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r-ruslan [8.4K]
3 years ago
12

Ms.arch is paid $1250 per week but is fined $100 each day she is late to work. Ms.arch wants to make at least $3,000 over the ne

xt three weeks so she can take a vacation. Over the next three weeks, what is the maximum number of days she can be late to work and still reach her goal of making at least $3,000?
Mathematics
1 answer:
Hatshy [7]3 years ago
6 0
She Can Miss No More Than 7 Days To Make 3000
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What is the radius and diameter of the following circle?
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1. Volume of a cone: 1. V = (1/3)πr2h 2. Slant height of a cone: 1. s = √(r2 + h2) 3. Lateral surface area of a cone: 1. L = πrs = πr√(r2 + h2) 4. Base surface area of a cone (a circle): 1. B = πr2 5. Total surface area of a cone: 1. A = L + B = πrs + πr2 = πr(s + r) = πr(r + √(r2 + h2))

Step-by-step explanation:

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maya wants to burs 600 calories, so far, she burned 584.3 calories. how many more calories must maya burn
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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

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3 years ago
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