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mariarad [96]
3 years ago
14

Select all the expressions that are equivalent to the polynomial below.

Mathematics
2 answers:
PIT_PIT [208]3 years ago
5 0
(4x+5)(-3x-1)=
-12x²-19x-5
So, you answers would be:
A. (-16x²+10x-3) + (4x²-29x-2)
D. (2x²-11x-9) - (14x²+8x-4)
Have a nice day! ♪
amm18123 years ago
5 0

Answer:

A. C. And D. Are same

Step-by-step explanation:

(4x+5)(-3x-1)

Applying distributive law

4x(-3x-1)+5(-3x-1)

Again applying distributive law in both the brackets

-12x^2-4x-15x-5

Adding the terms having same power of x

-12x^2-19x-5

1. -16x^2+10x-3+(4x^2-29x-2 )

Simplifying the terms containing the same power of x

-16x^2+10x-3+4x^2-29x-2

-12x^2-19x-5

Hence it is true

2. 3(x-5)-2(6x^2+9x+5)

   3x-15-12x^2-18x-10

  -12x^2-15x-25

  False

3. 2(x-1)-3(4x^2+7x+1)

   2x-2-12x^2-21x-3

  -12x^2-19x-5

True

4. 2x^2-11x-9-(14x^2+8x-4)

   2x^2-11x-9-14x^2-8x+4

  -12x^2-19x-5

True

5. -15x^2+9x-10+3x^2-10x-5

-12x^2+19x-15

False

6. 4x^2-13x-7-(16x^2+9x-5)

   4x^2-13x-7-16x^2-9x+5

  -12x^2-22x-2

False

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Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and list
Radda [10]

Answer:

P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}

Possible values of x: Any from 0 to 5.

P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807

P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015

P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087

P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323

P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835

P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this question:

n = 5, p = 0.3, q = 1 - p = 0.7

So

P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807

P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015

P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087

P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323

P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835

P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243

8 0
3 years ago
I need answers for the question below to end a debate. Explain why that is the answer too.
IRINA_888 [86]
PEMDAS. Parenthesis comes first. 3+2 = 5 which is then multiplied to 6/2 which is 3 so 3 x 5 is 15. 
7 0
3 years ago
Read 2 more answers
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