Answer:
1 9/16
Step-by-step explanation:
First, you do 25 - 16, and the answer is 9. After that you put a 1, as its a whole. Lastly, you put the 9 on top of the 16.
First you have to know, that the price the agent found is 70% of $2075
(100%-30%=70%)
70%=0,7
$2075 * 0,7 = $1452,5
So 70% 0f $2075 is 1452,5
The price the agant found is $1452,5.
Answer:
0.8762 or 87.62%
Step-by-step explanation:
Since our mean is μ=14.3 and our standard deviation is σ=3.7. If we're trying to figure out what percentage is P(10 ≤ x ≤ 26) equal to we must first calculate our z values as such:
Our x value ranges from 10 to 26 therefore let x=10 and we obtain:
If we look at our z-table we find that the probability associated with a z value of -1.16 is 0.1230 meaning 12.30%.
Now let's calculate the z value when x = 26 and so:
Similarly, we use the z-table again and find that the probability associated with a z value of 3.16 is 0.9992 meaning 99.92%.
Now we want to find the probability in between 10 and 26 so we will now subtract the upper limit minus the lower limit in P(10 ≤ x ≤ 26) therefore:
0.9992 - 0.1230 = 0.8762
or 87.62%
Answer:
x = 3
Step-by-step explanation:
The fractions need to have common denominators. So, we would make the common denominator (x + 3)(x - 3), or, x^2 - 9
Multiply the numerator by the same number or variable as the denominator and then combine.
Once their is x and a number left you would solve for x.
Answer:
The probability is
Step-by-step explanation:
We can divide the amount of favourable cases by the total amount of cases.
The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8,
For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.
Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.
We can conclude that the probability for 8 rooks not being able to capture themselves is