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2 years ago

Which graph shows the solution to the system of linear inequalities? 2x – 3y <12 y < –3

Mathematics

2 answers:

MrMuchimi2 years ago

6 0

Answer:

Graph.

2 x − 3 y < 12

y < − 3

Step-by-step explanation:

-Dominant- [34]2 years ago

6 0

Answer:

-3 please brainliest me PLease i expain it too

Step-by-step explanation:

Let's put the equation into point slope for

-3y < -2x + 12

Divide by -3

y < 2/3x - 4

The answer is the first one because the shades are pointing down.

y < -3 means the shade points down y because y is less than -3

Read more on Brainly.com - brainly.com/question/7589507#readmore

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Find the total cost to the nearest cent $43 dinner 18% gratuity

Norma-Jean [14]

43*1.18=$50.74
You pay 100% plus 18% tip

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2 years ago

From a recent company survey, it is known that the proportion of employees older than 55 and considering retirement is 8%. For a

Akimi4 [234]

Answer: 2.845

Step-by-step explanation:

The formula to find the standard deviation is given by :-

$\sigma=\sqrt{n(1-p)p}$ , where p is the probability of getting success in each trial and n is the sample size.

Given : Sample size : n=110

The proportion of employees older than 55 and considering retirement : p=0.08

Then, the standard deviation is given by :-

$\sigma=\sqrt{110(1-0.08)0.08}\approx2.845347079\approx2.845$

Hence, the standard deviation for the sampling distribution of the sample proportions is 2.845.

3 0

2 years ago

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

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2 years ago

Lelia says that 75% of a number will always be greater than 50% of a number. Complete the inequality to support Lelia's claim an

worty [1.4K]

Answer:

Let 'x' and 'y' be two different numbers.

Leila says that 75% of a number will always be greater than 50% of a number. The inequality that represents this statement is the following:

0.75x > 0.5y

Let x = 100 and y=200. We have that:

0.75(100) > 0.5(200)

75 > 100 ❌ INCORRECT ❌

Given that we found a case in which 75% of a number is not greater than 50% of a number, we can conclude that Leila's claim is incorrect.

3 0

2 years ago

How to decompose a rectangle into two or three smaller rectangles to demonstrate the distributive property?

koban [17]

Answer:

See explanation

Step-by-step explanation:

The distibutive property states that for all real numbers a, b and c

$(a+b)c=ac+bc$

Draw a rectangle with length of (a+b) units and width of c units. The area of this rectangle is

$(a+b)c\ un^2.$

Divide this rectangle into two rectangles: first rectangle with the length of a units and the width of c units and the second rectangle with the length of b units and the width of c units. The area of these rectangles are

$a\cdot c\ \text{and}\ b\cdot c$

These two rectangles together form initial rectangle, so the sum of the area of two smaller rectangles is the area of the bigger rectangle.

5 0

2 years ago