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2 years ago

Alexandra deposited $1,000 in

Mathematics

1 answer:

allochka39001 [22]2 years ago

4 0

$~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$1000\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ t=years\dotfill &5 \end{cases} \\\\\\ I = (1000)(0.035)(5)\implies I=175$

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Adrian's CD player can hold six disks at a time and shuffles all of the albums and their songs. If he has thirteen CD's, how man

Crank

Adrian can put a total of 1716 combinations in the player.

The formula to be used is as follows:

total combinations = n!/[r! *(n-r)!]
n = number of disks available = 13
r = number of disks that be held = 6

= 13! = (13• 12•11•10•9•8•7•6•5•4•3•2•1)/(6! • 7!)
=1716

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3 years ago

According to the National Vital Statistics, full-term babies' birth weights are Normally distributed with a mean of 7.5 pounds a

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Answer:

68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 7.5, \sigma = 1.1$

What is the probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

This is the pvalue of Z when X = 8.6 subtracted by the pvalue of Z when X = 6.4. So

X = 8.6

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8.6 - 7.5}{1.1}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 6.4

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6.4 - 7.5}{1.1}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

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3 years ago

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seraphim [82]

Answer:

Step-by-step explanation:

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