25, 34, 46, 37, 33, 42, 40, 37, 34, 49, 73, 46, 45, 45, 5. how many outliers do these data contain?
yKpoI14uk [10]
2: 73 and 5
...............................
4/3 x 7=9.3333
4/3 x 8=10.666
Can't go over 10, so you can only make 7 batches.
Hope I've helped!
Answer:
35 cm
Step-by-step explanation:
To find the area of the bottom portion, you would use the formula for finding out a triangle (B*H*1/2) which is:
(4+4)*5.75*1/2=<u>23</u>
Then, for the top portion, one would find the area of the triangles on the sides (with two marks going through). Since along the middle is 8cm, and along the top is 4, we can see that there is 2cm on either side, so that is the length of the base of the triangle. To solve for the top triangles, you would do almost the same thing as the last one:
2*2*1/2=2
But since there's two identical triangles on either side, we can multiply that by two, which would bring it to <u>4.</u>
That just leaves the rectangle that is left between the two triangles. To solve this, it's just B*H and luckily both of those are labeled for you already:
4*2=<u>8</u>
Now, to find the total area, all you have to do is add up the areas of the different sections:
23+4+8=35 cm
Hope this helps!
It would be A:303.45$ because 357-53.55 which would be 303.45$
Cos theta would be (adjacent side) / (hypotenuse), or 8/29.
Thus, the angle theta would be the inverse cosine of 8/29, which is
1.291 radians or 73.987 degrees.
Given this result, just take the sine of this angle.
