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3 years ago

Your company is hosting a benefit 5 kilometer race. How long will the race be to the nearest tenth of a mile?

2 answers:

7 0

1km=0.62 miles

so multiply 1km by 5 to get 5km
multiply 0.62 by 5 to get km to miles
0.62 times 5=3.1

answe ris 3.1 miles

julia-pushkina [17]3 years ago

6 0

Well, OK. My company never hosted anything, certainly
nothing with any benefit, so this scenario doesn't relate
very well to the real world. But here's the math:

(5 km) x (1 mile / 1.609344 km) = 3.106... mile

Rounded to the nearest tenth of a mile:

3.106... mile ===> 3.1 mile

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3 years ago

Helen got a 7% reduction in the price of pair jeans that are normally $32. What is the approximate percent decrease in the price

Colt1911 [192]

A percentage is a way to describe a part of a whole. The approximate percent decrease in the price is 21.87%, thus, the correct option is B.

<h3>What are Percentages?</h3>

A percentage is a way to describe a part of a whole. such as the fraction ¼ can be described as 0.25 which is equal to 25%.

To convert a fraction to a percentage, convert the fraction to decimal form and then multiply by 100 with the '%' symbol.

Given the reduction in the price is $7, while the original cost is $32. Therefore, the approximate percent decrease in the price is,

Percentage decrease = 7/32 × 100%

= 21.87%

Hence, the approximate percent decrease in the price is 21.87%, thus, the correct option is B.

The correct question is:

Helen got a $7 reduction in the price of a pair of jeans that are normally $32. What is the approximate percent decrease in the price?

The percent decrease is approximately 15%.

The percent decrease is approximately 21%.

The percent decrease is approximately 43%.

The percent decrease is approximately 78%

Learn more about Percentages:

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3 0

2 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

What is the solution of -x+4>2

Svetradugi [14.3K]

There is no single solution but there is a group of solutions also known as the interval.

$-x+4>2 \\-x>-2 \\x$