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3 years ago

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Jillian put $1200 into an account earning 6.5% interest compounded continuously. How long until she doubles her money?

1 answer:

VLD [36.1K]3 years ago

7 0

Answer:

$A= 2400\\P= 1200\\r = 6.5\% = 0.065\\n = 365 \\t = t\\A = P(1 + \frac{r}{n})^{nt}\\\\2400 = 1200(1 +\frac{0.065}{365} )^{365t}\\\\\frac{2400}{1200} = (1.00017808219)^{365t}\\\\(2)^{\frac{1}{365} } = (1.00017808219)^t\\\\\\Taking \ log \ on \ both \ sides\\log(1.00190083768) = t \times log(1.00017808219)\\\\t= \frac{log(1.00190083768)}{log(1.00017808219)} = 10.66\\\\t = 10.7years$

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Clayton is at most 2 meters above sea level. Using C for Clayton, which inequality correctly represents this statement?

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3 years ago

The playing time X of jazz CDs has the normal distribution with mean 52 and standard deviation 7; N(52, 7).

natita [175]

Answer:

68% of jazz CDs play between 45 and 59 minutes.

Step-by-step explanation:

<u>The correct question is:</u> The playing time X of jazz CDs has the normal distribution with mean 52 and standard deviation 7; N(52, 7).

According to the 68-95-99.7 rule, what percentage of jazz CDs play between 45 and 59 minutes?

Let X = <u>playing time of jazz CDs</u>

SO, X ~ Normal( $\mu=52, \sigma^{2} =7$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

Now, according to the 68-95-99.7 rule, it is stated that;

68% of the data values lie within one standard deviation points from the mean.

95% of the data values lie within two standard deviation points from the mean.

99.7% of the data values lie within three standard deviation points from the mean.

Here, we have to find the percentage of jazz CDs play between 45 and 59 minutes;

For 45 minutes, z-score is = $\frac{45-52}{7}$ = -1

For 59 minutes, z-score is = $\frac{59-52}{7}$ = 1

This means that our data values lie within 1 standard deviation points, so it is stated that 68% of jazz CDs play between 45 and 59 minutes.

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