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Norma-Jean [14]

3 years ago

9

Use experimental probability to make a prediction. John made 35 free throws and missed 15. Predict the number of free throws he

will make in the next 100 tries.
A.45

B.57

C.85

D.55

PLEASE HELP!!!!!!!!!

1 answer:

ss7ja [257]3 years ago

8 0

If you add 35 + 15 that mean in all he shot 50 times so maybe if he does it again .. blah ion feel like explaining i think its D.55

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Write each polynomial function in standard form , classify it by degree and determine the end behavior of its graph Y=12-x^4

Dmitry_Shevchenko [17]

In standard form, zero is always on the right:
y = 12 - x^4
y + x^4 - 12 = 0

End behavior:
As x approaches ∞, y approaches -∞
As x approaches -∞, t approaches -∞

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2 years ago

Which equation describes this line (1,10) (-3,2)

zubka84 [21]

Find the slope first:
m=y2-y-1/x2-x1

m=2-10/-3-1
m= -8/-4
m= 2

Select a point & put into y=mx+b to find b.
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2 years ago

What is the value of m?

Dima020 [189]

$\huge \underline \mathcal{Answer}$

The given angles forms linear pair, and we know the angles forming linear pair are supplementary,

Therefore,

Angle MHJ + Angle MHL = 180°

Let's solve :

$(5m + 100) \degree + (2 m + 10) \degree = 180 \degree$

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2 years ago

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

2 years ago

Read 2 more answers

Use the inequality to answer the following question.

katrin2010 [14]

Answer:

<h3>45,50,60</h3>

Step-by-step explanation:

<h3>pa correct nlng ty sa points.</h3>

3 0

2 years ago