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Katen [24]
2 years ago
14

Determine the intercepts of the line 2x+5y=-6?

Mathematics
1 answer:
SVETLANKA909090 [29]2 years ago
7 0

Answer:

The x intercept is (-3,0)

The y intercept is (0,-6/5)

Step-by-step explanation:

To find the x intercept, set y =0 and solve for x

2x+5y=-6

2x+0 = -6

Divide by 2

2x/2 = -6/2

x = -3

The x intercept is (-3,0)

To find the y intercept, set x =0 and solve for y

2x+5y=-6

0+5y = -6

Divide by 5

5y/5 = -6/5

y = -6/5

The y intercept is (0,-6/5)

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Simone takes dance lessons. Her shoes cost $150 per year, her tights and leotards cost $70 per year, her hair accessories cost $
KATRIN_1 [288]
Ok so all you do is add 150+70+15+1000. That equals to $1235
6 0
3 years ago
Total profit P is defined as total revenue R minus total cost​ C, and is given by the function P(x)=R(x)−C(x). Find the total pr
satela [25.4K]

Given:

The revenue function is

R(x)=126.98x-0.5x^2

The cost function is

C(x)=3809.40+0.7x^2

To find:

The profit function.

Solution:

We know that, the total profit P is defined as total revenue R minus total cost​ C.

So, the profit function is

P(x)=R(x)-C(x)

P(x)=126.98x-0.5x^2-(3809.40+0.7x^2)

P(x)=126.98x-0.5x^2-3809.40-0.7x^2

P(x)=-3682.42-1.2x^2

Therefore, the profit function is P(x)=-3682.42-1.2x^2.

5 0
3 years ago
Andre says 10x + 6 and 5x + 11 are equivalent because they both equal 16 when x is 1. Do you agree with Andre?​
Serga [27]

Answer:

No, They are not equivalent

Step-by-step explanation:

The two expressions are only said to be equivalent if they are the same regardless of whatever value the variable(s) is.

The variable in this question is x.

To prove this, we equate both expressions:

10x + 6 = 5x + 11

10x - 5x = 11 - 6\\\\\\5x = 5\\\\\\x = 1

This shows that they have the same value ONLY when x is 1.

This is why Andre says that they are equivalent, but they are not.

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.

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natima [27]

Answer:

1000000000000000000000

Step-by-step explanation:

just multiply it ;-; or add 2 zeros in 1,0000000000000000000

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