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ehidna [41]
3 years ago
5

Write the equation in standard form using factoring. f(x) = (2x + 1)(x – 4)

Mathematics
1 answer:
klemol [59]3 years ago
7 0

Answer:

2x^2-7x-4

Step-by-step explanation:

=2x^2-8x+x-4

=2x^2-7x-4

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Let M be the midpoint of LN. If LM = 9 inches what must be the length of LN? (draw a picture if needed!)
Pani-rosa [81]

Answer:

LN = 18

Step-by-step explanation:

hope that's the answer..

6 0
3 years ago
Let X1 and X2 be two random variables following Binomial distribution Bin(n1,p) and Bin(n2,p), respectively. Assume that X1 and
ryzh [129]

Answer:

a) X1+X2 have distribution Bi(n1+n2, p)

b)

P(X1+X2 = 1 | X2 = 0) =  np(1-p)ⁿ¹⁻¹

P(X1+X2 = 1| X2 = 1) = (1-p)ⁿ¹

P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

Step-by-step explanation:

Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).

b)

Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).

If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹

If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹

Thus,

P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

4 0
3 years ago
What is the value of X?
rodikova [14]

Answer:

102

Step-by-step explanation:

triangle area = 180 degrees

straight line = 180 degrees

180 - 127 = 53

49

49 + 53 + 3rd angle = 180

3rd angle = 78

180-78 = x

x = 102

4 0
2 years ago
Pls help me find the answer​
Korolek [52]
To what?? Please explain.
8 0
3 years ago
Which expression correctly displays the calculations to find the a^5b^4 term of (a+b)^8
LUCKY_DIMON [66]

Answer:

Step-by-step explanation:

THE BINOMIAL THEOREM shows how to calculate a power of a binomial -- (a + b)n -- without actually multiplying.

For example, if we actually multiplied out the 4th power of (a + b) --

(a + b)4 = (a + b)(a + b)(a + b)(a + b)

-- then on collecting like terms we would find:

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 .  .  .  .  (1)

Note:  The literal factors are all possible terms in a and b where the sum of the exponents is 4:  a4,  a3b,  a2b2,  ab3,  b4.

The degree of each term is 4.

The first term is actually a4b0, which is a4 · 1.

Thus to "expand" (a + b)5, we would anticipate the following terms, in which the sum of all the exponents is 5:

(a + b)5 =  ? a5 +  ? a4b +  ? a3b2 +  ? a2b3 +  ? ab4 +  ? b5

The question is, What are the coefficients?

They are called the binomial coefficients.  In the expansion of

(a + b)4, the binomial coefficients are

1  4  6  4  1

line (1) above.

 Note the symmetry:  The coefficients from left to right are the same right to left.

The answer to the question, "What are the binomial coefficients?" is called the binomial theorem.  It shows how to calculate the coefficients in the expansion of (a + b)n.

The symbol for a binomial coefficient is The binomial theorem.  The upper index n is the exponent of the expansion; the lower index k indicates which term, starting with k = 0.

For example, when n = 5, each term in the expansion of  (a + b)5  will look like this:

The binomial theorema5 − kbk

k will successively take on the values 0 through 5.

(a + b)5 = The binomial theorema5  +  The binomial theorema4b  +  The binomial theorema3b2  +  The binomial theorema2b3  +  The binomial theorem ab4  +  The binomial theoremb5

Note:  Each lower index is the exponent of b.  The first term has k = 0 because in the first term, b appears as b0, which is 1.

Now, what are these binomial coefficients, The binomial theorem ?

The theorem states that the binomial coefficients are none other than the combinatorial numbers, nCk .

The binomial theorem  =  nCk

 (a + b)5  =  5C0a5 + 5C1a4b + 5C2a3b2 + 5C3a2b3 + 5C4ab4 + 5C5b5

  =  1a5 + The binomial theorema4b + The binomial theorema3b2 + The binomial theorema2b3 + The binomial theoremab4 + The binomial theoremb5

  =  a5  +  5a4b  +  10a3b2  +  10a2b3  +  5ab4  +  b5

The binomial coefficients here are

1  5  10  10  5  1.

8 0
2 years ago
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