Answer:
given:
height[h]=8mm
radius [r]=5mm
Now,
Volume of cone=1/3 πr²h=⅓×3.14×5²×8=209mm³
<u>C) 209 </u><u>mm³</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>required</u><u> </u><u>answer</u><u>.</u>
Answer:
∠A = 30°
∠B = 60°
∠C= 90°
Step-by-step explanation:
This is a right triangle, you can see it mainly by the red square in C, and it is always used to mark 90 degrees.
Knowing that, you now know <em>∠C is 90°</em>
Now, to find ∠B, you should use the following equation:
This means that the sum of the three angles of a triangle gives 180. ALWAYS. So to find the missing angle, ∠B, do the following:
Fill the values of the equation with the angles you now know:
Solve the equation, passing the 30° and 90° to the other side of the equal sing with Inverse Operation:
<em>B = 60</em>
<em>Hope it helps!!</em>
The answer you put in is y = 1/2 + 4
7 = r^1/2 is the same as saying 7 = square root of r. So, to get rid of the square root sign on the r, you need to square it, and when you do something to one side of the equation, you have to do the same to the other side. So after squaring both sides, you get 49 = r.
(a)
Q1, the first quartile, 25th percentile, is greater than or equal to 1/4 of the points. It's in the first bar so we can estimate Q1=5. In reality the bar includes values from 0 to 9 or 10 (not clear which) and has around 37% of the points so we might estimate Q1 a bit higher as it's 2/3 of the points, say Q1=7.
The median is bigger than half the points. First bar is 37%, next is 22%, so its about halfway in the second bar, median=15
Third bar is 11%, so 70% so far. Four bar is 5%, so we're at the right end of the fourth bar for Q3, the third quartile, 75th percentile, say Q3=40
b
When the data is heavily skewed left like it is here, the median tends to be lower than the mean. The 5% of the data from 80 to 120 averages around 100 so adds 5 to the mean, and 8% of the data from the 60 to 80 adds another 5.6, 15% of the data from 40 to 60 adds about 7.5, plus the rest, so the mean is gonna be way bigger than the median of around 15.
