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ArbitrLikvidat [17]
3 years ago
6

What is the difference of the following polynomials?

Mathematics
2 answers:
Snowcat [4.5K]3 years ago
5 0
(6x^3-2x^2+4) <span>- (2x^3+4x^2-5)

Subtract the like terms

6x^3 - 2x^3 = 4x^3

-2x^2 - 4x^2 = -6x^2

4 - (-5) = 9

The final answer would be

4x^3 - 6x^2 + 9

</span>
Julli [10]3 years ago
4 0

Answer:

D) 4x^3 -6x^2 + 9

Step-by-step explanation:

Given: 6x^3-2x^2+4) - (2x^3+4x^2-5)

Distribute the negative sign, we get

6x^3 -2x^2 + 4 -2x^3 -4x^2 + 5

Now combine the like terms and simplify.

= 6x^3 -2x^3 -2x^2 -4x^2 + 4 +5

= 4x^3 -6x^2 + 9

Answer: D) 4x^3 -6x^2 + 9

Hope this will helpful.

Thank you.

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The difference of two numbers is 192, the sum is 3782
LenaWriter [7]

(3782 - 192) : 2 = 1795 (first number)

1795 + 192 = 1987 ( second number)

--------------------------

1987 - 1795 = 192

1987 + 1795 = 3782

4 0
3 years ago
An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba
madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C) The events "overweight" and "obese" exhaustive.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person is overweight

<em>Y</em> = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0
3 years ago
MATH<br><br> HELP<br><br> FIND THE EQUATION OF THE LINE
Dominik [7]

y=3/3x+-2

Step-by-step explanation:

it goes up 3 and over 3 so its 3/3 and it starts at -2

5 0
2 years ago
Read 2 more answers
Why does the shape of the distribution of the weights of russet potatoes tend to be symmetrical?
Sergio [31]

Answer:

Mean ≈ Median

6 0
3 years ago
Read 2 more answers
According to a 2010 study conducted by the Toronto-based social media analytics firm Sysomos, 71% of all tweets get no reaction.
klemol [59]

Answer:

a) the expected number of tweets with no reaction is 71 tweets

b) the variance is 20.59 tweets² ( ≈ 21 tweets )and the standard deviation is

4.537 tweets (≈ 5 tweets)

Step-by-step explanation:

we can use the binomial probability distribution

P(x, N, p ) = N!/[(N-x)!x!] * p^x * (1-p)^(N-x)

where

x = number of successful events

N = population total

p = probability for success for every individual and independent event

P = probability for x successful events

in our case p = 71% = 71/100 (probability of a tweet without reaction) , N = 100 tweets

a) the expected number of tweets

E(x) = N* p = 100 tweets * 71/100 = 71 tweets

b) the variance is

V(x) = N * p * (1-p)

V = 100 * 71/100 * (1-71/100) = 71* 29/100 = 20.59 tweets² ( units of variance are [N²] )

the standard deviation is

s = √V = √20.59 tweets² = 4.537 tweets

6 0
3 years ago
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