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juin [17]
3 years ago
10

If you play imvu send me 10k tyvm. my username is "shoinks"

Mathematics
1 answer:
bixtya [17]3 years ago
4 0

Answer:

lil shawty

Step-by-step explanation:

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35.4%

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(100/80) x 27.6

basically 1 of 80 is 1.25 and multiply 1.25 times the money jaylon spent sorry i am really bad at explaining only good at solving

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2 years ago
What is the difference? StartFraction x 5 Over x 2 EndFraction minus StartFraction x 1 Over x squared 2 x EndFraction StartFract
Troyanec [42]

The difference of the given fraction is expressed as \frac{x^2+4x-1}{x(x+2)}

Given the expression:

  • \frac{x+5}{x+2}-\frac{x+1}{x^2+2x} \\

Take the LCM of the fraction to have:

=\frac{x+5}{x+2}-\frac{x+1}{x(x+2)} \\=\frac{x(x+5)-(x+1)}{x(x+2)} \\=\frac{x^2+5x-x-1}{x(x+2)}\\=\frac{x^2+4x-1}{x(x+2)}

Hence the difference of the given fraction is expressed as \frac{x^2+4x-1}{x(x+2)}

Learn more on fractions here; brainly.com/question/78672

7 0
2 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
2 years ago
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