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Tems11 [23]
3 years ago
11

PLEASE HELP ME!!! Ms. Jones wants to install a stone border along the flower bed in her rectangular backyard. Which is closest t

o the value of F?
A: 16 ft
B: 25 ft
C: 30 ft
D: 53 ft

Mathematics
1 answer:
Nataly_w [17]3 years ago
3 0

Answer:

use Pythagoras theorem formula   C^{2} = A^{2} + B^{2}

Step-by-step explanation:

C^{2} =  24^{2} + 18^{2}

C^{2}  = 576 + 324

C^{2} = 900

C = \sqrt{900}

C = 30

Does that make sense?  :|

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Answeration:

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Melinda bought 6 bowls of for $13.20 what was the unit rate in dollars
dem82 [27]
Divide 13.20 by 6.

6/13.20 = 1/2.2

1 bowl is $2.20
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Solve the inequality and identify solution. Please help. <br><br> 5x-4&gt;-4
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Step-by-step explanation:

3 0
3 years ago
Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o
timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 1 the average number of aircraft arrival is:

\lambda t=8\times 1=8

The probability distribution of a Poisson distribution is:

P(X=x)=\frac{e^{-8}(8)^{x}}{x!}

Compute the value of P (X = 6) as follows:

P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

P(X\geq 6)=1-P(X

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

P(X\geq 10)=1-P(X

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 1.5=12

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

SD=\sqrt{\lambda t}=\sqrt{12}=3.464

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 2.5=20

Compute the value of P (X ≥ 20) as follows:

P(X\geq 20)=1-P(X

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0
3 years ago
Hey I really need help on this question please
Dmitry_Shevchenko [17]
A circle’s standard form of an equation is:
(x-h)^2 + (y-k)^2 = radius^2

Plug in h and k immediately because that is something you automatically know. H and k are derived from the center of the circle. The center of the circle is (h,k). Don’t get tripped up though, your center of a circle has negative coordinates. When you have two negatives, they become positive.

So now you have:
(x+4)^2 + (y-2)^2 = radius^2

So figure out what the radius is. Use the distance formula to find out. You have a change of 5 from -4 to 1 in x. You have a change of 2 from 2 to 4 in y. Distance formula has the distance as the square root of x distance squared and y distance squared. That would mean that the distance/radius is equal to the square root of (25 + 4). 5 squared is 25 while 2 squared is 4.

The radius of the circle is equal to the square root of (29). However, looking back at the circle equation the radius should be squared for the equation. Square root of 29 squared gets you 29.

Plug that in and you get:
(x+4)^2 + (y-2)^2 = 29

5 0
2 years ago
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