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Fiesta28 [93]
2 years ago
7

Krista is planning to mow her grandparents lawn the rectangular yard is 10 yards by 12 yards but there is a square patio that ha

s 12 foot sides in the middle of Krista kenmo 15 square feet per minute about how long will it take to finish the yard
Mathematics
1 answer:
kompoz [17]2 years ago
4 0
<span>First fine the total area of the yard 10 yards x 12 yards = 120 yards^2 Now find the area of the patio in yards 1 yard = 3 ft and each side of the patio is 12 ft --> 4 yards so the patio sides are 4 yds 4 yrds x 4 yrds = 16 yards To find the area of the yard only: 120 yards^2 - 16 yards^2 = 104 yards^2 Krista needs to mow 104 yards^2 total. And she can do 15 ft^2 per minute need the same units so convert the yards to feet 104 yards^2 x 3 ft/yard X 3 ft/yard = 936 ft^2 To find the minutes, divide the total by the amount she can do in one minute: 936 ft^2/15 ft^2 = 62.4 62.4 minutes</span>
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(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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