Given:
Uniform distribution of length of classes between 45.0 to 55.0 minutes.
To determine the probability of selecting a class that runs between 51.5 to 51.75 minutes, find the median of the given upper and lower limit first:
45+55/2 = 50
So the highest number of instances is 50-minute class. If the probability of 50 is 0.5, then the probability of length of class between 51.5 to 51.75 minutes is near 0.5, approximately 0.45. <span />
Answer:
Use the distance formula on both points AC and AB.
<em>Distance formula is this</em><em>:</em>
<em>\begin{gathered}d=\sqrt{(x2-x1)^2+(y2-y1)^2} \\\\d=\sqrt{(1--5)^2+(8--7)^2} \\\\d=\sqrt{(6)^2+(15)^2} \\\\d=\sqrt{36+225} \\\\d=\sqrt{261} \\\\\end{gathered}d=(x2−x1)2+(y2−y1)2d=(1−−5)2+(8−−7)2d=(6)2+(15)2d=36+225d=261</em>
Distance for AC is 16.16
Now do the same with the numbers for AB and get the distance of 5.39
2. To get the area, use the formula 1/2 x base x height
AB is the base and AC is the height.
1/2 x 16.16 x 5.39 = 43.55
the answer is 43.5
Answer:
16.34 square inches.
Step-by-step explanation:
Given:
Base of triangle, b = 3 inches
Height of triangle, h = 10.89 inches
Question asked:
What is the area of this triangle in square inches?
Round your answer to the nearest hundredths place .
Solution:
As we know,
As, we have to find the answer to the nearest hundredths place, hence the area of triangle will be 16.34 square inches.
