A student has a business selling snowballs during the summer. For each snowball the student sells, they makes a profit of $1.50. On the hottest day of the summer, the student sold 125 snowballs and made a profit of $187.50.Choose a model (equation) in point-slope form {y−y1=m(x−x1)} to represent the profit the student makes, y, based on the number of snowballs sold, x.

y−125=1.5(x−187.5) A


y+187.5=1.5(x+125) B


y−187.5=1.5(x−125) C

y+125=1.5(x+187.5) D

Answer:

a) the probability of waiting more than 10 min is 2/3 ≈ 66,67%

b) the probability of waiting more than 10 min, knowing that you already waited 15 min is 5/15 ≈ 33,33%

Step-by-step explanation:

to calculate, we will use the uniform distribution function:

p(c≤X≤d)= (d-c)/(B-A) , for A≤x≤B

where p(c≤X≤d) is the probability that the variable is between the values c and d. B is the maximum value possible and A is the minimum value possible.

In our case the random variable X= waiting time for the bus, and therefore

B= 30 min (maximum waiting time, it arrives 10:30 a.m)

A= 0 (minimum waiting time, it arrives 10:00 a.m )

a) the probability that the waiting time is longer than 10 minutes:

c=10 min , d=B=30 min --> waiting time X between 10 and 30 minutes

p(10 min≤X≤30 min) = (30 min - 10 min) / (30 min - 0 min) = 20/30=2/3 ≈ 66,67%

a) the probability that 10 minutes or more are needed to wait starting from 10:15 , is the same that saying that the waiting time is greater than 25 min (X≥25 min) knowing that you have waited 15 min (X≥15 min). This is written as P(X≥25 | X≥15 ). To calculate it the theorem of Bayes is used

P(A | B )= P(A ∩ B ) / P(A) . where P(A | B ) is the probability that A happen , knowing that B already happened. And P(A ∩ B ) is the probability that both A and B happen.

In our case:

P(X≥25 | X≥15 )= P(X≥25 ∩ X≥15 ) / P(X≥15 ) = P(X≥25) / P(X≥15) ,

Note: P(X≥25 ∩ X≥15 )= P(X≥25) because if you wait more than 25 minutes, you are already waiting more than 15 minutes

-   P(X≥25) is the probability that waiting time is greater than 25 min

c=25 min , d=B=30 min --> waiting time X between 25 and 30 minutes

p(25 min≤X≤30 min) = (30 min - 25 min) / (30 min - 0 min) = 5/30 ≈ 16,67%

-  P(X≥15) is the probability that waiting time is greater than 15 min --> p(15 min≤X≤30 min) = (30 min - 15 min) / (30 min - 0 min) = 15/30

therefore

P(X≥25 | X≥15 )= P(X≥25) / P(X≥15) = (5/30) / (15/30) =5/15=1/3  ≈ 33,33%

Note:

P(X≥25 | X≥15 )≈ 33,33% ≥ P(X≥25) ≈ 16,67%  since we know that the bus did not arrive the first 15 minutes and therefore is more likely that the actual waiting time could be in the 25 min - 30 min range (10:25-10:30).

Answer:

Midpoint = (  (x1+x2)/2 , (y1+y2)/2 )

Step-by-step explanation:

Please upload the line segment otherwise, you can use the equation above to solve for it.

Answer:

Her second lap was 8% better when compared to her first lap.

Step-by-step explanation:

Given:

Emily ran her first lap in 75 seconds she ran her second lap in 69 seconds.

Now, to find how much better was her second lap when compared to her first lap.

Emily ran her first lap in 75 seconds.

Emily ran her second lap in 69 seconds.

So, we get the difference in seconds:

Thus, in her second lap she was 6 seconds better than first.

Now, to get the percentage of her first lap better than second:

Therefore, her second lap was 8% better when compared to her first lap.