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Rina8888 [55]
3 years ago
10

For the month of October Lisa plan to go running every fifth day and walk every sixth day. If she begins on October 1, on what d

ay of the month will she run and walk during the same day?
Mathematics
1 answer:
Tomtit [17]3 years ago
4 0

Answer:

October 30 is the day of the month she will run and walk during the same day.

Step-by-step explanation:

To determine what day of the month will she run and walk during the same day, we will list the days she plans to go running and the days she plans to walk.

From the question,  Lisa plan to go running every fifth day and walk every sixth day, then

If she begins on October 1, the days she will run will be

October 5, October 10, October 15, October 20, October 25, and October 30;

while the days she will walk will be

October 6, October 12, October 18, October 24, and October 30.

Since October 30 occurred among the days she will run and walk, then October 30 is the day of the month she will run and walk during the same day.

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Jina wanted to study how the area of a rectangle changes with the length if it’s width is fixed. She computed the areas of sever
olganol [36]

Answer:

The domain and the range of the function are, respectively:

Dom\{f\} = [0\,m,5\,m]

Ran\{f\} = [0\,m^{2}, 10\,m^{2}]

Step-by-step explanation:

Jina represented a function by a graphic approach, where the length, measured in meters, is the domain of the function, whereas the area, measured in square meters, is its range.

Dom\{f\} = [0\,m,5\,m]

Ran\{f\} = [0\,m^{2}, 10\,m^{2}]

8 0
3 years ago
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
Yesterday, Julie’s bank account balance was –$10.28. She deposited some money into her account so that her balance is now $39.89
Tcecarenko [31]
10.28+39.89=50.17

hope this helps
5 0
3 years ago
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Mike put sports books on 1/3 of his bookshelf. He put history books on 1/4 of the bookshelf . What fraction of the bookshelf did
Shkiper50 [21]

Answer:

66.67% or 2/3

Step-by-step explanation:

He put 1/3 of his sports books on the shelf. Ignore the 1/4 and focus on how much space the sports books are not taking up which would be 1-1/3=2/3 or 66.67%.

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3 years ago
Which number would you divide the numerator and the denominator of the first fraction to yield the second fraction?
blondinia [14]

Answer:

the number I would to divide the numerator and denominator is 5/5

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3 years ago
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