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Mariulka [41]
2 years ago
11

Matthew invested $5000 in an account that earns 3.8% interest, compounded annually. The formula for compound interest is A(t) =

P(1 + i)t. How much did Matthew have in the account after 3 years?
A. $6900.00
B. $5594.37
C. $5570.00
D. $5591.93
Mathematics
2 answers:
Burka [1]2 years ago
6 0

Yup D Is Right Because if you divide by 3.8% * 5000 you get d

sweet [91]2 years ago
4 0

Answer:

D. $5591.93.

Step-by-step explanation:

We have been given that Matthew invested $5000 in an account that earns 3.8% interest, compounded annually.

We will use compound interest formula to solve our given problem.

A=P(1+\frac{r}{n})^{nt}, where,

A = Final amount after t years,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

Let us convert our given interest rate in decimal form.

3.8\%=\frac{3.8}{100}=0.038

Upon substituting our given values in above formula we will get,

A=\$5000(1+\frac{0.038}{1})^{1*3}

A=\$5000(1+0.038)^{3}

A=\$5000(1.038)^{3}

A=\$5000*1.118386872

A=\$5591.93436\approx \$5591.93

Therefore, Matthew will have an amount of $5591.93 is his account after 3 years and option D is the correct choice.

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Andre45 [30]
The sum of two consecutive even integers is a+(a+2) and divided by four is
(a+(a+2))/4 = 189.5
(2a+2)/4 = 189.5
2a+2 = 189.5 * 4
2a+2 = 758
2a = 758 - 2
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a = 756/2 = 378
first number is a = 378
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1 year ago
For Joe's illness, the doctor gave him a prescription of 28 pills. He needs to take 4 pills the first day, and then 2 pills each
kiruha [24]

Answer:

12 days.

Step-by-step explanation:

I looked at it this way:

He has to take 4/28 pills the first day, so it's 28-4=24. Then I divided 24 by 2.

In simpler terms:

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The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes"† proposed using the Poisson distribution
oksian1 [2.3K]

Answer:

a. P(X ≤ 5) = 0.999

b. P(X > λ+λ) = P(X > 2) = 0.080

Step-by-step explanation:

We model this randome variable with a Poisson distribution, with parameter λ=1.

We have to calculate, using this distribution, P(X ≤ 5).

The probability of k pipeline failures can be calculated with the following equation:

P(k)=\lambda^{k} \cdot e^{-\lambda}/k!=1^{k} \cdot e^{-1}/k!=e^{-1}/k!

Then, we can calculate P(X ≤ 5) as:

P(X\leq5)=P(0)+P(1)+P(2)+P(4)+P(5)\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\P(3)=1^{3} \cdot e^{-1}/3!=1*0.3679/6=0.061\\\\P(4)=1^{4} \cdot e^{-1}/4!=1*0.3679/24=0.015\\\\P(5)=1^{5} \cdot e^{-1}/5!=1*0.3679/120=0.003\\\\\\P(X\leq5)=0.368+0.368+0.184+0.061+0.015+0.003=0.999

The standard deviation of the Poisson deistribution is equal to its parameter λ=1, so the probability that X exceeds its mean value by more than one standard deviation (X>1+1=2) can be calculated as:

P(X>2)=1-(P(0)+P(1)+P(2))\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\\\P(X>2)=1-(0.368+0.368+0.184)=1-0.920=0.080

4 0
1 year ago
Stefanie is planning a 610-mile trip. Her car’s EPA rating is 34 mpg on the highway. How many gallons of gas will she require fo
agasfer [191]
It going to be 79 mile 
3 0
2 years ago
Every year the United States Department of Transportation publishes reports on the number of alcohol related and non-alcohol rel
Damm [24]

Complete question:

The line graph relating to the question was not attached. However, the line graph has can be found in the attachment below.

Answer:

17,209

Step-by-step explanation:

The line graph provides information about alcohol-related highway fatalities between year 2001 to 2010.

Determine the average number of alcohol-related fatalities from 2001 to 2006. Round to the nearest whole number.

The average number of alcohol related fatalities between 2001 - 2006 can be calculated thus :

From the graph:

Year - - - - - - - - - - Number of fatalities

2001 - - - - - - - - - - 17401

2002 - - - - - - - - -  17525

2003 - - - - - - - - -  17013

2004 - - - - - - - - - 16694

2005 - - - - - - - - - 16885

2006 - - - - - - - - - 17738

To get the average :

Sum of fatalities / number of years

(17401 + 17525 + 17013 + 16694 + 16885 + 17738) / 6

= 103256 / 6

= 17209.333

Average number of alcohol related fatalities is 17,209 (to the nearest whole number)

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2 years ago
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