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2 years ago

Whats the absolute value of -8 and 8

Mathematics

1 answer:

Alla [95]2 years ago

4 0

Answer:

Step-by-step explanation:

-8--------------------------0---------------------------8

imagine this is a straight line where you can measure the distances.

to get from -8 to 0 we have a distance of 8 units.

so, knowing this, now from 0 to 8 we have another distance of 8 units.

8+8=16

Distance is the absolute value, meaning that it'll always be positive.

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|-2x + 4| > or equal to 4

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4 years ago

Let f(x)=3x^2-x+2 and g(x)= 5x^2-1. find f(g(x))

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4 years ago

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GuDViN [60]

Answer:

The factors of 2q²-5pq-2q+5p are (2q-5p) (q-1)....

Step-by-step explanation:

The given expression is:

2q²-5pq-2q+5p

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(2q²-5pq) - (2q-5p)

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Now factor the polynomial by factoring out the G.C.F, 2q-5p

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3 years ago

One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst

grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let X = number of telephone calls.

The average number of calls per minute is, λ = 3.0.

The random variable X follows a Poisson distribution with parameter λ = 3.0.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...$

(a)

Compute the probability of X = 0 as follows:

$P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498$

Thus, the probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of X > 5 as follows:

P (X > 5) = 1 - P (X ≤ 5)

$=1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084$

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of X = 3 as follows:

$P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09$

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of X ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

$=\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578$

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

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3 years ago