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Setler79 [48]
3 years ago
13

If x= 3 what is the value of 50 - 2x?

Mathematics
2 answers:
Natali [406]3 years ago
4 0

50 - 2x; x = 3

____________________

50 - 2(3)

= 50 - 6

= 44

Vitek1552 [10]3 years ago
4 0
50-2(3)

50-6=44

Answer :

44
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Solve: -2x - 5 = -3 <br> plz help me hurry
lord [1]

Answer:

x=-1

Step-by-step explanation:

-2x - 5 = -3

add 5 to both sides

-2x = 2

divide both sides by -2

x = -1

8 0
3 years ago
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Name the coordinates for the reflection. Reflect point (4, -1) over the x-axis.
inysia [295]

Answer:

i just took the test its (4,1)

Step-by-step explanation:

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Can you help me with these questions<br>​
dlinn [17]

Answer:

see below (I hope this helps!)

Step-by-step explanation:

41. √20 = √4 * 5 = √4 * √5 = 2√5

43. The sum of interior angles in a triangle is 180°, therefore, x = 180 - (75 + 64) = 41°.

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7 0
3 years ago
Divide using a common factor of 6 to find an equivalent fraction for
Savatey [412]

5: six twelfths is one half

6: eight tenths is four fifths

7: four eighths is one half

8: ten tenths is two halves

9: two sixths is one third

10: one quarter, because you divide the top and bottom number by 3

method:

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7 0
3 years ago
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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
3 years ago
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