A test has the capability to detect a defect with 98% accuracy. It also tends to indicate a defect when there is none about 1% of the time. About 1 in 2000 test subjects has a defect. What is the probably that a subject chosen at random and is indicated to have a defect, really does have a defect? (0.25 points)

a. PMF: [tex]\(P(X=k) = \frac{1}{2^k}\) for \(k=0,1,2,3\); \(P(X > 3) = \frac{1}{2^3}\)[/tex]

b. [tex]\(E(X) = \frac{11}{4}\)[/tex]

a. To calculate the probability mass function (PMF) of X, we need to consider the different scenarios leading to the player's success or removal from the game. Let P(X = k) be the probability that the player misses \(k\) shots before succeeding or getting removed.

[tex]\[P(X = 0) = \frac{1}{2}\][/tex]

The player succeeds on the first shot.

[tex]\[P(X = 1) = \frac{1}{2} \cdot \frac{1}{3}\][/tex]

(The player misses the first shot, then succeeds on the second.)

[tex]\[P(X = 2) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4}\][/tex]

(The player misses the first two shots, then succeeds on the third.)

[tex]\[P(X = 3) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{1}{5}\][/tex]

(The player misses the first three shots, then succeeds on the fourth.)

[tex]\[P(X > 3) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5}\][/tex]

(The player misses the first four shots and is removed.)

b. To compute the expected value of X, we use the formula:

[tex]\[E(X) = \sum_{k=0}^{\infty} k \cdot P(X = k)\][/tex]

[tex]\[E(X) = 0 \cdot P(X = 0) + 1 \cdot P(X = 1) + 2 \cdot P(X = 2) + 3 \cdot P(X = 3) + \sum_{k=4}^{\infty} k \cdot P(X > 3)\][/tex]

Now plug in the values from part (a) into this formula and compute the sum. The result will give you the expected value of X.

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.

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:

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: 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.

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:

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:

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:

Answers may vary but will most likely be close to 2.

Step-by-step explanation

                 first test:38%

                 second test:76%

    SIMULATION FIRST TEST

Randomly select a 2-digit number.

If the digit is between 00 and 35 then you passed the test,else you did not pass the test.

SIMULATION SECOND TEST

Randomly select a 2-digit number.

if the digit is between 00 and 75 then you passed the test,else you did not pass the test.

SIMULATION TRIAL

Perform the simulation of the first test.if you did not pass the first test then perform the simulation of the second test.

Record the number of trials needed to pass the first or second test.

Repeat 20 times and take the average of the 20 recorded number of trials

(what is the sum of recorded values divided by 20).

Note:you will most likely obtain a result of about two trials needed.

The estimated average number of tests drivers take to get a license is 1.9

To estimate the average number of tests drivers take to get a license, we can simulate the process using a simple random experiment. We will assume that each test attempt is independent of the others and that the probabilities of passing are 38% on the first attempt and 76% on subsequent attempts. Here's how we can perform the simulation:

1. For each driver, simulate the number of tests they take by generating a random number between 0 and 1. If the number is less than or equal to 0.38 (38%), the driver passes on the first attempt. Otherwise, they proceed to a second attempt.

2. For the second attempt and any subsequent attempts, generate another random number. If the number is less than or equal to 0.76 (76%), the driver passes.

3. Repeat this process until the driver passes the test. Record the number of attempts it took.

4. Perform this simulation for 20 drivers to get a sample size that will help estimate the average number of tests taken.

5. Calculate the average number of tests by summing the number of tests taken by all drivers and dividing by the number of drivers (20).

Let's perform the simulation:

1. For the first attempt, generate a random number for each of the 20 drivers and check if it's less than or equal to 0.38.

2. For any subsequent attempts, generate a random number for each driver who did not pass on the first attempt and check if it's less than or equal to 0.76.

3. Record the number of attempts for each driver.

4. Sum the total number of attempts and divide by 20 to find the average.

Now, let's calculate the average number of tests based on the simulation: Assuming we have performed the simulation and recorded the number of attempts for each driver, we would have a list of numbers. For example, it might look something like this:

 Driver 1: 1 attempt

Driver 2: 2 attempts

Driver 3: 1 attempt

Driver 20: 3 attempts

Let's say after performing the simulation, we have the following counts for each number of attempts:

1 attempt: 8 drivers

2 attempts: 7 drivers

3 attempts: 4 drivers

4 attempts: 1 driver

Now, we calculate the average number of tests taken by these 20 drivers:

Average = (8 * 1 + 7 * 2 + 4 * 3 + 1 * 4) / 20

Average = (8 + 14 + 12 + 4) / 20

Average = 38 / 20

Average = 1.9

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:

Step-by-step explanation:

$97.50=15hours

$x=1hour

$15x= 97.50

X= 97.50/15

X= 6.5$

Therefore

1hour=6.5$

Therefore

130/6.5

=20hours

Answer:

  ai) 5.5

  aii) 22

  aiii) 4

  bi) 0

  bii) 0

  biii) undefined

Step-by-step explanation:

ai) The change in x is x2 -x1 = 7.5 -2 = 5.5 . . . . change in x

__

aii) The change in y is (4(7.5) +8) -(4(2) +8) = 4(7.5-2) = 22.0 . . . . change in y

__

aiii) The ratio of changes is (change in y)/(change in x) = 22/5.5 = 4

The change in y is 4 times the change in x.

___

bi) The difference is -5.1 -(-5.1) = 0 . . . . change in x

__

bii) The difference is (4(-5.1)+8) -(4(-5.1)+8) = 0 . . . . change in y

__

biii) The ratio of changes is (change in y)/(change in x) = 0/0 = undefined.

The multiplier of the change in x to get the change in y is undefined.

_____

Comment on part B

We know that the relative rates of change for x and y in this linear function are 1 : 4. However, we cannot compute that ratio directly when the change in x is 0. (The ratio holds for vanishingly small values of change in x, so is 4 in the limit as Δx → 0. That isn't what the problem asks.)

Final answer:

Over the given interval, x changes by 5.5 and y changes by 38. The ratio of the change in y to the change in x is approximately 6.91. Over the given interval, x remains constant at -5.1 and y does not change. Therefore, the ratio of the change in y to the change in x is undefined.

Explanation:

i. To find the change in x, we subtract the initial value from the final value: 7.5 - 2 = 5.5.

ii. To find the change in y, we substitute the initial and final values of x into the equation y = 4x + 8 and subtract: y_final - y_initial = (4 * 7.5 + 8) - (4 * 2 + 8) = 38.

iii. To find the ratio of the change in y to the change in x, we divide the change in y by the change in x: 38 / 5.5 ≈ 6.91.

iv. To find the change in x, we subtract the initial value from the final value: -5.1 - -5.1 = 0.

v. To find the change in y, we substitute the initial and final values of x into the equation y = 4x + 8 and subtract: y_final - y_initial = (4 * -5.1 + 8) - (4 * -5.1 + 8) = 0.

vi. To find the ratio of the change in y to the change in x, we divide the change in y by the change in x: 0 / 0.

Answer:

a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

Step-by-step explanation:

To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem

Normal probability distribution

Problems of normally distributed samples can be 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]\frac{\sigma}{\sqrt{n}}[/tex].

In this problem, we have that:

[tex]\mu = 6.3, \sigma = 2.1, n = 49, s = \frac{2.1}{\sqrt{49}} = 0.3[/tex]

A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?

This is the pvalue of Z when X = 7.1 subtracted by the pvalue of Z when X = 5.5.

By the Central Limit Theorem, the formula for Z is:

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

X = 7.1

[tex]Z = \frac{7.1 - 6.3}{0.3}[/tex]

[tex]Z = 2.67[/tex]

[tex]Z = 2.67[/tex] has a pvalue of 0.9962

X = 5.5

[tex]Z = \frac{5.5 - 6.3}{0.3}[/tex]

[tex]Z = -2.67[/tex]

[tex]Z = -2.67[/tex] has a pvalue of 0.0038

So there is a 0.9962 - 0.0038 = 0.9924 = 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.

This is the pvalue of Z when X = 6.7 subtracted by the pvalue of Z when X = 5.9

X = 6.7

[tex]Z = \frac{6.7 - 6.3}{0.3}[/tex]

[tex]Z = 1.33[/tex]

[tex]Z = 1.33[/tex] has a pvalue of 0.9082

X = 5.9

[tex]Z = \frac{5.9 - 6.3}{0.3}[/tex]

[tex]Z = -1.33[/tex]

[tex]Z = -1.33[/tex] has a pvalue of 0.0918.

So there is a 0.9082 - 0.0918 = 0.8164 = 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

Answer: 4.85%

Step-by-step explanation: productivity = parcel delivered / number of drivers.

At when parcel delivered was 103, 000 the average number of drivers was 84 thus making the productivity to be

103,000/ 84 = 1226.19

At when the parcel delivered was 112, 800 the average number of drivers was 96 thus making the productivity to be

112, 000/ 96 = 1166.67

Thus the change in productivity = 1226.19 - 1166.67 = 59.52.

Initial productivity = 1226.19

final productivity = 1166.67

% change in productivity = change in productivity / initial productivity * 100

% change in productivity = 59.52 /1226 * 100

% change in productivity = 4.85%

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:

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:

a. 74.63%

b. 33.63%

c. 87.67%

Step-by-step explanation:

If 59% (0.59) of the workers are married, then It means (100-59 = 41%) of the workers are not married.

If 43% (0.43) of the workers are College graduates, then it means (100-43= 57%) of the workers are not college graduates.

If 1/3 of college graduates are married, it means portion of graduate that are married = 1/3 * 43% = 1/3 * 0.43 = 0.1433.

For question a, Probability that the worker is neither married nor a college graduate becomes:

= (probability of not married) + (probability of not a graduate) - (probability of not married * not a graduate)

= 0.41 + 0.57 - (0.41*0.57) = 0.98 - 0.2337

= 0.7463 = 74.63%

For question b, probability that the worker is married but not a college graduate becomes:

=(probability of married) * (probability of not a graduate.)

= 0.59 * 0.57

= 0.3363 = 33.63%

For question c, probability that the worker is either married or a college graduate becomes:

=probability of marriage + probability of graduate - (probability of married and graduate)

= 0.59 + 0.43 - (0.1433)

= 0.8767. = 87.67%

The probability that a randomly chosen worker is neither married nor a college graduate is 0%. The probability that a worker is married but not a college graduate is approximately 34%. The probability that a worker is married or a college graduate is approximately 75%.

To calculate the probabilities, we need to use the formulas for conditional probability. Let's solve each part:

a) To find the probability that a randomly chosen worker is neither married nor a college graduate, we can subtract the probability that the worker is married and the probability that the worker is a college graduate from 1. The probability of being married is 59%, and the probability of being a college graduate is 43%. Using the formula, 1 - 0.59 - 0.43 = 0.

b) To find the probability that a worker is married but not a college graduate, we can multiply the probability of being married and the probability of not being a college graduate. The probability of being married is 59%, and the probability of not being a college graduate is 57% (100% - 43%). Using the formula, 0.59 * 0.57 = 0.3363 (approximately 34%).

c) To find the probability that a worker is married or a college graduate, we can add the probabilities of being married and being a college graduate, and then subtract the probability of being both married and a college graduate to avoid double counting. The probability of being married is 59%, the probability of being a college graduate is 43%, and the probability of being both married and a college graduate is 1/3 of the college graduates (1/3 * 43%). Using the formula, 0.59 + 0.43 - (1/3 * 0.43) = 0.745 (approximately 75%).

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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: John must make at least 44 tips to reach his goal.

Step-by-step explanation:

Let x represent the number of tips that John must make this week to reach his goal.

John works at a restaurant and gets paid $80 per week plus $5 per tip. This means that the total amount that John would earn in a week if he makes x tips would be

80 + 5x

This week, John wants to earn at least $300. This is expressed as

80 + 5x ≥ 300

5x ≥ 300 - 80

5x ≥ 220

x ≥ 220/5

x ≥ 44

Answer:

∑E(x[tex]_{i}[/tex]) = 13.49167 floors

Step-by-step explanation:

The expected number of floors no one get off = ∑E(x[tex]_{i}[/tex]) where i is from 0 to 23

and E(x[tex]_{i}[/tex]) = ∑x[tex]_{i}[/tex]P(x[tex]_{i}[/tex])

here x[tex]_{i}[/tex] is the indicator of floor where no one gets off, its value is 0 when  atleast one person get off on its floor and 1 when when no one gets off.

Now,

P(x[tex]_{i}[/tex]=1) = (22/23)¹²

P(x[tex]_{i}[/tex]=0) = [1-(22/23)¹²]

Now,

E(x[tex]_{i}[/tex]) = ∑x[tex]_{i}[/tex]P(x[tex]_{i}[/tex]) = 0* [1-(22/23)¹²] + 1*(22/23)¹² =0.586594704

For total number of floors where no one gets off

∑E(x[tex]_{i}[/tex]) = E(x₁)+E(x₂)+E(x₃)........................+E(x₂₃)

∑E(x[tex]_{i}[/tex]) = 23*0.586594704

∑E(x[tex]_{i}[/tex]) = 13.49167 floors

Final answer:

To answer, we calculate the expectation by considering the probability of a floor being skipped by all 12 people in a 24-floor building. The expected number of floors not chosen by anyone is around 8.

Explanation:

To find this, we can use the concept of probability. For any given floor (except the first), the probability that a single person does not choose it is (23/23) for the first choice, and the probability that they do choose another floor is (22/23), considering there are 23 floors above the first. Since each of the 12 people chooses independently, the probability that a given floor is not chosen by any of the 12 people is ((22/23)¹²).

With 23 possible floors (excluding the first) for people to exit on, the expected number of floors not chosen by anyone is 23 * ((22/23)¹²). A calculation reveals this to be approximately 7.9. Therefore, there's an expectation that around 8 floors will have no one exiting on them, under the given conditions.

Answer:

m arc DC=48°

Step-by-step explanation:

Angles in a Circle

We know [tex]\overline{BE}[/tex] and [tex]\overline{AC}[/tex] are diameters, so

[tex]m\angle BFA+m\angle AFE=180^o[/tex].

Since [tex]m\angle AFE=3x+12[/tex] and [tex]m\angle BFA=4x[/tex]

We set the equation

[tex]3x+12+4x=180^o[/tex]

Solving

[tex]7x=180-12=168[/tex]

[tex]x=24^o[/tex]

Thus

[tex]m\angle BFA=4(24)=96^o[/tex]

We also know

[tex]m\angle CFD\cong m\angle DFE[/tex]

Being [tex]\angle CFE[/tex] opposite to [tex]\angle BFA[/tex]

Then [tex]m\angle CFE=96^o[/tex]

It's divided in two equal angles [tex]\angle CFD\ and\ \angle DFE[/tex], thus m arc DC is half of 96°:

m arc DC=48°

First option

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:

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) 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:

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).

Answer:

a) 23%

Step-by-step explanation:

To find the price change as percentage :

Multiply 100 by 160 then divide it by 130

100 × 160 ÷ 130 = 123 approximately which means there's an additional 23% to the 100% price

Answer:

92.10% probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long.

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

[tex]\mu = 30.05, \sigma = 0.2[/tex]

What is the probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long?

This is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 29.75

X = 30.5

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

[tex]Z = \frac{30.5 - 30.05}{0.2}[/tex]

[tex]Z = 2.25[/tex]

[tex]Z = 2.25[/tex] has a pvalue of 0.9878

X = 29.75

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

[tex]Z = \frac{29.75 - 30.05}{0.2}[/tex]

[tex]Z = -1.5[/tex]

[tex]Z = -1.5[/tex] has a pvalue of 0.0668

So there is a 0.9878 - 0.0668 = 0.9210 = 92.10% probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long.

Answer:

49.34% probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers

Step-by-step explanation:

For each television, there are only two possible outcomes. Either they have defective speakers, or they do not. The probabilities of each television having defective speakers are independent from each other. So we use the binomial probability distribution to solve this problem.

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]p = 0.01, n = 500[/tex]

Find an approximate probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers

This is

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

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 5) = C_{500,5}.(0.01)^{5}.(0.99)^{495} = 0.1764[/tex]

[tex]P(X = 6) = C_{500,6}.(0.01)^{6}.(0.99)^{494} = 0.1470[/tex]

[tex]P(X = 7) = C_{500,7}.(0.01)^{7}.(0.99)^{493} = 0.1048[/tex]

[tex]P(X = 8) = C_{500,8}.(0.01)^{8}.(0.99)^{492} = 0.0652[/tex]

[tex]P(5 \leq X \leq 8) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.1764 + 0.1470 + 0.1048 + 0.0652 = 0.4934[/tex]

49.34% probability that 5 to 8 televisions (inclusive) in the shipment have defective speakers

The approximate probability that 5 to 8 out of 500 televisions have defective speakers is 0.49. This was calculated using the binomial distribution with n = 500 and p = 0.01. We summed P(X = 5), P(X = 6), P(X = 7), and P(X = 8) to find the result.

To find the approximate probability that 5 to 8 televisions out of 500 have defective speakers, we can use the binomial distribution.

The probability mass function for a binomial distribution is given by:

[tex]P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)[/tex]

where C(n, k) is the binomial coefficient.

We need to calculate P(5 ≤ X ≤ 8), which is P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8).

Using a binomial calculator or statistical software, we get:

Summing these probabilities gives:

P(5 ≤ X ≤ 8) ≈ 0.1755 + 0.1447 + 0.1040 + 0.0681

≈ 0.4923

Therefore, the approximate probability that 5 to 8 televisions have defective speakers is 0.49 (rounded to two decimal places)

Answer:

1. ΔXYZ is a right Δ with altitude YU.

Given

2. ΔXYZ ~ ΔYUZ

Right Triangle Altitude Similarity Theorem

3. VW || XY

Given

4. ∠VWZ ≅ ∠XYZ

Corresponding angles

5. ∠Z ≅ ∠Z

Reflexive property of congruence

6. ΔXYZ ~ ΔVWZ

AA Similarity postulate

7. ΔYUZ ~ ΔVWZ

Transitive property of similar triangles

Step-by-step explanation:

The first statement is given in the problem.  Since we know the altitude of a right triangle, we can use the Right Triangle Altitude Similarity Theorem to say that the triangles formed by the altitude are similar to each other and the original triangle.

Next, we are given in the problem statement that the lines VW and XY are parallel.  Therefore, ∠VWZ and ∠XYZ are corresponding angles, which makes them congruent.  And since ∠Z is equal to itself (by reflexive property), we can use AA similarity to say ΔXYZ and ΔVWZ are similar.

Finally, combining statements 2 and 6, we can use transitive property to say that ΔYUZ and ΔVWZ are similar.