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chubhunter [2.5K]
3 years ago
9

Hello, can someone help me with this. I had a brain fart and don’t remember how to do it.

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
7 0

Answer:

i believe its 36x but im not sure i just did the math..

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What’s the midpoint of the line segment joining A and B<br><br> A(2,-5);B(6,1)
IRISSAK [1]

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{2}~,~\stackrel{y_1}{-5})\qquad B(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{6+2}{2}~~,~~\cfrac{1-5}{2} \right)\implies \left( \cfrac{8}{4}~,~\cfrac{-4}{2} \right)\implies (4,-2)

7 0
2 years ago
Jasmine is a receptionist at a doctor's office she books 86 appointments each day four out of every 24 appointments are canceled
bonufazy [111]

Answer:

3.07

Step-by-step explanation:

(86.0/(4.0+24.0))

5 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
Help me please!!!!!!!!!!!!!!!!!!!!!!!!!!!
SVEN [57.7K]
82.2

Add all of them up and then divide by 10.
5 0
3 years ago
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Madaline wants to plant some spices in a cube-shaped planter box. If one edge of the box mcasures 7 inches, how much soil will b
Shkiper50 [21]

Answer:

B

Step-by-step explanation: B is the only reasonable answer

8 0
2 years ago
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