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Mekhanik [1.2K]
3 years ago
14

Write 0.851 as a fraction in simplest form.

Mathematics
1 answer:
Viktor [21]3 years ago
4 0

Answer:

\frac{851}{1000}

Step-by-step explanation:

First, we can simply multiply that number by 1000, and divide again by 1000 to get a base fraction:

.851\\\\= \frac{1000}{1000} \times .851\\\\= \frac{1000 \times .851}{1000}\\\\= \frac{851}{1000}

851 is a secondary prime, having only two factors, both of which are prime.  Those factors are 23 and 37, neither of which is a factor of 1000, so this is already in simplest form.

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Tricio mokes $9.00 an hour and works 40 hours a week. How much money does she earn in a year
nadezda [96]

Answer:

18,771 $

Step-by-step explanation:

there are 52.1429 weeks in one year with 40 hours of work each week and then 9 dollars made an hour this equals to 18,771 $

3 0
3 years ago
Find the angle measure C in the parallelogram ABCD.​
Kryger [21]

Answer:

135°

Step-by-step explanation:

In a parallelogram the opposite angles are congruent, so

∠BCD = ∠BAD = 135°

5 0
3 years ago
A square based prism and a cylinder both have
Anna11 [10]

Answer:

Parameter of base = 70.88 inch (Approx.)

Step-by-step explanation:

Given:

Volume of cylinder = 3768 cubic inch

Height of Prism = 12 inch

Find;

Parameter of base

Computation:

Volume of cylinder = Volume of Prism

3768 = (L)(B)(H)

3768 = (L)(B)(12)

(L)(B) = 314

L = B

So.

L = 17.72 inch

B = 17.72 inch

Parameter of base = 2[l+b]

Parameter of base = 2[17.72 + 17.72]

Parameter of base = 70.88 inch (Approx.)

8 0
3 years ago
sand falls from an overhead bin and accumulates in a conical pile with a radius that is always four times its height. suppose th
blagie [28]

Answer:

\frac{dv}{dt} =7239.168 cm/sec

Step-by-step explanation:

From the question we are told that:

Rate \frac{dh}{dt}=1cm

Height h=12cm

Radius r=4h

Generally the equation for Volume of Cone is mathematically given by

V=\frac{1}{3}\pi r^2h

V=\frac{1}{3}\pi (4h)^2h

Differentiating

\frac{dv}{dt} =\frac{16}{3}\pi3h^2\frac{dh}{dt}

\frac{dv}{dt} =\frac{16}{3}*3.142*3*12^2*1

\frac{dv}{dt} =7239.168 cm/sec

7 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
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