybar)}{sum(x-xbar)^2}[/tex]

b=-80/1.66

b=-48.193

Final answer:

The estimated slope parameter, which represents the rate of change in candy bar sales for every dollar change in price, is -3.810, indicating that there is a decrease in sales as the price increases. So the correct option is C.

Explanation:

To estimate the slope parameter of the regression line representing the relationship between candy bar price and sales, we can apply the formula for the slope (m) of the linear regression line:

m = Σ((Xi - μx) × (Yi - μy)) / Σ(Xi - μx)²

where Xi and Yi are the individual sample points and μx and μy are the mean values for the independent (X) and dependent (Y) variables, respectively.

Using the data provided, we can calculate the means (μx and μy) and then use the points to calculate the sum of the products of the deviations and the sum of the squared deviations. We substitute these into the slope formula to find the estimated slope parameter, which would provide the rate of change in candy bar sales concerning the price change.

After performing the calculations, we find that the estimated slope parameter is -3.810. Thus, the correct answer is (C).

Answer:

The probability that the woman selected does not have​ red/green color​ blindness is 0.9945.

Step-by-step explanation:

Let X = a woman has red/green color blindness.

It is provided that, in a certain group of women the rate of red/green color blindness is P (X) = 0.0055.

A complement of an event E, is defined as the event of not E.

The probability of complement of an event E is:

[tex]P(E^{c})=1-P(E)[/tex]

Compute the probability that the woman selected does not have​ red/green color​ blindness as follows:

[tex]P (X^{c})=1-P(X)\\=1-0.0055\\=0.9945[/tex]

Thus, the probability of the complement of the event of a woman having red/green color blindness is 0.9945.

The probability that the woman selected does not have​ red/green color​ blindness is 0.9945.

Since A certain group of women has a 0.55​% rate of​ red/green color blindness.

Here we assume a woman has red/green color blindness be X

So, here the probability should be

= 1-0.55%

= 0.9945

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Answer:

This statement can be made with a level of confidence of 97.72%.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.1 mm

Standard Deviation, σ = 0.5 mm

Sample size, n = 100

We are given that the distribution of thickness is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

Standard error due to sampling:

[tex]=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.5}{\sqrt{100}} = 0.05[/tex]

P(mean thickness is less than 8.2 mm)

P(x < 8.2)

[tex]P( x < 8.2)\\\\ = P( z < \displaystyle\frac{8.2 - 8.1}{0.05})\\\\ = P(z < 2)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x < 8.2) =0.9772 = 97.72\%[/tex]

This statement can be made with a level of confidence of 97.72%.

The z-score calculation reveals that the mean thickness is less than 8.2 mm with approximately 97.50% confidence.

To determine the level of confidence that the mean thickness is less than 8.2 mm, we can calculate the z-score of the sample mean and then find the corresponding percentile in the standard normal distribution. The z-score is calculated by the formula:

z = (x - μ) / (σ / √n)

where x is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size.

Using the given information, x = 8.1 mm, μ = 8.2 mm (the mean to test against), σ = 0.5 mm, and n = 100, we get:

z = (8.1 mm - 8.2 mm) / (0.5 mm / √100) = -0.1 / 0.05 = -2

The z-score corresponds to a percentile in the standard normal distribution, which can be found using a z-table or statistical software. A z-score of -2 corresponds to approximately the 2.5th percentile, which means that the sample mean is less than 8.2 mm with about 97.5% confidence.

Therefore, the statement that the mean thickness is less than 8.2 mm can be made with approximately 97.50% confidence.

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Answer:

We conclude that the value of the mean is 40.

Step-by-step explanation:

We know that the researcher records the following estimates: 40, 46, 30, 50, and 34. If the researcher removes the estimate of 40 (say, due to an experimenter error).  

Now we have the following estimates: 46, 30, 50, and 34.

We calculate  the value of the mean. We get:

[tex]x=\frac{46+30+50+34}{4}\\\\x=\frac{160}{4}\\\\x=40\\[/tex]

We conclude that the value of the mean is 40.

Answer:

the demand curve will shift upward by $20 and the price paid by buyers will decrease by $20.

Step-by-step explanation:

the reduction in the fixed amount of tax from $50 t0 $30 will bring about reduction  of $20 in the price of ticket. the reduction in the price of the ticket, other factors held constant, will brings about change in the demand curve.

The correct answer is D, as supply curve will shift downward by $20, and the effective price received by sellers will increase by $20.

This is so because by reducing the tax, the value that each airline will effectively increase. In this way, as they obtain greater income for each ticket, the airlines will in turn increase the offer of tickets.

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Answer:

[tex] \frac{dP}{\sqrt{P}} = k dt[/tex]

And if we integrate both sides we got:

[tex] 2 \sqrt{P} = kt +C[/tex]

Where C is a constant., we can rewrite the expression like this:

[tex] \sqrt{P} = \frac{1}{2} (kt +C)[/tex]

If we square both sides we got:

[tex] P = \frac{1}{4} (kt +C)^2 [/tex]

If we use the initial condition we have that:

[tex] P(0) = 100 = \frac{1}{4} (k*0 +C)^2 [/tex]

And we can solve for C like this:

[tex] 400 = C^2[/tex]

[tex] C = 20[/tex]

And now we can find the derivate of the function and we got:

[tex] P'(t) = 2* \frac{1}{4} (kt + 20) * k[/tex]

Using the condition [tex] P'(0) = 10 [/tex] we got:

[tex] 10 = \frac{1}{2} k (k*0 +20)[/tex]

[tex] 20 = 20 k[/tex]

k= 1

And then the model is defined as:

[tex] P = \frac{1}{4} (t +20)^2 [/tex]

And for t =12 months we have:

[tex] P(12) = \frac{1}{4} (12 +20)^2 = 256 [/tex]

Step-by-step explanation:

For this case we cna use the proportional model given by:

[tex] \frac{dP}{dt} = k \sqrt{P}[/tex]

Where k is a proportional constant, P the population and the represent the number of months

For this case we know the following initial condition [tex] P(0) =100[/tex] and [tex] P'(0) = 10 [/tex]

we can rewrite the differential equation like this:

[tex] \frac{dP}{\sqrt{P}} = k dt[/tex]

And if we integrate both sides we got:

[tex] 2 \sqrt{P} = kt +C[/tex]

Where C is a constant., we can rewrite the expression like this:

[tex] \sqrt{P} = \frac{1}{2} (kt +C)[/tex]

If we square both sides we got:

[tex] P = \frac{1}{4} (kt +C)^2 [/tex]

If we use the initial condition we have that:

[tex] P(0) = 100 = \frac{1}{4} (k*0 +C)^2 [/tex]

And we can solve for C like this:

[tex] 400 = C^2[/tex]

[tex] C = 20[/tex]

And now we can find the derivate of the function and we got:

[tex] P'(t) = 2* \frac{1}{4} (kt + 20) * k[/tex]

Using the condition [tex] P'(0) = 10 [/tex] we got:

[tex] 10 = \frac{1}{2} k (k*0 +20)[/tex]

[tex] 20 = 20 k[/tex]

k= 1

And then the model is defined as:

[tex] P = \frac{1}{4} (t +20)^2 [/tex]

And for t =12 months we have:

[tex] P(12) = \frac{1}{4} (12 +20)^2 = 256 [/tex]

a. Expectation is linear, so that

[tex]E[27X_1+125X_2+512X_3]=27E[X_1]+125E[X_2]+512E[X_3]=98,630[/tex]

The variance of a sum of independent random variables is equal to a weighted sum of the variances, with weights equal to the squares of the coefficients of the [tex]X_i[/tex]:

[tex]V[27X_1+125X_2+512X_3]=27^2V[X_1]+125^2V[X_2]+512^2V[X_3]=2,306,838[/tex]

b. If the [tex]X_i[/tex] were dependent on one another, we would have the same expectation, but now the variance of a sum of random variables becomes the sum of their covariances:

[tex]\displaystyle\sum_{i=1}^3V[\alpha_iX_i]=\sum_{i=1}^3\sum_{j=1}^3\mathrm{Cov}[X_i,X_j]=\sum_{i=1}^3{\alpha_i}^2V[X_i]+2\sum_{i\neq j}\alpha_i\alpha_j\mathrm{Cov}[X_i,X_j][/tex]

where

[tex]\mathrm{Cov}[X_i,X_j]=E[(X_i-E[X_i])(X_j-E[X_j])]=E[X_iX_j]-E[X_i]E[X_j][/tex]

When we assumed independence, we were granted [tex]E[X_iX_j]=E[X_i]E[X_j][/tex], but this may not be the case if the variables are dependent.

The expected value of the total volume shipped is 98630 ft3. The variance of the total volume shipped is 17423640 ft6. If the variables were not independent, only the expected value would remain correct.

To calculate the expected value of the total volume shipped, we use the linearity of the expectation:

E(Total Volume) = E(27X1 + 125X2 + 512X3)

E(Total Volume) = 27E(X1) + 125E(X2) + 512E(X3)

E(Total Volume) = 27(μ1) + 125(μ2) + 512(μ3)

E(Total Volume) = 27(220) + 125(250) + 512(120)

E(Total Volume) = 5940 + 31250 + 61440

E(Total Volume) = 98630 ft3

To calculate the variance of the total volume shipped, given that X1, X2, X3 are independent:

V(Total Volume) = V(27X1 + 125X2 + 512X3)

V(Total Volume) = 272V(X1) + 1252V(X2) + 5122V(X3)

V(Total Volume) = 729(σ12) + 15625(σ22) + 262144(σ32)

V(Total Volume) = 729(92) + 15625(132) + 262144(82)

V(Total Volume) = 729(81) + 15625(169) + 262144(64)

V(Total Volume) = 59049 + 2639375 + 16777216

V(Total Volume) = 17423640 ft6

If the Xi's were not independent, the expected value would still be correct but the variance would not be correct because the calculation of variance for non-independent variables includes covariance terms that are not present for independent variables.

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Answer:

a) 125 scoops b) 31 scoops

Step-by-step explanation:

The sink = 4000/3 pi

a) The volume of a Cone is 1/3 (pi)(h)(r)^2, meaning...

(1/3)(pi)(8)(2^2)=32/3 pi, and...

(4000pi/3)/(32pi/3) =

4000/32 =

125

b) The volume of a Cone is 1/3 (pi)(h)(r)^2, meaning...

(1/3)(pi)(8)(4^2) = 128pi/3, and...

(4000pi/3)/(128pi/3) =

4000/128 =

31.25, which is approx. 31

Answer:

a) 125

b) 31

Step-by-step explanation:

a) radius = 2

Volume of cup = ⅓×pi×r²×h

= 32pi/3

No. of scoops = 4000pi/3 ÷ 32pi/3

= 125

b) radius = 4

Volume of cup = ⅓×pi×r²×h

= 128pi/3

No. of scoops = 4000pi/3 ÷ 128pi/3

= 31.25

Nearest whole number: 31

Answer:

For part  ;a)

probability that the Atlanta Braves win the World Series=0.40

For part b ;b)

average number of games played regardless of winner =5.8

Step-by-step explanation:

using the  rand() function for the simulation of  the probabilities and  compare with given probabilites if less than above then Atlanta wins otherwise loses

a)

probability that the Atlanta Braves win the World Series=0.40

b)

average number of games played regardless of winner =5.8

Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; [tex]\mu[/tex] = 57,800  and  [tex]\sigma[/tex] = 750.

Let X = randomly selected element of the population

The z probability is given by;

           Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)  

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{58000-57800}{750}[/tex] ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{57000-57800}{750}[/tex] ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

                                                          = 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{750}{\sqrt{100} }[/tex] = 75

The z probability is given by;

           Z = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)  

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] <= [tex]\frac{58000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{57000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

Using the normal distribution and the central limit theorem, it is found that:

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

In this problem:

The probability that a single randomly selected element X of the population is between 57,000 and 58,000 is the p-value of Z when X = 58000 subtracted by the p-value of Z when X = 57000, thus:

X = 58000:

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

[tex]Z = \frac{58000 - 57800}{750}[/tex]

[tex]Z = 0.27[/tex]

[tex]Z = 0.27[/tex] has a p-value of 0.6064.

X = 57000:

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

[tex]Z = \frac{57000 - 57800}{750}[/tex]

[tex]Z = -1.07[/tex]

[tex]Z = -1.07[/tex] has a p-value of 0.1423.

0.6064 - 0.1423 = 0.4641.

0.4641 = 46.41% probability that a single randomly selected element X of the population is between 57,000 and 58,000.

For samples of size 100, [tex]n = 100[/tex], and then:

[tex]s = \frac{750}{\sqrt{100}} = 75[/tex]

For samples of size 100, the mean is of 57800 and the standard deviation is 75.

Then, the probability is:

X = 58000:

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

By the Central Limit Theorem

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

[tex]Z = \frac{58000 - 57800}{75}[/tex]

[tex]Z = 2.7[/tex]

[tex]Z = 2.7[/tex] has a p-value of 0.9965.

X = 57000:

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

[tex]Z = \frac{57000 - 57800}{75}[/tex]

[tex]Z = -10.7[/tex]

[tex]Z = -10.7[/tex] has a p-value of 0.

0.9965 - 0 = 0.9965.

0.9965 = 99.65% probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.

A similar problem is given at https://brainly.com/question/24663213

Answer:

Step-by-step explanation:

Hello!

Since the data on the text is missing, I've found a similar exercise with the same questions so I'm going to use the data of that one to explain how to calculate the probabilities.

We have a sample of 323000 high school students

154000 are girls, of those 47700 dropped out

16900 are boys, of those 10300 dropped out

a) To calculate this probability you have to divide the total number of girls by the total number of students:

P(G)= 154000/323000= 0.476 ≅ 0.48

b) In this item you have to calculate the probability of dropouts, the total of students that dropped out are the girls + boys that dropped out:

Dropouts: 47700+10300= 58000

Now you divide it by the total of students to reach the probability:

P(D)= 58000/323000= 0.179≅ 0.18

c) Now you have to calculate the probability of the intersection between the events "female" and "dropped out", symbolically:

P(G∩D)= P(G)*P(D)= 0.48*0.18= 0.0864

d) The probability of the student dropping out given that it is female is a conditional probability and you can calculate using the following formula:

P(D/G)= P(G∩D) = 0.0864 = 0.18

                 P(G)        0.48

e) Now you have to calculate the probability of the student being a femal, given that she dropped out of school, symbolically:

P(G/D)= P(G∩D) = 0.0864 = 0.48

                 P(D)        0.88

Note:

As you can see in d) P(D/G)=P(D)=0.18 and in e) P(G/D)=P(G)=0.48, this means that both events "the student is a girl" and "the student dropped out" are independent.

Remember two events are not independent when the occurrence of one modifies the probability of occurrence of the other.

I hope it helps!

Answer:

2xy

Step-by-step explanation:

10x²y² - 8xy

2xy(5xy - 4)

The greatest common factor (GCF) of the expressions 10x^2y^2 and 8xy is 2xy. The GCF for the numbers is 2, and for the variables x and y it's x and y, respectively.

The subject is the greatest common factor (GCF), applied to algebraic expressions. The GCF is the largest expression that can be multiplied by another expression to get the original expression. For the expressions 10x^2y^2 and 8xy, the numbers are 10 and 8. Their GCF is 2. The term x appears in both expressions, and the lowest power of x is 1. The term y also appears in both expressions, with the lowest power being 1. So, 2x and y are the GCFs for the numbers and the variables respectively. Combining them, we get 2xy, which is the GCF of 10x^2y^2 and 8xy.

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Answer:

A sampling method is  independent qualitative when an individual selected for one sample does not dictate which individual is to be in the second sample.

Step-by-step explanation:

individuals selected for one sample do not dictate which individuals are to be in the second sample, while dependent quantitative involves individuals selected to be in one sample are used to determine the individuals in the second sample.

Final answer:

An independent sampling method allows for selection of individuals for one sample without influencing the selection for another, maintaining the integrity of quantitative research.

Explanation:

A sampling method is considered independent when an individual selected for one sample does not dictate which individual is to be in the second sample. This type of sampling strategy ensures that the selection of participants for one sample does not influence the selection for another, which is crucial for maintaining the validity and reliability of research findings, particularly in quantitative research.

In research, particularly in studies involving statistical analysis, it is essential to use appropriate sampling methods to ensure that the results are representative of the wider population. Probability sampling methods are often employed in quantitative research as they provide a means to make generalizations from the sample to the population. Types of probability samples include simple random samples, stratified samples, and cluster samples.

Answer: It's C.5

Step-by-step explanation:

bro I hope someone can help you like an expert or something...

Answer:

Probability a team from national football conference wins the same year will be equal to 0.53

Step-by-step explanation:

Probability that a National League team wins baseball world series = 0.44

Probability that a National Football conference team wins the           Superbowl = 0.53

The two events are independent, since the sports played are different (baseball and football). This is a logical assumption to make.

Since these events are independent, the occurrence of one event will not change the probability of the other event. This means that it does not matter whether the national league team won or lost the world series. The probability of the football team winning the Superbowl will remain the same, which is 0.53.

Answer:

f'(x) is increasing in the open interval (-1/2, +∞), and it is decreasing in the interval (-∞, -1/2).

Step-by-step explanation:

f'(x) is a quadratic function with positive main coefficient, then it will be decreasing until the x-coordinate of its vertex and it will be increasing from there onwards. The x-coordinate of the vertex is given by the equation -b/2a = -1/2. Hence

- f'(x) is incresing in the interval (-1/2, +∞)

- f'(x) is decreasing in the interval (-∞, -1/2)

Final answer:

To find intervals where f'(x) is increasing or decreasing, locate the critical points by solving f'(x) = 0 and test the signs of f'(x) around these points. This process determines where the original function f(x) is increasing or decreasing.

Explanation:

To determine the open intervals in which the derivative of a function f(x), denoted as f'(x), is increasing or decreasing, one must:

In the scenario one would solve the equation for f'(x) = 0 to find the critical points. Once these are found, the signs of f'(x) to the left and right of these points can be tested to see where the function is increasing or always decreasing.

Answer:

(a) The probability that there are no dandelions in a randomly selected area of 1 square meter in this region is 0.1353.

(b) The probability that there are at least one dandelion in a randomly selected area of 1 square meter in this region is 0.8647.

Step-by-step explanation:

Let X = number of dandelions per square meter.

The average number of dandelions per square meter is, λ = 2.

The random variable X follows a Poisson distribution with parameter λ = 2.

The probability mass function of X is:

[tex]P(X=x)=\frac{e^{-2}2^{x}}{x!};\ x=0,1,2,3...[/tex]

(a)

Compute the value of P (X = 0) as follows:

[tex]P(X=0)=\frac{e^{-2}2^{0}}{0!}=\frac{0.13534\times 1}{1}=0.1353[/tex]

Thus, the probability that there are no dandelions in a randomly selected area of 1 square meter in this region is 0.1353.

(b)

Compute the value of P (X ≥ 1) as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             [tex]=1-0.1353\\=0.8647[/tex]

Thus, the probability that there are at least one dandelion in a randomly selected area of 1 square meter in this region is 0.8647.

To find the probability of no dandelions, use the Poisson distribution formula with the mean value of 2. To find the probability of at least one dandelion, use the complement rule.

To find the probability of no dandelions in a randomly selected area of 1 square meter in this region, we need to use the Poisson distribution. The mean number of dandelions per square meter is 2. The probability of no dandelions is given by the formula: P(X=0) = e^(-2) * (2^0 / 0!). Calculating this gives us approximately 0.1353.

To find the probability of at least one dandelion, we can use the complement rule. The probability of at least one dandelion is equal to 1 minus the probability of no dandelions. So, P(X >= 1) = 1 - P(X=0). Calculating this gives us approximately 0.8647.

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Answer:

Probability that the sample contains at least five Roman Catholics = 0.995 .

Step-by-step explanation:

We are given that In Quebec, 90 percent of the population subscribes to the Roman Catholic religion.

The Binomial distribution probability is given by;

 P(X = r) = [tex]\binom{n}{r}p^{r}(1-p)^{n-r}[/tex] for x = 0,1,2,3,.......

Here, n = number of trials which is 8 in our case

         r = no. of success which is at least 5 in our case

         p = probability of success which is probability of Roman Catholic of

                 0.90 in our case

So, P(X >= 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

= [tex]\binom{8}{5}0.9^{5}(1-0.9)^{8-5} + \binom{8}{6}0.9^{6}(1-0.9)^{8-6} + \binom{8}{7}0.9^{7}(1-0.9)^{8-7} + \binom{8}{8}0.9^{8}(1-0.9)^{8-8}[/tex]

= 56 * [tex]0.9^{5} * (0.1)^{3}[/tex] + 28 * [tex]0.9^{6} * (0.1)^{2}[/tex] + 8 * [tex]0.9^{7} * (0.1)^{1}[/tex] + 1 * [tex]0.9^{8}[/tex]

= 0.995

Therefore, probability that the sample contains at least five Roman Catholics is 0.995.

The probability that a random sample of eight Quebecois contains at least five Roman Catholics is approximately 0.9950 or 99.50%.

Given:

- The probability of success (subscribing to the Roman Catholic religion) p = 0.9 .

- The number of trials (sample size)  n = 8 .

- We want to find the probability of getting at least 5 successes.

The binomial probability formula is:

[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]

Where:

- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient [tex]\(\frac{n!}{k!(n-k)!} \)[/tex],

- k is the number of successes,

- p is the probability of success,

- (1-p)  is the probability of failure.

We need to find [tex]\( P(X \geq 5) \)[/tex], which is the sum of the probabilities of getting exactly 5, 6, 7, and 8 successes.

[tex]\[ P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) \][/tex]

Calculating each term individually:

1.

[tex]\( P(X = 5) \):\[ P(X = 5) = \binom{8}{5} (0.9)^5 (0.1)^3 \]\[ \binom{8}{5} = \frac{8!}{5!3!} = 56 \]\[ P(X = 5) = 56 \times (0.9)^5 \times (0.1)^3 \]\[ P(X = 5) = 56 \times 0.59049 \times 0.001 \]\[ P(X = 5) = 56 \times 0.00059049 \]\[ P(X = 5) \approx 0.0331 \][/tex]

2.  P(X = 6) :

[tex]\[ P(X = 6) = \binom{8}{6} (0.9)^6 (0.1)^2 \]\[ \binom{8}{6} = \frac{8!}{6!2!} = 28 \]\[ P(X = 6) = 28 \times (0.9)^6 \times (0.1)^2 \]\[ P(X = 6) = 28 \times 0.531441 \times 0.01 \]\[ P(X = 6) = 28 \times 0.00531441 \]\[ P(X = 6) \approx 0.1488 \][/tex]

3.  P(X = 7) :

[tex]\[ P(X = 7) = \binom{8}{7} (0.9)^7 (0.1)^1 \]\[ \binom{8}{7} = \frac{8!}{7!1!} = 8 \]\[ P(X = 7) = 8 \times (0.9)^7 \times 0.1 \]\[ P(X = 7) = 8 \times 0.4782969 \times 0.1 \]\[ P(X = 7) = 8 \times 0.04782969 \]\[ P(X = 7) \approx 0.3826 \][/tex]

4. P(X = 8) :

[tex]\[ P(X = 8) = \binom{8}{8} (0.9)^8 (0.1)^0 \]\[ \binom{8}{8} = 1 \]\[ P(X = 8) = 1 \times (0.9)^8 \times 1 \]\[ P(X = 8) = (0.9)^8 \]\[ P(X = 8) = 0.43046721 \]\[ P(X = 8) \approx 0.4305 \][/tex]

Now sum these probabilities:

[tex]\[ P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) \]\[ P(X \geq 5) \approx 0.0331 + 0.1488 + 0.3826 + 0.4305 \]\[ P(X \geq 5) \approx 0.9950 \][/tex]

So, the probability that the sample contains at least five Roman Catholics is approximately 0.9950 or 99.50%.

Answer:

Step-by-step explanation:

Hello!

You have 4 events A, B, C and D

With probabilities:

P(A∪B)= 5/8

P(A)= 3/8

P(C∩D)= 1/3

P(C)= 1/2

A and B are disjoint events, this means that there are no shared elements between then and their intersection is void, symbolically A∩B= ∅, in consequence, these events are mutually exclusive.

C and D are independent events, this means that the occurrence of one of them does not affect the probability of occurrence of the other one in two consecutive repetitions.

a.

i. P(A∩B)= 0

⇒ Since A and B are disjoint events, the probability of their intersection is zero.

ii. A and B are mutually exclusive events, this means that P(A∪B)= P(A)+P(B)

⇒ From this expression, you can clear the probability of b as P(B)= P(A∪B)-P(A)= 5/8-3/8= 1/4

iii. If Bc is the complementary event of B, its probability would be P(Bc)= 1 - P(B)= 1 - 1/4= 3/4. If the events A and B are mutually exclusive and disjoint, it is logical to believe that so will be the events A and Bc, so their intersection will also be void:

P(A∩Bc)= 0

vi.P(A∪Bc)= P(A) + P(Bc)= 3/8+3/4= 9/8

b.

If A and B are independent then the probability of A is equal to the probability of A given B, symbolically:

P(A)= P(A/B)

[tex]P(A/B)= \frac{P(AnB)}{P(B)}= \frac{0}{1/4}= 0[/tex]

P(A)= 3/8

P(A) ≠ P(A/B) ⇒ A and B are not independent.

c.

i. P(D) ⇒ Considering C and D are two independent events, then we know that P(C∩D)= P(C)*P(D)

Then you can clear the probability of D as:

P(D)= P(C∩D)/P(C)= (1/3)/(1/2)= 2/3

ii. If Dc is the complementary event of D, then its probability is P(Dc)= 1 - P(D) = 1 - 2/3= 1/3

P(C∩Dc)= P(C)*P(Dc)= (1/2)*(1/3)= 1/6

iii. Now Cc is the complementary event of C, its probability is P(Cc)= 1 - P(C)= 1 - 1/2= 1/2

P(Cc∩Dc)= P(Cc)*P(Dc)= (1/2)*(1/3)= 1/6

vi. and e.

[tex]P(C/D)= \frac{P(CnD)}{P(D)} = \frac{1/3}{2/3} = 1/2[/tex]

P(C)=1/2

As you can see the P(C)=P(C/D) ⇒ This fact proves that the events C and D are independent.

d.

i. P(C∪D)= P(C) + P(D) - P(C∩D)= 1/2 + 2/3 - 1/3= 5/6

ii. P(C∪Dc)= P(C) + P(Dc) - P(C∩Dc)= 1/2 + 1/3 - 1/6= 2/3

I hope it helps!

Complete Question

The complete question is shown on the first uploaded image

Answer:

p(X> 6 ) = 0.01132 + 0.01672 = 0.0280

Step-by-step explanation:

The probability that the US woman would have more than six children is equal to the probability that she would have children that are equal to 7 or that she would that she would have children equal to 8 or more as shown on the table in the question

Now representing this mathematically we have

                     p(X > 6) = p(x =7 ) + p(x = 8 or more)

    From the table the p(x = 7 ) [tex]= \frac{523}{46165} = 0.01132[/tex]

     And the p(x= 8 or more ) [tex]= \frac{772}{46165} = 0.01672[/tex]

Hence  p(X> 6 ) = 0.01132 + 0.01672 = 0.0280

An estimation of the probability that a U.S. woman has more than six children using the Empirical Method involves counting how many women in the sample data have more than six children divided by the total number of women in the sample. The specific calculation can't be provided without the concrete data.

To estimate the probability that a U.S. woman has more than six children using the Empirical Method, you would need specific data from the sample. However, since this data hasn't been provided, it's important to explain the process.

First, list the number of women in your sample data who have more than six children. Let's call this number 'X'. Then, calculate the total number of women in your sample data. Let's call this number 'Y'. The empirical probability is then calculated by dividing the number of successful outcomes (X) by the total number of outcomes (Y).

So, if we had data, the equation would be: Probability = X/ Y. Remember to round your answer to four decimal places.

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