Answer:
A. 3561 ft²
Step-by-step explanation:
πr² = the area of a circle
base · height = volume
9²π = 254.47
254.47 · 14 = 3561
To get the inverse
of any function f(x) [ just make sure that it passes the horizontal line test]
follow the steps:
- Solve for x, i.e separate x in one side
and the other terms in the other side, included f(x).
- Swap x and f(x).
- The result is the inverse of the original
function 
.
-------------------------
Apply that on your function
Answer:
Since the calculated z=40 falls in the critical region this indicates that the true obesity rate for children in Marion County is different from the national average at the 0.05 significance level. We reject the null hypothesis that population proportion is 0.17.
Step-by-step explanation:
The national average is 17% .The z proportional hypothesis test is used.
1) Let the null and alternate hypothesis be
H0: p =0.17
against the claim
Ha: p ≠ 0.17
Choose the significance level ∝= 0.05
The critical region is z > 1.96 and Z <- 1.96 because it is two tailed test.
Computing
z= p^-p / sqrt [pq/n]
Z= 0.22-0.17/ √0.17*(1- 0.17)/90147
z= 39.965
Since the calculated z=40 falls in the critical region this indicates that the true obesity rate for children in Marion County is different from the national average at the 0.05 significance level. We reject the null hypothesis that population proportion is 0.17.
Completed question:
In the game of tic-tac-toe, if all moves are performed randomly the probability that the game will end in a draw is 0.127. Suppose six random games of tic-tac-toe are played. What is the probability that at least one of them will end in a draw?
Answer:
0.557
Step-by-step explanation:
For each game, the probability of not end in a draw is 1 - 0.127 = 0.873. Thus, because each game is independent of each other, the probability of all of them not end in a draw is the multiplication of the probability of each one:
0.873x0.873x0.873x...x0.873 = 0.873⁶ = 0.443
Thus, the probability that at least one of them end in a draw is the total probability (1) less the probability that none of them en in a draw:
1 - 0.443
0.557
If the letters are considered distinct, then the number of permutations is
.
If we count either C as the same character, then we would be double-counting - to correct this, we would simply divided by the number of ways we can choose C from the available characters, or
.
