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luda_lava [24]
3 years ago
12

12v-38=⅛*(4v-3) I need this answered plz

Mathematics
2 answers:
lorasvet [3.4K]3 years ago
8 0
Answer:

12v-38=1/8

We move all terms to the left:

12v-38-(1/8)=0

We add all the numbers together, and all the variables

12v-38-(+1/8)=0

We get rid of parentheses

12v-38-1/8=0

We multiply all the terms by the denominator


12v*8-1-38*8=0

We add all the numbers together, and all the variables

12v*8-305=0

Wy multiply elements

96v-305=0

We move all terms containing v to the left, all other terms to the right

96v=305

v=305/96

v=3+17/96

(Hope that helps)
Mrrafil [7]3 years ago
7 0

Answer:

v = 3.2

Step-by-step explanation:

12v - 38 = (4v-3)/8

multiply each side by 8 to get:

96v - 304 = 4v - 3

92v - 304 = -3

92v = 301

v = 3.27

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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
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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

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