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Cerrena [4.2K]
3 years ago
13

Isaac invested $1,800 in an account paying an interest rate of 4.7% compounded continuously. Assuming no deposits or withdrawals

are made, how much money, to the nearest hundred dollars, would be in the account after 15 years?
Mathematics
1 answer:
valkas [14]3 years ago
7 0

Answer:

First, convert R as a percent to r as a decimal

r = R/100

r = 4.7/100

r = 0.047 rate per year,

Then solve for A,

              A = P\cdot e^{rt}\\\\A= 1800 \cdot (2.71828)^{(.047)(15)}, \ where \ e = 2.71828\\A = 3642.92  

The amount he receives after 15years $3,642.92

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Find a1 if Sn = 89,800, r = 3.4, and n = 10. Round to the nearest hundredth if necessary.
xxMikexx [17]

Answer:

a_1=1.04

Step-by-step explanation:

We have a geometric  sequence with:

Sn = 89,800, r = 3.4, and n = 10

Where

Sn is the sum of the sequence

r is the common ratio

a_1 is the first term in the sequence

n is the number of terms in the sequence

The formula to calculate the sum of a finite geometric sequence is:

S_n=\frac{a_1(1-r^n)}{1-r}

Then:

89,800=\frac{a_1(1-(3.4)^{10})}{1-3.4}

Now we solve for a_1

89,800(1-3.4)=a_1(1-(3.4)^{10})

a_1=\frac{89,800(1-3.4)}{1-(3.4)^{10}}\\\\a_1=1.04

4 0
3 years ago
Simplify the expression. x2/3 · x7/2
WITCHER [35]

Answer:

x^2/3 x7/2

=1/6 x^9

Step-by-step explanation:

try using mathpapa

6 0
3 years ago
Helpppppppppp plzzzzz
Ratling [72]

so do all the question like this types

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4 0
3 years ago
3x + 2y = 2
iragen [17]
A) (-2,4) hope this helps 
6 0
3 years ago
Suppose that the population mean for income is $50,000, while the population standard deviation is 25,000. If we select a random
Fudgin [204]

Answer:

Probability that the sample will have a mean that is greater than $52,000 is 0.0057.

Step-by-step explanation:

We are given that the population mean for income is $50,000, while the population standard deviation is 25,000.

We select a random sample of 1,000 people.

<em>Let </em>\bar X<em> = sample mean</em>

The z-score probability distribution for sample mean is given by;

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

where, \mu = population mean = $50,000

            \sigma = population standard deviation = $25,000

            n = sample of people = 1,000

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

So, probability that the sample will have a mean that is greater than $52,000 is given by = P(\bar X > $52,000)

  P(\bar X > $52,000) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{52,000-50,000}{\frac{25,000}{\sqrt{1,000} } } ) = P(Z > 2.53) = 1 - P(Z \leq 2.53)

                                                                    = 1 - 0.9943 = 0.0057

<em>Now, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.</em>

Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.

5 0
3 years ago
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