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ANTONII [103]
3 years ago
7

For breakfast, Sarah can choose eggs, granola or oatmeal as a main course, and orange juice or milk for a drink. Sarah says that

the sample space for choosing one of each contains 32 = 9 outcomes. What is her error? Explain. Because there are choices for the first item and for the second, there are possible outcomes.
Mathematics
1 answer:
svlad2 [7]3 years ago
4 0

Answer:

The explanation is given below

Step-by-step explanation:

Her error is that she multiplied the possibilities of her meal that involves eggs with milk, granola & oatmeal but the possibilities involves orange juice with egg and granola & oatmeal

Now

If she choose milk so she can select eggs, granola, and oatmeal so there are 3 possibilities also with the orange juice there is 3 possibilities

So, the sample space is

= 3 + 3

= 6 outcomes  

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jocelyn bought 2 sweaters for the same price.she paid $23.56, including sales tax of $1.36 and a $5.00 discount coupon.what was
k0ka [10]
<span>Jocelyn bought 2 sweaters for the same price.
She paid $23.56, including sales tax of $1.36 and a $5.00 discount coupon.
Problem: Find the price of 1 sweater before the tax and coupon
=> 2 sweaters = 23.56 dollars included tax and discount
=> remove the tax and discount first
=> 23.56 + 5.00 = 28.56 dollars since this is a discount, we need to add this to the amount
=> 28.56 – 1.36 =  27.2 dollars is the price of 2 sweaters
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4 0
3 years ago
Find the limit
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]

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\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}

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Answer:

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Step-by-step explanation:

<u>Equation of the line in intercept form</u>

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Where:

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