Hector is competing in a 42 mile bicycle race. He has already completed 18 miles of the race and is traveling at a constant spee

Answer:

A) The estimated proportion of women whose platelet counts fall within 2 standard deviations of the mean, or between 118.5 and 377.3, is 95%.

B) The estimated percentage of women with platelet counts ranging from 53.8 to 442.0 is 99.7%.

Step-by-step explanation:

Provided data:

mean;μ = 247.9

standard deviation;σ = 64.7

A) We seek to find the estimated percentage of women with platelet counts within 2 standard deviations from the mean, which translates to values between 118.5 and 377.3.

Based on the attached image, the empirical curve indicates that the likelihood within 1 standard deviation of the mean is (34% + 34%) = 68%.

In contrast, the likelihood within 2 standard deviations from the mean is (13.5% + 34% + 34% + 13.5%) = 95%

Therefore, the estimated percentage of women having platelet counts within 2 standard deviations of the mean, or ranging from 118.5 to 377.3, equals 95%

B) Next, we want to determine the estimated percentage of women with platelet counts between 53.8 and 442.0.

The values of 53.8 and 442.0 correspond to 3 standard deviations from the mean.

Let’s verify that.

Since mean;μ = 247.9

standard deviation;σ = 64.7;

μ = 247.9

σ = 64.7

μ + 3σ = 247.9 + 3(64.7) = 442

Also;

μ - 3σ = 247.9 - 3(64.7) = 53.8

From the attached empirical curve, it can be seen that at 3 standard deviations from the mean, the probability percentage is;

(2.35% + 13.5% + 34% + 34% + 13.5% + 2.35%) = 99.7%

Response:

Step-by-step breakdown:

Kevin has already gathered five and a half gallons of water for his trip

He understands that he requires a minimum of 20 gallons of water for the journey.

The water is packaged in 32-fluid ounce (quarter-gallon) containers.

1 fluid ounce equals 0.0078125 gallons

32-fluid ounce

Let x represent the number of 32-fluid ounce (quarter-gallon) containers needed to collect at least 20 gallons of water for the trip.

One container holds 0.25 gallons of water

Therefore, x containers hold 0.25x gallons of water

Thus, Kevin's total gallons of water =

Since it is given that he needs at least 20 gallons of water for the trip.

Hence,

Thus, the algebraic inequality representing this scenario is

Answer:

The triangulation technique may not always yield a precise result (aside from any errors in measurement) for the following reasons:

It employs a variety of data sources, different researchers, and multiple theories or viewpoints.

Step-by-step explanation:

Research triangulation integrates various methods, data sources, diverse investigators, and assorted theories to develop a deeper insight into the situations being studied. This process fortifies qualitative research by incorporating data from multiple sources, perspectives, and methodologies.

Answer:

A. P = 0.73

B. P(A∩B∩C') = 0.22

C. P(B/A) = 0.5

P(A/B) = 0.75

D. P(A∩B/C) = 0.4

E. P(A∪B/C) = 0.85

Step-by-step explanation:

Denote A as the event of a student having a Visa card, B as the event of holding a MasterCard, and C as the event of owning an American Express card. Additionally, let A' indicate the event of not having a Visa card, B' signify not having a MasterCard, and C denote the event of not possessing an American Express card.

Thus, with the given probabilities, we can determine the following probabilities:

P(A∩B∩C') = P(A∩B) - P(A∩B∩C) = 0.3 - 0.08 = 0.22

Here, P(A∩B∩C') refers to the chance that a student has both a Visa and MasterCard but does not own an American Express, P(A∩B) indicates the probability that a student possesses both a Visa and a MasterCard, and P(A∩B∩C) represents the likelihood that a student has a Visa, MasterCard, and American Express. Similarly, we can find:

P(A∩C∩B') = P(A∩C) - P(A∩B∩C) = 0.15 - 0.08 = 0.07

P(B∩C∩A') = P(B∩C) - P(A∩B∩C) = 0.1 - 0.08 = 0.02

P(A∩B'∩C') = P(A) - P(A∩B∩C') - P(A∩C∩B') - P(A∩B∩C)

                   = 0.6 - 0.22 - 0.07 - 0.08 = 0.23

P(B∩A'∩C') = P(B) - P(A∩B∩C') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.4 - 0.22 - 0.02 - 0.08 = 0.08

P(C∩A'∩A') = P(C) - P(A∩C∩B') - P(B∩C∩A') - P(A∩B∩C)

                   = 0.2 - 0.07 - 0.02 - 0.08 = 0.03

A. The likelihood that the selected student holds at least one of the three card types is calculated as follows:

P = P(A∩B∩C) + P(A∩B∩C') + P(A∩C∩B') + P(B∩C∩A') + P(A∩B'∩C') +              

     P(B∩A'∩C') + P(C∩A'∩A')

P = 0.08 + 0.22 + 0.07 + 0.02 + 0.23 + 0.08 + 0.03 = 0.73

B. The probability that the chosen student possesses both a Visa and a MasterCard without an American Express card can be represented as P(A∩B∩C') equaling 0.22

C. P(B/A) represents the chance that a student holds a MasterCard provided they have a Visa. This is calculated as:

P(B/A) = P(A∩B)/P(A)

By substituting in the values, we find:

P(B/A) = 0.3/0.6 = 0.5

In a similar manner, P(A/B) represents the probability a student has a Visa given they possess a MasterCard, calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. For a student with an American Express card, the likelihood they also hold both a Visa and a MasterCard is expressed as P(A∩B/C), calculated as:

P(A∩B/C) = P(A∩B∩C)/P(C) = 0.08/0.2 = 0.4

E. If the student has an American Express card, the probability they possess at least one of the other two card types is denoted as P(A∪B/C), computed as:

P(A∪B/C) = P(A∪B∩C)/P(C)

Where P(A∪B∩C) = P(A∩B∩C)+P(B∩C∩A')+P(A∩C∩B')

Consequently, P(A∪B∩C) equals 0.08 + 0.07 + 0.02 = 0.17

Ultimately, P(A∪B/C) equals:

P(A∪B/C) = 0.17/0.2 =0.85