Euclidean distance can be used to calculate the dissimilarity between two observations. Let u = (25, $350) correspond to a 25-year-old customer that spent $350 at Store A in the previous fiscal year. Let v = (53, $420) correspond to a 53-year-old customer that spent $4,100 at Store A in the previous fiscal year. Calculate the dissimilarity between these two observations using Euclidean distance.

a. 66.21
b. 88.57
c. 72.28
d. 75.39

Answer:

70 combinations

Step-by-step explanation:

Given:

- The total number of designated location n = 8

- The number of location the ointment is to be applied r = 4

Find:

How many different choices were there for the four locations to apply the new ointment?

Solution:

- For this question we will apply combinations to the given problem. We have total of 8 available places out of which we have to "choose" 4 locations to apply the new ointment. We will use combinations " C " operator.

- Out of 8 "choose" 4 = 8_C _4 = 70 possible combinations.

- Since there is requirement of "order" so its only limited to the selection process and not an arrangement.

Answer:

1) 0.375

2) 0.5

3) 0.875

4) 0.5                  

Step-by-step explanation:

We are given the following in the question:

Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

[tex]\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}[/tex]

1. The probability of getting exactly one tail

P(Exactly one tail)

Favorable outcomes ={HHT, HTH, THH}

[tex]\text{P(Exactly one tail)} = \dfrac{3}{8} = 0.375[/tex]

2. The probability of getting a head on the first toss

P(head on the first toss)

Favorable outcomes ={HHH, HHT, HTH, HTT}

[tex]\text{P(head on the first toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]

3. The probability of getting at least one head

P(at least one head)

Favorable outcomes ={HHH, HHT, HTH, HTT, THH, THT, TTH}

[tex]\text{P(at least one head)} = \dfrac{7}{8} = 0.875[/tex]

4. The probability of getting at least two heads

P(Exactly one tail)

Favorable outcomes ={HHH, HHT, HTH,THH}

[tex]\text{P(Exactly one tail)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]

Final answer:

The question involves calculating probabilities based on the outcomes of tossing a coin three times. The probabilities for getting exactly one tail, a head on the first toss, at least one head, and at least two heads are found to be 3/8, 1/2, 7/8, and 1/2 respectively, by counting the favorable outcomes within the complete sample space.

Explanation:

The question asks for the probability of various outcomes when tossing a coin three times, given the sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Let's solve each part step-by-step.

Probability of getting exactly one tail: There are three outcomes (HTT, THT, TTH) out of eight that satisfy this condition, so the probability is 3/8.

Probability of getting a head on the first toss: Four outcomes (HHH, HHT, HTH, HTT) satisfy this, so the probability is 4/8 or 1/2.

Probability of getting at least one head: Every outcome except TTT includes at least one head, so the probability is 7/8.

Probability of getting at least two heads: There are four outcomes (HHH, HHT, HTH, THH) that satisfy this condition, leading to a probability of 4/8 or 1/2.

Each probability is based on the fundamental principle of counting favorable outcomes over the total number of outcomes.

Answer:

87.26 mm Hg is the reading that represents the 80th percentile of the distribution.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 78 mm Hg

Standard Deviation, σ = 11 mm Hg

We are given that the distribution of diastolic blood pressure is a bell shaped distribution that is a normal distribution.

Formula:

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

We have to find the value of x such that the probability is 0.80

P(X < x)  

[tex]P( X < x) = P( z < \displaystyle\frac{x - 78}{11})=0.8[/tex]  

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

[tex]P(z < 0.842) = 0.8[/tex]

[tex]\displaystyle\dfrac{x - 78}{11} = 0.842\\\\x = 87.262 \approx 87.26[/tex]

87.26 mm Hg is the reading that represents the 80th percentile of the distribution.

To find the DBP reading representing the 80th percentile, we can use the z-score formula. The DBP reading is 86.8 mmHg.

To find the diastolic blood pressure (DBP) reading that represents the 80th percentile of the distribution, we can use the z-score formula. The z-score is calculated by subtracting the mean from the desired value and dividing it by the standard deviation. Then, we can use the z-score table or calculator to find the corresponding percentile. In this case, since we want the 80th percentile, we need to find the z-score that corresponds to an area of 0.8 to the left of it.

The formula for the z-score is: z = (x - mean) / standard deviation

Using the given values for Canada (mean = 78 mmHg, standard deviation = 11 mmHg), we can substitute them into the formula and solve for x:

0.8 = (x - 78) / 11

Multiplying both sides by 11, we get:

8.8 = x - 78

Adding 78 to both sides, we get:

x = 86.8 mmHg

Therefore, the DBP reading that represents the 80th percentile of the distribution is 86.8 mmHg.

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Here is the complete question.

1). What are the odds for rolling a sum of 5 in a single roll of two fair dice?

2). If you bet $1 that a sum of 5 will turn up, what should the house pay (plus returning your $1 bet) if a sum of 5 turns up in order for the game to be fair?

Answer:

1).  [tex]\frac{1}{8}[/tex]

2).  $8.

Step-by-step explanation:

1).

The sample space (S) for rolling two fair dice is given as the following parameters illustrated below:

[tex]\left[\begin{array}{cccccc}(1,2)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\end{array}\right] \left[\begin{array}{cccccc}(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\end{array}\right][/tex]

Let F represent the event that the sum of both dice turns up is 5.

F = [(1,4),(2,3),(3,2),(4,1)}

∴ The probability for an event F  is illustrated below as:

[tex]\frac{P(F)}{P(F')}[/tex]  [tex]=\frac{\frac{4}{36} }{1-\frac{4}{36} }[/tex]

[tex]=\frac{\frac{4}{36} }{\frac{32}{36} }[/tex]

[tex]={\frac{4}{36}}*\frac{36}{32}[/tex]

= [tex]\frac{1}{8}[/tex]

2). From above, we can see that the probability for rolling a sum of 5 are 1 to 8. Therefore, if you roll a sum of 5, in order for the game to be fair, the house is required to pay $8.

Answer:

0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5)y' + (ln t)y = 6t

and the condition y(1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [(ln t)/(t - 5)]y = 6t/(t - 5)

(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).

6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.

Answer:

The probability that a person has HIV given that the test negative is 0.0033.

Step-by-step explanation:

Denote the events as follows:

X = a person in Uganda had HIV

Y = a person in Uganda was tested positive for HIV.

The information provided is:

[tex]P(Y^{c}\cap X)=\frac{4}{1517}\\[/tex]

[tex]P(Y\cap X)=\frac{166}{1517}\\[/tex]

[tex]P(Y\cap X^{c})=\frac{129}{1517}\\[/tex]

[tex]P(Y^{c}\cap X^{c})=\frac{1218}{1517}\\[/tex]

The probability that a person has HIV given that he/she was tested negative is:

[tex]P(X|Y^{c})=\frac{P(Y^{c}\cap X)}{P(Y^{c})}[/tex]

Compute the probability of a person not having HIV as follows:

[tex]P(Y^{c})=P(Y^{c}\cap X)+P(Y^{c}\cap X^{c})=\frac{4}{1517}+\frac{1218}{1517} =\frac{4+1218}{1517} =\frac{1222}{1517}[/tex]

Compute the value of [tex]P(X|Y^{c})[/tex] as follows:

[tex]P(X|Y^{c})=\frac{P(Y^{c}\cap X)}{P(Y^{c})}=\frac{4}{1517}\times\frac{1517}{1222}=0.0033[/tex]

Thus, the probability that a person has HIV given that he/she was tested negative is 0.0033.

Final answer:

The probability that a randomly selected person from this population who tests negative actually has HIV is approximately 0.327%. This accounts for the possibility of false negatives. However, confirmatory testing is necessary to accurately diagnose HIV.

Explanation:

To calculate the probability that a randomly selected person from the population who tested negative for HIV actually has the virus, we need to focus on the false negatives and true negatives provided in the figures from Uganda.

Out of 1517 tested individuals, there were 4 false negatives and 1218 true negatives.

Thus, for someone who tests negative, to find out the probability that they actually have HIV (false negative), we use the formula: Probability = Number of false negatives / (Number of false negatives + Number of true negatives).

In this case, we have Probability = 4 / (4 + 1218) = 4 / 1222. When we calculate this probability, we get approximately 0.00327, or 0.327%.

Answer:

546.94 meters is the height of the mountain.

Step-by-step explanation:

Pressure at the top of mountain = [tex]P_1=93.8 kPa[/tex]

Pressure at the bottom of the mountain = [tex]P_2=100.5 kPa[/tex]

Pressure difference  =[tex]P_2-P_1=100.5kPa-93.8kPa=6.7 kPa[/tex]

6.7 kPa = 6.7 × 1000 Pa (1 kPa= 1000 pa)

Density of the air = d = [tex]1.25 kg/m^3[/tex]

Acceleration due to gravity = g = [tex]9.8 m/s^2[/tex]

Height of the mountain = h

[tex]P=h\times d\times g[/tex]

[tex]h=\frac{P}{d\times g}=\frac{6.7\times 1000 Pa}{1.25 kg/m^3\times 9.8 m/s^2}[/tex]

[tex]h=546.94 m[/tex]

546.94 meters is the height of the mountain.

$ 320

41 miles ............ 1 gallon

4700 miles .......x gallon

x = 4700×1/41 = 114.63 gallons

114.63 gallons×$2.79 ≈  $ 320

To find the cost of gasoline for a 4,700-mile automobile trip, divide the total distance by the car's miles per gallon, then multiply by the price of gas per gallon, finally we get a total cost of $318.65.

To find the cost of gasoline for a 4,700-mile automobile trip, we need to divide the total distance by the car's miles per gallon to get the number of gallons of gas needed.

Then, we multiply the number of gallons by the price of gas per gallon to get the total cost.

In this case, the car gets 41 miles per gallon, so the number of gallons needed is 4700 miles / 41 miles per gallon = 114.63 gallons.

Multiplying this by the average price of gas, $2.79 per gallon, we get a total cost of $318.65.

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Answer:

20

Step-by-step explanation:

The variance of the number of children in a randomly chosen family is approximately 9.4148.

To determine the variance of the number of children in a randomly chosen family, we can use the properties of the Poisson distribution, assuming that the number of children in a family follows a Poisson distribution with a mean of 2.3.

The Poisson distribution has a variance equal to its mean, so if the mean number of children per family is 2.3, then the variance is also 2.3.

However, we can also calculate the variance directly from the given information.

Let ( X ) be the random variable representing the number of children in a family.

Given:

- The mean number of children in a family, [tex]\( \mu = 2.3 \)[/tex]

- Each child has, on average, 1.6 siblings.

To find the variance, we use the formula:

\[ \text{Variance} = \text{E}(X^2) - [\text{E}(X)]^2 \]

[tex]First, we find \( \text{E}(X^2) \), the expected value of \( X^2 \):\[ \text{E}(X^2) = \text{Variance} + [\text{E}(X)]^2 \]\[ \text{E}(X^2) = 1.6^2 \times 2.3 + (2.3)^2 \]\[ \text{E}(X^2) = 4.096 \times 2.3 + 5.29 \]\[ \text{E}(X^2) = 9.4148 + 5.29 \]\[ \text{E}(X^2) = 14.7048 \][/tex]

Then, using the formula for variance:

[tex]\[ \text{Variance} = \text{E}(X^2) - [\text{E}(X)]^2 \]\[ \text{Variance} = 14.7048 - (2.3)^2 \]\[ \text{Variance} = 14.7048 - 5.29 \]\[ \text{Variance} = 9.4148 \][/tex]

So, the variance of the number of children in a randomly chosen family is approximately 9.4148.

Question:

In a town there are on average 2.3 children in a family and a randomly chosen child has on average 1.6 siblings. Determine the variance of the number of children in a randomly chosen family.

Answer:

a) 8 units

b) 3 units

c) 4 units

d) [tex]4\sqrt{5}\text{ units}[/tex]

e) [tex]\sqrt{73}\text{ units}[/tex]

f) 5 units

Step-by-step explanation:

We are given the following:

Point (3, −4, 8)

We have to find the distance of the point from the following:

Distance formula:

[tex](x_1,y_1,z_1),(x_2,y_2,z_2)\\\\d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}[/tex]

(a) the xy-plane

We have to find the distance from (3, −4, 8) to (3, −4, 0)

[tex]d = \sqrt{(3-3)^2 + (-4+4)^2 + (0-8)^2} = 8\text{ units}[/tex]

(b) the yz-plane

We have to find the distance from (3, −4, 8) to (0, −4, 8)

[tex]d = \sqrt{(0-3)^2 + (-4+4)^2 + (8-8)^2} = 3\text{ units}[/tex]

(c) the xz-plane

We have to find the distance from (3, −4, 8) to (3, 0, 8)

[tex]d = \sqrt{(3-3)^2 + (0+4)^2 + (8-8)^2} = 4\text{ units}[/tex]

(d) the x-axis

We have to find the distance from (3, −4, 8) to (3, 0, 0)

[tex]d = \sqrt{(3-3)^2 + (0+4)^2 + (0-8)^2} = \sqrt{80} = 4\sqrt{5}\text{ units}[/tex]

(e) the y-axis (0,−4,0)

We have to find the distance from (3, −4, 8) to (0, -4, 0)

[tex]d = \sqrt{(0-3)^2 + (-4+4)^2 + (0-8)^2} = \sqrt{73}\text{ units}[/tex]

(f) the z-axis (0,0,8)

We have to find the distance from (3, −4, 8) to (0, 0, 8)

[tex]d = \sqrt{(0-3)^2 + (0+4)^2 + (8-8)^2} = \sqrt{25} = 5\text{ units}[/tex]

Answer:

a) The the approximate number of babies between 2363 and 3234 are:

n = 0.683*621= 423.95 and that's approximately 424 babies

b) The approximate number of babies between 1492 and 4976 are:

n = 0.955*621= 592.74 and that's approximately 593 babies

Step-by-step explanation:

Assuming this problem:"A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 621 babies born in New York. The mean weight was 3234 grams with a standard deviation of 871 grams. Assume that birth weight data are approximately bell-shaped.

Estimate the number of newborns who weighed between 2363 grams and 4105 grams. Round to the nearest whole number.

The number of newborns who weighed between 1492 grams and 4976 grams is . "

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(3234,871)[/tex]  

Where [tex]\mu=3234[/tex] and [tex]\sigma=871[/tex]

We select a sample size of n = 621. And we want to find this probability:

[tex] P(2363 < X < 4105) [/tex]

[tex]P(2363<X<4105)=P(\frac{2363-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4105-\mu}{\sigma})=P(\frac{2363-3234}{871}<Z<\frac{4105-3234}{871})=P(-1<z<1)[/tex]

And we can find this probability taking this difference:

[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.841-0.159=0.683 [/tex]

And the the approximate number of babies between 2363 and 3234 are:

n = 0.683*621= 423.95 and that's approximately 424 babies

Part b

[tex] P(1492 < X < 4976) [/tex]

[tex]P(1492<X<4976)=P(\frac{1492-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4976-\mu}{\sigma})=P(\frac{1492-3234}{871}<Z<\frac{4976-3234}{871})=P(-1<z<1)[/tex]

And we can find this probability taking this difference:

[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.977-0.0228=0.955 [/tex]

And the the approximate number of babies between 1492 and 4976 are:

n = 0.955*621= 592.74 and that's approximately 593 babies

Answer:

-2

1

3125

Step-by-step explanation:

Answer:

The complete question is stated below:

An obstetrician knew that there were more live births during the week than on weekends. She wanted to determine whether the mean number of births was the same for each of the five days of the week. She randomly selected eight dates for each of the five days of the week and obtained the following data:

Monday:      Tuesday:    Wednesday:    Thursday:        Friday:

10,456            11,621         11,084             11,171                 11,545

10,023            11,944         11,570            11,745                12,321

10,691             11,045         11,346            12,023               11,749

10,283             12,927        11,875            12,433               12,192

10,265             12,577         12,193           12,132                12,422

11,189                11,753         11,593           11,903                11,627

11,198                12,509        11,216            11,233                11,624

11,465                13,521         11,818            12,543              12,543

a. Write the null and alternative hypotheses.

b. State the requirements that must be satisfied to use the one-way ANOVA procedure.

c On which day or dates are there more births?

Answer:

a. Null Hypothesis: There is no statistically significant difference between the mean number of babies born alive on Mondays, Tuesdays, Wednesdays, Thursdays and Fridays.

Alternative Hypothesis: The mean number of babies born alive on each weekday from Monday to Friday are not the same.

b) 1. The dependent variable should be measured at ratio or interval level

2. The independent variable should contain at least two groups that are categorical and independent

3. There must independence of observations between and within the various groups

4. No significant outliers should exist

5. for each group of the independent variable, the dependent variable should be approximately normally distributed.

6. the variances should be homogeneous

c. There are more births on Tuesdays and Fridays

Step-by-step explanation:

a. A Null Hypothesis is one that states that no statistical significance exist between the two variables in the hypothesis. It is the null hypothesis that the researcher tries to disprove. While the alternative hypothesis is simply the opposite of the null hypothesis, it hypothesizes that there is indeed a statistically significant relationship between the variables in question.

b.  1. The dependent variable should be measured at ratio or interval level; among the various levels of measurements such as ordinal, nominal, ratio or interval, the dependent variable must be either on the interval or ratio level.

2. The independent variable should contain at least two groups that are categorical and independent; the independent variable, in this case, the number of live births, should be two groups or more, and they should be categorical.

3. There must independence of observations between and within the various groups; the values observed within each group should be independent of the other. For example, no one pregnant woman should be involved in more than one group.

4. No significant outliers should exist; outliers are single data points within the measured variable that do not follow the usual point pattern of the other values, either values that are too high or too low. they reduce the accuracy of the one-way ANOVA.

5. for each group of the independent variable, the dependent variable should be approximately normally distributed; violations of normality causes the result to be invalid. It can only hold little variations to produce valid results.

6. the variances should be homogeneous; the variances (distribution or spread around the mean) of the two or more test groups must be considered equal.

c. Tuesday with a mean livebirth of 12,237 births and Friday with an average livebirth of 12,002 births have the most number of births.

Answer:

242,784,360 hands.

Step-by-step explanation:

The number of hands that contain at least one hearts card is given by the total possible number of hands subtracted by the number of hands with no hearts card.

The total possible number of hands is:

[tex]N_T=\frac{52!}{(52-5)!}=\frac{52!}{(47)!} =52*51*50*49*48\\N_T=311,875,200[/tex]

Since there are 13 hearts cards in a deck, the number of possible hands with no hearts card is:

[tex]N_{H=0} =\frac{52-13!}{(52-13-5)!}=\frac{39!}{(34)!} =39*38*37*36*35\\N_{H=0}=69,090,840[/tex]

The number of hands with at least one hearts card is:

[tex]N_{H>0} = N_T-N{H=0}\\N_{H>0}=311,875,200-69,090,840\\N_{H>0}=242,784,360[/tex]

242,784,360 hands contain at least one hearts card.

Answer:

a. Mean = 4.5, Standard Deviation = 1.775

b. 0.0152

Step-by-step explanation:

Given

n = 15 purchasers

p = Success = 30%

p = 0.3

q =Failure = 70%

q = 0.7

a.

Mean = np

Mean = 15 * 0.3

Mean = 4.5

Standard Deviation = √Variance

Variance = npq

Variance = 15 * 0.3 * 0.7

Variance = 3.15

Standard Deviation = √3.15

Standard Deviation = 1.774823934929884

Standard Deviation = 1.775 ---------- Approximated

b.

The probability that the number who want new copies is more than two standard deviations away from the mean value

Standard Deviation = 1.775

Mean = 4.5

2 Standard Deviation and Mean = 2 * 1.775 + 4.5

= 3.55 + 4.5

= 8.05

P(X>8.05) = P(9) + P(10) +........+ P(15)

Using the binomial distribution

(p + q) ^ n where p = 0.3, q = 0.7 , n = 15

Expanding (p+q)^n where n = 15 and r > 8

We have

15C9 p^9 q^6 + 15C10 p^10 q^5 + 15C11 p^11 q⁴ + 15C12 p^12 q³ + 15C13 p^13 q² + 15C14 p^14 q + p^15

= 5005 (0.3)^9 (0.7)^6 + 3003 (0.3)^10 (0.7)^5 + 1365 (0.3)^11 (0.7)⁴ + 455 (0.3)^12 (0.7)³ + 105 (0.3)^14 (0.7)² + 15 (0.3)^14 (0.7) + 0.3^15

=0.015234

Answer:

The probability that the coins are thrown more than three times to show the same face is 0.3164.

Step-by-step explanation:

The problem is related to Geometric distribution.

The Geometric distribution defines the probability distribution of X failures before the first success.

The probability distribution function is:

[tex]P(X=k)=(1-p)^{k}p;\ k = 0, 1, 2, ...[/tex]

First compute the probability that in the [tex]i^{th}[/tex] throw all the three coins will show the same face.

P (All the 3 coins shows the same face) = P (All the three coins shows Heads) + P (All the three coins shows Tails)

                      [tex]=(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} )+(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} )\\=\frac{1}{8}+\frac{1}{8}\\ =\frac{2}{8} \\=\frac{1}{4}[/tex]

Now compute the probability that it takes more than 3 throws for the coins to show the same face.

P (X > 3) = 1 - P (X ≤ 3)

[tex]=1-[P(X=1)+P(X=2)+P(X=3)]\\=1-[[(1-\frac{1}{4} )^{0}\times\frac{1}{4}]+[(1-\frac{1}{4} )^{1}\times\frac{1}{4}]+ [(1-\frac{1}{4} )^{2}\times\frac{1}{4}]+[(1-\frac{1}{4} )^{3}\times\frac{1}{4}]]\\=1-[0.2500+0.1875+0.1406+0.1055]\\=1-0.6836\\=0.3164[/tex]

Thus, the probability that it takes more than 3 throws for the coins to show the same face is 0.3164.

The probability that it takes more than 3 throws for the coins to show the same face is 0.3164.

Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen. It is the ratio of the favorable event to the total number of events.

Given

Three identical fair coins are thrown simultaneously until all three show the same face.

Then the total event will be

[tex]\rm Total \ event = 2^3 = 8[/tex]

All the possibilities.

[tex]\rm (HHH),( HHT ),(HTH),( THH),(HTT),(THT),(TTH),(TTT)[/tex]

For getting the same face, then the probability will be

[tex]\rm P(same\ face) = \dfrac{2}{8} = \dfrac{1}{4}[/tex]

Now compute the probability that it takes more than 3 throws for the coins to show the same face.

[tex]\rm P (X > 3) = 1 - P (X ≤ 3)\\\\P (X > 3) = 1 -[P(x=1) +P(x=2) + P(x=3)]\\\\P (X > 3) =1 -\{[(1-\dfrac{1}{4})^0 * \dfrac{1}{4}] +[(1-\dfrac{1}{4})^1 * \dfrac{1}{4}] +[(1-\dfrac{1}{4})^2 * \dfrac{1}{4}] +[(1-\dfrac{1}{4})^3 * \dfrac{1}{4}] \}\\\\P (X > 3) =1 - 0.6836\\\\P (X > 3) =0.3164[/tex]

Thus, the probability that it takes more than 3 throws for the coins to show the same face is 0.3164.

More about the probability link is given below.

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Answer:

Total distance is 8 meters.

Step-by-step explanation:

First, let's take the equation and analyse it:

v(t) = 6 + t²

The delta t = 2

To find the acceleration, we need to differentiate v with respect to t like this:

dv/dt = d/dt (6 + t²)

         = 2t

At t = 2, the change in velocity (dv/dt) = 2×2 = 4 m/s². This is the acceleration,a

The final velocity is given by v = 6 + (2)²

                                                   = 10 m/s

The distance is given by the formula:

s = ut + 1/2at²

initial velocity, u = 0

Therefore, the equation becomes s = 1/2at²

                                                            = 1/2× 4×4

                                                             = 8 m Ans

The upper estimate of the distance traveled is 150 meters, the lower estimate is 12 meters, and the average of these estimates is 81 meters.

This math problem involves the concept of approximating the distance traveled using velocity and small intervals of time. The distance traveled, in approximation, can be calculated by taking the velocity at a certain time, and multiplying it by the change in time, which is given as 2 seconds in this case.

To find the upper estimate, we calculate using the endpoint of the interval, t = 6. To find the lower estimate, we calculate using the beginning of the interval, t = 0.

For upper estimate: v(6) x 2 = [6 +  (6^2)] x 2 = 150 meters

For lower estimate: v(0) x 2 = [6 + (0^2)] x 2 = 12 meters

The average of the two estimates is then found by adding them together and dividing by 2, which results in 81 meters.

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To calculate the probabilities, we can use the binomial distribution formula. For each event, substitute the appropriate values into the formula and calculate the probabilities using the combination symbol and the probabilities of winning and losing a game. For part c, sum the probabilities of winning at least 7 games.

To find the probabilities for the given events, we can use the binomial distribution formula. Let p be the probability of winning a game for both the computer and the human, and q be the probability of losing a game. Since they are evenly matched, p = q = 0.5.

a. The probability that they each win five games is P(X = 5), where X follows a binomial distribution with n = 10 (number of games) and p = 0.5. We can calculate this probability using the formula P(X = 5) = C(10, 5) * (0.5)^5 * (0.5)^5.

b. The probability that the computer wins seven games is P(X = 7), where X follows a binomial distribution with n = 10 and p = 0.5. We can calculate this probability using the formula P(X = 7) = C(10, 7) * (0.5)^7 * (0.5)^3.

c. The probability that the human chess champion wins at least seven games can be calculated by summing the probabilities of winning exactly 7, 8, 9, and 10 games: P(X >= 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10).

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Answer:

1. 12/52 or 0.2308

2. 10/13 or 0.7692

3.  3/26 or 0.1154

4. 3/26 or 0.1154

5. 9/52 or 0.1731

Step-by-step explanation:

1.

The face card are king,queen and jack.

There are 12 face card in a deck of 52 cards i.e. 4 king, 4 queen and 4 jack.

P(FC)=12/52 or 0.2308

So, the probability that the card is a face card is 0.2308.

2.

The card is not a face card.

P(FC')=1-P(FC)=1-(12/52)=(52-12)/52=40/52=10/13

P(FC')=10/13 or 0.7692

So, the probability that the card is not a face card is 0.7692.

3.

The card is a red face card

There are 6 red face card in a deck of 52 cards i.e. 2 king, 2 queen and 2 jack.

P(RFC)=6/52

P(RFC)=3/26 or 0.1154

So, the probability that the card is a red face card is 0.1154.

4.

The card is a 6 or lower.

There are 6 cards in a deck of 52 cards that are 6 or lower i.e. 6,5,4,3,2 and ace.

P(6 or lower)=6/52

P(6 or lower)=3/26 or 0.1154

So, the probability that the card is a 6 or lower is 0.1154.

5.

The card is a 9 or lower.

There are 9 cards in a deck of 52 cards that are 9 or lower i.e. 9, 8, 7, 6, 5, 4, 3, 2 and ace.

P(9 or lower)=9/52

P(9 or lower)=9/52 or 0.1731

So, the probability that the card is a 9 or lower is 0.1731.

In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e. spades , hearts , diamonds  and clubs.  King, Queen and Jack  are face cards. So, there are 12 face cards in the deck of 52 playing cards.

In a Deck of 52 playing cards,

Number of face card = 12

number of non face card = 40

Number of red face card = 6

Number of Ace cards = 4

So, Probability of the card is face card is , = 12/52 =3/13

Probability of the card is not face card ,= 40/52 =  10/13

Probability of the card is a red face card is,=6/52 =  3/26

The probability of  card is a 6 or lower is, = 24/52 =  6/13

The probability of  card is a 9 or lower is, = 36/52 =  9/13.

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Answer:

a) [tex]183-1.984\frac{20}{\sqrt{100}}=179.032[/tex]    

[tex]183+1.984\frac{20}{\sqrt{100}}=186.968[/tex]    

So on this case the 95% confidence interval would be given by (179.032;186.968)    

b) [tex]190-1.984\frac{23}{\sqrt{100}}=185.437[/tex]    

[tex]190+1.984\frac{23}{\sqrt{100}}=194.563[/tex]    

So on this case the 95% confidence interval would be given by (185.437;194.563)    

c) For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance

For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Part a : Summer

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=100-1=99[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,99)".And we see that [tex]t_{\alpha/2}=1.984[/tex]

Now we have everything in order to replace into formula (1):

[tex]183-1.984\frac{20}{\sqrt{100}}=179.032[/tex]    

[tex]183+1.984\frac{20}{\sqrt{100}}=186.968[/tex]    

So on this case the 95% confidence interval would be given by (179.032;186.968)    

Part b: Winter

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=100-1=99[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,99)".And we see that [tex]t_{\alpha/2}=1.984[/tex]

Now we have everything in order to replace into formula (1):

[tex]190-1.984\frac{23}{\sqrt{100}}=185.437[/tex]    

[tex]190+1.984\frac{23}{\sqrt{100}}=194.563[/tex]    

So on this case the 95% confidence interval would be given by (185.437;194.563)    

Part c

For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance

For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance

Final answer:

Using statistical methods, we have constructed 95% confidence intervals for the mean summer and winter weights of Frontier Airlines passengers, finding that the new FAA recommendations align with our calculated intervals but are positioned conservatively towards the upper limits.

Explanation:

To answer the student's question about constructing and interpreting 95% confidence intervals for the mean summer and winter weights of Frontier Airlines passengers, and then commenting on the new FAA recommendations, we'll follow statistical methods.

Part A: Summer Weight

For the summer weight, the mean is 183 lb, the standard deviation is 20 lb, and the sample size (n) is 100 passengers. The formula for a 95% confidence interval is: mean ± (z*standard error), where the standard error is (standard deviation/√n). Using z=1.96 for 95% confidence, we calculate the interval as 183 ± (1.96*(20/√100)), which simplifies to 183 ± 3.92. Thus, the 95% confidence interval for summer is ±183 lb ± 3.92 lb.

Part B: Winter Weight

Similar calculations for the winter weight (mean = 190 lb, standard deviation = 23 lb, n = 100) yield a 95% confidence interval of 190 ± (1.96*(23/√100)), or 190 ± 4.51. Therefore, the 95% confidence interval for winter is ±190 lb ± 4.51 lb.

Part C: Comment on FAA Recommendations

The new FAA recommendations are 190 lb for summer and 195 lb for winter. These fall within our confidence intervals, but lean towards the upper limit, suggesting a conservative approach by the FAA, possibly to ensure safety by accommodating potential underestimations of passenger weights.

Answer:

[tex] P(D|P)= \frac{0.11}{0.59}=0.186[/tex]

Step-by-step explanation:

For this case we have defined the following events:

D= "An international flight leaving the United States is delayed in departing"

P="An international flight leaving the United States is a transpacific flight "

And we have defined the probabilities:

[tex] P(D)= 0.29 , P(P) = 0.59[/tex]

And for the event: "an international flight leaving the U.S. is a transpacific flight and is delayed in departing" [tex] D \cap P [/tex] we know the probability:

[tex] P(D \cap P) =0.11[/tex]

We want to find this probability:

What is the probability that an international flight leaving the United States is delayed in departing given that the flight is a transpacific flight

So we want this probability:

[tex] P(D|P)[/tex]

And we can use the conditional formula from the Bayes theorem given two events A and B:

[tex] P(A|B) = \frac{P(A \cap B)}{P(B)}[/tex]

And if we use this formula for our case we have:

[tex] P(D|P)= \frac{P(D \cap P)}{P(P)}[/tex]

And if we replace the values we got:

[tex] P(D|P)= \frac{0.11}{0.59}=0.186[/tex]

Answer:

Try Photomath !

Step-by-step explanation:

The given series is an alternating harmonic series. It is convergent because the sequence of absolute values decreases to zero.

The given series is: 1/2 , 3/4 , 1/8 , 3/16 , 1/32 , 3/64 . . . As you can see, the numerator alternates between 1 and 3, while the denominator doubles each time. This type of series is known as an alternating harmonic series.

To determine if an alternating series converges or diverges, we can apply the Alternating Series Test. This test says that an alternating series converges if the sequence of absolute values decreases to zero. In our case, the absolute value of the terms based solely on the denominator is decreasing to zero, hence, the series converges.

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Answer:

76,050 ft²

Step-by-step explanation:

If the area must be rectangular, let L be the length of the side opposite to the creek, and S be the length of the remaining two sides.

The perimeter of the fencing and the area of the pasture are:

[tex]780 = L+2S\\A= LS\\\\L=780-2S\\A=-2S^2+780S[/tex]

The value of S for which the derivate of the area function is zero is the length of S that maximizes the area of pasture:

[tex]\frac{dA}{dS}=0=-4S+780\\S= 195\\L=780-(2*195)=390[/tex]

The maximum possible area is:

[tex]A_{MAX}=390*195=76,050\ ft^2[/tex]

Answer:

The answer to your question is z ≥ -159

Step-by-step explanation:

Process

1.- Write the inequality

                                   130 ≥ - 29 - z

2.- Add 29 to both sides of the inequality

                              130 + 29 ≥ - 29 + 29 - z

3.- Simplify

                                    159 ≥   - z

4.- Divide by -1 and change the inequality

                                    159/-1 ≤ -z/-1

5.- Simplify

                                   -159 ≤ z

Answer: Tara earned $775 from her summer job.

Step-by-step explanation:

Let x represent the amount of money that Tara earned from her summer job.

Lewis earned $1800 from his summer job at the grocery store. This is $250 more than twice what his friend Tara earned. The equation would be

2x + 250 = 1800

Subtracting 250 from the left hand side and the right hand side of the equation, it becomes

2x + 250 - 250 = 1800 - 250

2x = 1550

Dividing the left hand side and the right hand side of the equation by 2, it becomes

2x/2 = 1550/2

x = 775

The student has 210 ways to schedule the remaining six hours of classes across the week, after meeting specific daily minimum class hour requirements.

The student wants to schedule twelve hours of classes across a week while meeting minimum class hour requirements on specific days.

Monday requires at least three hours, Tuesday at least two, and Friday at least one hour of classes.

Considering the constraints, we must calculate the number of ways she can distribute her remaining class hours across the week.

Steps to Calculate the Distribution:

Therefore, the student has 210 ways to schedule the remaining six hours of classes across the week.

Answer:

P(A∪B)=17/20 or 0.85

P(A∪B')=2/5 or 0.4

P(A'∪B')=4/5 or 0.8

Step-by-step explanation:

There are four font colors so each color had equal chance and thus,

P(A)=1/4

There are 5 font sizes and so not the smallest fonts are 4.Thus,

P(B)=4/5

P(A∪B)=P(A)+P(B)-P(A∩B)

The design is generated randomly so event A and event B are independent.

P(A∩B)=P(A)*P(B)

P(A∩B)=1/4(4/5)=1/5

P(A∪B)=P(A)+P(B)-P(A∩B)

P(A∪B)=1/4+4/5-1/5=1/4+3/5

P(A∪B)=17/20 or 0.85

P(A∪B')=P(A)+P(B')-P(A∩B')

P(B')=1-P(B)=1-4/5=1/5

P(A∩B')=P(A)*P(B')=1/4*1/5=1/20

P(A∪B')=P(A)+P(B')-P(A∩B')

P(A∪B')=1/4+1/5-1/20=9/20-1/20=8/20

P(A∪B')=2/5 or 0.4

P(A'∪B')=P(A∩B)'

P(A'∪B')=1-P(A∩B)

P(A'∪B')=1-1/5=4/5

P(A'∪B')=4/5 or 0.8

The probability of event A or B occurring can be calculated using the addition rule. The probability of A occurring is 1/4, the probability of B occurring is 4/5, and the probability of both A and B occurring is 1/5. Using the addition rule, the probability of A or B occurring is 13/20.

To calculate the probability of events A or B occurring, we can use the addition rule. The addition rule states that the probability of A or B occurring is equal to the sum of their individual probabilities minus the probability that both A and B occur together.

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Answer:

It does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

Step-by-step explanation:

We are given the following data in the question:

75, 88, 51, 73, 49, 31, 69, 74, 72, 59, 72, 81, 99, 101, 73

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{1067}{15} = 71.13[/tex]

Sum of squares of differences = 4739.733

[tex]S.D = \sqrt{\frac{4739.733}{14}} = 18.39[/tex]

Population mean, μ = 60 minutes

Sample mean, [tex]\bar{x}[/tex] = 71.13 minutes

Sample size, n = 15

Alpha, α = 0.10

Sample standard deviation, s = 18.39 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 60\text{ minutes}\\H_A: \mu \neq 60\text{ minutes}[/tex]

We use Two-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{71.13 - 60}{\frac{18.39}{\sqrt{15}} } = 2.34[/tex]

Calculating the p-value from the table, we have,

P-value = 0.034354

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, we conclude that it does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

Option A is the correct set of hypotheses for testing if students can estimate one minute. The test statistic and p-value are needed to make a decision, typically using a t-test. The conclusion depends on the p-value relative to the significance level.

The null and alternative hypotheses you are working with for testing whether students are good at estimating one minute are:

To determine if students are good at estimating one minute, we have option A as correct, which is H0: \\(\mu = 60\\) and Ha: \\(\mu \not= 60\\).

The test statistic and p-value should be calculated using appropriate statistical methods (typically a t-test assuming we do not know the population standard deviation, which would require the mean, standard deviation, and sample size). After calculating the test statistic and p-value, you will compare the p-value to the significance level (alpha). If the p-value is less than alpha, you reject H0; otherwise, you fail to reject H0.

Based on the decision to reject or fail to reject H0, you can make a conclusion regarding the original claim about whether students are reasonably good at estimating one minute.

Answer:

Interquartile range: Difference between the third and the first quartile.

Range: Difference between the highest and the lowest value of a dataset.

Since the range is not in the middle of a set(like the quartiles), it is more sensitive to extreme values.

Step-by-step explanation:

The interquartile range is the difference between the third quartile and the first quartile.

The first quartile(Q1) separates the lower 25% from the upper 75% of a set. So 25% of the values in a data set lie at or below the first quartile, and 75% of the values in a data set lie at or above the first quartile.

The third quartile(Q3) separates the lower 75% from the upper 25% of a set. So 75% of the values in a data set lie at or below the third quartile, and 25% of the values in a data set lie at or the third quartile.

The range is the difference between the highest and the lowest value of a dataset.

Lets see two different sets of cardinality 8

Set 1: 5,5,6,7,8,9,13,13

Range = 13-5 = 8

Q1 = Element at the position 0.25*8 = 2 = 5

Q3 = Element at the position 0.75*8 = 6 = 9

IQR = Q3 - Q1 = 9 - 5 = 4

Now, replacing the first and the last element by extreme values

Set 1: 0,5,6,7,8,9,13,20

Range = 20-0 = 20

Q1 = Element at the position 0.25*8 = 2 = 5

Q3 = Element at the position 0.75*8 = 6 = 9

IQR = Q3 - Q1 = 9 - 5 = 4

The range changes and the IQR does not. Since extreme values are not in the middle of the distribution(in which the quartiles are), the range is more sensitive to extreme values, as this example showed.

Final answer:

The interquartile range measures the spread of the middle 50% of data and is less sensitive to extreme values compared to the range, which measures the spread of all data points by looking at the difference between the maximum and minimum values.

Explanation:

Difference Between Interquartile Range and Range in Data Analysis:

The difference between the interquartile range (IQR) and the range is that while the range measures the spread of all data points by subtracting the smallest value from the largest, the interquartile range measures the spread of the middle 50% of data, between the 25th and 75th percentiles (Q1 and Q3 respectively). To calculate the range in a dataset, if the highest score is 100 and the lowest is 70, the range is 100 - 70 = 30. On the other hand, if Q1 is 72.25 and Q3 is 86.75, the interquartile range would be 86.75 - 72.25 = 14.50.

The range is more sensitive to extreme values or outliers because it takes into account the most extreme scores in the dataset. In contrast, the IQR is less sensitive to extreme values since it focuses on the central portion of the data set and ignores the extremes. This is why IQR is often used to identify potential outliers, which are suspected if they lie more than 1.5 times the IQR below Q1 or above Q3.

Therefore, the IQR provides a more robust measure of data variability, especially when dealing with skewed distributions or datasets with outliers.