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3 years ago

Jennifer is saving money to buy a bike. The bike costs $223. She has $124 saved, and each week she adds $11 to her savings.

Mathematics

1 answer:

Elan Coil [88]3 years ago

4 0

233-124 is 109.

109/11 is nearly 10.

answer is 10 weeks

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four cards numbered 1, 5, 8, and 9 are placed in a bag. A card is drawn at random and then replaced. Then a card is drawn at ran

Vadim26 [7]

Answer:

1/16

Step-by-step explanation:

Four cards numbered (1, 5, 8,9)

Given the sample space S= 4

1.The first selection with replacement

Probability of picking a 9= 1/4

2. Second selection, probability of picking a 9= 1/4

Hence the probability that both cards drawn have the number 9.

Is = (1/4)*(1/4)= 1/16

6 0

3 years ago

What’s the Value of x?

denpristay [2]

X will be 8.49
the other side is 6 too

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3 years ago

What points are on the graph -3x+6y+5=-7

Simora [160]

The slope 1/2 y -intercept -2

8 0

3 years ago

The residents of a certain dormitory have collected the following data: People who live in the dorm can be classified as either

Ad libitum [116K]

Answer:

The steady state proportion for the U (uninvolved) fraction is 0.4.

Step-by-step explanation:

This can be modeled as a Markov chain, with two states:

U: uninvolved

M: matched

The transitions probability matrix is:

$\begin{pmatrix} &U&M\\U&0.85&0.15\\M&0.10&0.90\end{pmatrix}$

The steady state is that satisfies this product of matrixs:

$[\pi] \cdot [P]=[\pi]$

being π the matrix of steady-state proportions and P the transition matrix.

If we multiply, we have:

$(\pi_U,\pi_M)*\begin{pmatrix}0.85&0.15\\0.10&0.90\end{pmatrix}=(\pi_U,\pi_M)$

Now we have to solve this equations

$0.85\pi_U+0.10\pi_M=\pi_U\\\\0.15\pi_U+0.90\pi_M=\pi_M$

We choose one of the equations and solve:

$0.85\pi_U+0.10\pi_M=\pi_U\\\\\pi_M=((1-0.85)/0.10)\pi_U=1.5\pi_U\\\\\\\pi_M+\pi_U=1\\\\1.5\pi_U+\pi_U=1\\\\\pi_U=1/2.5=0.4 \\\\ \pi_M=1.5\pi_U=1.5*0.4=0.6$

Then, the steady state proportion for the U (uninvolved) fraction is 0.4.

4 0

3 years ago

Please solve equations by completing square

BlackZzzverrR [31]

$x^2-4x+1=0\\
x^2-4x+4-3=0\\
(x-2)^2=3\\
|x-2|=\sqrt3\\
x-2=\sqrt3 \vee x-2=-\sqrt3\\
x=2+\sqrt3 \vee x=2-\sqrt3$

$x^2+6x-9=0\\
x^2+6x+9-18=0\\
(x+3)^2=18\\
|x+3|=\sqrt{18}=3\sqrt2\\
x+3=3\sqrt2 \vee x+3=-3\sqrt2\\
x=-3+3\sqrt2 \vee x=-3-3\sqrt2\\$

$x^2+20x+3=0\\
x^2+20x+100-97=0\\
(x+10)^2=97\\
|x+10|=\sqrt{97}\\
x+10=\sqrt{97} \vee x+10=-\sqrt{97}\\
x=-10+\sqrt{97} \vee x=-10-\sqrt{97}$

$x^2-2x-5=0\\
x^2-2x+1-6=0\\
(x-1)^2=6\\
|x-1|=\sqrt6\\
x-1=\sqrt6 \vee x-1=-\sqrt6\\
x=1+\sqrt6 \vee x=1-\sqrt6$

8 0

3 years ago