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Advocard [28]
3 years ago
14

How do i draw two number lines that show 0.20 and 1/5 are equivalent?

Mathematics
1 answer:
leonid [27]3 years ago
6 0
The frist line may be one with the numbers 0, 1 and 2 and 10 divisions (marks at equal distance) between each integer. Each division will be equal to 0.1 units and then you can mark the second division from the zero point to the right as the 0.20 mark.

The other line must have the same integers, 0 , 1 and 2 placed in identical form as the first line. Then
   - draw an inclined straight line since the point zero,
   - mark 5 points in the inclined lined equally spaced over the line.
   - draw a sttraight line from the 5th point to the point with the mar 1 over the base number line.
   -  draw a parallel line to the previous one passing trhough the second point of the inclined line and mark the point where this parallel touchs the base number line. This point shall be at the same distance from zero than the 0.2 mark was in the first number line, meaning that 0.2 and 1/5 are equivalent.
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Find a simplified expression to represent the area of the triangle. The area formula for a triangle is bh, where b is the base a
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Will improving customer service result in higher stock prices for the companies providing the better service? "When a company’s
Orlov [11]

Question:

Company                           2007 Score          2008 Score

Rite Aid                                73                          76

Expedia                                75                          77

J.C. Penney                          77                          78

a. For Rite Aid, is the increase in the satisfaction score from 2007 to 2008 statistically  significant? Use α= .05. What can you conclude?

b. Can you conclude that the 2008 score for Rite Aid is above the national average of  75.7? Use α= .05.

c. For Expedia, is the increase from 2007 to 2008 statistically significant? Use α= .05.

d. When conducting a hypothesis test with the values given for the standard deviation,

sample size, and α, how large must the increase from 2007 to 2008 be for it to be statistically  significant?

e. Use the result of part (d) to state whether the increase for J.C. Penney from 2007 to  2008 is statistically significant.

Answer:

a. There is sufficient statistical evidence to suggest that the increase in satisfaction score for Rite Aid from 2007 to 2008 is statistically significant

b. There is sufficient statistical evidence to suggest that the 2008 Rite Aid score, is above the national average of 75.7

c. The statistical evidence support the claim of a significant increase from 2007 to 2008

d. 1.802 and above is significant

e. The increase of J. C. Penney from 2007 is not statistically significant.

Step-by-step explanation:

Here we have

n = 60

σ = 6

μ₁ = 73

μ₂ = 76

We put H₀ : μ₁ ≥ μ₂ and

Hₐ : μ₁ < μ₂

From which we have;

z=\frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}+\frac{\sigma _{2}^{2}}{n_{2}}}} = \frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{2\sigma_{}^{2} }{n_{}}}}}

Plugging in the values we have

z =  \frac{(73-76)}{\sqrt{\frac{2\times 6^{2} }{60_{}}}}} = -2.7386

The probability, P from z function computation gives;

P(Z < -2.7386) = 0.0031

Where we have P < α, we reject the null hypothesis meaning that there is sufficient statistical evidence to suggest that the increase in satisfaction score for Rite Aid from 2007 to 2008 is statistically significant

b. To test here, we have

H₀ : μ ≤ 75.7

Hₐ : μ > 75.7

The test statistic is given as follows;

z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}} = \frac{76-75.7 }{\frac{6 }{\sqrt{60}}} = 0.3873

Therefore, we have the probability, P given as the value for the function at z = 0.3873 that is we have;

P = P(Z > 0.3873) = P(Z < -0.3873) = 0.3493

Therefore, since P > α which is 0.05, we fail to reject the null hypothesis, that is there is sufficient statistical evidence to suggest that the 2008 Rite Aid score, is above the national average of 75.7

c. Here we put

Null hypothesis H₀ : μ₁ ≥ μ₂

Alternative hypothesis Hₐ : μ₁ < μ₂

The test statistic is given by the following equation;

z=\frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}+\frac{\sigma _{2}^{2}}{n_{2}}}} = \frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{2\sigma_{}^{2} }{n_{}}}}}

Plugging in the values we have

z =  \frac{(75-77)}{\sqrt{\frac{2\times 6^{2} }{60_{}}}}} = -1.8257

The probability, P from z function computation gives;

P(Z < -1.8257) = 0.03394

The statistical evidence support the claim of a significant increase

d. For statistical significance at 0.05 significant level, we have z = -1.644854

Therefore, from;

z=\frac{(\bar{x_{1}}-\bar{x_{2}})-(\mu_{1}-\mu _{2} )}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}. we have;

z \times \sqrt{\frac{\sigma_{1}^{2} }{n_{1}}+\frac{\sigma _{2}^{2}}{n_{2}}} + (\mu_{1}-\mu _{2} )}{} ={(\bar{x_{1}}-\bar{x_{2}})

Which gives

{(\bar{x_{1}}-\bar{x_{2}}) = z \times \sqrt{\frac{2\sigma_{}^{2} }{n_{}}}} + (\mu_{1}-\mu _{2} )}{}  = -1.644854 \times \sqrt{\frac{2\times 6_{}^{2} }{60_{}}}} + 0 = -1.802

Therefore an increase of 1.802 and above is significant

e. Based on the result of part d. we have for J.C. Penney from 2007 to 2008 an increase of 1  which is less than 1.802 at 5% significant level, is not significant.

5 0
3 years ago
V=1/3 pi r squared h what is h?
Sveta_85 [38]

Step-by-step explanation:

V = 1/3 pi r^2 h

root(V ÷ 1/3 pi) = h

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3 years ago
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