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BARSIC [14]
3 years ago
10

Please help sorry if it’s blurry

Mathematics
1 answer:
Sveta_85 [38]3 years ago
8 0

Answer:

a

Step-by-step explanation:

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Suppose the number of messages that an inbox receives may be modeled by a Poisson distribution. If the average number of message
docker41 [41]

Answer:

0.36427

Step-by-step explanation:

Mean = λ = 18 messages per hour

P(X = x) = (e^-λ)(λ⁻ˣ)/x!

P(X ≤ x) = Σ (e^-λ)(λ⁻ˣ)/x! (Summation From 0 to x)

But the probability required is that the messages thay come in an hour is between 15 and 20, that is, P(15 < X < 20)

P(15 < X < 20) = P(X < 20) - P(X ≤ 15)

These probabilities will be evaluated using a cumulative frequency calculator.

P(X < 20) = 0.65092

P(X ≤ 15) = poissoncdf(18, 15) = 0.28665

P(15 < X < 20) = P(X < 20) - P(X ≤ 15) = 0.65092 - 0.28665 = 0.36427.

You can use the Poisson distribution calculator here

https://stattrek.com/online-calculator/poisson.aspx

4 0
3 years ago
May someone please help mee????
Arlecino [84]

Answer:

width of rectangle is 87

Step-by-step explanation:

because perimeter of rectangle = 2(l+w)

3 0
3 years ago
How much would something cost if it was $20 but was 1/5 percent off??
QveST [7]
4$ would be the cost of the item
6 0
3 years ago
Read 2 more answers
Consider x^2 + (b/a)x = (-c/a). Which variable cannot have a value of zero? Justify your answer.
GREYUIT [131]
The answer is C because if something is divided by zero, it is undefined.
5 0
3 years ago
Money Flow  The rate of a continuous money flow starts at $1000 and increases exponentially at 5% per year for 4 years. Find the
77julia77 [94]

Answer:

Present value =  $4,122.4

Accumulated amount = $4,742

Step-by-step explanation:

Data provided in the question:

Amount at the Start of money flow = $1,000

Increase in amount is exponentially at the rate of 5% per year

Time = 4 years

Interest rate = 3.5%  compounded continuously

Now,

Accumulated Value of the money flow = 1000e^{0.05t}

The present value of the money flow = \int\limits^4_0 {1000e^{0.05t}(e^{-0.035t})} \, dt

= 1000\int\limits^4_0 {e^{0.015t}} \, dt

= 1000\left [\frac{e^{0.015t}}{0.015} \right ]_0^4

= 1000\times\left [\frac{e^{0.015(4)}}{0.015} -\frac{e^{0.015(0)}}{0.015} \right]

= 1000 × [70.7891 - 66.6667]

= $4,122.4

Accumulated interest = e^{rt}\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t}} \, dt

= e^{0.035\times4}\times4,122.4

= $4,742

8 0
3 years ago
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