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allsm [11]
3 years ago
8

Let f(x) = (x^2+3x-5)/(x^2-8x+15) What is the domain of f(x) ?

Mathematics
1 answer:
Lyrx [107]3 years ago
8 0

Answer:

So, Domain = {set of all the real numbers except 3, 5}

Interval notation = {x ∈ set of all real numbers/ x ≠3 and x ≠ 5}

Step-by-step explanation:

The given function is f(x) = \frac{x^2 + 3x - 5}{x^2 - 8x + 15}

Here we need to find the domain.

To find the domain, let's find the restricted domain.

If we set the denominator equal to zero and solve for x, we can find the restricted domain.

Here the denominator is x^2 -8x + 15 = 0

Now we can factorize and solve for x.

(x - 3)(x - 5) = 0

(x - 3) or (x - 5) = 0

x -3 = 0  and x - 5 = 0

Solving the above equations, we get

x = 3 and x = 5

The restricted domains are x = 3 and x = 5.

Now we can find the domain.

Except 3 and 5, all the real numbers can be domain of this function.

So, Domain = {set of all the real numbers except 3, 5}

= {x ∈ set of all real numbers/ x ≠3 and x ≠ 5}.

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The population of Portland Oregon has recently grown to more than 500,000. What is the order of magnitude for the number of peop
miss Akunina [59]

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Order is 6

Step-by-step explanation:

The population of Portland Oregon has recently grown to more than 500,000.

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P=500,000

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5 0
3 years ago
At a shelter, 15% of the dogs are puppies their are 60 dogs
devlian [24]

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9 puppies

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6 0
3 years ago
Two coworkers commute from the same building. They are interested in whether or not there is any variation in the time it takes
jenyasd209 [6]

Answer:

Step-by-step explanation:

Hello!

You have two samples.

Sample 1 (time that takes one commute of worker 1)

n₁= 20

S₁²= 12.2

Sample 2 (time that takes one commute of worker 2)

n₂= 20

S₂²= 16.9

The hypothesis is that the first coworker is more consistent with his commute times, this means that the variability of his commute times is less than the variability of the commute times of worker 2. With this in mind, the hypothesis for a variance ratio test is:

H₀: σ₁² ≥ σ₂²

H₁:: σ₁² < σ₂²

α: 0.10

The statistic for this test is:

F= (S₁²/S₂²) * (σ₁²/σ₂²) ~ F_{n1-1;n2-1}

The test is one-tailed (left) with just one critical value:

F_{n_1-1;n_2-1;\alpha } = \frac{1}{F_{n_2-1;n_1-1;1-\alpha }} = \frac{1}{F_{19;19;0.90}} = \frac{1}{1.82} =0.549

(The F-table I use only has the values for high probabilities so I've used a conversion method to obtain the value for the small probability)

Your decision rule is:

If F ≤ 0.549, you reject the null hypothesis.

if F > 0.549, you support the null hypothesis.

The calculated value:

F= (S₁²/S₂²) * (σ₁²/σ₂²) = (12.2/16.9) * 1= 0.722

Since the calculated F-value is greater than the critical value, the decision is to not reject the null hypothesis. So you can say that the variability of the commute times of worker 1 is less than the variability of the commute times of worker 2.

I hope you have a SUPER day!

7 0
3 years ago
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