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3 years ago

Please help me I am struggling

Mathematics

2 answers:

Otrada [13]3 years ago

8 0

Answer:

$48 Option 1

Step-by-step explanation:

$24/3 hours = $x/6 Hours

24/3=8 so it is means 8 dollars per hour

and 8*6 is 48 so x would equal 48.

Also since its like fractions or something like that the ratio is 1:2 and there fore 2 times the number of the left side.

The other way is looking at the fractions and seeing that 6 is 2 times larger than 3 so you would multiply 24 by 2 to get the demoators equal

Oxana [17]3 years ago

3 0

Answer:

it's 48

Step-by-step explanation:

Because 8x3=24 and you multiply 6×8=48

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Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

4 years ago

Find the geometric proof

GrogVix [38]

Answer:

See Explanation

Step-by-step explanation:

Line l is bisector of MO... (given)

$\therefore MN = NO.... (1)$

$\because MN + NO = MO..(M-N-O)$

$\therefore NO + NO = MO..[From \:(1)]$

$\therefore 2NO = MO$

$\therefore NO = \frac{1}{2} MO$

$\therefore NO = \frac{1}{2}\times 24..(\because MO=24)$

$\therefore NO =12$

Thus Proved

6 0

3 years ago

Read 2 more answers

When principal = Rs x, Time = y years and Rate = z% per year, calculate the simple interest.

Marianna [84]

Answer:

X×Y×Z%/100

Step-by-step explanation:

using formula

si=p×t×r%/100

3 0

3 years ago

Is (4,3) a solution to the inequality graph below? Why or why not? I need help plz don’t upload files

Studentka2010 [4]

Answer:

yes

Step-by-step explanation:

Go to x=4 and y= 3

It is in the shaded area to it is a solution

7 0

3 years ago

A share of ABC stock was worth $70 in 2011 and only

Bezzdna [24]

Answer:

5v + 32y = 350

Step-by-step explanation:

Let, the year counting starts in 2011 and represented by y.

So, the share of ABC stock value was $70 at y = 0. And in the year 2016 i.e. y = (2016 - 2011) = 5, the share worth $38.

If there is a linear relationship between the number of years since 2011 i.e. y and the value of the share of ABC stock i.e. v, then we can write the equation of this situation as

$\frac{v - 70}{70 - 38} = \frac{y - 0}{0 - 5}$

⇒ 5v - 350 = - 32y

⇒ 5v + 32y = 350 (Answer)

6 0

3 years ago