Answer:
33.33% probability that it takes Isabella more than 11 minutes to wait for the bus
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability that we find a value X lower than x is given by the following formula.

For this problem, we have that:
Uniformly distributed between 3 minutes and 15 minutes:
So 
What is the probability that it takes Isabella more than 11 minutes to wait for the bus?
Either she has to wait 11 or less minutes for the bus, or she has to wait more than 11 minutes. The sum of these probabilities is 1. So

We want P(X > 11). So

33.33% probability that it takes Isabella more than 11 minutes to wait for the bus
The correct answer is:
How to find X if P percent of it is Y. Use the percentage formula Y/P% = X
Convert the problem to an equation using the percentage formula: Y/P% = X
Y is 26,400, P% is 16%, so the equation is 26,400/16% = X
Convert the percentage to a decimal by dividing by 100.
Converting 16% to a decimal: 16/100 = 0.16
Substitute 0.16 for 16% in the equation: 26,400/0.16 = X
Do the math: 26,400/0.16 = X
X = 165,000
So $ 165,000 is the price of the home.
I hope this helps you!!
Answer:
Step-by-step explanation:
Angle sum property of triangle: Sum of all the angles of a triangle is 180
Alternate interior angles: When two parallel lines are intersected by a transversal, the pair of angles on the inner side of each of these lines but on the opposite side of the transversal are called alternate interior angles
In ΔABC,
∠1 + 90 + 38 = 180 {angle sum property of triangle}
∠1 + 128 = 180
∠1 = 180 - 128

AB // CD and AC is transversal.
{Alternate interior angles are equal}
In ΔACD,
∠2 + ∠3 + 63 = 180 {angle sum property of triangle}
∠2 + 38 +63 = 180
∠2 + 101 =180
∠2 = 180 - 101

Answer:
1:1
Step-by-step explanation:
They look exactly the same so they would equal each other
High positive correlation<span>
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