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3 years ago

Simplify:

Mathematics

1 answer:

stepan [7]3 years ago

3 0

Answer:

simplify question is 3a^2 +6a+9ab

Step-by-step explanation:

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Find the absolute maximum and minimum values of f(x.y)=x^2+y^2-2x-2y on the closed region bounded by the triangle with vertices

dem82 [27]

Try this suggested solution, note, 'D' means the region bounded by the triangle according to the condition. It consists of 6 steps.

Answers are underlined with red colour.

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3 years ago

Solve 6x - 3y = 12 for y A. y = 6x + 4 B. y = 2x - 4 C. y = -6x + 12 D. y = 2x - 12

DedPeter [7]

Answer: y = 2x − 4

The correct answer is B) y = 2x − 4

Step-by-step explanation:

Move all terms that don't contain y to the right side and solve:

y = −4 + 2x

Then reorder −4 and 2x:

y = 2x − 4

Final answer is:

B) y = 2x − 4

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2 years ago

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She has 30 children........

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3 years ago

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Definition of radius

Lina20 [59]

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2 years ago

A data set includes data from student evaluations of courses. The summary statistics are nequals83, x overbarequals3.32, sequa

Andrews [41]

Answer:

Hypothesis

Null hypothesis: $\mu =3.50$

Alternative hypothesis: $\mu \neq 3.50$

Statistic

$t=\frac{3.32-3.50}{\frac{0.56}{\sqrt{83}}}=-2.928$

P value

The degrees of freedom are given by:

$df=n-1=83-1=82$

And the p value would be given by:

$p_v =2*P(t_{(82)}$

Conclusion

Since the p value is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 3.5

Step-by-step explanation:

Information given

$\bar X=3.32$ represent the sample mean

$s=0.56$ represent the sample deviation

$n=83$ sample size

$\mu_o =3.5$ represent the value to test

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the true mean is equal to 3.5, the system of hypothesis are:

Null hypothesis: $\mu =3.50$

Alternative hypothesis: $\mu \neq 3.50$

Statistic

The statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing we got:

$t=\frac{3.32-3.50}{\frac{0.56}{\sqrt{83}}}=-2.928$

P value

The degrees of freedom are given by:

$df=n-1=83-1=82$

And the p value would be given by:

$p_v =2*P(t_{(82)}$

Conclusion

Since the p value is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 3.5

8 0

2 years ago