Answer:
x^2 + 5x - 14
Step-by-step explanation:
We can assume that if there is 2 options for x, this is a quadratic equation that equals 0.
Therefore, we can put this into brackets as one bracket must equal 0.
(x - 2) (x + 7)
After this we expand, which is a process easier to search up
The answer is:
x^2 + 7x - 2x - 14
We can simplify this to:
x^2 + 5x - 14
x^2 represents x squared in case you were wondering
Therefore, the answer is:
x^2 + 5x - 14
Answer:
Answers A (1/3), C (4/15) and D (5/6) are all repeating decimals.
Step-by-step explanation:
Repeating decimals are decimals in which a number or sequence of numbers is repeated over and over when dividing the numerator by the denominator of a fraction. For example, when you divide 1 by 3 (1/3), you will get a repeating decimal of 0.3333...since 3 goes into 10 three times with a remainder of 1 and will keep going. Likewise, when you divide 4/15, you get an initial value of 0.2, with a repeating 6, or 0.266666... Lastly, when you divide 5/6, you will get a repeating decimal of 0.833333.... The other answers will all be terminating decimals when you divide the numerator by the denominator. Terminating decimals mean the stop at some point and don't continue.
Any of the 4 people can take the first seat
since it gets occupied by one person , the next seat can be occupied by any of the remaining 3
similarly, the next one has 2 possibilities and the last seat can only be occupied by the man who is left
so by the principal of multiplication, no. of ways equals 4! = 4×3×2×1 = 24 ways
If <em>X</em> is uniformly distributed on the interval (0, 12), then its PDF is

or simply

and the zero-elsewhere case is assumed.
Whether you include 0 and 12 in the domain is irrelevant, since the probability that <em>X</em> = 0 or <em>X</em> = 12 are both zero.