Math question 4:) please and thanks

20 grams of KCl to moles of

Irina-Kira [14]

0.013413582325191 got it from google

4 0

3 years ago

Find the amount of time. <br> 8. I = $2000, P = $5000, r = 8%

algol [13]

Answer:

t = 5 years

Step-by-step explanation:

I = $ 2000

P = $ 5000

r = 8%

$I = P * \frac{r}{100}*t\\\\t =\frac{ I * 100}{P*r}\\\\t =\frac{2000*100}{5000*8}\\\\ = 5$

5 0

2 years ago

A library has five copies of a certain textbook on reserve of which two copies (1 and 2) are hardcover books and the other three

Gre4nikov [31]

Answer and Step-by-step explanation:

(a) Since the student have to stop when he gets to w softcover book, then the sample space should always stop at either 3 or 4 or 5. If the students examines the books randomly and stops at a softcover book, here are the possible ways he could carry this out,

S = {3, 4, 5, 13, 14, 15, 23, 24, 25, 123, 124, 125, 213, 214, 215}

2. Since only one book is to be examined then the student must check only one softcover book and end the examining. To do this, either he examines just 3 or 4 or 5. Therefore,

A = {3, 4, 5}

3. We need to pick the possibilities from the sample space that shows that the examining ended at book 5. Therefore,

B = {5, 15, 25, 125, 215}

4. We need to pick possibilities from the sample space that shows that book 1 was not examined. Therefore,

C = {3, 4, 5, 23, 24, 25}

5 0

2 years ago

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose

Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

In which

x is the number of sucesses

$
e = 2.71828$ is the Euler number

$\mu$ is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson mean:

$\mu = 0.2$

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

$P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819$

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017$

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so $p = 0.181$

We want to find P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671$

0.671 = 67.1% probability that neither contains a missing pulse

8 0

2 years ago

Y= -1/2(x+10)^2+14 Find the range

Andre45 [30]

Answer:

Step-by-step explanation:

y=-1/2(x+10)^2+14

2y=-(x+10)^2+28

(x+10)^2=28-2y

L.H.S. is positive.

so 28-2y≥0

28≥2y

or 2y≤28

or y≤ 14

Range is (-∞,14]

8 0

2 years ago