Ask question

Ask question

All categories

2 years ago

10

You work as a cashier for a bookstore and earn $6 per hour. You also baby sit and earn $6 per hour. You want to earn at least $6

0 per week, but would like to work no more than 12 hours per week. Which system of inequalities, along with y ≥ 0 and x ≥ 0, would you use to solve the real-world problem?

1 answer:

klemol [59]2 years ago

3 0

For this case we first define varials:
x: number of hours working as cashier
y: number of hours working as a baby sit
We now write the system of equations:
6x + 6y ≥ 60
x + y ≤ 12
Answer:
to solve the real-world problem the system of inequalities is:
6x + 6y ≥ 60
x + y ≤ 12

You might be interested in

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo

Shkiper50 [21]

Answer:

a) $P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335$

b) $P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452$

c) $P(X \geq 8) = 1-P(X$

d) $P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that $X \sim Poisson(\lambda=4)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda=4$ , $Var(X)=\lambda=2$ , $Sd(X)=2$

a. Compute both P(X≤4) and P(X<4).

$P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183$

$P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733$

$P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465$

$P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646$

$P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692$

b. Compute P(4≤X≤ 8).

$P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595$

$P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298$

$P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452$

c. Compute P(8≤ X).

$P(X \geq 8) = 1-P(X$

$P(X \geq 8) = 1-P(X$

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

4 0

3 years ago

PLEASE HELP QUICK I CANT FAIL THIS TEST

Finger [1]

Answer:

24+24x I believe is the answer. Good luck on your test :)

5 0

3 years ago

Read 2 more answers

Pleasee help asap! will give brainiest

AfilCa [17]

Answer:x=-0.8

5x - 7 =3

-7 -7

5/5x = -4/5

x = -0.8

8 0

3 years ago

87+25<br><img src="https://tex.z-dn.net/?f=12%20%2B%2010%20%20" id="TexFormula1" title="12 + 10 " alt="12 + 10 " align="absmid

Butoxors [25]

Answer:

Seriously?

U don't know the answer

3 0

2 years ago

Find the equation of the tangent line. y=(9-x^2)^1/3 at (1,2)

Schach [20]

Get the derivative:

<em>y</em> = (9 - <em>x</em>²)¹ʹ³

d<em>y</em>/d<em>x</em> = 1/3 (9 - <em>x</em>²)⁻²ʹ³ d/d<em>x</em> [9 - <em>x</em>²]

d<em>y</em>/d<em>x</em> = 1/3 (9 - <em>x</em>²)⁻²ʹ³ (-2<em>x</em>)

d<em>y</em>/d<em>x</em> = -2/3 <em>x</em> (9 - <em>x</em>²)⁻²ʹ³

Evaluate it at <em>x</em> = 1 :

d<em>y</em>/d<em>x</em> (1) = -2/3 • 8⁻²ʹ³

Since 8 = 2³, we have

8⁻²ʹ³ = 1 / 8²ʹ³ = 1 / (2³)²ʹ³ = 1 / 2² = 1/4

Then the tangent line has equation

<em>y</em> - 2 = 1/4 (<em>x</em> - 1) → <em>y</em> = 1/4 <em>x</em> + 7/4

7 0

2 years ago