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Sidana [21]
3 years ago
15

I need help with this!

Mathematics
1 answer:
Viefleur [7K]3 years ago
4 0

Answer:

the answer is option E.

Step-by-step explanation:

it is in the form f/g (x).

we know f and g from the equation;

we can rewrite it as, (√(9-x^2)/(3x-1))

if you notice you cannot put any number less than -3 or greater than 3 in the numeror because if you do you get a negative root which is false. for instance if you put 4 or -4 in the numerator you get 9 - (4 or -4 square ) which is 9- 16 which is a negative number and you cannot take root of a negative number.

on the numerator if you put 1/3 as the value for x you will get zero in the denominator. and any number divided by zero is undefined so that cannot be.

this means that option E is the right one that satisfies the condition. it means the domain is [-3, 1/3) U (1/3, 3] .

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We need to convert this equation to slope-intercept form first.

We can do that by solving for y.

x - 5y = 15

<em><u>Add 5y to both sides.</u></em>

x = 5y + 15

<em><u>Subtract 15 from both sides.</u></em>

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<em><u>Divide both sides by 5.</u></em>

y = 1/5x - 3

We now know the slope is 1/5.

The slope of the line perpendicular to the line with a slope of 1/5 is -5.

The slope of a perpendicular line is the negative reciprocal of the original slope.

Using a graphing calculator, we know the y-intercept of the line that is perpendicular to the original line must have a y-intercept of -6 to run through the points (-2, 5).

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Answer: m=-1

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  2m + 2  =   2 • (m + 1)

2.1      Solve :    2   =  0

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Answer:

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Step-by-step explanation:

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Answer:

(a) 0.14%

(b) 2.28%

(c) 48%

(d) 68%

(e) 34%

(f) 50%

Step-by-step explanation:

Let <em>X</em> be a random variable representing the prices paid for a particular model of HD television.

It is provided that <em>X</em> follows a normal distribution with mean, <em>μ</em> = $1600 and standard deviation, <em>σ</em> = $100.

(a)

Compute the probability of buyers who paid more than $1900 as follows:

P(X>1900)=P(\frac{X-\mu}{\sigma}>\frac{1900-1600}{100})

                   =P(Z>3)\\=1-P(Z

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid more than $1900 is 0.14%.

(b)

Compute the probability of buyers who paid less than $1400 as follows:

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*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid less than $1400 is 2.28%.

(c)

Compute the probability of buyers who paid between $1400 and $1600 as follows:

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*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1400 and $1600 is 48%.

(d)

Compute the probability of buyers who paid between $1500 and $1700 as follows:

P(1500

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*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1500 and $1700 is 68%.

(e)

Compute the probability of buyers who paid between $1600 and $1700 as follows:

P(1600

                              =P(0

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1700 is 34%.

(f)

Compute the probability of buyers who paid between $1600 and $1900 as follows:

P(1600

                              =P(0

*Use a <em>z</em>-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1900 is 50%.

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Answer: 20/33

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Let E be the event,

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