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avanturin [10]
2 years ago
14

-2x + y = 1 -4x + y = -1 Solve the following system of equation

Mathematics
1 answer:
andre [41]2 years ago
4 0

Answer:

x = 2 and y = 5

Step-by-step explanation:

2x = 2

x = 2

-4 + y = 1

y = 5

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Find the value of x and y<br> 3х<br> 20<br> 25<br> 18<br> бу -2<br> Х=<br> ү=
sweet-ann [11.9K]

Answer: first one =109360044xy−36453348x−2, 2nd,=3037778998,last one,=1968480790

Step-by-step explanation:Evaluate for x=3x,y=6y−2

33x(2025186)(6y−2)−2

33x(2025186)(6y−2)−2

=109360044xy−36453348x−2

Evaluate for x=20,y=25

(3)(20)(2025186)(25)−2

(3)(20)(2025186)(25)−2

=3037778998

because 18 is by itself i just did Evaluate for x=18,y=18

(3)(18)(2025186)(18)−2

(3)(18)(2025186)(18)−2

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2 years ago
I need answer ASAP! Please!!
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3 years ago
90 divided by 5 in distributive property
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7 0
3 years ago
Read 2 more answers
The sum of three numbers is 26. The second number is twice the first and the third number is 6 more than the second. Find the nu
inessss [21]
The numbers are 4,8, and 14 because 4+8+14=26 an 4x2=8 and 8+6=14
7 0
3 years ago
Suppose a basketball player has made 231 out of 361 free throws. If the player makes the next 2 free throws, I will pay you $31.
statuscvo [17]

Answer:

The expected value of the proposition is of -0.29.

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either the player will make it, or he will miss it. The probability of a player making a free throw is independent of any other free throw, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Suppose a basketball player has made 231 out of 361 free throws.

This means that p = \frac{231}{361} = 0.6399

Probability of the player making the next 2 free throws:

This is P(X = 2) when n = 2. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{2,2}.(0.6399)^{2}.(0.3601)^{0} = 0.4095

Find the expected value of the proposition:

0.4095 probability of you paying $31(losing $31), which is when the player makes the next 2 free throws.

1 - 0.4059 = 0.5905 probability of you being paid $21(earning $21), which is when the player does not make the next 2 free throws. So

E = -31*0.4095 + 21*0.5905 = -0.29

The expected value of the proposition is of -0.29.

3 0
2 years ago
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