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2 years ago

If y = 124 and x = 12, find y when x = -24.

Mathematics

1 answer:

tatyana61 [14]2 years ago

7 0

Answer:

y = -248

Step-by-step explanation:

Hi there!

Set up a proportion:

$\displaystyle\frac{124}{12} =\frac{y}{-24}$

Cross-multiply:

$12y=(-24)(124)\\12y=-2976\\y=-248$

I hope this helps!

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Answer:

Step-by-step explanation:

20x2.5=50

51.79-50=1.79

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2 years ago

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Nathan has just bought a new car. He models the value, V, in dollars, of the car after t years as V(t) = 21,000(0.861)t. Based o

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Answer:

13.9%

Step-by-step explanation:

The value of car is modeled as:

$V(t)=21,000(0.861)t$

Here we can see that, each year Nathan has considered 0.861 of the previous year value or we can say that Nathan has considered 86.1% of the previous year value. So,

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The value of new car is decreasing by 13.9% each year.

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3 years ago

A forester studying the effects of fertilization on certain pine forests in the Southeast is interested in estimating the averag

ahrayia [7]

Answer:

$P(-2 \leq \bar X -\mu \leq 2)$

If we divide both sides by $\frac{\sigma}{\sqrt{n}}$ we got:

$P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})$

And we can use the normal distribution table or excel to find the probabilites and we got:

$P(-1.5 \leq Z \leq 1.5)= P(Z$ 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".

Solution to the problem

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

$X \sim N(M,4)$

Where $\mu=M$ and $\sigma=4$

We select a a sample of n =4 and since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

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

And we want to find this probability:

$P(-2 \leq \bar X -\mu \leq 2)$

If we divide both sides by $\frac{\sigma}{\sqrt{n}}$ we got:

$P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})$

And we can use the normal distribution table or excel to find the probabilites and we got:

$P(-1.5 \leq Z \leq 1.5)= P(Z$

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2 years ago

2) 1 - 3p+2p - 9<br> Your<br> answer<br> This is a required question

Pavlova-9 [17]

Answer:

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Step-by-step explanation:

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2 years ago

(a) Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the powe

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Answer:

a) $\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}$

b) See Below for proper explanation

Step-by-step explanation:

a) The objective here is to Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the power series expansion of the given function.

The function is $e^x + 3 \ cos \ x$

The expansion is of $e^x$ is $e^x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ...$

The expansion of cos x is $cos \ x = 1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...$

Therefore; $e^x + 3 \ cos \ x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ... 3[1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...]$

$e^x + 3 \ cos \ x = 4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} + \dfrac{x^3}{3!}+ ...$

Thus, the first three terms of the above series are:

$\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} ...}$

The series for $e^x + 3 \ cos \ x$ is $\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!} + 3 \sum \limits^{\infty}_{x=0} ( -1 )^x \dfrac{x^{2x}}{(2n)!}$

let consider the series; $\sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!}$

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Thus it converges for all value of x

Let also consider the series $\sum \limits^{\infty}_{x=0}(-1)^x\dfrac{x^{2n}}{(2n)!}$

It also converges for all values of x

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3 years ago