2.82 For married couples living in a certain suburb, the probability that the husband will vote on a bond referendum is 0.21, the probability that the wife will vote on the referendum is 0.28, and the probability that both the husband and the wife will vote is 0.15. What is the probability that (a) at least one member of a married couple will vote? (b) a wife will vote, given that her husband will vote? (c) a husband will vote, given that his wife will not vote?

Answer:c

Step-by-step explanation:because the arc length is x that is variable and multiplying it by the the formula gives the right answers.

Answer:

Part a: The equation is [tex]N=N_{0}e^{-kt}[/tex]

Part b: The half life of the material is 2.4 days.

Step-by-step explanation:

Part a

The relation is given as

[tex]\frac{dN}{dt}=-kN[/tex]

Rearranging the equation gives

[tex]\frac{dN}{N}=-kdt[/tex]

Integrating and simplifying the equation as

                 [tex]\int \frac{dN}{N}=\int (-kdt)\\ln N=-kt+lnC\\N=e^{-kt+lnC}\\N=Ce^{-kt}[/tex]

This is the equation of radioactive decay of the radioactive material. For estimation of C consider following IVP where t=0,N=N_o so the equation becomes

[tex]N=Ce^{-kt}\\N_0=Ce^{-k(0)}\\N_0=Ce^{0}\\N_0=C[/tex]

Now substituting the value of C in the equation gives

[tex]N=N_{0}e^{-kt}[/tex]

This is the relation of concentration of unstable radioactive material at a given time .

Part b

From the given data the equation becomes as

                                       [tex]N=N_{0}e^{-kt}\\9=12\times e^{-k*1}\\e^{-k}=\frac{9}{12}[/tex]

Now half like is defined as the time when the quantity is exact half, i.e. N/N_o =0.5 so

                                      [tex]N=N_{0}e^{-kt}\\6=12\times e^{-k*t}\\e^{-kt}=\frac{6}{12}\\{e^{-k}}^t}=0.5\\(0.75)^t=0.5\\So\, t\, is\, given\, as \\t=\frac{ln 0.5}{ln 0.75}\\t=2.4 \, days[/tex]

So the half life of the material is 2.4 days.

Answer:

The p-value for the test statistic is 0.1587.

Step-by-step explanation:

As the sample size for both the airlines are large, according to the central limit theorem the sampling distribution of sample proportion follows a normal distribution.

The test statistic for the difference between two proportions is:

[tex]z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{P(1-P)\frac{1}{n_{1}}+\frac{1}{n_{2}} } }[/tex]

The test statistic value is, z = -1.00

The p-value of the test statistic is:

[tex]P (Z<-1.00)=1-P(Z<1.00)=1-0.8413=0.1587[/tex]

**Use the standard normal table for the probability.

Thus, the p-value for the test statistic is 0.1587.

Answer:

The correct option is (d) 11.070

Step-by-step explanation:

The test statistic for Goodness of fit test for k observations is:

[tex]\chi^{2}=\sum\frac{(Observed-Expected)^{2}}{Expected}[/tex]

This statistic follows a Chi-square distribution with (k - 1) degrees of freedom and α level of significance.

Here α = 0.05 and degrees of freedom is, k - 1 = 6 - 1 = 5 d.f.

Use the chi-square table for the critical value.

[tex]\chi^{2}_{(5)}=11.070[/tex]

Thus, the correct option is (d).

The velocity vector of the bee is simply the rate of change of the position of the bee.

The bee is located at [tex]\mathbf{ (3,-6,10)}[/tex] at 9 seconds.

The given parameters are:

[tex]\mathbf{r'(t) = (3,-6,10)}[/tex], when t = 0

A vector is represented as:

[tex]\mathbf{r'(t) = xi + yj + zk}[/tex]

From the question, we have:

[tex]\mathbf{\int\limits^9_0 {r'(u)} \, du = 0}[/tex]

Integrate

[tex]\mathbf{r(u)|\limits^9_0 = 0}[/tex]

Expand

[tex]\mathbf{r(9) - r(0) = 0}[/tex]

Rewrite as:

[tex]\mathbf{r(9) = r(0) }[/tex]

Recall that: [tex]\mathbf{r'(t) = (3,-6,10)}[/tex]

So, we have:

[tex]\mathbf{r(9) = 3i -6j + 10k }[/tex]

Also, we have:

[tex]\mathbf{xi + yj + zk = 3i -6j + 10k }[/tex]

By comparison;

[tex]\mathbf{x = 3}[/tex]

[tex]\mathbf{y = -6}[/tex]

[tex]\mathbf{z = 10}[/tex]

So, the bee is located at [tex]\mathbf{ (3,-6,10)}[/tex] at 9 seconds.

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After 9 seconds, the bee returns to its initial position (3, -6, 10) because the integral of the velocity function from 0 to 9 seconds is zero, indicating that the net displacement of the bee in these 9 seconds is zero.

The question is essentially asking where the bee is located based on its velocity vector, initial position, and the change in time. To answer this, we will use the concept of integrals from calculus. The integral of the velocity function, r′(t), over a period of time gives the displacement of the bee. However, it's given that ∫9 0 r′(u) du = 0. This tells us that the net displacement of the bee in that 9 seconds is zero.

Displacement refers to the change in position of an object, so if the net displacement in these 9 seconds is zero, then the bee is at the same position at t=9 as at t=0. In other words, after travelling around for 9 seconds, the bee returns to the same starting point. Therefore, at time t=9, the bee is located at the same initial position, that is, (3, -6, 10).

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Answer:

B. stratification

Step-by-step explanation:

Stratification of data is a process in which data is agrupated into some groups or strats depending on its value.

This action is helpful when the person has a lot of data that can be clasified on several variables that will make their analysis easier to manage.

Answer:convenience sampling

Step-by-step explanation:

This is the type of sampling where the first available primary data source will be used without additional requirements.

a) A nonzero vector orthogonal to the plane through the points P, Q, and R is [tex]PQ= < 5, 5, -3 >[/tex] and [tex]PR= < 5, 6, 1 >[/tex].

b) The area of the triangle PQR is 32.09 square units.

Given points are P(0, -4, 0), Q(5, 1, -3), T(5, 2, 1)

a) [tex]PQ= < 5, 1, -3 > - < 0, -4, 0 >[/tex]

[tex]PQ= < 5-0, 1-(-4), -3-0 >[/tex]

[tex]PQ= < 5, 5, -3 >[/tex]

[tex]PR= < 5, 2, 1 > - < 0, -4, 0 >[/tex]

[tex]PR= < 5-0, 2-(-4), 1-0 >[/tex]

[tex]PR= < 5, 6, 1 >[/tex]

b) The cross product of PQ and PR is

[tex]PQ\times PR=\left[\begin{array}{ccc}i&j&k\\5&5&-3\\5&6&1\end{array}\right][/tex]

[tex]PQ\times PR=(5+18)i-(5+15)j+(30-25)k[/tex]

[tex]PQ\times PR=23i-20j+10k[/tex]

The magnitude of [tex]PQ\times PR[/tex] is

[tex]|PQ\times PR| =|23i-20j+10k|[/tex]

[tex]|PQ\times PR| =\sqrt{23^2+(-20)^2+10^2}[/tex]

[tex]|PQ\times PR| =\sqrt{529+400+100}[/tex]

[tex]|PQ\times PR| =\sqrt{1029}[/tex]

[tex]|PQ\times PR| =32.09[/tex]

Therefore,

a) A nonzero vector orthogonal to the plane through the points P, Q, and R is [tex]PQ= < 5, 5, -3 >[/tex] and [tex]PR= < 5, 6, 1 >[/tex].

b) The area of the triangle PQR is 32.09 square units.

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Final answer:

To find a nonzero vector orthogonal to the plane through points P, Q, and R, calculate the cross product of two vectors in the plane. The area of triangle PQR can be found by dividing the length of the cross product vector by 2.

Explanation:

To find a nonzero vector orthogonal to the plane through points P, Q, and R, we can calculate the cross product of two vectors in the plane. Let's take vectors PQ and PR. Using the formula for cross product, we have:

PQ x PR = (5 - 0, 1 - (-4), -3 - 0) = (5, 5, -3)

Therefore, the vector (5, 5, -3) is orthogonal to the plane through points P, Q, and R.

To find the area of triangle PQR, we can use the length of the cross product vector and divide it by 2. The length of the cross product vector PQ x PR is:

|PQ x PR| =√(5² + 5² + (-3)²) = sqrt(59)

Dividing by 2, we get the area of triangle PQR as:

Area = √(59) / 2

Answer:

0.0369 = 3.69% probability that all six cells of the selected cells are able to replicate.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes

The combinations formula is important in this problem:

[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]

Desired outcomes

31 cells

19 are able to replicate.

We pick six.

The order is not important, that is, if it had been 2 cells, cell A and cell B would be the same outcome as cell B and cell A. So we use the combinations formula to find the number of desired outcomes.

It is a combination of 6 from 19(cells who are able to replicate).

[tex]D = C_{19,6} = \frac{19!}{6!(19 - 6)!} = 27132[/tex]

Total outcomes

31 cells

6 are picked.

[tex]T = C_{31,6} = \frac{31!}{6!(31 - 6)!} = 736281[/tex]

What is the probability that all six cells of the selected cells are able to replicate?

[tex]P = \frac{D}{T} = \frac{27132}{736281} = 0.0369[/tex]

0.0369 = 3.69% probability that all six cells of the selected cells are able to replicate.

Answer:

The answer would be 98.8

Step-by-step explanation:

Resin being you must round simply to the nearest hundreds

Answer: 1/2

Step-by-step explanation:

Since it's a fair coin, the probability of tossing head, P(H) = probability of tossing tail, P(T) = 1/2.

Hence, for the event of A being exactly 2 tails, then we have a possibility of:

A= [ HTT, THT, TTH]

For the event of B being first two tosses resulting in tail, it becomes:

B= [TTH)

For event of C being that the three tosses are tail, it becomes:

C=[TTT]

Hence, union of Event A, B and C is given as:

AuBuC= [HTT, THT, TTH, TTT]

Prob [AuBuC] = [ (1/2 * 1/2 * 1/2)] * 4

P[AuBuC] = (1/8) *4

P(AuBuC) = 1/2.

Answer:

1.8413 is the radian measure of angle at the center of circle.

Step-by-step explanation:

We are given the following in the question:

Radius,r = 70.0 cm

Arc length,s = 127 cm

Formula:

[tex]\theta = \dfrac{s}{r}[/tex]

where [tex]\theta[/tex] is the angle measure in radians, s is the intercepted arc and r is the radius of the circle.

Putting the values, we get,

[tex]\theta = \dfrac{127}{70} = 1.8143[/tex]

Thus, 1.8413 is the radian measure of angle at the center of circle.

The radian measure of an angle in a circle, given an arc length of 127 cm and radius of 70 cm, is approximately 1.814. This is computed by dividing the arc length by the radius.

In mathematics, the measure of an angle in radians in a circle is given by the ratio of the length of the arc that it subtends and the radius of the circle in which this occurs. This is represented by the formula: θ = s / r where s is the arc length and r is the radius.

Given the radius r = 70 cm and arc length s = 127 cm, you can substitute these values into the formula to get the radian measure of the angle. Therefore, by substitution we get: θ = 127 cm / 70 cm which simplifies to approximately 1.814 radians.

To note, a radian is the standard unit of angular measure, used in many areas of mathematics. An angle's measurement in radians is numerically equal to the length of a corresponding arc of a unit circle, hence why this method works.

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Answer:

a) 95% of the data falls between $135,000 and $175,000.

b) 81.5% of new homes priced between $135,000 and $165,000.

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 deviations of the mean.

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

We also have that:

50% of the measures are below the mean and 50% of the measures are above the mean.

34% of the measures are between 1 standard deviation below the mean and the mean, and 34% of the measures are between the mean and 1 standard deviations above the mean.

47.5% of the measures are between 2 standard deviations below the mean and the mean, and 47.5% of the measures are between the mean and 2 standard deviations above the mean.

49.85% of the measures are between 3 standard deviations below the mean and the mean, and 49.85% of the measures are between the mean and 3 standard deviations above the mean.

In this problem, we have that:

Mean = $155,000.

Standard deviation = $10,000.

(a) Between what two values do about 95% of the data fall?

By the Empirical Rule, 95% of the values fall within 2 standard deviations of the mean.

So

155000 - 2*10000 = 135,000

155000 + 2*10000 = 175,000

95% of the data falls between $135,000 and $175,000.

(b) Estimate the percentage of new homes priced between $135,000 and $165,000?

We have to find how many fall between $135,000 and the mean($155,000) and how many fall between the mean and $165,000

$135,000 and the mean

$135,000 is two standard deviations below the mean.

By the empirical rule, 47.5% of the measures are between 2 standard deviations below the mean and the mean.

So 47.5% of the measures are between $135,000 and the mean

Mean and $165,000

$165,000 is one standard deviation above the mean.

By the empirical rule, 34% of the measures are between the mean and 1 standard deviations above the mean.

So 34% of the measures are between the mean and $165,000.

$135,000 and $165,000

47.5% + 34% = 81.5% of new homes priced between $135,000 and $165,000.

Final answer:

The answer explains how to determine the range of values where 95% of data falls based on mean and standard deviation and estimates the percentage of new homes within a specific price range.

Explanation:

The mean price for new homes from a sample of houses is $155,000 with a standard deviation of $10,000.

(a) Between what two values do about 95% of the data fall?

(b) Estimate the percentage of new homes priced between $135,000 and $165,000?

Answer:

a) 0.5

b) 0.309

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 70 inches

Standard Deviation, σ = 4 inches

Formula:

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

Height of George Washington = 6 feet = [tex]6\times 12 = 72\text{ inches}[/tex]

x = 72

a) z-score

[tex]z_{score} = \displaystyle\frac{72-70}{4} = 0.5[/tex]

Interpretation:

This, mean that George Washington is 0.5 standard deviation more than the mean height of for men.

b) P(individual will be as tall or taller than George Washington)

[tex]P( x \geq 72) = P( z \geq 0.5)[/tex]

[tex]= 1 - P(z < 0.5)[/tex]

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

[tex]P(x \geq 72) = 1 - 0.691 = 0.309 = 30.9\%[/tex]

Interpretation:

0.309 is the probability  that a randomly selected individual will be as tall or taller than George Washington.

Answer:

the equation[tex]5(\frac{d^{2}x }{dt^{2} }) +4(\frac{dx}{dt})+9x=2cos3t[/tex] is a partial differential equation(PDE) because it contains unknown multi variables and their derivatives. This is a PDE of order 2.

The independent variable is x while the dependent variable is t.

The PDE is Linear.

Step-by-step explanation:

Partial Differential Equation (PDE): This is a differential equation that contains multi variables and their derivatives.

Ordinary Differential Equation (ODE): this is a differential equation containing a function of one independent variable and its derivatives.

Answer:

Step-by-step explanation:

given that a computer sends a packet of information along a channel and waits for a return signal acknowledging that the packet has been received. If no acknowledgment is received within a certain time, the packet is re-sent

Let X represent the number of times the packet is sent. Assume that the probability mass function of X is given by

p(x)={ cx for x=1,2,3,4, or 5

{ 0 otherwise

where c is a constant.

Let us write pdf as follows

x             1              2            3            4           5              Total

p             c             2c           3c         4c         5c              15c

Since total probability = 1

we get

[tex]c=\frac{1}{15}[/tex]

P(X=2)=2c = [tex]\frac{2}{15}[/tex]

Answer:

b. False

See explanation below

Step-by-step explanation:

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:  

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

For this case we know that r =0.87. But we can't conclude that the linear correlation coefficient represent a cause-effect relationship, since the linear correlation coefficient measures the linear dependency between two variables, but we can have other types of association between the variables and that's not measured by the Pearson correlation coeffcient.

So for this case the answer woudl be

b. False

Final answer:

The claim that a correlation of 0.87 between time in prison and aggression implies causation is false because correlation does not equal causation. There may be other factors influencing aggression, and the correlation observed might be spurious.

Explanation:

The statement that a correlation of 0.87 between the length of time a person is in prison and the amount of aggression displayed means that spending a longer time in prison causes people to become more aggressive is false.

Correlation does not imply causation. Just because two variables are correlated does not mean that one variable causes the change in the other. There could be other factors or variables that contribute to the aggression, which are not accounted for in simply observing the correlation.

Negative experiences do increase aggression, as people are more prone to aggress when experiencing negative emotions such as frustration, pain, or being in a bad mood. In a prison context, the aggression could also stem from the social structure of prison life and not merely the length of time served.

The Stanford prison experiment by Philip Zimbardo is a notable study highlighting how quickly people can turn to aggressive behaviors based on the roles they are assigned within a prison environment.

Answer:

The balance will be $2,731.82.

Step-by-step explanation:

This problem can be solved by the following formula:

[tex]A = P(1 + r)^{t}[/tex]

In which A is the final amount(balance), P is the principal(the deposit), r is the interest rate and t is the time, in years.

In this problem, we have that:

[tex]P = 2500, r = 0.03 t = 3[/tex]

We want to find A

So

[tex]A = P(1 + r)^{t}[/tex]

[tex]A = 2500(1 + 0.03)^{3}[/tex]

[tex]A = 2,731.82[/tex]

The balance will be $2,731.82.

Answer:

2731.8

Step-by-step explanation:

2500 deposited at 3 percent interest annually equation form

2500(1+(3/100))

for 3 years power the bract by 3

2500(1+(3/100))^3=2731.81

aka 2732

To find the 73rd percentile of the given data, we need to arrange the weights in ascending order and calculate the rank. Then, we interpolate to find the weight at the desired percentile.

To find the weight that represents the 73rd percentile, we need to follow these steps:

Therefore, the weight that represents the 73rd percentile is 8.02.

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Answer:

i dont know

Step-by-step explanation:

just working

Answer:

x=4

y=-3

Step-by-step explanation:

3x+4y=0

2x+5y=7

Change 3x+4y=0 into y= -3/4x

so substitute for y

2x-15/4x=7

-1 3/4x = 7

x=-4

substitue x for y and get

-3/4 * 4

y=-3

Answer: x = - 4

y = 3

Step-by-step explanation:

The given system of simultaneous equations is expressed as

3x+4y=0 - - - - - - - - - - - - -1

2x+5y=7- - - - - - - - - - - - - - - 2

From equation 1, we would make x the subject of the formula. Firstly, we would subtract 4y from the Left hand side and the right hand side of the equation. It becomes

3x +4y - 4y = 0 - 4y

3x = - 4y

We would divide the Left hand side and the right hand side of the equation by 3. It becomes

3x/3 = - 4y/3

x = - 4y/3

Substituting x = - 4y/3 into equation 2, it becomes

2 × - 4y/3 +5y = 7

- 8y/3 + 5y = 7

(- 8y + 15y)/3 = 7

7y/3 = 7

Cross multiplying, it becomes

7y = 21

Dividing the left hand side and the right hand side of the equation by 7, it becomes

7y/7 = 21/7

y = 3

Substituting y = 3 into x = - 4y/3, it becomes

x = - 4× 3/3

x = - 4

Answer:

a)

A={RRR,SSS,LLL}

b)

B={RSL,SLR,LRS,RLS,SRL,LSR}

c)

C={RRL,RLR,LRR,RRS,SRR,RSR}

d)

D={RRL, RRS, RLR, RLL, RSR, RSS, LRR, LRL, LLR, LLS, LSL, LSS, SRR, SRS, SLL, SLS, SSR, SSL}

e)

D'={RRR, RLS, RSL, LRS, LLL, LSR, SRL, SLR, SSS}

Step-by-step explanation:

Sample space=S= {RRR, RRL, RRS, RLR, RLL, RLS, RSR, RSL, RSS, LRR, LRL, LRS, LLR, LLL, LLS, LSR, LSL, LSS, SRR, SRL, SRS, SLR, SLL, SLS, SSR, SSL, SSS}

a)

Let A be the event that all three vehicles go in same direction. It means that the all three vehicles go to right or left or straight. Thus, event A can be represented as

A={RRR,SSS,LLL}

b)

Let B be the event that all three vehicles go in different direction. It means that the three vehicles can go to

1. Right, Straight, Left

2. Straight, Left, Right

3. Left, Right, Straight

4. Right, Left, Straight

5. Straight, Right, Left

6.  Left, Straight, Right

Thus, event B can be represented as

B={RSL,SLR,LRS,RLS,SRL,LSR}

c)

Let C be the event that exactly two of three vehicles turn right. It means that the three vehicles can go to

1. Right, Right , Left

2. Right , Left, Right

3. Left, Right, Right

4. Right, Right , Straight

5. Straight, Right,Right

6.  Right , Straight, Right

Thus, event C can be represented as

C={RRL,RLR,LRR,RRS,SRR,RSR}

d)

Let D be the event that exactly two of three vehicles go in same direction. It means that the three vehicles can go to

1. Right,  Right, Left

2. Right, Right, Straight

3. Right, Left,  Right

4. Right, Left, Left

5. Right, Straight, Right

6. Right, Straight, Straight

7.  Left,  Right, Right

8. Left,  Right, Left

9. Left, Left, Right

10. Left, Left, Straight

11. Left, Straight, Left

12. Left, Straight, Straight

13. Straight, Right, Right

14. Straight, Right, Straight

15. Straight, Left, Left

16. Straight, Left, Straight

17. Straight, Straight, Right

18.Straight,Straight, Left

Thus, event D can be represented as

D={RRL, RRS, RLR, RLL, RSR, RSS, LRR, LRL, LLR, LLS, LSL, LSS, SRR, SRS, SLL, SLS, SSR, SSL}

e)

D'=S-D= {RRR, RRL, RRS, RLR, RLL, RLS, RSR, RSL, RSS, LRR, LRL, LRS, LLR, LLL, LLS, LSR, LSL, LSS, SRR, SRL, SRS, SLR, SLL, SLS, SSR, SSL, SSS} -{RRL, RRS, RLR, RLL, RSR, RSS, LRR, LRL, LLR, LLS, LSL, LSS, SRR, SRS, SLL, SLS, SSR, SSL}

D'={RRR, RLS, RSL, LRS, LLL, LSR, SRL, SLR, SSS}

Final answer:

For a freeway exit where vehicles can go right, left, or straight, we determine the combination of directions for various events: Event A with all vehicles going the same direction has 3 outcomes; Event B with all vehicles taking different directions has 6 outcomes; Event C with two vehicles turning right has 6 outcomes; Event D with two vehicles going in the same direction has several outcomes depending on combinations; Event D' coincides with Event B.

Explanation:

When analyzing the outcomes of three successive vehicles taking different actions at a freeway exit, we can approach the given scenarios methodically:

To further explain, for (a), since there are three options for direction and all must be the same, we have exactly three outcomes when they are all identical. For (b), all vehicles must take a unique direction, and since there are three vehicles and three directions, we have 3! (factorial) permutations of the outcome, resulting in 6 possibilities. For (c), two vehicles must turn right and the third vehicle must take either of the two remaining directions, which can be arranged in 3 choose 2 (3C2 = 3) ways for each of the two non-right directions, making 6 outcomes total. For (d), it's similar to (c), but we include all same-direction pairs, resulting in 3C2 combinations for each pair of directions. Lastly, (e) is simply the case where no two outcomes are the same, hence it is the same as (b).

Answer:

You would expect 95% of your customers to stay in the store from 43.1 minutes to 57.5 minutes.

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 = 50.3 minutes

Standard deviation = 3.6 minutes

Within what range would you expect 95% of your customers to stay in your store?

Within 2 standard deviations of the mean.

So from

50.3 - 2*3.6 = 43.1 minutes

To

50.3 + 2*3.6 = 57.5 minutes

You would expect 95% of your customers to stay in the store from 43.1 minutes to 57.5 minutes.

Answer:

(A) 0.04

(B) 0.25

(C) 0.40

Step-by-step explanation:

Let R = drawing a red chips, G = drawing green chips and W = drawing white chips.

Given:

R = 8, G = 10 and W = 2.

Total number of chips = 8 + 10 + 2 = 20

[tex]P(R) = \frac{8}{20}=\frac{2}{5}\\P(G)= \frac{10}{20}=\frac{1}{2}\\P(W)= \frac{2}{20}=\frac{1}{10}[/tex]

As the chips are replaced after drawing the probability of selecting the second chip is independent of the probability of selecting the first chip.

(A)

Compute the probability of selecting a white chip on the first and a red on the second as follows:

[tex]P(1^{st}\ white\ chip, 2^{nd}\ red\ chip)=P(W)\times P(R)\\=\frac{1}{10}\times \frac{2}{5}\\ =\frac{1}{25} \\=0.04[/tex]

Thus, the probability of selecting a white chip on the first and a red on the second is 0.04.

(B)

Compute the probability of selecting 2 green chips:

[tex]P(2\ Green\ chips)=P(G)\times P(G)\\=\frac{1}{2} \times\frac{1}{2}\\ =\frac{1}{4}\\ =0.25[/tex]

Thus, the probability of selecting 2 green chips is 0.25.

(C)

Compute the conditional probability of selecting a red chip given the first chip drawn was white as follows:

[tex]P(2^{nd}\ red\ chip|1^{st}\ white\ chip)=\frac{P(2^{nd}\ red\ chip\ \cap 1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\=\frac{P(2^{nd}\ red\ chip)P(1^{st}\ white\ chip)}{P (1^{st}\ white\ chip)} \\= P(R)\\=\frac{2}{5}\\=0.40[/tex]

Thus, the probability of selecting a red chip given the first chip drawn was white is 0.40.

The probability of drawing a white chip on the first draw and a red chip on the second draw with replacement is 0.04. The probability of drawing two green chips is 0.25. The probability of drawing a red chip on the second draw, given that a white chip was drawn on the first draw, is 4.

To find the probability of different events, we need to use the concept of probability. Probability is the likelihood of an event occurring. In this case, we have an urn with 8 red chips, 10 green chips, and 2 white chips.

(A) The probability of drawing a white chip on the first draw and a red chip on the second draw with replacement is:

P(white first and red second) = P(white first) * P(red second) = (2/20) * (8/20) = 16/400 = 0.04.

(B) The probability of drawing two green chips is:

P(green first and green second) = P(green first) * P(green second) = (10/20) * (10/20) = 100/400 = 0.25.

(C) The probability of drawing a red chip on the second draw, given that a white chip was drawn on the first draw, is:

P(red second | white first) = P(red second and white first) / P(white first) = (8/20) / (2/20) = 8/2 = 4.

Answer:

Expression: N = C·L·l(t)· T + 20

The initial value problem and solution are expressed as a first order differential equation.

Step-by-step explanation:

First, gather the information:

total population, N = 2 000

Proportionality constant, C = 0.0002

l(t) number of infected individuals = l(t)

healthy individuals = L

The equation is given as follows:

N = C·L·l(t)

However, there is a change with time, so the expression will be:

[tex]\frac{dN}{dt}[/tex] = C·L·l(t)

multiplying both sides  by dt gives:

dN  =   C·L·l(t)

Integrating both sides gives:

[tex]\int\limits^a_b {dN} \, dt[/tex] = [tex]\int\limits^a_b {CLl(t)} \, dt[/tex]

N = C·L·l(t)· T + K

initial conditions:

T= 0, N₀ = (0.01 ₓ 2 000)  = 20

to find K, plug in the values:

N₀ = K

20 = K

At any time T, the expression will be:

N = C·L·l(t)· T + 20 Ans

Answer:

0.74141

Step-by-step explanation:

There are 52 cards in total

since each card is different,

the probability = number of favorable cards / total number of outcomes

P(C₁) = number of favorable cards / total number of outcomes

     = [tex]\frac{52}{52}[/tex]

P(C₂, C₁) = number of favorable cards / total number of outcomes

   = [tex]\frac{51}{52}[/tex]

P(C₃,C₁ ∩ C₂) = [tex]\frac{50}{52}[/tex]

P(C₄,C₁ ∩ C₂ ∩ C₃) = [tex]\frac{49}{52}[/tex]

P(C₅,C₁ ∩ C₂ ∩ C₃ ∩ C₄) = [tex]\frac{48}{52}[/tex]

P(C₆,C₁ ∩ C₂ ∩ C₃ ∩ C₄ ∩ C₅) = [tex]\frac{47}{52}[/tex]

General multiplication rule

P(A) =P(C₁ ∩ C₂ ∩ C₃ ∩ C₄ ∩ C₅ ∩ C₆)

         = [tex]\frac{52}{52}. \frac{51}{52}. \frac{50}{52}. \frac{49}{52}. \frac{48}{52} .\frac{47}{52}[/tex]

         = 8,808,975 / 11,881,376

         = 0.74141

Answer:

0.3506 is the required probability.

Step-by-step explanation:

We are given the following in the question:

A: Weather on a given day is cloudy

B: Weather on a given day is snowy

[tex]P(A) = 0.77\\P(A\cap B) = 0.27[/tex]

We have to find the probability that the randomly selected day in May will be snowy if it is cloudy.

That is we have to evaluate P(B|A)

[tex]P(B|A) = \dfrac{P(B\cap A)}{P(A)}\\\\P(B|A) = \dfrac{0.27}{0.77}\\\\P(B|A) = 0.3506[/tex]

Thus, 0.3506 is the probability hat a randomly selected day in May will be snowy if it is cloudy.

Answer:

y=25600

y=100(2)^{2t}

y=252

Step-by-step explanation:

A bacteria culture starts with 100 bacteria and doubles in size every half hour.

[tex]y=100(2)^{2x}[/tex]

where x represents the number of hours

(a) t= 4 hours

Plug in 4 for t  and find out the number of bacteria y

[tex]y=100(2)^{2 \cdot 4}\\y=25600[/tex]

25600 bacteria

[tex]y=100(2)^{2t}[/tex]

convert 40 seconds into hour

40 divide by 60 =2/3

[tex]y=100(2)^{2\frac{2}{3} }\\y=100(2)^{\frac{4}{3} }\\\\y=252[/tex]

The graph is attached below

Answer:

a) probability of choosing heads= 1/2 (50%)

b)  probability of choosing the fair coin knowing that it showed heads is= 1/3 (33.33%)

Step-by-step explanation:

Since the unfair coin can have 2 heads or 2 tails , and assuming both are equally possible . then

probability of choosing the fair coin  (named A)= 1/2

probability of choosing an unfair coin with 2 heads (named B)= (1-1/2)*1/2= 1/4

probability of choosing an unfair coin with 2 tails (named C)= (1-1/2)*(1-1/2)= 1/4

then

probability of choosing heads= probability of choosing A * probability of getting heads from A + probability of choosing B * probability of getting heads from B + probability of choosing C * probability of getting heads from C =

1/2*1/2 + 1/4*1 + 1/4*0 = 2/4 = 1/2

the probability of choosing the fair coin knowing that it showed heads is

P(A/B) = P(A∩B)/P(B)

denoting event A= the coin is fair and event B= the result is heads

P(A∩B) = 1/2*1/2 = 1/4

but since we know now that that the unfair coin is not possible , the probability of choosing heads is altered:

P(B)=probability of choosing heads= probability of choosing A * probability of getting heads from A + probability of choosing B * probability of getting heads from B  = 1/2*1/2+1/2*1 = 3/4

then

P(A/B) = P(A∩B)/P(B)  = (1/4)/(3/4) = 1/3

then the probability is 1/3

Answer:

Part A:

[tex]v_w=53.9\ m/s[/tex]

Part B:

Wrench is thrown away from the spaceship

Step-by-step explanation:

This the problem related to conservation of momentum.

According to the conservation of momentum:

Initial Momentum=Final Momentum

[tex]m_av_a=m_wv_w+(m_a-m_v)v[/tex]

where:

[tex]m_a[/tex] is the mass of astronaut

[tex]m_w[/tex] is the mass of wrench

[tex]m_a-m_w[/tex] is the mass when wrench is thrown

[tex]v_a[/tex] is the speed of astronaut

[tex]v_w[/tex] is the speed of wrench

v is the speed acquired

Part A:

(+ve sign for away from ship),( -ve sign for towards ship)

v= -0.10 m/s

[tex]90*0.2=0.5*v_w+(90-0.5)(-0.1)\\v_w=53.9\ m/s[/tex]

Part B:

Since velocity is +ve as calculated above and according to conditions:

Wrench is thrown away from the spaceship

Answer:

(a) The probability that a selected joint was judged to be defective by neither of the two inspectors is 0.906.

(b) The probability that a selected joint was judged to be defective by inspector B but not by inspector A is 0.0213.

Step-by-step explanation:

The sample of joints randomly selected is, n = 10,000.

Number of joints judged defective by inspector A is, n (A) = 727.

The probability that a joint is judged defective by inspector A is:

[tex]P(A)=\frac{n(A)}{n} =\frac{727}{10000} =0.0727[/tex]

Number of joints judged defective by inspector B is, n (B) = 756.

The probability that a joint is judged defective by inspector B is:

[tex]P(B)=\frac{n(B)}{n} =\frac{756}{10000} =0.0756[/tex]

Number of joints judged defective by at least one of the inspectors is,

n (At least 1) = 940.

The probability that a joint is judged defective by at least one of the inspectors is:

[tex]P(At\ least\ 1)=\frac{n(At\ least\ 1)}{n} =\frac{940}{10000}=0.094[/tex]

(a)

Compute the probability that a selected joint was judged to be defective by neither of the two inspectors as follows:

P (At least 1) = 1 - P (Less than 1)

                    = 1 - P (None)

     P (None) = 1 - P (At least 1)

                    [tex]=1-0.094\\=0.906[/tex]

Thus, the probability that a selected joint was judged to be defective by neither of the two inspectors is 0.906.

(b)

Compute the probability that a selected joint was judged to be defective by inspector B but not by inspector A as follows:

P (B but not A) = P (At least 1) - P (A)

                        [tex]=0.094-0.0727\\=0.0213[/tex]

Thus, the probability that a selected joint was judged to be defective by inspector B but not by inspector A is 0.0213.

a) The values of a, b and c are [tex]\frac{173}{4}, 41[/tex] and [tex]1547[/tex] respectively.

b) The number of craft beer breweries in 2016 will be 4643.

The quadratic equation holds an important significance in mathematics and is used in many situations, like calculating area or finding profit. When two linear expressions get multiplied, the resultant expression is a quadratic expression.

The number of craft beer breweries is modeled by a quadratic equation [tex]f(t)=at^2+bt+c[/tex].

In 2008, t=0.

a) Here, it is given that [tex]f(0)=1547[/tex], [tex]f(2)=1802[/tex] and [tex]f(4)=2403[/tex].

Now, use these values to find a, b and c as follows:

[tex]f(0)=a(0)^2+b(0)+c[/tex]

[tex]1547=c[/tex]

[tex]f(2)=a(2)^2+2b+c[/tex]

[tex]1802=4a+2b+1547[/tex]

[tex]255=4a+2b[/tex]

[tex]f(4)=a(4)^2+b(4)+c[/tex]

[tex]2403=16a+4b+1547[/tex]

[tex]856=16a+4b[/tex]

[tex]214=4a+b[/tex]

Now, setting the equations [tex]4a+2b=255[/tex] and [tex]4a+b=214[/tex] equal to each other:

[tex]4a=255-2b[/tex]

[tex]4a=214-b[/tex]

[tex]255-2b=214-b[/tex]

[tex]b=255-214[/tex]

[tex]b=41[/tex]

Now plugging the value of b into the equation [tex]4a+b=214[/tex], we get

[tex]4a+b=214[/tex]

[tex]4a=214-b[/tex]

[tex]a=\frac{214-b}{4}[/tex]

[tex]a=\frac{214-41}{4}[/tex]

[tex]a=\frac{173}{4}[/tex]

So, we have [tex]a=\frac{173}{4}[/tex], [tex]b=41[/tex] and [tex]c=1547[/tex]

b) Using the values calculated in part (a), the quadratic equation becomes

[tex]f(t)=\frac{173}{4}t^2+41t+1547[/tex]

Now, in 2016, the value of t=8

So, the number of craft beer breweries in 2016 will be calculated by [tex]f(8)[/tex]:

[tex]f(8)=\frac{173}{4}(8)^2+41(8)+1547[/tex]

[tex]= 2768+328+1547[/tex]

[tex]= 4643[/tex]

Therefore, the number of craft beer breweries in 2016 will be 4643.

Learn more about the quadratic function here:

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We first find the constants by substituting in the provided points into the quadratic equation to create a series of equations. Solving them gives us a = 61.5, b = 47, and c = 1547. To predict the number of breweries in 2016, we substitute t = 8 into the function to get approximately 3455 breweries.

To find the constants a, b, and c, we will substitute the given points (0, 1547), (2, 1802), and (4, 2403) into the equation f(t) = at² + bt + c. This will give us three equations:

Solving these gives us a = 61.5, b = 47, and c = 1547. Therefore the quadratic function is: f(t) = 61.5t² + 47t + 1547.

For (b), we want to find the number of breweries in 2016, which is 8 years from 2008. Substituting t = 8 into our quadratic function gives us f(8) = 61.5(8²) + 47(8) + 1547 which is approximately 3455. So, the model estimates there will be around 3455 craft-beer breweries in 2016.

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