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3 years ago

A person runs 1 4/5 miles in 1/5 of an hour determine how many miles they are running per hour

Mathematics

2 answers:

asambeis [7]3 years ago

8 0

Multiply 1 4/5 by 5
9/5 * 5/1 = 45/5 = 9 miles

AveGali [126]3 years ago

8 0

Answer:

Step-by-step explanation: There are different methods to solve this, but I chose to change these into decimals- 1.8 and .20, to make it easier. Since miles per hour is speed, and speed is distance divided by time, all you have to do is fill in the blanks.

Since 1.8 / .20= 9, the answer is 9 miles per hour.

To take it a step farther, you could multiply this by any number of hours, or convert it to minutes. For example, to figure out how far they'd get in 10 minutes, I could divide the unit rate by 60, since there are 60 minutes in an hour, and get 0.15. So, in 10 minutes, they'd get 0.15 miles away.

Just in the mood for math, ig.

Ew, did I just say that? >.<

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You buy items costing $1900 and finance the cost with a fixed installment loan for 24 months at 8% simple interest per year.

ddd [48]

9514 1404 393

Answer:

$304
$91.83

Step-by-step explanation:

1. The finance charge is found from the simple interest formula;

I = Prt

where P is the principal amount, r is the annual rate, and t is the number of years.

24 months is 2 years, so the interest charged is ...

I = $1900×0.08×2 = $304

The finance charge is $304.

2. The monthly payment will be the total amount due, divided by the number of months.

payment = ($1900 +304)/24 = $2204/24 ≈ $91.83

The monthly payment is $91.83.

6 0

3 years ago

Please help math

Naddika [18.5K]

$\displaystyle\\
15(2a - 2) = 5(a^2 -1) \\ \\ 
30a - 30 = 5a^2-5\\\\
-5a^2 + 30a -30 + 5 =0\\\\
-5a^2 + 30a -25 =0 ~~~~~\Big| \times (-1) \\\\
5a^2 - 30a +25 = 0~~~~~\Big| : 5 \\\\
a^2 - 6a +5 = 0 \\ \\ 
a_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}= \frac{6\pm \sqrt{36-20}}{2}= \frac{6\pm \sqrt{16}}{2}=\frac{6\pm 4}{2}= \boxed{3\pm 2} \\ \\ 
a_1 = 3+ 2 = \boxed{5}\\\\
a_2 = 3-2 = \boxed{1}$

8 0

3 years ago

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The formula for the circumference of a circle is c=2r,where r is the radios

Sedbober [7]

PI x diameter squared

6 0

3 years ago

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:

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

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3 years ago

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A music teacher budgets $150 for a new book. The minimum cost of a new book is $12. How many books can she buy? Is this a minimu

andreyandreev [35.5K]

She can buy a maximum of 12 books for her music class

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3 years ago