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3 years ago

I need help with this 1 problem on my homework

Mathematics

1 answer:

Andru [333]3 years ago

4 0

Step-by-step explanation:

10-5y=3x

1. add 10 to both sides

-5y=3x+10

2. divide everything by -5

y=-3/5x-2

3. your point would start at -2 on the y axis and you would go down 3 and over 5 to get to the next point

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3 years ago

Learning Theory Students in a psychology experiment were given an exam and then retested monthly with equivalent exams. The aver

forsale [732]

Answer:

Step-by-step explanation:

The average scores s (on a 100 point scale) for the students can be modeled by

s = 75 - 6 In(t + 1), 0 < t < 12

where t is the time in months.

a) Since the students were given an exam and then retested monthly with equivalent exams, then,

For the original exam, t = 0

Therefore,

s = 75 - 6 In(0 + 1) = 75 - 6 In1

s = 75 - 6 × 0 = 75

b) the average score after 4 months, t = 4

Therefore,

s = 75 - 6 In(4 + 1) = 75 - 6 In5

s = 75 - 9.66 = 65.34

c) s = 60

Therefore,

60 = 75 - 6 In(t + 1)

6 In(t + 1) = 75 - 60 = 15

In(t + 1) = 15/6 = 2.5

t + 1 = e^2.5 = 12.18

t = 12.18 - 1 = 11.18

t = 11 approximately

8 0

3 years ago

A player tosses a die 6 times.If gets a number 6 Atleast two times he wins 2 dollars ,otherwise he looses 1 dollar.. Find the ex

swat32

Answer:

$E(x)=-0.2101$

Step-by-step explanation:

The expected value for a discrete variable is calculated as:

$E(x)=x_1*p(x_1)+x_2*p(x_2)$

Where $x_1$ and $x_2$ are the values that the variable can take and $p(x_1)$ and $p(x_2)$ are their respective probabilities.

So, a player can win 2 dollars or looses 1 dollar, it means that $x_1$ is equal to 2 and $x_2$ is equal to -1.

Then, we need to calculated the probability that the player win 2 dollars and the probability that the player loses 1 dollar.

If there are n identical and independent events with a probability p of success and a probability (1-p) of fail, the probability that a events from the n are success are equal to:

$P(a)=nCa*p^a*(1-p)^{n-a}$

Where $nCa=\frac{n!}{a!(n-a)!}$

So, in this case, n is number of times that the player tosses a die and p is the probability to get a 6. n is equal to 6 and p is equal to 1/6.

Therefore, the probability $p(x_1)$ that a player get at least two times number 6, is calculated as:

$p(x_1)=P(x\geq2)=1-P(0)-P(1) \\\\P(0) =6C0*(1/6)^{0}*(5/6)^{6}=0.3349\\P(1)=6C1*(1/6)^{1}*(5/6)^{5}=0.4018\\\\p(x_1)=1-0.3349-0.4018\\p(x_1)=0.2633$

On the other hand, the probability $p(x_2)$ that a player don't get at least two times number 6, is calculated as:

$p(x_2)=1-p(x_1)=1-0.2633=0.7367$

Finally, the expected value of the amount that the player wins is:

$E(x)=x_1*p(x_1)+x_2*p(x_2)\\E(x)=2*(0.2633)+ (-1)*0.7367\\E(x)=-0.2101$

It means that he can expect to loses 0.2101 dollars.

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2 years ago

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