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3 years ago

Find the simple interned earned in savings account that pay 5% interest if $500 is deposited for 4 years

Mathematics

1 answer:

Anuta_ua [19.1K]3 years ago

4 0

The answer is 100 dollars.

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There are 8 members on a board of directors. if they must form a subcommittee of 6 members, how many different subcommittees are

Shtirlitz [24]

1. Given a group of n people. There are C(n, r) ways of forming groups of r out of n.

2. Where C(n, r)= $\frac{n!}{r!(n-r)!}$

3. For example, given {Andy, John, Julia}. We want to pick 2 people to give a gift: we can pick {(Andy, John), (Andy, Julia), (John, Julia)}, so there are 3 ways. So we can list and count.

4. Or we could do this with the formula C(3, 2)= $\frac{3!}{2!(1)!}= \frac{3*2*1}{2*1}=3$

5. C(8, 6)= $\frac{8!}{2!6!}= \frac{8*7*6!}{2*6!}= \frac{8*7}{2}= 4*7=28$

So there are C(8,6)=28 ways of chosing 6 out of 8 people to form the subcommittees.

5 0

3 years ago

True or False The Solutions, Roots, X-intercepts and zeros of a quadratic equation are all the same thing? why

Zepler [3.9K]

Yes, solutions, roots, x-intercepts, and zeros are the same thing.

<h3>What is a quadratic equation?</h3>

The general quadratic equation is given by:

a*x^2 + b*x + c = 0

So the solutions are the values of x such that the above thing is zero.

On another hand, a parabola or a quadratic function is given by:

a*x^2 + b*x + c = y

The roots, zeros, or x-intercepts (these represent the same thing) are given by:

a*x^2 + b*x + c = 0

Zero or Root means that when you evaluate the function in that value the outcome is zero.
X-intercept means that for that value of x, the function intercepts the x-axis, so the function is equal to zero.

So these are the values of x such that the function becomes equal to zero, so these are exactly the same thing as the solutions of a quadratic equation.

Concluding, yes, solutions, roots, x-intercepts, and zeros are the same thing.

If you want to learn more about quadratic functions, you can read:

brainly.com/question/1214333

8 0

2 years ago

Given the following data set 2,3,1,6,1,1,1,0,2,4,5,1,2,2,3 Mean Median Range Mid range

igomit [66]

Answer:

mean: 34/15=2.27

median: 2

range: highest-lowest...... 6-0=6

mid range: high + low divided by 2

6+0=6/2=3

8 0

3 years ago

Read 2 more answers

Find the measure of exterior angle A.

max2010maxim [7]

Answer:

40* degress angled

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Find the mass and the center of mass of a wire loop in the shape of a helix (measured in cm: x = t, y = 4 cos(t), z = 4 sin(t) f

Sholpan [36]

Answer:

Mass

$\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

Center of mass

Coordinate x

$\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate y

$\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate z

$\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Step-by-step explanation:

Let W be the wire. We can consider W=(x(t),y(t),z(t)) as a path given by the parametric functions

x(t) = t

y(t) = 4 cos(t)

z(t) = 4 sin(t)

for 0 ≤ t ≤ 2π

If D(x,y,z) is the density of W at a given point (x,y,z), the mass m would be the curve integral along the path W

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt$

The density D(x,y,z) is given by

$D(x,y,z)=x^2+y^2+z^2=t^2+16cos^2(t)+16sin^2(t)=t^2+16$

on the other hand

$||W'(t)||=\sqrt{1^2+(-4sin(t))^2+(4cos(t))^2}=\sqrt{1+16}=\sqrt{17}$

and we have

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt=\\\\\sqrt{17}\displaystyle\int_{0}^{2\pi}(t^2+16)dt=\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

The center of mass is the point $(\bar x,\bar y,\bar z)$

where

$\bar x=\displaystyle\frac{1}{m}\displaystyle\int_{W}xD(x,y,z)\\\\\bar y=\displaystyle\frac{1}{m}\displaystyle\int_{W}yD(x,y,z)\\\\\bar z=\displaystyle\frac{1}{m}\displaystyle\int_{W}zD(x,y,z)$

We have

$\displaystyle\int_{W}xD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}t(t^2+16)dt=\\\\=\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)$

$\bar x=\displaystyle\frac{\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}yD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}4cos(t)(t^2+16)dt=\\\\=16\sqrt{17}\pi$

$\bar y=\displaystyle\frac{16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}zD(x,y,z)=4\sqrt{17}\displaystyle\int_{0}^{2\pi}sin(t)(t^2+16)dt=\\\\=-16\sqrt{17}\pi$

$\bar z=\displaystyle\frac{-16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

3 0

2 years ago