Answer:
can I see the graphs
Step-by-step explanation:
I cannot see the graph
Margin of error, e = Z*SD/Sqrt (N), where N = Sample population
Assuming a 95% confidence interval and substituting all the values;
At 95% confidence, Z = 1.96
Therefore,
0.23 = 1.96*1.9/Sqrt (N)
Sqrt (N) = 1.96*1.9/0.23
N = (1.96*1.9/0.23)^2 = 262.16 ≈ 263
Minimum sample size required is 263 students.
Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
Answer:
Infinite solutions
Step-by-step explanation:
All solutions are possible.
Answer:
25 – 4.33 = 20.67
7 + (1 - 100) = -92
Step-by-step explanation:
First one) Simply use your knowledge of integers and decimals to solve this. You can even start by drawing a number line and practising with that before beginning to do harder problems.
Second one) Just like the first one but use PEDMAS, BEDMAS or BODMAS and make sure to solve the brackets first.
