All categories

3 years ago

Help plsssssssssssssssss

Mathematics

2 answers:

Helen [10]3 years ago

8 0

Answer: A .0115

Step-by-step explanation:

There are 1000 mg in a gram. So to convert to grams, you would divide by 1000

erastova [34]3 years ago

6 0

Answer:

A number is the Right answer

0.0115g is the Right answer

Thank you

You might be interested in

Please help, thanks!<br><br> Factor completely<br><br> Sx²- 13x² + 15x²

Andru [333]

Answer:

x^2(S+2)

Step-by-step explanation:

7 0

3 years ago

The weekly earnings of students in one age group are normally distributed with a standard deviation of 47 dollars. A researcher

ELEN [110]

Answer:

Option D) 340

Step-by-step explanation:

We are given the following in the question:

Alpha, α = 0.05

Population standard deviation, σ = $47

Margin of error = 5

95% Confidence Interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

$\text{Margin of error} = z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$5 = 1.96(\dfrac{47}{\sqrt{n}} )\\\\\sqrt{n} = \dfrac{47\times 1.96}{5}\\\\n = 339.443776\\\Rightarrow n \approx 340$

Thus, the correct answer is

Option D) 340

7 0

3 years ago

Select all expressions that are equivalent to 3x - 2 (x - 4) - 1

tigry1 [53]

Answer:

x+7

Step-by-step explanation:

Remove Parentheses

Collect like terms

Calculate

8 0

2 years ago

Read 2 more answers

Choose the correct simplification of the expression (c2)9.

Lapatulllka [165]

The correct simplification of the expression is 9c^2 i think.

5 0

3 years ago

Read 2 more answers

A new shopping mall is considering setting up an information desk manned by one employee. Based upon information obtained from s

quester [9]

Answer:

a) $P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

b) $p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

c) $L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

d) $L_q =\frac{20^2}{30(30-20)}=1.333 people$

e) $W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

f) $W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

Step-by-step explanation:

Notation

P represent the probability that the employee is idle

$p_x$ represent the probability that the employee is busy

$L_s$ represent the average number of people receiving and waiting to receive some information

$L_q$ represent the average number of people waiting in line to get some information

$W_s$ represent the average time a person seeking information spends in the system

$W_q$ represent the expected time a person spends just waiting in line to have a question answered

This an special case of Single channel model

Single Channel Queuing Model. "That division of service channels happen in regards to number of servers that are present at each of the queues that are formed. Poisson distribution determines the number of arrivals on a per unit time basis, where mean arrival rate is denoted by λ".

Part a

Find the probability that the employee is idle

The probability on this case is given by:

In order to find the mean we can do this:

$\mu = \frac{1question}{2minutes}\frac{60minutes}{1hr}=\frac{30 question}{hr}$

And in order to find the probability we can do this:

$P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

Part b

Find the proportion of the time that the employee is busy

This proportion is given by:

$p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

Part c

Find the average number of people receiving and waiting to receive some information

In order to find this average we can use this formula:

$L_s= \frac{\lambda}{\lambda -\mu}$

And replacing we got:

$L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

Part d

Find the average number of people waiting in line to get some information.

For the number of people wiating we can us ethe following formula"

$L_q =\frac{\lambda^2}{\mu(\mu-\lambda)}$

And replacing we got this:

$L_q =\frac{20^2}{30(30-20)}=1.333 people$

Part e

Find the average time a person seeking information spends in the system

For this average we can use the following formula:

$W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

Part f

Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

For this case the waiting time to answer a question we can use this formula:

$W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

6 0

2 years ago

Read 2 more answers