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3 years ago

Subtract (3m^3-6)-(-2m^3+m^2+3)

1 answer:

5 0

The answer you are looking for is: 
5 $m^{3}$ - $m^{2}$ -9 
Hope that helps!! 
Have a wonderful day!!

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Find the quotient 7/9 divided by 5/6

Blizzard [7]

7/9 divided by 5/6 = 14/15
in decimal for it would be 0.93

3 0

3 years ago

Indicate in standard form the equation of the line passing through the given points G (4,6) H (1,5)

Delicious77 [7]

First, find the slope (m) = $\frac{y2 - y1}{x2 - x1}$ = $\frac{5 - 6}{1 - 4}$ = $\frac{-1}{-3}$ = $\frac{1}{3}$

Now plug in ONE of the points and the slope into the point-slope equation:

y - y₁ = m(x - x₁); where (x₁, y₁) is the chosen point.

y - 5 = $\frac{1}{3}$ (x - 1) (I used (1,5) as the chosen point)

3(y - 5) = x - 1

3y - 15 = x - 1

3y -14 = x

-14 = x - 3y → x - 3y = -14

Answer: x - 3y = -14

4 0

3 years ago

A farmer uses 1/3 of his land to plant cassava, 2/5 of the remaining and to plant maize and the rest vegetables.if vegetables co

zimovet [89]

The answer is 75 because if u multiply 75 by 1/3 you get 25 then 25 times 2/5 u get 10 75 is your answer

7 0

3 years ago

PLEASE HELP FAST The Booster Club raised $30,000 for a sports fund. No more money will be placed into the fund. Each year the fu

zloy xaker [14]

Answer:

c. $24,435

Step-by-step explanation:

$30000(1 - .05)^{4} = 24,435.19$

3 0

3 years ago

A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator.

AVprozaik [17]

Answer:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean is also normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean". The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z$

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is also normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

8 0

3 years ago