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Keith_Richards [23]
3 years ago
11

Convert the following into millilitres a)13 litres​

Mathematics
2 answers:
skelet666 [1.2K]3 years ago
7 0

Answer:

13,000ml

Step-by-step explanation:

1000ml = 1L.

We can rewrite this forumla as:

yL x 1000=ml. y=the number of liters.

In this problem, y=13. Thus:

13Lx1000= 13,000ml

I hope this helps!

I am Lyosha [343]3 years ago
4 0

Answer:

13000

Step-by-step explanation:

Kilo                                         Base is whatever u are using liter, meter, ect.

Hecto                                     The amount goes up by ten when u move down                          

Deka                                       the metric system and vice versa.

Base 13

Deci  130

Centi 1300

Milli   13000

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Answer:

-\frac{10}{3}ft/s

Step-by-step explanation:

We are given that

Height of man=5 foot

\frac{dy}{dt}=-10ft/s

Height of street light=20ft

We have to find the rate of change of the length of his shadow when he is 25 ft form the street light.

ABE and CDE are similar triangle because all right triangles are similar.

\frac{20}{5}=\frac{x+y}{x}

4=\frac{x+y}{x}

4x=x+y

4x-x=y

3x=y

3\frac{dx}{dt}=\frac{dy}{dt}

\frac{dx}{dt}=\frac{1}{3}(-10)=-\frac{10}{3}ft/s=-\frac{10}{3}ft/s

Hence, the rate of change of the length of his shadow when he is 25 ft from the street light=-\frac{10}{3}ft/s

7 0
3 years ago
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Romashka-Z-Leto [24]
Just take each number in the first table and divide it by the total then x by 100

Eg the first one 24/120 x 100 = 20%
6 0
2 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.

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lana66690 [7]

x=125/100


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Step-by-step explanation:

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