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KiRa [710]
1 year ago
14

1.) Your 3 year investment of $20,000 received 5.2% interested compounded semi annually. What is your total return? ASW

Mathematics
1 answer:
Stella [2.4K]1 year ago
4 0

Let's begin by listing out the information given to us:

Principal (p) = $20,000

Interest rate (r) = 5.2% = 0.052

Number of compounding (n) = 2 (semi annually)

Time (t) = 3 years

The total return is calculated as shown below:

A = p(1 + r/n)^nt

A = 20000(1 + 0.052/2)^2*3 = 20000(1 + 0.026)^6

A = 20000(1.1665) = 23,330

A = $23,330

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Steve travels 5 times as fast as Kevin. Traveling in opposite directions, they are 390 miles apart after 5 hours. Find their rat
Kaylis [27]

Answer:

steve=65 mph

kevin=13 mph

Step-by-step explanation:

steve=s

kevin=k

s=5

k=1

1=5=6

390 divided by 6 equals s 65

divide further by 5 which equals k 13

3 0
2 years ago
we believe that 42% of freshmen do not visit their counselors regularly. For this year, you would like to obtain a new sample to
Maksim231197 [3]

Answer:

A sample of 1077 is required.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

42% of freshmen do not visit their counselors regularly.

This means that \pi = 0.42

98% confidence level

So \alpha = 0.02, z is the value of Z that has a pvalue of 1 - \frac{0.02}{2} = 0.99, so Z = 2.327.

How large of a sample size is required?

A sample size of n is required, and n is found when M = 0.035. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.035 = 2.327\sqrt{\frac{0.42*0.58}{n}}

0.035\sqrt{n} = 2.327\sqrt{0.42*0.58}

\sqrt{n} = \frac{2.327\sqrt{0.42*0.58}}{0.035}

(\sqrt{n})^2 = (\frac{2.327\sqrt{0.42*0.58}}{0.035})^2

n = 1076.8

Rounding up:

A sample of 1077 is required.

3 0
3 years ago
Use x = 3 to identify the value of each expression.<br><br> Exponent Value of expression
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5 0
3 years ago
The principal of the school is looking over grade reports, specifically scores on math quizzes.
guapka [62]

Answer:

Range = 45

Median = 75

Step-by-step explanation:

Based on the attached box and whisker plot :

From the plot :

Maximum value = 95

Minimum value = 50

The median value can be obtained from the box and whisker plot. The median value from the plot is ; the green dotted point inside the box :

This point is marked 75.

Hence, median value = 75

Range = maximum - minimum

Range =. 95 - 50 = 45

8 0
3 years ago
A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th
Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let S(t) be the amount of sugar in the tank at time t. Then S(0)=25.

Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of

\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}

Separate variables, integrate, and solve for <em>S</em>.

\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt

\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt

-100\ln\left|P-\dfrac S{100}\right|=t+C

\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t

P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}

\dfrac S{100}=P-Ce^{-100t}

S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}

Use the initial value to solve for <em>C</em> :

S(0)=25\implies 25=100P-C\implies C=100P-25

\implies S(t)=100P-(100P-25)e^{-100t}

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick <em>P</em> such that

S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}

5 0
3 years ago
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