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Serjik [45]
2 years ago
11

2. How long does this lease last?

Mathematics
1 answer:
Tatiana [17]2 years ago
7 0

Answer:

akeijd

Step-by-step explanation:

kahshsarihaieie

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−104=4x−4(2x+16)<br> I need a whole number i keep messing up
kotykmax [81]

Answer:

x=10

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Can somebody explain to me how they got 3^4(x) = 3^3(x+2)? Then explain what happened to the 3 in 3^4 and the 3 in 3^3? Please a
musickatia [10]

Answer:

x=6

Step-by-step explanation:

81^x = 27 ^(x+2)

81 = 3^4    and 27 =3^3  so replace 81 and 27 in the equation

3^4^x = 3^3^(x+2)

When we have a power to a power we can multiply the exponents

a ^b^c = a^(b*c)

3^(4x) = 3^(3*(x+2))

Since the bases are the same, the exponents  have to be the same

a^b = a^c   means b=c

4x = 3(x+2)

Now we can solve for x

Distribute

4x = 3x+6

Subtract 3x from each side

4x-3x = 3x-3x+6

x = 6

3 0
3 years ago
Find the point B on AC such that the ratio of AB to BC is 2:1.
Softa [21]

Answer: B=-2,2

Step-by-step explanation:

I got it wrong so you can get it right

3 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
Please help !! Simplify the following expression
puteri [66]
8x^4 + x^3 - 4x^2 +1. I believe the answer is B. Please correct me if I am wrong.

5 0
3 years ago
Read 2 more answers
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