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aivan3 [116]
3 years ago
13

Which of the following numbers are less than nine over four

Mathematics
1 answer:
allochka39001 [22]3 years ago
5 0

Answer:

option B and C

Step-by-step explanation:

option B and C

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Solve 3x-4= 7 you may use the flowchart to help you if you wish
laiz [17]

Answer:

x = 3 2/3

Step-by-step explanation:

Given

3x - 4 = 7

Add 4 to both sides of the equation

3x - 4 + 4 = 7 + 4

3x = 11

Divide both sides by 3

3x/3 = 11/3

x = 3 2/3

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3 years ago
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Please help with this problem
Paraphin [41]

Answer:

Step-by-step explanation:

See attachment.

The change from x to (x-4) moved the function to the right by 4.  The - 5 moved it down by 5 (as measured from the crossover/inflection point.

7 0
2 years ago
Find a formula for the linear function with slope -9 and x-intercept 2.
gregori [183]
Hello : 
<span> formula for the linear function with slope -9 and x-intercept 2 is : 
y - 0 = -9(x-2)
y= -9x +18

</span>
4 0
4 years ago
What are the multiples of 88
NARA [144]
88 * 2 = 176 ;
88 * 3 = 264 ;
88 * 4 = 362 ;
...................

8 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
3 years ago
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