Which equation has infinitely many solutions?

An arithmetic sequence has this recursive formula:

Marianna [84]

Answer:

Option D is the correct option.

please see the attached picture for full solution..

Hope it helps...

Good luck on your assignment...

5 0

3 years ago

A company makes rubber rafts. 12% of them develop cracks within the first month of operation. 27 new rafts are randomly sampled

rewona [7]

The probability that the number of tested rafts that develop cracks is no more than 3 is .00006.

The true proportion, p for the population is given to 0.12.

Thus, the mean, μ, for the sample = np = 27*0.12 = 3.24.

The sample size, n, given to us is 27.

Thus, the standard deviation, s, for the sample can be calculated using the formula, s = √{p(1 - p)}/n.

s = √{0.12(1 - 0.12)}/27 = √0.003911 = 0.0625389.

We are asked to calculate the probability that the number of tested rafts that develop cracks is no more than 3, that is, we need to calculate P(X ≤3).

P(X ≤ 3)

= P(Z ≤ {(3 - 3.24)/0.0625389) {Using the formula z = (x - μ)/s}

= P(Z ≤ -3.8376114706)

= .00006 {From table}.

Thus, the probability that the number of tested rafts that develop cracks is no more than 3 is .00006.

Learn more about sampling distributions at

brainly.com/question/15507495

#SPJ4

8 0

2 years ago

Domain and range of 10x - 2y = 4

Anna71 [15]

Answer:

domain: (-∞,∞)

range: (-∞,∞)

Step-by-step explanation:

You need to convert this into a linear equation (y=mx+b)

So we add 2y and divide by 2 to get y by itself

that gives us 5x-2=y

If you pictures this graph in your head you'll see that the domain and range are all real numbers

5 0

3 years ago

A square building with an area of 225 m2 has a garden surrounding it that has an equal width on all sides. The area of the garde

xenn [34]

Answer:

The width of the garden is 1.16 m

Step-by-step explanation:

step 1

Find the area of the garden

To find out the area of the garden multiply by 1/3 the area of the building

$225(\frac{1}{3})=75\ m^2$

step 2

Find the length side of the square building

The area of a square is

$A=b^2$

where

b is the length side of the square

we have

$A=225\ m^2$

so

$b^2=225\\b=15\ m$

step 3

Find the width of the garden

Let

x ----> the width of the garden

we know that

The area of the building plus the area of the garden is equal to

$(15+2x)^2=225+75$

solve for x

$225+60x+4x^2=225+75\\4x^2+60x-75=0$

Solve the quadratic equation by graphing

using a graphing tool

The solution is x=1.16 m

see the attached figure

therefore

The width of the garden is 1.16 m

Find the exact value

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$4x^2+60x-75=0$

so

$a=4\\b=60\\c=-75$

substitute in the formula

$x=\frac{-60\pm\sqrt{60^{2}-4(4)(-75)}} {2(4)}$

$x=\frac{-60\pm\sqrt{4,800}} {8}$

$x=\frac{-60\pm40\sqrt{3}} {8}$

$x=\frac{-60+40\sqrt{3}} {8}\ m$ ----> exact value

8 0

3 years ago

Find the perimeter of the figure

nalin [4]

Answer:

The perimeter is 66.9 units

Step-by-step explanation:

See the attached figure with letters to better understand the problem

step 1

In the right triangle ABD

Find the length side AD

$tan(33^o)=\frac{8}{AD}$ ----> by TOA (opposite side divided by the adjacent side)

$AD=\frac{8}{tan(33^o)}=12.3\ units$

Find the length side BD

$sin(33^o)=\frac{8}{BD}$ ---> by SOH (opposite side divided by the hypotenuse)

$BD=\frac{8}{sin(33^o)}=14.7\ units$

step 2

In the right triangle CDE

Find the length side CE

$cos(64^o)=\frac{CD}{CE}$ --> by CAH (adjacent side divided by the hypotenuse)

we have

$CD=BD/2=14.7/2=7.35\ units$

substitute

$CE=\frac{7.35}{cos(64^o)}=16.8\ units$

Find the length side DE

$tan(64^o)=\frac{DE}{CD}$ --> by TOA (opposite side divided by the adjacent side)

$DE=tan(64^o){7.35}=15.1\ units$

step 3

Find the perimeter of the figure

The perimeter is equal to

$P=AB+AD+BD+CE+DE$

substitute the values

$P=8+12.3+14.7+16.8+15.1=66.9\ units$

5 0

3 years ago