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Murljashka [212]
2 years ago
6

Solve the equation for x in terms of c.

Mathematics
1 answer:
Alex2 years ago
6 0

Answer:

  • x = 29 / (8c)

Step-by-step explanation:

  • (2/3)(cx + 1/2) - 1/4 = 5/2
  • (2/3)cx + 1/3 - 1/4 = 5/2
  • (2/3)cx + 1/12 = 5/2                            Multiply all terms by 12
  • 8cx + 1 = 30
  • 8cx = 29
  • x = 29 / (8c)

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1. A parachutist is 800 feet above the ground when she opens her parachute. She then falls at a constant rate of 5 feet per seco
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Answer:

h = -5*t +800

Step-by-step explanation:

The initial value is 800 ft

She is falling at a rate of 5 ft per second

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y = mx+b

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A real estate agent has 19 properties that she shows. She feels that there is a 30% chance of selling any one property during a
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Answer:

P(X \geq 5)=1-P(X

We can find the individual probabilities:

P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114

P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092

P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358

P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869

P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491

And replacing we got:

P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=19, p=0.3)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

And we want to find this probability:

P(X \geq 5)

And we can use the complement rule:

P(X \geq 5)=1-P(X

We can find the individual probabilities:

P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114

P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092

P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358

P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869

P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491

And replacing we got:

P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178

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