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Andrews [41]
3 years ago
14

Select the vertical asymptote(s) of the function

7B%28x-2%29%28x%2B6%29%7D" id="TexFormula1" title="f(x)=\frac{(x+6)(x-1)}{(x-2)(x+6)}" alt="f(x)=\frac{(x+6)(x-1)}{(x-2)(x+6)}" align="absmiddle" class="latex-formula">
A. x=2
B. x=-6
C. x=2, x=-6
D. x=-2
Mathematics
2 answers:
Bess [88]3 years ago
4 0

Answer:

As the described function, we want to find vertical asymtotes, we find the value of x so that the denominator is equal to 0.

Here, (x - 2)(x + 6) = 0

=> x = 2, x =-6

=> Option C is correct.

Hope this helps!

:)

Julli [10]3 years ago
4 0

Answer:

A. x = 2

Step-by-step explanation:

(x+6)(x-1) ÷ (x-2)(x+6)

Since x+6 is common to both, numerator and denominator, it will get cancelled out

And there will be a hole at x = -6

The simplified form would be

(x-1)/(x-2)

There will be a vertical asymptote at x = 2, because the denominator becomes zero at x = 2

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3 years ago
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It is desired to compare the hourly rate of an entry-level job in two fast-food chains. Eight locations for each chain are rando
notka56 [123]

Answer:

It can be concluded that at 5% significance level that there is no difference in the amount paid by chain A and chain B for the job under consideration

Step by Step Solution:

The given data are;

Chain A 4.25, 4.75, 3.80, 4.50, 3.90, 5.00, 4.00, 3.80

Chain B 4.60, 4.65, 3.85, 4.00, 4.80, 4.00, 4.50, 3.65

Using the functions of Microsoft Excel, we get;

The mean hourly rate for fast-food Chain A, \overline x_1 = 4.25

The standard deviation hourly rate for fast-food Chain A, s₁ = 0.457478

The mean hourly rate for fast-food Chain B, \overline x_2 = 4.25625

The standard deviation hourly rate for fast-food Chain B, s₂ = 0.429649

The significance level, α = 5%

The null hypothesis, H₀:  \overline x_1 = \overline x_2

The alternative hypothesis, Hₐ:  \overline x_1 ≠ \overline x_2

The pooled variance, S_p^2, is given as follows;

S_p^2 = \dfrac{s_1^2 \cdot (n_1 - 1) + s_2^2\cdot (n_2-1)}{(n_1 - 1)+ (n_2 -1)}

Therefore, we have;

S_p^2 = \dfrac{0.457478^2 \cdot (8 - 1) + 0.429649^2\cdot (8-1)}{(8 - 1)+ (8 -1)} \approx 0.19682

The test statistic is given as follows;

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{S_{p}^{2} \cdot \left(\dfrac{1 }{n_{1}}+\dfrac{1}{n_{2}}\right)}}

Therefore, we have;

t=\dfrac{(4.25-4.25625)}{\sqrt{0.19682 \times \left(\dfrac{1 }{8}+\dfrac{1}{8}\right)}} \approx -0.028176

The degrees of freedom, df = n₁ + n₂ - 2 = 8 + 8 - 2 = 14

At 5% significance level, the critical t = 2.145

Therefore, given that the absolute value of the test statistic is less than the critical 't', we fail to reject the null hypothesis and it can be concluded that at 5% significance level that chain A pays the same as chain B for the job under consideration

3 0
3 years ago
Castel and Kali are selling wrapping paper for a school fundraiser. Customers can buy rolls of plain wrapping paper and rolls of
True [87]

Answer:

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Step-by-step explanation:

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Kali: 9p + 10h = 213

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