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bearhunter [10]
3 years ago
9

Richard and teo have a combined age of 42. Richard is 9 years older that twice teos age. How old are Richard and Teo

Mathematics
1 answer:
satela [25.4K]3 years ago
6 0

Step-by-step explanation:

Richard and teo age is 51 years old

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Which of the following is not the same as positive 7?
bekas [8.4K]
Seven units to the left of the 0 on a number line would = -7 thus answer 3 being the correct answer
3 0
3 years ago
Read 2 more answers
3/8 x 5 as a mixed number​
leva [86]

Answer:

1 7/8

Step-by-step explanation:

3/8 * 5/1 = 15/8

<-- First multiply 3/8 by 5

----------------------------------------------

15/8 = 1 with a remainder of 7

<-- Then convert 15/8 to a mixed number by dividing 15 by 8

15/8 = 1 7/8

8 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
Noah wants to put $1,000 in a savings account with a 1.5% annual interest rate. How much more money will he have after one year
larisa86 [58]
Assuming that 1.5% annual interest is converted to monthly basis with the same amount, then the monthly interest should be: 1.5%/12= 0.125%.

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If you put $1000 for monthly interest, the saving account would become: ($1000*(100%+0.125%)= $1000*1.0151035559= $1015.10

Then, the money difference should be:  $1015.10-$1015= $0.10
6 0
3 years ago
angelo rode his bike around a bike trail that was 1/4 of a mile long.he rode his bike around thetrail 8 times.Angelo says he rod
forsale [732]
Hi there!

From this problem we can pull out some key information and remove any excess.

- The trail is 1/4 of a mile long.
- Angelo rode around the trail 8 times.
- Angelo claims he rode 8/4 miles.
- Teresa claims he rode 2 miles.

From that we know that if Angelo rode around the trail 8 times and the trail is 1/4 a mile, he in total rode 8/4 of a mile.

However, Teresa claims he rode 2 miles!

And the answer is... they're both correct!

Angelo did indeed ride 8/4 miles however that is equivalent to 2 miles so they are both in the right.
7 0
4 years ago
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