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3 years ago

How to write the ratio 8:30 as a fraction in simplest form

1 answer:

5 0

4/15 because you make it into a fraction: 8/30, you find the GCF which is 2, so then you divide both the numerator and denominator by 2 and get 4/15 at it's simplest term

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If f'(x)=2/x and f(sqrt(e))=5, then f(e)=

docker41 [41]

$f'(x)=\dfrac2x$
$\implies f(x)=\displaystyle\frac2x\,\mathrm dx=2\ln|x|+C$

$f(\sqrt e)=5\implies 5=2\ln|\sqrt e|+C$
$\implies 5=\dfrac22\ln e+C$
$\implies C=4$
$\implies f(x)=2\ln|x|+4$
$\implies f(e)=2\ln|e|+4=6$

3 0

3 years ago

The radius of a circle is 16 cm. Find its area in terms of pi

anzhelika [568]

Area is 804.25 if the radius is 16cm

8 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.

Step-by-step explanation:

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

The arrival rate is, λt = 8 per hour.

(a)

For t = 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 t = 90 minutes = 1.5 hour, the value of λ, 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 t = 2.5 the value of λ, 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

2 years ago

Raquel has created a program for a robot. The robot travels 10 meters and stops. The program is designed for wheels that are 2.5

Grace [21]

Answer:

Raquel should adjust the number of rotations to 115.81 rotations

Step-by-step explanation:

step 1

Find the circumference of the wheels that are 2.5 cm in diameter

The circumference is equal to

$C=2\pi r$

we have

$r=2.5/2=1.25\ cm$

assume

$\pi =3.14$

substitute

$C=2(3.14) (1.25)$

$C=7.85\ cm$

step 2

Find the number of rotations

Divide 10 meters by the circumference

$10\ m=1,000\ cm$

$1,000/7.85=127.39\ rotations$

step 3

For a diameter of 2.75 cm find how should she adjust the number of rotations so that the robot travels the same distance

Find the circumference of the wheels

The circumference is equal to

$C=2\pi r$

we have

$r=2.75/2=1.375\ cm$

assume

$\pi =3.14$

substitute

$C=2(3.14) (1.375)$

$C=8.635\ cm$

step 4

Find the number of rotations

Divide 10 meters by the circumference

$10\ m=1,000\ cm$

$1,000/8.635=115.81\ rotations$

3 0

2 years ago

Help ASAP, will give BRAINLIEST

vlada-n [284]

Answer:

B: 39 degrees

Step-by-step explanation: The angles in a triangle add up to 180 degrees. 180 - 30 - 24 = 126.

126 = (3x-9).

126 - 9 = 117.

117 / 3 = 39

4 0

2 years ago

Read 2 more answers