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3 years ago

PLEASE HELP ASAP!!!

Mathematics

1 answer:

Dafna1 [17]3 years ago

6 0

Answer:

Step-by-step explanation:

Any time you have compounding more than once a year (which is annually), unless we are talking about compounding continuously, you will use the formula

$A(t)=P(1+\frac{r}{n})^{(n)(t)}$

Here's what we have:

The amount after a certain time that she has in the bank is 4672.12; that's A(t).

The interest rate in decimal form is .18; that's r.

The number of times the interest compounds is 12; that's n

and the time that the money is invested is 3.5 years; that's t.

Filling all that into the formula:

$4672.12=P(1+\frac{.18}{12})^{(12)(3.5)}$ Simplifying it down a bit:

$4672.12=P(1.015)^{42}$ Raise 1.015 to the 42nd power to get

4672.12 = P(1.868847115) and divide to get P alone:

P = 2500.00

She invested $2500.00 initially.

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The heights of baby giraffe are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 100 b

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Answer:

$P(\bar X$

And we can solve this using the following z score formula:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we use this formula we got:

$z = \frac{63-63.6}{\frac{2.5}{\sqrt{100}}}= -2.4$

So we can find this probability equivalently like this:

$P( Z$

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 heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(63.6,2.5)$

Where $\mu=63.6$ and $\sigma=2.5$

We select n =100. Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

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

We want this probability:

$P(\bar X$

And we can solve this using the following z score formula:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we use this formula we got:

$z = \frac{63-63.6}{\frac{2.5}{\sqrt{100}}}= -2.4$

So we can find this probability equivalently like this:

$P( Z$

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All mental math no work needed.

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