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3 years ago

you deposit 200 into an account that pays 7% interest compounded quarterly. how much will you have in 5 years

1 answer:

8 0

First, convert R percent to r a decimal
r = R/100
r = 7%/100
r = 0.07 per year,

Then, solve our equation for A
A = P(1 + r/n)nt
A = 200.00(1 + 0.005833333/12)(12)(5)
A = $ 283.53

Summary:
The total amount accrued, principal plus interest,
from compound interest on an original principal of
$ 200.00 at a rate of 7% per year
compounded 12 times per year
over 5 years is $ 283.53.

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Joanna set a goal to drink more water daily. The number of ounces of water she drank each of the last seven days is shown below.

hammer [34]

Answer:

The median amount of water is higher when the eighth day's amount is included in the data.

Explanation:

Given the ounces of water drank is 60, 58, 64, 64, 68, 50, 57

Arranging in ascending order: 50, 57, 58, 60, 64, 64, 68

Calculating the median of the above data

Median (when n is odd) = $\left(\frac{n+1}{2}\right)^{t h} \text { term }$

where n =7

Median = 4th term = 60

When the eight day is added,

Series will be: 50, 57, 58, 60, 64, 64, 68, 82

Median (when n is even) = $\frac{\left(\frac{n}{2}\right)^{t h} t e r m+\left(\frac{\pi}{2}+1\right)^{t h} t e r m}{2}$

Median = $\frac{60+64}{2}=62$

So, here we can say that median has increased after adding 82 in the data

3 0

3 years ago

Read 2 more answers

Solve the following compound inequality. 5x+7<=-3 or 3x-4>=11

Kobotan [32]

Answer:

$x\leq -2$ or $x\geq 5$

Step-by-step explanation:

$5x+7\leq -3$ or $3x-4\geq 11\\$

$5x+7\leq -3\\5x\leq -10\\x\leq -2$

$3x-4\geq 11\\3x\geq 15\\ x\geq 5$

4 0

3 years ago

PLease help The segments shown below could form a triangle True or false ?

grigory [225]

Answer:

true

Step-by-step explanation:

for a right triangle no

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3 years ago

What is the solution to the following system?

notsponge [240]

Answer:

the answer is c (2,1,0)

Step-by-step explanation:

4 0

3 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

2 years ago