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3 years ago

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.

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

3 years ago