Amanda exercised for 10 minutes every day in the first week, 20 minutes in the second week, 30 minutes in the third week, and 40 minutes in the fourth week.Billy exercised for 5 minutes every day in the first week, 10 minutes in the second week, 20 minutes in the third week, and 40 minutes in the fourth week.

Which statement best describes the methods used by Amanda and Billy to increase the time they spent exercising?


A) Amanda's method is linear because the number of minutes increased by an equal number every week.
B) Billy's method is linear because the number of minutes increased by an equal factor every week.
C) Both Billy's and Amanda's methods are exponential because the number of minutes increased by an equal factor every week.
D) Both Billy's and Amanda's methods are exponential because the number of minutes increased by an equal number every week.

To find the distance between the points (a, a) and (b, b), we can use the distance formula. The distance formula calculates the distance between two points in a coordinate plane.

The distance between two points (x1, y1) and (x2, y2) is given by the formula:

Distance = √((x2 - x1)² + (y2 - y1)²)

In this case, since the points are (a, a) and (b, b), we substitute a for both x1 and y1, and b for both x2 and y2:

Distance = √((b - a)² + (b - a)²)

Simplifying this expression further:

Distance = √((b - a)² + (b - a)²) = √(b² - 2ab + a² + b² - 2ab + a²) = √(2a² + 2b² - 4ab)

Therefore, the distance between the points (a, a) and (b, b) is √(2a² + 2b² - 4ab).

Answer:

• E(X) = 60*1/6 = 10

• sd(X) = √8.666 = 2.886

• E(Y) = 600*1/6 = 100

• sd(Y) = √86.666 = 9.1287

Step-by-step explanation:

Lets call X the amount of aces obtained in 60 rolls, and Y the amount of aces obtained in 600 rolls.

Note that both X and Y are obtained from counting the amount of successful tries from repetitions of independent experiments that have 1/6 of probability of success. Thus, both X and Y are random variables with binomial distribution, with n = 60 and 600 respectively and probability 1/6.

Remember that if Z is a random variable, Z ≈ Bi(n,p), then

• E(Z) = np, where E(Z) denotes the expected value of Z
• V(Z) = np(1-p), where V(Z) denotes the variance of Z. Hence, the standard deviation is the square root of V(Z), √(np(1-p)).

As a result

• E(X) = 60*1/6 = 10
• V(X) = 10*(1-1/6) = 50/6 ≅ 8.666
• sd(X) = √8.666 = 2.886
• E(Y) = 600*1/6 = 100
• V(Y) = 100*(1-1/6) = 500/6 ≅ 86.666

• sd(Y) = √86.666 = 9.1287

The observed amount of aces is more likely to be closer from the expected value with 60 rolls because, since we have less rolls, it is more difficult to obtain spread results.

You can also notice that X and Y can be obtained by summing independent variables with distribution BI(1,p) (also called Bernoulli(p) ). When you sum independent variables with the same distribution you have this property:

• E(r1+r2+...+rn) = n*E(r1)
• V(r1+r2+...+rn) = n*V(r1)

• sd(r1+r2+...+rn) = √n*sd(r1)

X can be obtained by summing 60 independent variables r1, ...., r60 with mean 1/6 and variance 1/6*(5/6) = 5/36. So we obtain that V(X) = 60*5/36, and sd(X) = √60 * √(5/36). While for the same argument sd(Y) = √600*√(5/36). The higher the number of rolls, the more spread the results are.

I hope this helped you!

The expected number of aces from 60 rolls of a fair die is 10 with a standard deviation of approximately 3.72. For 600 rolls, the expected number is 100 with a standard deviation of about 11.79. The observed count of aces is more likely to be closer to the expected value with fewer rolls due to the smaller standard deviation relative to the number of trials.

The expected value for the number of aces in a fair die roll is computed by multiplying the probability of rolling an ace by the number of rolls. For 60 rolls, the expected number is . For 600 rolls, the expected number is

The standard deviation for the number of aces is calculated using the formula for the standard deviation of a binomial distribution, , where n is the number of trials, p the probability of success ( for an ace). For 60 rolls, it is . For 600 rolls, it's .

When you roll the die 60 times, the chances of the observed count of aces being within 2 from the expected value (10) is higher because the standard deviation is smaller relative to the number of trials than when you roll the die 600 times.

As the number of trials increases, the expected standard deviation grows larger, and the observed count is more likely to be within a wider range from the expected value (100).

Learn more about standard deviation here:

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