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svetlana [45]
2 years ago
6

What is -3y(y-8)(2y+1)=0

Mathematics
1 answer:
Marta_Voda [28]2 years ago
7 0

Step-by-step explanation:

(-3y^2+24y)(-6y^2-3y)

open the bracket and collect the like terms

-3y^2-6y^2+24y-3y

-9y^2+21y

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If a student (represented by initials) was chosen at random, find P(HH|C').
valentina_108 [34]

Answer:

3/5

Step-by-step explanation:

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Which of the following equations has a graph PARALLEL to the line y = 4x + 7
Kobotan [32]

Answer:

y = 4x - 2

Step-by-step explanation:

This equation has the same slope as the equation given which means that they are parallel lines. This equation also has a y-intercept of -2.

I graphed both equations below to show you that they are parallel.

If this answer is correct, please make me Brainliest!

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3 years ago
Suppose we roll a fair six-sided die 20 times and draw ten cards from a standard 52-card deck. Let X be the number of "6"s rolle
Lera25 [3.4K]

Answer:

a) Expected value = 6.406

Variance = 4.905

Standard deviation = 2.45

b) The probability is 0.08547

Step-by-step explanation:

a) Let's suppose that:

X₁ = number of 6´s

X₂ = number of Jack, Queen, King or Aces

The mean of X₁ is:

MeanX₁ = n * p = 20 * (1/6) = 3.33

The variance of X₁ is:

Var-X_{1} =np(1-p)=3.33(1-(1/6))=2.775

The mean of X₂ is:

MeanX₂ = 10 * (16/52) = 3.076

The variance of X₂ is:

Var-X_{2} =3.076(1-(16/52))=2.13

The expect value of X is:

Xexp = MeanX₁ + MeanX₂ = 3.33 + 3.076 = 6.406

The variance of X is:

VarX = VarX₁ + VarX₂ = 2.775 + 2.13 = 4.905

The standard deviation is:

Xdevi = 4.905/2 = 2.45

b) The probability of drawing at least five six out of 20 rolls is equal to:

∑(1/6)ˣ(5/6)²⁰⁻ˣ = 0.231 with x = 5

The probability of at least 4 Jack, Queen, Kings or Aces is:

∑(16/52)ˣ(1-(16/52))¹⁰⁻ˣ = 0.37 with x = 4

The probability of given event is equal to:

P = 0.231 * 0.37 = 0.08547

5 0
3 years ago
The table shows a function. Is the function linear or nonlinear?<br> x y<br> 7 20<br> 8 12<br> 9 7
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Answer:

Non-linear because none of them repeat.

Step-by-step explanation:

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Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
grandymaker [24]
(2x+1)^{\cot x}=\exp\left(\ln(2x+1)^{\cot x}\right)=\exp\left(\cot x\ln(2x+1)\right)=\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)

where \exp(x)\equiv e^x.

By continuity of e^x, you have

\displaystyle\lim_{x\to0^+}\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)

As x\to0^+ in the numerator, you approach \ln1=0; in the denominator, you approach \tan0=0. So you have an indeterminate form \dfrac00. Provided the limit indeed exists, L'Hopital's rule can be used.

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Now the numerator approaches \dfrac21=2, while the denominator approaches \sec^20=1, suggesting the limit above is 2. This means

\displaystyle\lim_{x\to0^+}(2x+1)^{\cot x}=\exp(2)=e^2
7 0
2 years ago
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