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2 years ago

Can y’all help me on this graph work asap pls

Mathematics

2 answers:

Alenkasestr [34]2 years ago

7 0

Answer:i know it its 8 sorry

Step-by-step explanation:

nataly862011 [7]2 years ago

7 0

Just divide one of number of tickets by which ever of the ticket prices that goes on a line, like for example $24 divided by 3 equals $8 a ticket

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A bag contains tiles of different colors. Roshan randomly selects a tile from the bag and returns the tile to the bag after reco

sweet [91]

Answer:

6/41

Step-by-step explanation:

6 + 7 + 20 + 8 = 41 so that is number of tiles he took. 6 of those were 41 so therefore, the chance is 6/41

7 0

2 years ago

Read 2 more answers

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

3 years ago

Look at the picture below

Lelu [443]

Answer: Phillip is correct. The triangles are not congruent.

How do we know this? Because triangle ABC has the 15 inch side between the two angles 50 and 60 degrees. The other triangle must have the same set up (just with different letters XYZ). This isn't the case. The 15 inch side for triangle XYZ is between the 50 and 70 degree angle.

This mismatch means we cannot use the "S" in the ASA or AAS simply because we don't have a proper corresponding pair of sides. If we knew AB, BC, XZ or YZ, then we might be able to use ASA or AAS.

At this point, there isn't enough information. So that means John and Mary are incorrect, leaving Phillip to be correct by default.

Note: Phillip may be wrong and the triangles could be congruent, but again, we don't have enough info. If there was an answer choice simply saying "there isn't enough info to say either if the triangles are congruent or not", then this would be the best answer. Unfortunately, it looks like this answer is missing. So what I bolded above is the next best thing.

7 0

3 years ago

Read 2 more answers

High grade steel consists of 85% iron and 15% magnese. Low grade steel consists of 67% iron and 33% mag ese. NASA orders 500 ton

Mamont248 [21]

Answer:

$361\dfrac{1}{9}$ tons of high steel and $138\dfrac{8}{9}$ tons of low steel