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Komok [63]
3 years ago
14

Simplify square root 3y diviede by square root of y​

Mathematics
1 answer:
Nataly [62]3 years ago
5 0

Answer:

Step-by-step explanation:

√3y/√y

= √(3y/y)

= √3

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Which set of ordered pairs could be generated by an exponential function?
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Answer: Third option.

Step-by-step explanation:

By definition, Exponential functions have the following form:

y = ab^x

Where "b" is the base (b > 0 and b\neq 1), "a" is a coefficient (a\neq 0) and "x" is the exponent.

 It is importat to remember that the "Zero exponent rule" states that  any base with an exponent of 0 is equal to 1.

Then, for an input value 0 (x=0) the output value (value of "y") of the set of ordered pairs that could be generated by an exponential function must be 1 (y=1).

You can observe in the Third option shown in the image that when x=0, y=1

 Therefore, the set of ordered pairs that could be generated by an exponential function is the set shown in the Third option.

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T-8 in a algebric expression
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4 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
3 years ago
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