1/10 of an hour (D) is correct. Hope this helped :)
Answer:
24.5 unit²
Step-by-step explanation:
Area of ∆
= ½ | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |
= ½ | (-1)(3 -(-4)) + 6(-4 -3) + (-1)(3 - 3) |
= ½ | -7 - 42 |
= ½ | - 49 |
= ½ (49)
= 24.5 unit²
<u>Method 2:</u>
Let the vertices are A, B and C. Using distance formula:
AB = √(-1-6)² + (3-3)² = 7
BC = √(-6-1)² + (-4-3)² = 7√2
AC = √(-1-(-1))² + (4-(-3))² = 7
Semi-perimeter = (7+7+7√2)/2
= (14+7√2)/2
Using herons formula:
Area = √s(s - a)(s - b)(s - c)
here,
s = semi-perimeter = (14 + 7√2)/2
s - a = S - AB = (14+7√2)/2 - 7 = (7 + √2)/2
s - b = (14+7√2)/2 - 7√2 = (14 - 7√2)/2
s - c = (14+7√2)/2 - 7 = (7 + √2)/2
Hence, on solving for area using herons formula, area = 49/2 = 24.5 unit²
First problem:
The answer is C because theoretically it should have a 50% of landing
heads but instead it lands heads 56% of the time. Thus this is 6% higher
than 50%. It's not D because there is not specified detail of how the
person got this data so you can assume that the person did a fair
survey/data collection.
Second problem:
So in a data size of 490, 140 of them we trout. This means that 140/490 or 28.57% of the fish are trout. This means that in a sample size of 5000 fish, 5000*0.2857 or 1428.57 are trout.
Third problem:
The Science students seem to have a higher average score because the average score of Math students are:
(32+33+34+34+35+37+39+40+40)/9=36
average score of Science students are:
(41+42+43+43+46+46+47+49+49)/9=45.111
Answer:
y= 1/5x + 0.8
Step-by-step explanation:
Hope this helps
Answer:
.................
Step-by-step explanation:
