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joja [24]
3 years ago
6

Wasani was playing Jeopardy. In this game of Jeopardy, players earn 50 points for a correct response and lose 50 points for an i

ncorrect response. So far, Wasani has answered 4 questions and his score is 0. Which of the following could be true?
Mathematics
2 answers:
nata0808 [166]3 years ago
6 0

Answer:

he answered 2 right and 2 wrong

Step-by-step explanation:

all you have to do is say 50+50 is 100 then he had he lost 50 points which would put him back to 0 so how much points would he lose if he answered 4 questions well 100

Nady [450]3 years ago
5 0
He got two correct and two wrong.
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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.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 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 <em>X</em> = 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 <em>X</em> > 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 <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

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 <em>X</em> ≤ 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.

5 0
3 years ago
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UNO [17]
X both sides by 11
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3 years ago
−2/3a+1/8−1/6a−3/4 PLEASE HELP
prohojiy [21]

Answer:

Evaluate:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Adding to the checkbook those debits and credits not already there brings its balance to ...

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Adjusting the bank's balance by the deposits and checks not already there brings its balance to ...

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Use Structure Tyler explained how he multiplies
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Answer:

The property that I would use is that however many decimal points are in the equation is the number of decimal places in the answer

Step-by-step explanation:

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