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IRISSAK [1]
3 years ago
10

Answer these 2 questions please

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
7 0
\cfrac{5.0*10^8}{2.0*10^4} =2.5*10^{8-4}=2.5*10^4 \\  \\  \\  \\ (2.35*10^5)(5.92*10^7)=13.912*10^{12}=1.3912*10^{13}
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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
According to the manufacturer of the candy Skittles, 20% of the candy produced are red. If we take a random sample of 100 bags o
igomit [66]

Answer:

Probability that the proportion in our sample of red candies will be less than 20% is 0.5 .

Step-by-step explanation:

We are given that 20% of the candy produced are red. A random sample of 100 bags of Skittles is taken.

The distribution we can use here is;

               \frac{\hat p - p}{\sqrt{\frac{\hat p (1- \hat p)}{n} } } ~ N(0,1)

where, p = 0.20 and n = 100

Let \hat p = proportion of red candies in our sample

So, P(\hat p < 0.20) = P(\frac{\hat p - p}{\sqrt{\frac{\hat p (1- \hat p)}{n} } } < \frac{0.20 - 0.20}{\sqrt{\frac{0.2 (1- 0.2)}{100} } } ) = P(Z < 0) = 0.5

Therefore, probability that the proportion in our sample of red candies will be less than 20% is 0.5 .

5 0
3 years ago
$75 dress shirt; 20% discounts
kherson [118]
Is it adding ,subracting,mulipty,or divide
7 0
3 years ago
Read 2 more answers
3n+75=50+2n please help righting more
alexandr1967 [171]

Answer: n = -25

Step-by-step explanation:

you would need to get N by itself so you would subtract 2n from both sides and that would bring you to n+75=50 and then you would need to subtract 75 from both sides and that leads you to n=-25

5 0
3 years ago
Imagine that a traffic intersection has a stop light that repeatedly cycles through the normal sequence of traffic signals (gree
Nutka1998 [239]

Answer:

Your answer is 3.61%.

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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