What is the area,in square yards of the rectangle shown down below 12yd 15yd Answer choices are 225,54,180,27
2 answers:
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Answer: Any real number x as long as and
In other words, anything but 0 or -2/3 is valid.
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Explanation:
Set the denominator equal to zero and solve for x
2(3x^2 + 2x) = 0
3x^2 + 2x = 0
x(3x + 2) = 0
x = 0 or 3x+2 = 0 .... zero product property
x = 0 or 3x = -2
x = 0 or x = -2/3
If either x = 0 or x = -2/3, then the denominator 2(3x^2 + 2x) will be zero. But recall that we cannot have zero in the denominator. Dividing by zero is not allowed. The expression is undefined when we divide by zero.
Therefore, we must exclude x = 0 and x = -2/3 from the domain. Any other real number is valid as an x input.
The probability of an event is the measurement of the chance of that event's occurrence. The probabilities of considered events are:
- P(At least 8 have the disease) ≈ 0.4378
- P(At most 4 have the disease) ≈ 0.0342
Binomial distributions consist of n independent Bernoulli trials. Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))
Suppose we have random variable X pertaining to a binomial distribution with parameters n and p, then it is written as
X = B(n,p)
The probability that out of n trials, there'd be x successes is given by
P(X=x) =
Since 10 people can be either diseased or not and they be so independent of each other (assuming them to be selected randomly) , thus, we can take them being diseased or not as outputs of 10 independent Bernoulli trials.
Let we say
Success= Probability of a diseased person tagged as diseased by the clinic
Failure = Probability of a diseased person tagged as not diseased by the clinic.
Then,
P(Success) = p = 72% = 0.72 (of a single person)
P(Failure) = q = 1-p = 0.28
Let X be the number of people diagnosed diseased by the clinic out of 10 diseased people. Then we have: X ≈ B(n+10,P=0.73)
Calculating the needed probabilities, we get:
a) P(At leased 8 have disease) = P(X≥8) =P(X=8) + P(X=9) + P(X=10)
P(X≥8) =
P(X≥8) ≈ 0.2548 + 0.1456 + 0.0374 ≈ 0.4378
b) P(At most 4 have the disease) = P(X≤4) = P(X=0) + P(X=1)+P(X=2)+P(X=3)+P(X=4)
P (X ≤ 4) =
P (X ≤ 4) = 0.000003 + 0.000076+0.00088+0.00604+0.02719
P (X ≤ 4) = 0.0342
Thus,
The probabilities of considered events are:
- P(At leased 8 have disease) = 0.4378 approx
- P(At most 4 have the disease) = 0.0342 approx
Learn more about binomial distribution here:
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